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r Editor Tim Major and Editor Editor CalifCalifCalifCalifCalifororororornia Special Needs esa J visor::::: TTTTTerererereresa J esa J visor nia Special Needs AdAdAdAdAdvisor visor nia Special Needs nia Special Needs esa J..... Miller Miller Miller Miller esa J visor nia Special Needs Miller Teresa is a Mathematics Teacher in Capistrano Unified School District. She has a master’s degree in Education with an emphasis on Special Education. She holds two teaching credentials: Foundation Level Mathematics and Resource Specialist. ames Schierhierhierhierhierererererer ames Sc visor::::: J J J J James Sc ames Sc visor General al al al al AdAdAdAdAdvisor visor Gener Gener Gener ames Sc visor Gener James is a High School Mathematics Teacher for the King City Joint Union High School District. He has a bachelor’s degree in History. He holds two teaching credentials: Mathematics and History. visor::::: Daisy Lee Daisy Lee Daisy Lee visor General al al al al AdAdAdAdAdvisor visor Gener Gener Daisy Lee Daisy Lee visor Gener Gener Daisy is a High School Mathematics Teacher in Los Angeles Unified School District. She has a bachelor’s degree in Mathematics. terial: terial: ditional Ma AdAdAdAdAdditional Ma ditional Ma ence E. Da Da Da Da Davisvisvisvisvis,,,,, Ph.D Ph.D Ph.D ence E. ence E. Clar Clar terial: Clar ditional Material: Ph.D Clarence E. terial: ditional Ma Ph.D ence E. Clar Clarence is an educational consultant for Middle Grade Mathematics and was an Assistant Professor of Mathematics Education. AdAdAdAdAdditional Ma les Seraaaaaphinphinphinphinphin les Ser les Ser osemarie Ing terial: R R R R Rosemarie Ing osemarie Ing terial: terial: ditional Ma ditional Ma osemarie Ingles Ser ditional Material: les Ser osemarie Ing terial: ditional Ma Rosemarie is a Middle School and High School Mathematics Teacher, Fairfax County, Virginia. ting Editors:s:s:s:s: ting Editor ting Editor Suppor Suppor Supporting Editor Suppor ting Editor Suppor Chris Dennett Sarah Hilton Kate Houghton Sharon Keeley Ali Palin Alan Rix Glenn Rogers Emma Stevens Ami Snelling Claire Thompson Julie Wakeling Tim Wakeling eading: Judith Curran Buck, Jennifer Heller, Amanda Jones, Rafiq Ladhani, Betsey McCarty and Cathy Podeszwa. eading: PrPrPrPrProofroofroofroofroofreading: eading: eading: phic Design: Caroline Batten, Russell Holden, Jane Ross and Ash Tyson. phic Design: GrGrGrGrGraaaaaphic Design: phic Design: phic Design: Mathematics Content Standards for California Public Schools reproduced by permission, California Department of Education, CDE Press, 1430 N Street, Suite 3207, Sacramento, CA 95814. ISBN 13: 978 1 60017 014 0 website: www.cgpeducation.com Wakeling Publishing Services: www.wakelingpublishingservices.co.uk Printed by Elanders Hindson Ltd, UK and Johnson Printing, Boulder, CO Clipart sources: CorelDRAW and VECTOR. Text, design, layout, and illustrations © CGP, Inc. 2007 All rights reserved. xxxxx Chapter 1 Working with Real Numbers Section 1.1 Sets and Expressions ............................... 2 Section 1.2 The Real Number System ....................... 14 Section 1.3 Exponents, Roots, and Fractions ............ 37 Section 1.4 Mathematical Logic ................................. 55 Investigation Counting Collections ............................... 65 11111 TTTTTopicopicopicopicopic 1.1.11.1.1 1.1.11.1.1 1.1.1 Section 1.1 The Basics of Sets The Basics of Sets The Basics of Sets The Basics of Sets The Basics of Sets The Basics of Sets The Basics of Sets The Basics of Sets The Basics of Sets The Basics of Sets California Standards: 1.0:1.0:1.0:1.0:1.0: Students identify Students identify Students identify Students identify and Students identify use the arithmetic properties integggggererererersssss and inte inte subsets of subsets of of subsets of subsets of inte inte subsets of rational, irrational, and real numbers, including closure properties for the four basic arithmetic operations where applicable. What it means for you: You’ll learn about sets and subsets. Key words: set element subset universal set empty set Check it out: In this course the universal set consists of all real numbers (see Topic 1.1.2), unless otherwise specified. Sets are a really useful way of being able to say whether numbers or variables have something in common. There’s some new notation here, but the math isn’t too hard at all. Elements Elements e Collections of e Collections of Sets ar Sets ar Elements e Collections of Elements Sets are Collections of Elements e Collections of Sets ar Sets ar A set is a collection of objects. Each object in the set is called an element or a member. You use the symbol Œ to show that something is a member of a set — you read it as “is an element of ” or “is a member of.” The symbol œ means the opposite — you read it as “is not an element of ” or “is not a member of.” Sets are usually named by capital letters. The elements of a set are enclosed in braces { }. For example, A = {1, 3, 5, 7} is a set containing 4 elements — the numbers 1, 3, 5, and 7. The universal set, denoted by the symbol U, is the set of all objects under consideration (so in a math course, the universal set often consists of all numbers). Example Example Example Example Example 11111 Given that set A = {1, x, 2, b}, determine whether each of the following statements is true or false. a) x Œ A b) 2 Œ A c) 1 œ A d) 3 Œ A Solution Solution Solution Solution Solution a) x is an element of A, so the statement is true. b) 2 is an element of A, so the statement is true. c) 1 is an element of A (1 Œ A), so the statement is false. d) 3 is not an element of A, so the statement is false. Guided Practice Given that set A = {x, 2, 4, y}, determine whether each of the following statements is true or false. 1. y Œ A 2. 2 Œ A 3. 6 œ A 4. 4 œ A 22222 Section 1.1 Section 1.1 Section 1.1 — Sets and Expressions Section 1.1 Section 1.1 The Empty (or Null) Set has No Elements The Empty (or Null) Set has No Elements The Empty (or Null) Set has No Elements The Empty (or Null) Set has No Elements The Empty (or Null) Set has No Elements An empty set (or null set) is a set without any elements or members. It’s denoted by Δ or { }. e Contained WWWWWithin Other Sets ithin Other Sets ithin Other Sets e Contained e Contained Subsets ar Subsets ar ithin Other Sets Subsets are Contained ithin Other Sets e Contained Subsets ar Subsets ar A subset is a set whose elements are also contained in another set. The symbol Ã means “is a subset of.” Any set is a subset of itself — and the empty set is a subset of any set. Example Example Example Example Example 22222 Let A = {all odd numbers} and B = {1, 3, 5, 7}. Is B a subset of A? Solution Solution Solution Solution Solution Work through the elements of B one by one: 1 Œ A, 3 Œ A, 5 Œ A, and 7 Œ A, because they are all odd numbers. All elements of B are also elements of A — so B ÃÃÃÃÃ A. Example Example Example Example Example 33333 Let A = {0, 1, 2}. Determine all the subsets of set A. Solution Solution Solution Solution Solution List the empty set first because it’s a subset of any set. Then write all the subsets with 1 element, then all the subsets with 2 elements, and so on until finally you finish with the whole set. So the subsets of A are: ΔΔΔΔΔ, {0}, {1}, {2}, {0, 1}, {0, 2}, {1, 2}, and {0, 1, 2} Guided Practice Let A = {a, 2, 3, b, c} and B = {1, 2, c, d}. Use sets A and B to answer questions 5 and 6: 5. Explain whether B Ã A. 6. List all the subsets of set B. Let C = {all prime numbers less than 13 but greater than 7}. 7. List set C and all its subsets. For exercises 8 and 9, let M = {all real numbers b such that b = 3x – 1}. 8. List the members of set M if x Œ {1, 2, 3, 5}. 9. List the subsets of set M if x Œ {1, 2, 3, 5}. Section 1.1 Section 1.1 Section 1.1 — Sets and Expressions Section 1.1 Section 1.1 33333 Check it out: Another way of saying that set A and B are equal is that A is a subset of B... ...and B is a subset of A. Equality of Sets Equality of Sets Equality of Sets Equality of Sets Equality of Sets Two sets are equal if they have all of the same elements in them. So if A = {1, 3, 5} and B = {3, 5, 1}, then A and B are equal. Or more mathematically, two sets A and B are said to be equal if every element in set A is in set B, and every element in set B is in set A. Note also that all empty sets are equal — because they’re exactly the same. That’s why you say the empty set, not an empty set. Independent Practice 1. Determine the set H whose elements are all the multiples of both 2 and 3 that are less than 30 but greater than 12. 2. Let U = {all letters of the alphabet}. Determine set V, whose elements are the vowels. 3. Write down G = {all prime numbers greater than 11 but less than 13}. For exercises 4 and 5, let F = {red, yellow, blue, purple}, and let G Ã F and H Ã F. 4. Determine set G, whose elements are 3-letter words in F. 5. Determine set H, whose elements are 5-letter words in F. For exercises 6 and 7, let A = {all even numbers} and B = {2, 4, 6, 8}. 6. Explain whether B Ã A. 7. List all the subsets of B. For exercises 8 and 9, let M = {3, a, 9, b, 15} and N = {3, 6, 9}. 8. Explain whether N Ã M. 9. List all the subsets of N. 10. Let C = {all even numbers less than 10 but greater than 2}. List set C and all its subsets. 11. Let J = {all real numbers y such that y = 2x + 3}. List the members of set J if x Œ {0, 1, 2, 3}. 12. Let A = {1, 3, 5, 7, 9} and B = {all odd numbers less than 10 but greater than 0}. Explain whether A = B. ound Up ound Up RRRRRound Up ound Up ound Up RRRRRound Up ound Up ound Up ound Up ound Up Sets sound a little odd but they’re just a way of grouping together types of numbers or variables. You’ve seen a lot of the stuff in this Topic in earlier grades, but in Algebra I you’ve got to treat everything formally and give the proper names for things like the empty set. 44444 Section 1.1 Section 1.1 Section 1.1 — Sets and Expressions Section 1.1 Section 1.1 TTTTTopicopicopicopicopic 1.1.21.1.2 1.1.
21.1.2 1.1.2 California Standards: 1.0:1.0:1.0:1.0:1.0: Students identify Students identify Students identify Students identify and Students identify use the arithmetic properties integggggererererers and s and s and inte inte subsets of subsets of subsets of inte of subsets of s and inte subsets of s and rrrrraaaaational, and realealealealeal and r and r tional, tional, tional, ir ir ir ir irrrrrraaaaational, tional, tional, tional, and r and r tional, tional, umbersssss, including closure umber umber nnnnnumber umber properties for the four basic arithmetic operations where applicable. What it means for you: You’ll learn about the set of real numbers and its subsets. Key words: real numbers natural numbers whole numbers integers rational numbers irrational numbers set element subset Check it out: The natural numbers (N) are sometimes called the counting numbers, because they’re the numbers you use in everyday life to count things. Check it out: The symbol À means “is not a subset of.” eal Numbersssss eal Number eal Number the R the R Subsets of Subsets of the Real Number Subsets of the R eal Number the R Subsets of Subsets of eal Numbersssss eal Number eal Number the R the R Subsets of Subsets of the Real Number Subsets of the R the R Subsets of eal Number Subsets of This Topic is about one set of numbers that is really important. You’ll be referring to the real numbers throughout Algebra I. s Has Subsets s Has Subsets eal Number he Set of R R R R Real Number eal Number he Set of TTTTThe Set of he Set of s Has Subsets eal Numbers Has Subsets s Has Subsets eal Number he Set of In Algebra I, the sets used are usually subsets of the real numbers. R — Real Numbers The set of real numbers consists of all positive numbers, zero, and all negative numbers. The following are subsets of R: N — Natural numbers W — Whole numbers Z — Integers N = {1, 2, 3, …} W = {0, 1, 2, 3, …} Z = {..., –2, –1, 0, 1, 2, ...} Example Example Example Example Example 11111 Explain why N Ã W. Solution Solution Solution Solution Solution The set N of natural numbers is a subset of the set W of whole numbers because every member of set N is a member of set W. Example Example Example Example Example 22222 Explain why Z À N. Solution Solution Solution Solution Solution To show this, you need to find an element in Z that is not in N. (You can use any negative integer or zero.) Since –1 Œ Z, but –1 œ N, the set Z is not a subset of N (Z À N). Guided Practice 1. Determine the subset of N (natural numbers) whose elements are multiples of 2 that are greater than 10 but less than 20. 2. Determine the subset of Z (integers) whose elements are greater than –1 but less than 5. 3. Give an example of an element in R (real numbers) that is not in Z (integers). 4. Explain why R (real numbers) À Z (integers). Section 1.1 Section 1.1 Section 1.1 — Sets and Expressions Section 1.1 Section 1.1 55555 Check it out: Notice that since q Œ N, q cannot equal zero. e Subsets of R e Subsets of e Subsets of o Mor tionals — TTTTTwwwwwo Mor o Mor tionals — tionals and Irrrrrraaaaationals — tionals — tionals and Ir RRRRRaaaaationals and Ir tionals and Ir o More Subsets of e Subsets of o Mor tionals — tionals and Ir A rational number is a number that can be expressed in the form p q , where p is an integer and q is a natural number. Rational Numbers: = all numbers that can be expressed as fractions ⎧ ⎪⎪ ⎨ ⎩⎪⎪ , where p q∈ , ∈ p q ⎫ ⎪⎪ ⎬ ⎭⎪⎪ For example, 3.5 is a rational number — it can be expressed as 7 2 . An irrational number is a number that cannot be expressed in the form p q , where p is an integer and q is a natural number. Irrational Numbers: I = all numbers that cannot be expressed as fractions ⎧ ⎪⎪ ⎨ ⎩⎪⎪ , wher ee p q∈ , ∈ p q ⎫ ⎪⎪ ⎬ ⎭⎪⎪ For example, ÷2 = 1.4142... is an irrational number — it can’t be expressed in the form p q . Both Q and I are subsets of the real numbers, R. Guided Practice 5. Explain why N Ã Q. 6. Explain why Z À I. 7. Determine the subset, C, of Q whose elements are also members of I. Independent Practice Remember that R = {real numbers}, N = {natural numbers}, W = {whole numbers}, Z = {integers}, Q = {rational numbers}, and I = {irrational numbers}. Classify each of the following statements as true or false. 2. Δ À N 1. N Ã W 3. R Ã Q { 4. Z À Q 5. 2 3 . 0 4 7 25 8 6 , }Ã Q . , , , ound Up ound Up RRRRRound Up ound Up ound Up ound Up ound Up RRRRRound Up ound Up ound Up “Real numbers” is just a formal name for all positive and negative numbers and zero — there’s nothing more to it than a new name. You need to know all the common labels for the subsets. 66666 Section 1.1 Section 1.1 Section 1.1 — Sets and Expressions Section 1.1 Section 1.1 TTTTTopicopicopicopicopic 1.1.31.1.3 1.1.31.1.3 1.1.3 California Standards: 1.0:1.0:1.0:1.0:1.0: Students identify Students identify Students identify Students identify and Students identify use the arithmetic properties integggggererererers and s and s and inte inte subsets of subsets of subsets of inte of subsets of s and inte subsets of s and rrrrraaaaational, and realealealealeal and r and r tional, tional, tional, ir ir ir ir irrrrrraaaaational, tional, tional, tional, and r and r tional, tional, umbersssss, including closure umber umber nnnnnumber umber properties for the four basic arithmetic operations where applicable. What it means for you: You’ll learn about unions and intersections of sets. Key words: union intersection set element Check it out: A » B is said “A union B.” sections sections Unions and Inter Unions and Inter sections Unions and Intersections sections Unions and Inter Unions and Inter sections sections Unions and Inter Unions and Inter Unions and Intersections sections Unions and Inter sections Unions and Inter This Topic is all about two symbols that represent ways of grouping elements in sets — unions and intersections. The Union of Sets is Anything in Either or Both Sets The Union of Sets is Anything in Either or Both Sets The Union of Sets is Anything in Either or Both Sets The Union of Sets is Anything in Either or Both Sets The Union of Sets is Anything in Either or Both Sets The union of sets A and B, denoted A »»»»» B, is the set of all elements in either A or B or both sets. Example Example Example Example Example 11111 Let A = {4, 6, 8, 20} and B = {6, 8, 9, 15}. Find A » B. Solution Solution Solution Solution Solution Elements are not counted twice — they are either “in or out.” So 8, which is in both A and B, only appears once in A » B. So A »»»»» B = {4, 6, 8, 9, 15, 20}. Guided Practice Let A = {1, 2, 4, 5}, B = {0, 3, 6}, C = {2, 4, 6, 8}, D = {1, 3, 5, 7}. 1. Find A » B. 2. Find C » D. 3. Find A » C. 4. Find B » D. Sets is AnAnAnAnAnything in Both Sets ything in Both Sets ything in Both Sets Sets is Sets is section of section of he Inter TTTTThe Inter he Inter ything in Both Sets section of Sets is he Intersection of ything in Both Sets Sets is section of he Inter The intersection of sets A and B, denoted A ««««« B, is the set of all elements that are in both set A and set B. Check it out: A « B is said “A intersect B” (or “A intersection B”). Example Example Example Example Example 22222 Let A = {4, 6, 8, 20} and B = {6, 9, 15}. Find A « B. Solution Solution Solution Solution Solution Only the number 6 is in both A and B. So A ««««« B = {6}. Example Example Example Example Example 33333 Let A = {all even numbers} and B = {all odd numbers}. Find A « B. Solution Solution Solution Solution Solution There are no elements in both A and B, so A ««««« B = ΔΔΔΔΔ. Section 1.1 Section 1.1 Section 1.1 — Sets and Expressions Section 1.1 Section 1.1 77777 Guided Practice Let A = {1, 2, 4, 5}, B = {0, 3, 6}, C = {2, 4, 6, 8}, D = {1, 3, 5, 7}. 5. Find A « B. 6. Find A « C. 7. Find C « D. 8. Find B « D. sections for or or or or TTTTTwwwwwo Sets o Sets o Sets sections f sections f ou Can Find Unions and Inter YYYYYou Can Find Unions and Inter ou Can Find Unions and Inter o Sets ou Can Find Unions and Intersections f o Sets sections f ou Can Find Unions and Inter If you have any two sets, you can always work out the union and intersection. Example Example Example Example Example 44444 Let A = {2, 4, 6, 8, 10, 12} and B = {3, 6, 9, 12, 15}. Find A « B and A » B. Solution Solution Solution Solution Solution A « B is the set of all the elements that appear in both A and B. So A ««««« B = {6, 12}. A » B is the set of all the elements that appear in A or B, or both sets. So A »»»»» B = {2, 3, 4, 6, 8, 9, 10, 12, 15}. Independent Practice 1. Explain why A « B is a subset of A. If A = {1, 2, 3, 4, 5, 11, 12}, B = {2, 4, 6, 8, 10, 11, 20}, and C = {5, 10, 15, 20, 24}, write down the following sets. 2. A « B 3. (A « B) » C 4. (A « B) « C Let U = {all b Œ N such that b £ 20}, M = {all b in U such that b is a multiple of 2}, V = {all b in U such that b is a multiple of 3}, and H = {all b in U such that b is a multiple of 4}. 5. Determine the set M « V. 6. Determine the set (M « V) « H. 7. Determine the set (M » V) » H. Let A = {prime numbers}, B = {square numbers}, C = {even numbers}, D = {odd numbers}, and E = {natural numbers less than 10}. 8. Copy and complete the following expressions using sets A to E: a. A « ...... = Δ, b. ...... « ...... = {2}, c. ...... « D « E = {1, 9} d. A « ...... « ...... = {3, 5, 7} 9. Write down the number of subsets for each set a–d in exercise 8. 10. Use your answer to exercise 9 to write down a formula for the number of subsets of a set with n members. ound Up ound Up RRRRRound Up ound Up ound Up ound Up ound Up RRRRRound Up ound Up ound Up The big thing to remember here is that the union sign is » and the intersection sign is « . Make sure you understand the definitions of each way of grouping elements. 88888 Section 1.1 Section 1.1 Section 1.1 — Sets and Expressions Section 1.1 Section 1.1 TTTTTopicopicopicopicopic 1.1.41.1.4 1.1.41.1.4 1.1.4 California Standards: 1.1: Students use 1.1: Students use 1.1: Students use 1.1: Students use 1.1: Students use p
rprprprproper umbersssss to umber ties of n n n n number umber ties of ties of oper oper operties of umber ties of oper demonstrate whether assertions are true or false. What it means for you: You’ll deal with simple expressions that contain numbers and unknowns. Key words: expression numeric expression algebraic expression variable aic and Numeric aic and Numeric AlgAlgAlgAlgAlgeeeeebrbrbrbrbraic and Numeric aic and Numeric aic and Numeric AlgAlgAlgAlgAlgeeeeebrbrbrbrbraic and Numeric aic and Numeric aic and Numeric aic and Numeric aic and Numeric essions essions ExprExprExprExprExpressions essions essions ExprExprExprExprExpressions essions essions essions essions Expressions are mathematical statements. This Topic is all about two kinds of expressions — ones that contain only numbers, and ones that contain both numbers and unknown values. lude Numbersssss lude Number lude Number ust Inc ust Inc essions J essions J Numeric Expr Numeric Expr ust Include Number essions Just Inc Numeric Expressions J lude Number ust Inc essions J Numeric Expr Numeric Expr A numeric expression contains only numbers — for example, 2.54 × 2 is a numeric expression. Since there are no unknown quantities, you can always work out the value of a numeric expression — the value of 2.54 × 2 is 5.08. Numeric expressions with the same value are like different names for the same thing. All these expressions have the same value: 2.54 × 2 5 + 0.08 6 – 0.92 The value of each of these numeric expressions is 55555.08.08.08.08.08. So if you saw (2.54 × 2) somewhere, you could write (5 + 0.08) instead, since the two expressions have the same value — they’re describing the same thing. Guided Practice Calculate the value of each of these numeric expressions: 1. 18.4 + 8.23 2. 37.82 – 11.19 3. 716 ÷ 2 4. 1790 ÷ 5 5. 37 ÷ 37 6. 19284 ÷ 19284 For each of the following, write a numeric expression that contains the number 100 and has the same value as the expression given. 7. 465 – 253 8. 3850 ÷ 5 9. 15.6 × 5 10. 24.2 + 1.5 Section 1.1 Section 1.1 Section 1.1 — Sets and Expressions Section 1.1 Section 1.1 99999 Check it out: When you see a number and a letter written next to each other (like 2.54x), it means they are multiplied together. essions Contain VVVVVariaariaariaariaariabbbbbleslesleslesles essions Contain essions Contain aic Expr AlgAlgAlgAlgAlgeeeeebrbrbrbrbraic Expr aic Expr aic Expressions Contain essions Contain aic Expr A lot of the time in math, you can use letters to stand for unknown amounts. So if you were not sure how many inches long your desk was, you could just call the length x inches. When you see letters in equations, they’re just numbers that you don’t know the value of yet. Letters that represent unknown numbers are called variables. Example Example Example Example Example 11111 The length of a desk is x inches. How long is this in centimeters? To convert a length in inches to a length in centimeters, you multiply by 2.54. Solution Solution Solution Solution Solution You don’t know the length of the desk, so it’s been called x. If the desk was 10 inches long, it would be 2.54 × 10 = 25.4 cm long. If the desk was 50 inches long, it would be 2.54 × 50 = 127 cm. In the same way, a desk that is x inches long would be 2.54 × x = 2.54x cm. You don’t have enough information to write a numerical value, so leave x in the solution. 2.54x is called an algebraic expression, as it contains an unknown quantity. The variable x represents the unknown quantity — the number of inches to be converted to centimeters. An algebraic expression always contains at least one variable (and very often it contains numbers as well). Example Example Example Example Example 22222 Write an algebraic or numeric expression for each of the following: a) The total cost of an item whose price is $15.75 plus sales tax of $1.30. b) The number of cents in d dollars. c) Three more than twice the length, l. Solution Solution Solution Solution Solution a) $15.75 + $1.30 = $17.05 (since there are no unknowns, this is a numeric expression) b) 100d (since the number of cents is always the number of dollars multiplied by 100) c) 2l + 3 (since twice the length l is 2 × l = 2l, and three more than this is 2l + 3) 1010101010 Section 1.1 Section 1.1 Section 1.1 — Sets and Expressions Section 1.1 Section 1.1 Guided Practice Write an algebraic expression representing each of the following. 11. The sum of twice m and 7. 12. The sum of three times x and 12. 13. The difference between the total cost c, and the down payment d. 14. The number of notebooks you can buy for $75 if each notebook costs $n. 15. The difference between (3x + 7) and t. 16. Two more than five times d. 17. Three less than the product of x and t. 18. The age of someone six years ago who is 2y years old now. Independent Practice Determine whether the following expressions are numeric or algebraic. 1. 3.14 × 2 2. 189x ÷ 2 3. 2w + 2l 4. 656.4 – 33 Write an algebraic or numeric expression for each of the following. 5. The difference between x and 3 6. The cost of 5 CDs at $x 7. The difference between the original price of a pair of jeans, p, and the sale price, s 8. The product of 2 and 4, plus 3 9. Three times the difference between t and 2 10. One-half of 200 times a number 11. The number of feet in x yards (3 feet = 1 yard) 12. The perimeter of a rectangle with length, l, and width, w. ound Up ound Up RRRRRound Up ound Up ound Up RRRRRound Up ound Up ound Up ound Up ound Up Don’t worry if you come across an expression that contains variables. Variables are just letters or symbols that represent unknown numbers, and expressions containing variables follow all the same rules as numeric expressions. Section 1.1 Section 1.1 Section 1.1 — Sets and Expressions Section 1.1 Section 1.1 1111111111 TTTTTopicopicopicopicopic 1.1.51.1.5 1.1.51.1.5 1.1.5 California Standards: 1.1: Students use 1.1: Students use 1.1: Students use 1.1: Students use 1.1: Students use prprprprproper umbersssss to umber ties of n n n n number umber ties of ties of oper oper operties of umber ties of oper demonstrate whether assertions are true or false. What it means for you: You’ll find out what coefficients are, and you’ll calculate the value of expressions. Key words: coefficient expression numeric expression algebraic expression variable ficients and Evaluaaluaaluaaluaaluatingtingtingtingting ficients and Ev ficients and Ev CoefCoefCoefCoefCoefficients and Ev ficients and Ev CoefCoefCoefCoefCoefficients and Ev ficients and Evaluaaluaaluaaluaaluatingtingtingtingting ficients and Ev ficients and Ev ficients and Ev You’ve just learned about variables — coefficients are just another element that make up a full expression. In this Topic you’ll learn about evaluating expressions, which just means calculating the value. y a y a VVVVVariaariaariaariaariabbbbblelelelele y a y a Amount to Multipl Amount to Multipl ficient is the ficient is the A Coef A Coef Amount to Multiply a ficient is the Amount to Multipl A Coefficient is the Amount to Multipl ficient is the A Coef A Coef In the algebraic expression 2.54x, the number 2.54 is the coefficient of x. A coefficient is a number that multiplies a variable. Example Example Example Example Example 11111 Find the coefficients in these expressions: a) the coefficient of x in the algebraic expression 2x b) the coefficient of v in v + 5 c) the coefficient of y in 7 – 6y d) the coefficient of k in 5 – k e) the coefficient of m in 5(3 – 2m) Solution Solution Solution Solution Solution a) x is multiplied by 2, so the coefficient is 2. b) v is the same as 1v, so the coefficient is 1. c) The minus sign is part of the coefficient as well — so the coefficient is –6. d) There’s a minus sign here too — the coefficient is –1. e) First multiply out the parentheses: 5(3 – 2m) = 15 – 10m Now you can see that the coefficient is actually –10. Guided Practice Find the coefficient of the variable in each of these expressions: 1. 4a – 15 2. 15 – 4z 3. x – 3 4. 2 – k 5. 2(3b + 1) 6. 4(2 – m) 1212121212 Section 1.1 Section 1.1 Section 1.1 — Sets and Expressions Section 1.1 Section 1.1 heir Simplest Fororororormmmmm heir Simplest F essions in TTTTTheir Simplest F heir Simplest F essions in essions in rite Expr AlAlAlAlAlwwwwwaaaaays ys ys ys ys WWWWWrite Expr rite Expr rite Expressions in heir Simplest F essions in rite Expr Different numeric expressions can have the same value. The expressions 4 + 3 and 14 ÷ 2 both have the same value of 7. You can think of these expressions as different names for the number 7. When you simplify an expression, you replace it with its simplest name. So you could simplify “4 + 3” or “14 ÷ 2” by writing “7” instead. Guided Practice Simplify each expression below. 9. 24 ÷ 6 7. 11 + 4 10. 10a + 4a 8. 10 × 3.14 11. 12a + 2 – 8a – 5 12. 28b ÷ 28 essions essions aic Expr ting AlgAlgAlgAlgAlgeeeeebrbrbrbrbraic Expr aic Expr ting EvEvEvEvEvaluaaluaaluaaluaaluating ting essions aic Expressions essions aic Expr ting You’ve already seen algebraic expressions — they’re expressions containing variables. But if you know the value of the letter, you can evaluate the expression — that means you calculate its value. For example, the algebraic expression 2.54x contains the variable x. When x = 33, the expression 2.54x is equal to 2.54 × 33 = 83.82. That’s what is meant by “evaluating an algebraic expression” — finding the value of the expression when any variables are replaced by specific numbers. Guided Practice Don’t forget: Variables are letters that represent unknown quantities. Evaluate each of these expressions when a = 2, b = 7, and c = –4. 13. a + b 14. b × a 15. c ÷ a 16. a + b + c 17. b – a 18. b – c Independent Practice 1. Find the coefficient of x in the algebraic expression 8 – 5x. 2. Evaluate 8 – 5x when x = 3. 3. Find the coefficient of k in 4.18k + 2. 4. Evaluate 10x + 2 when x = 0. 5. Evaluate 2x + 3y when x = 10 and y = –1. 6. Evaluate 2x + 3y – z when x = 5, y = 2, and z = 4. 7. The formula C = 5 9
degrees Fahrenheit (°F) to degrees Celsius (°C). What is 131 °F in °C? (F – 32) is used to convert temperatures from ound Up ound Up RRRRRound Up ound Up ound Up ound Up ound Up RRRRRound Up ound Up ound Up A coefficient is just the number in front of a variable. It doesn’t matter whether an expression includes just numbers, or numbers and variables — you always follow the same rules when you simplify. Section 1.1 Section 1.1 Section 1.1 — Sets and Expressions Section 1.1 Section 1.1 1313131313 TTTTTopicopicopicopicopic 1.2.11.2.1 1.2.11.2.1 1.2.1 California Standards: 1.0: Students identify and 1.0: Students identify and 1.0: Students identify and 1.0: Students identify and 1.0: Students identify and use the arithmetic use the arithmetic use the arithmetic use the arithmetic use the arithmetic prprprprproper ties of subsets of ties of ties of oper oper operties of ties of oper integers and rational, umbersssss,,,,, umber umber eal n eal n irrational, and rrrrreal n eal number umber eal n incincincincincluding c losureeeee losur losur luding c luding c luding closur losur luding c prprprprproper our basic our basic or the f or the f ties f ties f oper oper or the four basic our basic ties for the f operties f our basic or the f ties f oper tions wherherherherhereeeee tions w tions w arithmetic operaaaaations w arithmetic oper arithmetic oper arithmetic oper tions w arithmetic oper pplicabbbbblelelelele..... pplica aaaaapplica pplica pplica What it means for you: You’ll learn some of the rules that the real numbers always follow. Key words: binary operation equality real numbers substitution The Number System The Number System The Number System The Number System The Number System The Number System The Number System The Number System The Number System The Number System You’ve already met the real numbers in Section 1.1. Now it’s time to look at the properties of the real numbers in detail. eal Number System is Based on Simple Rulesulesulesulesules eal Number System is Based on Simple R TTTTThe Rhe Rhe Rhe Rhe Real Number System is Based on Simple R eal Number System is Based on Simple R eal Number System is Based on Simple R The rules of the real number system are based on the existence of a set of numbers, plus two binary operations. A binary operation allows you to combine two numbers in some way to produce a third number. In Algebra I, the set of numbers used is the real numbers, R, and the two binary operations are addition and multiplication. Equality AlAlAlAlAlwwwwwaaaaays Hold ys Hold ys Hold Equality Equality ties of ties of oper oper he Pr TTTTThe Pr he Pr ys Hold ties of Equality operties of he Proper ys Hold Equality ties of oper he Pr The following statements about equality hold true for any real numbers a, b, and c: For any number a Œ R: a = a This is the reflexive property of equality. For any numbers a, b Œ R: if a = b then b = a This is the symmetric property of equality. For any numbers a, b, c Œ R: if a = b and b = c, then a = c This is the transitive property of equality. Guided Practice Name the property of equality being used in each statement. 1. If a = 3 then 3 = a. 2. If a = 3 and 3 = b, then a = b. 3. 3 = 3 4. If 3x = 2 and 2 = 2y, then 3x = 2y. 5. If 3x = 2y then 2y = 3x. Check it out: For addition, the inputs are called adadadadaddends dends dends dends, and for dends multiplication they are called fffffactor actor actor actorsssss. actor tions tions y Operaaaaations y Oper y Oper e Binar e Binar tion ar tion ar dition and Multiplica dition and Multiplica AdAdAdAdAddition and Multiplica tions e Binary Oper tion are Binar dition and Multiplication ar tions y Oper e Binar tion ar dition and Multiplica The operations of addition and multiplication are called binary operations. In order to carry either of them out you need to have two “inputs.” What this basically means is that you can combine two numbers to produce a third number. A set of numbers is said to be closed under a given binary operation if, when you perform that operation on any two members of the set, the result is also a member of the set. 1414141414 Section 1.2 Section 1.2 Section 1.2 — The Real Number System Section 1.2 Section 1.2 Check it out: Compare this to the set of irrational numbers, I, which is not closed under multiplication, since 2 × 2 = 2, where 2 Œ I, but 2 œ I. Check it out: Multiplication of a and b can be written as ab, a·b, or a × b. They all mean the same thing. The set of real numbers, R, is closed under both addition and multiplication — adding and multiplying real numbers always produces other real numbers. If a Œ R and b Œ R, then (a + b) ŒŒŒŒŒ R and (a × b) ŒŒŒŒŒ R. Example Example Example Example Example 11111 Use the fact that 10 Œ R and 6 Œ R to explain why 16 Œ R and 60 Œ R. Solution Solution Solution Solution Solution 10 Œ R and 6 Œ R, so you can add the two numbers to produce another number that is a member of R. So 10 + 6 = 16 Œ R. You can also multiply the two numbers to produce another number that is a member of R. So 10 × 6 = 60 Œ R. operty)ty)ty)ty)ty) oper oper he Substitution Principle (or Substitution Pr TTTTThe Substitution Principle (or Substitution Pr he Substitution Principle (or Substitution Pr he Substitution Principle (or Substitution Proper oper he Substitution Principle (or Substitution Pr For any real numbers a and b, the number a may be substituted for the number b if a = b. This principle means that if two expressions have the same value, then one expression can be substituted for (written instead of) the other. For example, the expression 23 can be replaced by the expression (10 + 13) or by (17 + 6). Similarly, the expression 2.3 × 4.5 can be replaced with 10.35. Independent Practice Determine whether each of the following statements is true or false. If false, rewrite the statement so that it is true. 1. If k = l then l = k. This is the symmetric property. 2. If k = l and l = 3, then k = 3. This is the reflexive property. In Exercises 3–6, demonstrate the closed nature of the set under both addition and multiplication using the numbers given. 3. R using 3 and 5 4. R using –10 and 11 5. N using 6 and 8 6. Q using 1.5 and 0.3 ound Up ound Up RRRRRound Up ound Up ound Up ound Up ound Up RRRRRound Up ound Up ound Up These rules are a bit abstract at the moment, but don’t worry. You’ll see how useful the real number properties are in the rest of the Section, and you’ll use them all the way through Algebra I. Section 1.2 Section 1.2 Section 1.2 — The Real Number System Section 1.2 Section 1.2 1515151515 TTTTTopicopicopicopicopic 1.2.21.2.2 1.2.21.2.2 1.2.2 California Standards: 2.0:2.0:2.0:2.0:2.0: Students under stand stand Students under Students under Students understand stand stand Students under h operaaaaations as tions as tions as h oper h oper and use suc and use suc tions as and use such oper and use suc tions as h oper and use suc taking the opposite,,,,, finding finding finding taking the opposite taking the opposite finding taking the opposite finding taking the opposite ocal ocal ecipr ecipr the r the r ocal, taking a root, eciprocal the recipr the r ocal ecipr the r and raising to a fractional power. They understand and use the rules of exponents. What it means for you: You’ll find out about identities, which don’t change your original value, and inverses, which take the opposite. Key words: identity inverse Check it out: The additive inverse is sometimes called a negative, or an opposite. Identities and Invvvvvererererersessessessesses Identities and In Identities and In Identities and In Identities and In Identities and Invvvvvererererersessessessesses Identities and In Identities and In Identities and In Identities and In The numbers 0 and 1 are special numbers, since they are the identities of addition and multiplication. dition and Multiplicationtiontiontiontion dition and Multiplica e Identities of AdAdAdAdAddition and Multiplica dition and Multiplica e Identities of e Identities of 0 and 1 ar 0 and 1 ar 0 and 1 are Identities of dition and Multiplica e Identities of 0 and 1 ar 0 and 1 ar In the real number system, there’s an additive identity. This means that there’s a number (called zero) which you can add to any number without changing the value of that number. For any number a Œ R There’s also a multiplicative identity. This is a number (called one) which any number can be multiplied by without its value being changed. For any number a Œ R to Make 0e 0e 0e 0e 0 d to Mak se is WWWWWhahahahahat t t t t YYYYYou ou ou ou ou AdAdAdAdAdd to Mak d to Mak se is TTTTThe he he he he AdAdAdAdAdditiditiditiditiditivvvvve Ine Ine Ine Ine Invvvvverererererse is se is d to Mak se is Every real number has an additive inverse — “additive inverse” is just a more mathematical term for the “negative” of a number. When you add a number to its additive inverse, the result is 0 (the additive identity). Or, more formally: For every real number m, there is an additive inverse written –m. When you add a number to its additive inverse, you get 0 — which is the additive identity. So, for any number m Œ R: –m + m = 0 = m + (–m) The inverse of a negative number is a positive number — that is, –(–m) = m, for any m. And if you take the opposite of a number, you change it into its additive inverse. Guided Practice Find the additive inverse of the following: 1. 6 2. 81 3. –4 4. y 5. –k 6. a + 5 7. –5a – 2 8. 2ab 9. State the value of x if 51 + x = 0. 10. State the value of x if x – 5 = 0. 1616161616 Section 1.2 Section 1.2 Section 1.2 — The Real Number System Section 1.2 Section 1.2 Check it out: The multiplicative inverse is sometimes called the reciprocal. Check it out: See Topic 1.2.5 for more information about not being able to divide by zero. se to Make 1e 1e 1e 1e 1 se to Mak y the Multiplicatititititivvvvve Ine Ine Ine Ine Invvvvverererererse to Mak se to Mak y the Multiplica Multiply by by by by by the Multiplica y the Multiplica Multipl Multipl se to Mak y the Multiplica Multip
l Multipl Multiplication has a property that’s similar to the property of addition: Every real number also has a multiplicative inverse (or “reciprocal”). When you multiply a number by its multiplicative inverse, the result is 1 — the multiplicative identity. However, there’s an important exception: zero has no reciprocal — its reciprocal cannot be defined. That means that you can’t divide by zero. More formally this becomes: For every nonzero real number m, there is a multiplicative inverse written m–1. When you multiply a number by its multiplicative inverse, you get 1 — which is the multiplicative identity. So, for any number m Œ R, m π 0: m–1 × m = 1 = m × m–1 If m is a nonzero real number then its reciprocal is m–1, which is given by m–1 = 1 m . And the reciprocal of m–1 is m — that is, ( )m− − 1 1 = m, 1 − = m, or 1m ( 1 1 ) = m. m which means that Example Example Example Example Example 11111 Find the reciprocals of: a) 3 b) 12 c) x d) 3–1 Solution Solution Solution Solution Solution 3 1− = 1 3 a) b) 12 1− = 1 12 c) x− =1 1 x d) (3–1)–1 = 3 Guided Practice Find the multiplicative inverse: 12. –31 11. 2 a 14. − 1 b 16 15. 16. Find the value of x if 3x = 1. Independent Practice 13. 1 8 State the additive and multiplicative inverses of each expression below: 1. 6 1 8 4. 2. –11 5. x 3. 1 3 6. (a + b) ound Up ound Up RRRRRound Up ound Up ound Up ound Up ound Up RRRRRound Up ound Up ound Up Identity and inverse are just math names for quite simple things. The identity doesn’t change your value, and inverses just mean taking the opposite, or finding the reciprocal. Section 1.2 Section 1.2 Section 1.2 — The Real Number System Section 1.2 Section 1.2 1717171717 TTTTTopicopicopicopicopic 1.2.31.2.3 1.2.31.2.3 1.2.3 California Standards: 1.1: Students use 1.1: Students use 1.1: Students use 1.1: Students use 1.1: Students use prprprprproper umbers s s s s to umber ties of n n n n number umber ties of ties of oper oper operties of umber ties of oper demonstrate whether assertions are true or false. What it means for you: You’ll use the number line to show real numbers, and you’ll describe them in terms of absolute values. Key words: real numbers absolute value The Number Line The Number Line The Number Line The Number Line The Number Line The Number Line The Number Line The Number Line The Number Line The Number Line alues alues Absolute VVVVValues Absolute Absolute and and alues and Absolute alues Absolute and and Absolute VVVVValues alues alues Absolute Absolute and and and Absolute alues Absolute and alues and The number line is a useful way of representing numbers visually. This Topic also includes information about absolute values, which show how far from zero numbers are on the number line. oints on a Line oints on a Line s as P s as P eal Number Number Lines Show Rw Rw Rw Rw Real Number eal Number Number Lines Sho Number Lines Sho oints on a Line s as Points on a Line eal Numbers as P oints on a Line s as P eal Number Number Lines Sho Number Lines Sho –6 –5 –4 –3 –2 –1 0 All real numbers can be found on the number line, no matter how big or small they are, and no matter whether they are rational or irrational. And for every point on the number line, there is a real number. 1,000,000,000 –4.5 ÷2 1 2 3 5 1 6 4 2 So there’s a one-to-one correspondence between the real numbers and the points on the number line. Check it out: A “graph” is just a type of “picture” of something mathematical. For each point on the number line, the corresponding real number is called the coordinate of the point. And for each real number, the corresponding point is called the graph of the number. Don’t forget: The “sign” of a number refers to whether it is positive or negative. The numbers to the left of zero on the number line are all negative — they’re less than zero. The numbers to the right of zero are positive — they’re greater than zero. Zero is neither negative nor positive. Guided Practice 1. “Only integers can be found on the number line.” Is this statement true or false? 2. Identify the corresponding real numbers of points A–E on the number line below. A B C D E –5 –4 –3 –2 –. Draw the graph of 6 on a number line. 4. Draw the graph of –2 on a number line. 1818181818 Section 1.2 Section 1.2 Section 1.2 — The Real Number System Section 1.2 Section 1.2 tions tions or Calcula or Calcula he Number Line Can Be Useful f TTTTThe Number Line Can Be Useful f he Number Line Can Be Useful f tions or Calculations he Number Line Can Be Useful for Calcula tions or Calcula he Number Line Can Be Useful f If you think of the number line as a road, then you can think of coordinates as movement along the road — either to the left or to the right, depending on the coordinate’s sign. For example, –5 would indicate a movement of 5 units to the left, while 4 would mean 4 units to the right. Example Example Example Example Example 11111 Find 3 – 2 + 4. Solution Solution Solution Solution Solution Rewriting this as 3 + (–2) + 4, you can interpret this as: “Start at 3, move 2 to the left (to reach 1) and then 4 to the right (to reach 5).” So 3 – 2 + 4 = 5. om 0 om 0 a Number is its Distance fr a Number is its Distance fr alue of Absolute VVVVValue of alue of Absolute TTTTThe he he he he Absolute Absolute om 0 a Number is its Distance from 0 alue of a Number is its Distance fr om 0 a Number is its Distance fr alue of Absolute The opposite of a real number c (that is, –c) lies an equal distance from zero as c, but on the other side of zero. So the opposite of 4 (which lies 4 units to the right of zero) is –4 (which lies 4 units to the left of zero). And the opposite of –7 (which lies 7 units to the left of zero) is 7 (which lies 7 units to the right of zero). The distance from zero to a number is called the number’s absolute value. It doesn’t matter whether it is to the left or to the right of zero — so absolute value just means the “size” of the number, ignoring its sign. The absolute value of c is written \|c\|. More algebraically... = c ⎧ c ⎪⎪⎪⎪ ⎪⎪⎪⎪ c if ⎨ c if 0 − c ⎩ c if > 0 = 0 < 0 The absolute value of a number can never be negative. Section 1.2 Section 1.2 Section 1.2 — The Real Number System Section 1.2 Section 1.2 1919191919 Example Example Example Example Example 22222 Find: a) \|6\| b) \|0\| c) \|–23\| Solution Solution Solution Solution Solution a) 6 is positive, so \|6\| = 6. b) \|0\| = 0 (by definition) c) –23 is negative, so \|–23\| = –(–23) = 23. Guided Practice Find the following absolute values. 5. \|6\| 7. \|–3\| 9. \|0\| 11. \| 2 \| 6. \|15\| 8. \|–8\| 10. \| 1 2 \| 12. − 1 3 Independent Practice 1. Choose the correct word from each pair to complete this sentence. On a number line, positive numbers are found to the (left/right) of zero and negative numbers are found to the (left/right) of zero. 2. On a single number line, draw the graphs with the following coordinates: 5 –3.5 0.5 In exercises 3–11, find x. 3. \|3\| = x 6. \|3.14\| = x 9. \|465\| = x 4. \|–10.5\| = x 7. \|–2.17\| = x 10. \|–465\| = x 11. \|x\| = 465 (Hint: Look at exercises 9 and 10) 12. What is wrong with the equation \|x\| = –1? 5. \|–2\| = x 8. \|x\| = 0 ound Up ound Up RRRRRound Up ound Up ound Up RRRRRound Up ound Up ound Up ound Up ound Up You’ve seen the number line plenty of times in earlier grades, but it’s always useful. You don’t always need to draw it out, but you can imagine a number line to work out which direction an operation will move a number. 2020202020 Section 1.2 Section 1.2 Section 1.2 — The Real Number System Section 1.2 Section 1.2 TTTTTopicopicopicopicopic 1.2.41.2.4 1.2.41.2.4 1.2.4 California Standards: 1.0: Students identify and 1.0: Students identify and 1.0: Students identify and 1.0: Students identify and 1.0: Students identify and use the arithmetic use the arithmetic use the arithmetic use the arithmetic use the arithmetic prprprprproper ties of subsets of ties of ties of oper oper operties of ties of oper integers and rational, umbersssss,,,,, umber umber eal n eal n irrational, and rrrrreal n eal number umber eal n incincincincincluding c losureeeee losur losur luding c luding c luding closur losur luding c prprprprproper our basic our basic or the f or the f ties f ties f oper oper or the four basic our basic ties for the f operties f our basic or the f ties f oper tions wherherherherhereeeee tions w tions w arithmetic operaaaaations w arithmetic oper arithmetic oper arithmetic oper tions w arithmetic oper pplicabbbbblelelelele..... pplica aaaaapplica pplica pplica What it means for you: You’ll see how adding or multiplying positive and negative numbers affects the sign of the result. Key words: sum product addend Don’t forget: You don’t need to write the sign of a positive number — you can write +12 as 12. dition and Multiplicationtiontiontiontion dition and Multiplica dition and Multiplica AdAdAdAdAddition and Multiplica dition and Multiplica AdAdAdAdAddition and Multiplica dition and Multiplicationtiontiontiontion dition and Multiplica dition and Multiplica dition and Multiplica All real numbers fall into one of the following three categories: positive, zero, or negative. This is also true for the results of adding or multiplying real numbers (since R is closed under addition and multiplication). d Numbersssss hen YYYYYou ou ou ou ou AdAdAdAdAdd Number d Number d Number hen ou Get WWWWWhen hen ou Get he Sum is WWWWWhahahahahat t t t t YYYYYou Get ou Get he Sum is TTTTThe Sum is he Sum is d Number hen ou Get he Sum is When you add two real numbers a and b (find the sum a + b), the result can be positive, zero, or negative, depending on a and b. The sum of any two real numbers a and b is: 1. Positive if both addends are positive. This means that (a + b) > 0 if a > 0 and b > 0. This is the same as saying that the set of positive real numbers is closed under addition — so if you add two positive real numbers, you always get another positive real number. 2. Negative if both addends are negative. This means that (a + b) < 0 if a < 0 and b < 0. This is the same as saying that the
set of negative real numbers is closed under addition — so if you add two negative real numbers, you always get another negative real number. 3. Positive or Negative or Zero if a and b have opposite signs. In this case, the sign of (a + b) is the same as the sign of the addend with the larger absolute value. For example, (–5) + 7 = 7 + (–5) = 2 — positive, since \|7\| > \|–5\|. (The positive addend has a larger absolute value, so the sum is positive.) However, (–7) + 5 = 5 + (–7) = –2 — negative, since \|–7\| > \|5\|. (The negative addend has a larger absolute value, so the sum is negative.) Guided Practice State with a reason the sign of each sum, then find the sum. 1. 3 + 9 2. –3 + (–9) 3. –10 + (–5) 4. –3 + 9 5. 3 + (–9) 6. 10 + (–15) Section 1.2 Section 1.2 Section 1.2 — The Real Number System Section 1.2 Section 1.2 2121212121 Check it out: The numbers being multiplied together are called factors. ou Multiplyyyyy hen YYYYYou Multipl ou Multipl ou Multipl hen ou Get WWWWWhen hen ou Get oduct is WWWWWhahahahahat t t t t YYYYYou Get ou Get oduct is oduct is he Pr TTTTThe Pr he Pr he Product is ou Multipl hen ou Get oduct is he Pr In a similar way, the sign of the product of any real numbers depends on the signs of the numbers being multiplied (the factors). The rules for the sign of the product of any two real numbers are as follows: Signs of the factors: Sign of the product: + + – + – – + + – For example, 5 × 2 = 10 — positive. The set of positive real numbers is closed under multiplication. For example, –5 × –2 = 10 — positive. The set of negative real numbers is not closed under multiplication. For example, 5 × –2 = –10 — negative. So the product of any real numbers is: 1. Positive if the expression contains an even number of negative factors (or only positive factors). For example, 5 × (–2) × 2 × (–3) × (–1) × (–4) = 240 This is positive, since the expression has four negative factors — an even number. 2. Negative if the expression contains an odd number of negative factors. For example, (–2) × 3 × (–4) × (–2) = –48 This is negative, since the expression has three negative factors — an odd number. 2222222222 Section 1.2 Section 1.2 Section 1.2 — The Real Number System Section 1.2 Section 1.2 Guided Practice State with a reason the sign of each product, then find the product. 7. 2 × 6 8. 7 × (–7) 9. (–7) × (–7) 10. 9 × 10 11. (–11) × (–3) 12. (–1) × 3 × 4 Independent Practice State with a reason the signs of the following expressions. 1. (–5) + 3 2. 2 + (–8) 3. 5 × 3 4. (–10) × (–2) 5. (–2) × (–1) × (–8) × 6 × (–2) Evaluate the following numerical expressions. 6. –8 + (–2) 7. 8 + (–2) 8. 6 × (–4) 9. –8 + 2 10. –8 + 10 11. (–2) × (–5) × 4 12. (–5) + 2 + 6 13. (–3) × (–5) × (–2) State the signs of the following expressions. 14. a3 where a < 0. 15. –3a2b2 where a < 0 and b > 0. 16. a2b3c7 where a > 0, b < 0, and c < 0 ound Up ound Up RRRRRound Up ound Up ound Up RRRRRound Up ound Up ound Up ound Up ound Up You’ve done addition and multiplication a lot in previous grades — but a lot of Algebra I is about making formal rules for math methods. After a while you’ll carry out the rules of changing sign without really thinking about them — but for now you need to make sure you remember them. Section 1.2 Section 1.2 Section 1.2 — The Real Number System Section 1.2 Section 1.2 2323232323 TTTTTopicopicopicopicopic 1.2.51.2.5 1.2.51.2.5 1.2.5 California Standards: 1.0: Students identify and 1.0: Students identify and 1.0: Students identify and 1.0: Students identify and 1.0: Students identify and use the arithmetic use the arithmetic use the arithmetic use the arithmetic use the arithmetic prprprprproper ties of subsets of ties of ties of oper oper operties of ties of oper integers and rational, umbersssss,,,,, umber umber eal n eal n irrational, and rrrrreal n eal number umber eal n incincincincincluding c losureeeee losur losur luding c luding c luding closur losur luding c prprprprproper our basic our basic or the f or the f ties f ties f oper oper or the four basic our basic ties for the f operties f our basic or the f ties f oper tions wherherherherhereeeee tions w tions w arithmetic operaaaaations w arithmetic oper arithmetic oper arithmetic oper tions w arithmetic oper pplicabbbbblelelelele..... pplica aaaaapplica pplica pplica 2.0:2.0:2.0:2.0:2.0: Students under stand stand Students under Students under Students understand stand stand Students under h operaaaaations as tions as tions as h oper h oper and use suc and use suc tions as and use such oper and use suc tions as h oper and use suc taking the opposite,,,,, finding finding finding taking the opposite taking the opposite finding taking the opposite finding taking the opposite ocal ocal ecipr ecipr the r the r ocal, taking a root, eciprocal the recipr the r ocal ecipr the r and raising to a fractional power. They understand and use the rules of exponents. What it means for you: You’ll see how adding or multiplying positive and negative numbers affects the sign of the result. Key words: difference quotient reciprocal commutative Check it out: See Topic 1.2.7 for more on what “commutative” means. vision vision action and Di action and Di Subtr Subtr vision action and Division Subtraction and Di vision action and Di Subtr Subtr vision vision action and Di action and Di Subtr Subtr action and Division Subtraction and Di vision action and Di Subtr vision Subtr After addition and multiplication you can probably guess what’s next. This Topic gives more formal rules for subtraction and division that you’ve seen in earlier grades. ding the AdAdAdAdAdditiditiditiditiditivvvvve Ine Ine Ine Ine Invvvvvererererersesesesese ding the action Means AdAdAdAdAdding the ding the action Means action Means Subtr Subtr Subtraction Means ding the action Means Subtr Subtr Subtraction is the inverse operation of addition. It is defined as the addition of the opposite of a number. So to subtract a from b (that is, to find b – a), you add –a to b. For any a, b Œ R: b – a = b + (–a) Finding the difference between two numbers means you subtract them. For example, the difference between a and b would be a – b or b – a. Since subtraction is a type of addition, and R is closed under addition, R must be closed under subtraction. Subtraction is not commutative (meaning that a – b π b – a). Example Example Example Example Example 11111 Show that a – b π b – a for any a π b. Solution Solution Solution Solution Solution Write both subtractions as additions: a – b = a + (–b) b – a = b + (–a) = (–a) + b The two subtractions “a – b” and “b – a” contain different addends when they’re written as additions — so a – b π b – a. Subtracting a negative number is the same as adding a positive number (since subtracting means adding the opposite, and the opposite of a negative number is a positive number). For any a, b Œ R: a – (–b) = a + b 2424242424 Section 1.2 Section 1.2 Section 1.2 — The Real Number System Section 1.2 Section 1.2 Example Example Example Example Example 22222 a) Rewrite 11 – 7 as an addition. b) Simplify –7 – (–5). c) Find the sum –3 + (–8). Solution Solution Solution Solution Solution a) 11 – 7 = 11 + (–7) b) As the opposite of –5 is 5, subtracting –5 means adding 5. So –7 – (–5) = –7 + 5 = –2. c) –3 + (–8) = –(3 + 8) = –11 Guided Practice Rewrite the following subtractions as additions: 1. 9 – 6 2. 10 – 4 3. –4 – (–10) Evaluate the following: 4. 6 – (–5) 5. –3 + (–10) 6. –a – (–a) ocal ocal ecipr y a Ry a Recipr ecipr y a Ry a R ying b ying b vision Means Multipl DiDiDiDiDivision Means Multipl vision Means Multipl ocal eciprocal ying by a R vision Means Multiplying b ocal ecipr ying b vision Means Multipl Check it out: The reciprocal is another name for the multiplicative inverse. Division is the inverse operation of multiplication. It’s defined as multiplication by a reciprocal. So to divide b by a (that is, to find b ÷ a), you multiply b by a–1. For any a, b (a π 0) Œ R1 1 a One number divided by another is called a quotient. Division is not commutative (meaning that a ÷ b π b ÷ a). Example Example Example Example Example 33333 Show that a ÷ b π b ÷ a for any a π b, a, b π 0. Solution Solution Solution Solution Solution Write both divisions as multiplications: a ÷ b = a × (b–1) b ÷ a = b × (a–1) The two divisions “a ÷ b” and “b ÷ a” contain different factors when they’re written as multiplications — so a ÷ b π b ÷ a. Section 1.2 Section 1.2 Section 1.2 — The Real Number System Section 1.2 Section 1.2 2525252525 Dividing by a number’s reciprocal is the same as multiplying by the number itself. For any a, b (b π 0) Œ R ) And lastly, you cannot divide by 0 — since 0 has no reciprocal. Example Example Example Example Example 44444 a) Rewrite 25 ÷ 5 as a multiplication. b) Simplify 7 ÷ 5–1. Solution Solution Solution Solution Solution a) 25 ÷ 5 = 25 × 1 5 (or 25 × 0.2) b) The reciprocal of 5–1 is 5, so dividing by 5–1 means multiplying by 5. Therefore 7 ÷ 5–1 = 7 × 5 = 35. Guided Practice Rewrite these expressions as multiplications. 7. 24 ÷ 6 8. 36 ÷ 3 Evaluate the following expressions. 10. 10 ÷ 1 5 11. 16 ÷ 2–1 9. 2 ÷ 1 8 ⎛ 12. 16 ÷ 1 ⎜⎜⎜ ⎝ 8 − ⎞ 1 ⎟⎟⎟ ⎠ Independent Practice Simplify. 1. 12 – 15 4. –4 – 2 7. 15 ÷ 3–1 2. 18 + (–3) 3. 36 – (–4) 5. 20 ÷ (–2) 8. 18 ÷ ⎛ ⎜⎜⎜ ⎝ 1 6 − ⎞ 1 ⎟⎟⎟ ⎠ 6. 10 ÷ 1 2 9. Use examples to demonstrate that subtraction and division are not commutative. ound Up ound Up RRRRRound Up ound Up ound Up ound Up ound Up RRRRRound Up ound Up ound Up Check over the part on reciprocals until you’re sure you understand it. Although the notation is tricky, the actual ideas behind it should make sense if you read through it carefully. 2626262626 Section 1.2 Section 1.2 Section 1.2 — The Real Number System Section 1.2 Section 1.2 TTTTTopicopicopicopicopic 1.2.61.2.6 1.2.61.2.6 1.2.6 California Standards: 1.1: Students use 1.1: Students use 1.1: Students use 1.1: Students use 1.1: Students use prprprprproper umbers s s s s to umber ties of n n n n number umber ties of ties of o
per oper operties of umber ties of oper demonstrate whether assertions are true or false. What it means for you: You’ll see why it’s important to have a set of rules about the order in which you have to deal with operations. Key words grouping symbols parentheses brackets braces exponents tions tions Operaaaaations Oper Oper der of der of OrOrOrOrOrder of tions der of Oper tions Oper der of Operaaaaations OrOrOrOrOrder of tions tions Oper Oper der of der of der of Oper tions Oper der of tions It’s important that all mathematicians write out expressions in the same way, so that anyone can reach the same solution by following a set of rules called the “order of operations.” k Out Firststststst k Out Fir t to WWWWWororororork Out Fir k Out Fir t to ouping Symbols Show w w w w YYYYYou ou ou ou ou WWWWWhahahahahat to t to ouping Symbols Sho GrGrGrGrGrouping Symbols Sho ouping Symbols Sho k Out Fir t to ouping Symbols Sho If you wanted to write a numerical expression representing “add 4 and 3, then multiply the answer by 2,” you might be tempted to write 4 + 3 × 2. But watch out — this expression contains an addition and a multiplication, and you get different answers depending on which you do first. If you do the addition first, you get the answer 7 × 2 = 14. If you do the multiplication first, you get the answer 4 + 6 = 10. You might know the addition has to be done first, but somebody else might not. To be really clear which parts of a calculation have to be done first, you can use grouping symbols. Some common grouping symbols are: parentheses ( ), brackets [ ], and braces { }. Example Example Example Example Example 11111 Write an expression representing the phrase “add 4 and 3, then multiply the answer by 2.” Solution Solution Solution Solution Solution You need to show that the addition should be done first, so put that part inside grouping symbols: The expression should be (4 + 3) × 2. Guided Practice Write numeric expressions for these phrases: 1. Divide 4 by 8 then add 3. 2. Divide 4 by the sum of 8 and 3. 3. From 20, subtract the product of 8 and 2. 4. From 20, subtract 8 and multiply by 2. Evaluate the following sums and differences: 5. (3 – 2) – 5 6. 3 – (2 – 5) 7. 6 – (11 + 7) 8. (7 – 8) – (–3 – 8) – 11 9. (5 – 9) – (3 – 10) – 2 10. 9 + (5 – 3) – 4 Section 1.2 Section 1.2 Section 1.2 — The Real Number System Section 1.2 Section 1.2 2727272727 Nested grouping symbols are when you have grouping symbols inside other grouping symbols. When you see nested grouping symbols, you always start from the inside and work outwards. Example Example Example Example Example 22222 Evaluate {5 – [11 – (7 – 2)]} + 34. Solution Solution Solution Solution Solution Start from the inside and work outwards: {5 – [11 – (7 – 2)]} + 34 = {5 – [11 – 5]} + 34 = {5 – 6} + 34 = –1 + 34 = 33 Guided Practice Evaluate the following: 11. 8 + [10 + (6 – 9) + 7] 12. 9 – {[(–4) + 10] + 7} 13. [(13 – 12) + 6] – (4 – 2) 14. 14 – {8 + [5 – (–2)]} – 6 15. 13 + [10 – (4 + 5)] – (11 + 8) 16. 10 + {[7 – (–2)] – (3 – 1)} + (–14) Check it out See Topic 1.3.1 for more about exponents. Check it out When you’re simplifying expressions within grouping symbols, follow steps 2–4 (see Example 3). te Firststststst te Fir t to Evaluaaluaaluaaluaaluate Fir te Fir t to Ev About WWWWWhahahahahat to Ev t to Ev About About ules ules e Other R TTTTTherherherherhere are are are are are Other R e Other R ules About e Other Rules te Fir t to Ev About ules e Other R This order of operations is used by all mathematicians, so that every mathematician in the world evaluates expressions in the same way. 1. First calculate expressions within grouping symbols — working from the innermost grouping symbols to the outermost. 2. Then calculate expressions involving exponents. 3. Next do all multiplication and division, working from left to right. Multiplication and division have equal priority, so do them in the order they appear from left to right. 4. Lastly, do any addition or subtraction, again from left to right. Addition and subtraction have the same priority, so do them in the order they appear from left to right too. 2828282828 Section 1.2 Section 1.2 Section 1.2 — The Real Number System Section 1.2 Section 1.2 Example Example Example Example Example 33333 Simplify {4(10 – 3) + 32} × 5 – 11. Solution Solution Solution Solution Solution Work through the expression bit by bit: {4(10 – 3) + 32} × 5 – 11 = {4 × 7 + 32} × 5 – 11 inner ouping symbols firststststst ouping symbols fir ouping symbols fir most grrrrrouping symbols fir most g most g inner inner innermost g ouping symbols fir most g inner Now you have to calculate everything inside the remaining grouping symbols: = {4 × 7 + 9} × 5 – 11 = {28 + 9} × 5 – 11 = 37 × 5 – 11 xponent xponent k out the e k out the e firfirfirfirfirst wst wst wst wst wororororork out the e xponent k out the exponent xponent k out the e ultiplicationtiontiontiontion ultiplica ultiplica then do the m then do the m then do the multiplica ultiplica then do the m then do the m dition dition then do the ad then do the ad dition then do the addition dition then do the ad then do the ad Now there are no grouping symbols left, so you can do the rest of the calculation: = 185 – 11 = 174 tion firststststst tion fir tion fir ultiplica ultiplica do the m do the m ultiplication fir do the multiplica tion fir ultiplica do the m do the m action action y the subtr y the subtr and finall and finall action y the subtraction and finally the subtr action y the subtr and finall and finall Guided Practice Evaluate the following: 17. 24 ÷ 8 – 2 18. 32 – 4 × 2 19. 21 – {–3[–5 × 4 + 32] – 9 × 23} + 17 ) 4 − − ( 2 3 3 ⎡ + ÷ + −( 4 10 ⎣ ⎡ 12 ⎣⎢ 3 ⎤ ) ⎦ ⎤ ⎦⎥ 3 20. 21. 8 + {10 ÷ [11 – 6] × (–4)} 22. [17 + {(–33 + 4) × 8 – 17}] ÷ 8 Independent Practice Evaluate the following. 1. 14 – [9 – 4 × 2] 2. 11 + (9 – 3) – (4 ÷ 2) 3. (–1) × (7 – 10 + 12) 4. 12 + 9 × [(1 + 2) – (6 – 14)] 5. [(11 – 8) + (7 – 2)] × 3 – 13 6. [(10 + 9 + 5) ÷ 2] – (6 – 12) Insert grouping symbols in each of the following statements so that each statement is true: 7. 12 + 4 2 × 24 – 18 ÷ 3 = 44 8. 20 + 3 2 – 14 – 12 × 6 = 517 9 = 1662 ound Up ound Up RRRRRound Up ound Up ound Up ound Up ound Up RRRRRound Up ound Up ound Up You really need to learn the order of operation rules. You’ll be using them again and again in Algebra I so you might as well make sure you remember them right now. Section 1.2 Section 1.2 Section 1.2 — The Real Number System Section 1.2 Section 1.2 2929292929 TTTTTopicopicopicopicopic 1.2.71.2.7 1.2.71.2.7 1.2.7 California Standards: 1.0: Students identify and 1.0: Students identify and 1.0: Students identify and 1.0: Students identify and 1.0: Students identify and use the arithmetic use the arithmetic use the arithmetic use the arithmetic use the arithmetic prprprprproper ties of subsets of ties of ties of oper oper operties of ties of oper integers and rational, umbersssss,,,,, umber umber eal n eal n irrational, and rrrrreal n eal number umber eal n including closure properties for the four basic arithmetic operations where applicable. What it means for you: You’ll learn about the commutative, associative, and distributive laws of the real numbers. Key words commutative associative distributive binary operation eal Numbersssss eal Number eal Number ties of R R R R Real Number ties of ties of oper oper PrPrPrPrProper operties of eal Number ties of oper ties of R R R R Real Number PrPrPrPrProper eal Numbersssss eal Number eal Number ties of ties of oper oper operties of ties of oper eal Number In this Section you’ve already seen lots of real number properties — and now it’s time for some more. These rules are really important because they tell you exactly how to deal with real numbers. tter tter t Ma der Doesn’’’’’t Ma t Ma der Doesn der Doesn ws — the Or CommCommCommCommCommutautautautautatititititivvvvve Lae Lae Lae Lae Laws — the Or ws — the Or tter t Matter ws — the Order Doesn tter t Ma der Doesn ws — the Or It doesn’t matter which order you add two numbers in — the result is the same. This is called the commutative property (or law) of addition. For any a, b Œ R: a + b = b + a The same is true of multiplication — this is the commutative property (or law) of multiplication. For any a, b Œ R: a × b = b × a For example: 2 + 3 = 3 + 2 — commutative property of addition 2 × 3 = 3 × 2 — commutative property of multiplication oup Numbers s s s s AnAnAnAnAny y y y y WWWWWaaaaayyyyy oup Number oup Number ou Can Gr ws — YYYYYou Can Gr ou Can Gr ws — Associatititititivvvvve Lae Lae Lae Lae Laws — ws — Associa Associa ou Can Group Number oup Number ou Can Gr ws — Associa Associa Addition and multiplication are binary operations — you can only add or multiply two numbers at a time. So to add three numbers, for example, you add two of them first, and then add the third to the result. However, it doesn’t matter which two you add first: For any a, b, c Œ R: (a + b) + c = a + (b + c) Again, the same is true of multiplication: For any a, b, c Œ R: (a × b) × c = a × (b × c) These are called the associative properties (laws) of addition and multiplication. For example: (2 + 3) + 4 = 2 + (3 + 4) — associative property of addition (2 × 3) × 4 = 2 × (3 × 4) — associative property of multiplication 3030303030 Section 1.2 Section 1.2 Section 1.2 — The Real Number System Section 1.2 Section 1.2 Check it out In c), it’s fine to leave the parentheses out. It doesn’t matter if you do (3x + 3y) + 3z or 3x + (3y + 3z) — the result is the same. y Out Parararararentheses entheses entheses y Out P y Out P ws — Multipl Distributiutiutiutiutivvvvve Lae Lae Lae Lae Laws — Multipl ws — Multipl Distrib Distrib entheses ws — Multiply Out P entheses y Out P ws — Multipl Distrib Distrib The distributive law of multiplication over addition defines how multiplication and addition combine. When you multiply by a sum in parentheses, it’s the distributive law that you’re using. For any a, b, c Œ R: a × (b + c) = (a × b)
+ (a × c) So a factor outside parentheses multiplies every term inside. Example Example Example Example Example 11111 Expand: a) 3(x + y) b) x(3 + y) c) 3[(x + y) + z] Solution Solution Solution Solution Solution a) 3(x + y) = 3x + 3y b) x(3 + y) = x·3 + xy = 3x + xy (using the commutative law of multiplication) c) 3[(x + y) + z] = 3(x + y) + 3z = (3x + 3y) + 3z = 3x + 3y + 3z Independent Practice Identify the property that makes each of the following statements true. 1. 8 × 3 = 3 × 8 2. 2(x + 3) = 2x + 6 3. 9 + 2 = 2 + 9 4. (9 + 2) + 1 = 9 + (2 + 1) 5. 4(p + 1) = 4p + 4 6. (9 × 4) × 5 = 9 × (4 × 5) Expand: 7. a(b + c) 8. 6(t + 5) 9. 12t(s – r) ound Up ound Up RRRRRound Up ound Up ound Up RRRRRound Up ound Up ound Up ound Up ound Up The thing with these laws is that you’ve been using them for years without thinking about them, so it might seem strange to be taught them now. But when you rearrange equations, you’re using these laws. In Algebra I you can use these laws to justify anything you do to an equation. Section 1.2 Section 1.2 Section 1.2 — The Real Number System Section 1.2 Section 1.2 3131313131 TTTTTopicopicopicopicopic 1.2.81.2.8 1.2.81.2.8 1.2.8 California Standards: 2.0:2.0:2.0:2.0:2.0: Students under stand stand Students under Students under Students understand stand stand Students under tions as tions as h operaaaaations as h oper h oper and use suc and use suc tions as and use such oper and use suc tions as h oper and use suc taking the opposite,,,,, finding taking the opposite taking the opposite taking the opposite taking the opposite the reciprocal, taking a root, and raising to a fractional power. They understand and use the rules of exponents. What it means for you: You’ll learn about multiplicative opposites, and numbers that don’t change a value when you multiply by them. Key words inverse identity reciprocal Check it out See Topic 1.2.5 for more on additive inverses. Check it out A minus sign outside parentheses changes the sign of everything inside the parentheses. e on Multiplicationtiontiontiontion e on Multiplica e on Multiplica MorMorMorMorMore on Multiplica e on Multiplica MorMorMorMorMore on Multiplica e on Multiplicationtiontiontiontion e on Multiplica e on Multiplica e on Multiplica You already dealt with changing signs during multiplication in Topic 1.2.4. This Topic is all about opposites in multiplication, and numbers that don’t change a value if you multiply by them. y –1 to Find the AdAdAdAdAdditiditiditiditiditivvvvve Ine Ine Ine Ine Invvvvverererererse or Opposite se or Opposite se or Opposite y –1 to Find the Multiply by by by by by –1 to Find the y –1 to Find the Multipl Multipl se or Opposite se or Opposite y –1 to Find the Multipl Multipl Every real number has an additive inverse — a number that will give zero when added to that number. So for every number m, there is an opposite number (or negative), –m. In particular, the number 1 has the additive inverse –1. And if you multiply any number by –1, you get that number’s inverse. For any m Œ R: –m = –1 × m = m × –1 So the opposite of 4 is –1 × 4 = –4. And the opposite of x is –1 × x = –x. Example Example Example Example Example 11111 Find the opposite of (2a + 3). Solution Solution Solution Solution Solution The question is asking you to find –(2a + 3). Find the opposite by multiplying by –1: –(2a + 3) = –1 × (2a + 3) = [(–1) × (2a)] + [(–1) × (3)] = [(–1 × 2) × a)] + (–3) = –2a + (–3) = –2a – 3 Guided Practice Find the opposite of each of these expressions. 1. t 2. 3 3. t + 3 4. a – 2 5. –x + 1 6. –y – 4 Anything Multiplied by 0 is 0 Anything Multiplied by 0 is 0 Anything Multiplied by 0 is 0 Anything Multiplied by 0 is 0 Anything Multiplied by 0 is 0 Zero is a special number — it’s the additive identity. This means that for any number m. But zero has another useful property as well... 3232323232 Section 1.2 Section 1.2 Section 1.2 — The Real Number System Section 1.2 Section 1.2 The product of any real number and 0 equals 0. For any m Œ R Check it out See Topic 1.2.5 for the definition of division. One result of this is that zero doesn’t have a reciprocal, since there is no number that you can possibly multiply zero by in order to get 1 — the multiplicative identity. What this means in practice is that you can’t divide by zero. Check it out See Topic 7.1.1 for more examples of solving equations using this rule. s and Get Zerooooo s and Get Zer s and Get Zer o Number o Nonzererererero Number o Number o Nonz t Multiply y y y y TTTTTwwwwwo Nonz o Nonz t Multipl ou Can’’’’’t Multipl t Multipl ou Can YYYYYou Can ou Can o Numbers and Get Zer s and Get Zer o Number o Nonz t Multipl ou Can If nonzero real numbers are multiplied together, the result is never zero. The product of two (or more) nonzero real numbers cannot be zero. If x, y Œ R, and xy = 0, then either x = 0 or y = 0 (or both x = 0 and y = 0). This has practical uses when it comes to solving equations. Example Example Example Example Example 22222 Find two possible solutions to the equation x(x – 1) = 0. Solution Solution Solution Solution Solution There are two expressions multiplied together to give zero. Either one or the other must equal 0, so x = 0 or x – 1 = 0 — that is, x = 0 or x = 1. (You need to write “or,” since x can’t be both 0 and 1 at the same time.) Independent Practice Find the opposite. 1. –4 2. –a 3. g + 5 Solve each equation. 6. x + 1 = 0 7. y – 4 = 0 4. t – 6 5. –b + 8 8. y(y + 2) = 0 9. t(t – 3) = 0 ound Up ound Up RRRRRound Up ound Up ound Up ound Up ound Up RRRRRound Up ound Up ound Up These rules might just sound like common sense — but it’s important to write statements in formal math-speak to prove that they’re true. You’ll use these rules throughout Algebra I. Section 1.2 Section 1.2 Section 1.2 — The Real Number System Section 1.2 Section 1.2 3333333333 TTTTTopicopicopicopicopic 1.2.91.2.9 1.2.91.2.9 1.2.9 California Standards: 1.1: Students use 1.1: Students use 1.1: Students use 1.1: Students use 1.1: Students use ties of n n n n number prprprprproper umbers tos tos tos tos to umber umber ties of ties of oper oper operties of umber ties of oper hether hether demonstraaaaate wte wte wte wte whether demonstr demonstr hether demonstr hether demonstr ue or falsealsealsealsealse..... ue or f tions are tre tre tre tre true or f ue or f tions ar tions ar asser asser assertions ar ue or f tions ar asser asser What it means for you: You’ll see that the properties of real numbers that you learned in this section are useful to prove whether math statements are true. Key words theorem axiom postulate closure identity inverse commutative associative distributive Axioms Axioms eal Number eal Number TTTTThe Rhe Rhe Rhe Rhe Real Number Axioms eal Number Axioms Axioms eal Number TTTTThe Rhe Rhe Rhe Rhe Real Number Axioms Axioms eal Number eal Number eal Number Axioms Axioms eal Number Axioms Most of what has been covered in this Section has been about the axioms (or postulates) of the real number system. This Topic gives a summary of the axioms, and shows how you can use them. Assumptions Assumptions e Fundamental e Fundamental Axioms ar Axioms ar Assumptions e Fundamental Assumptions Axioms are Fundamental Assumptions e Fundamental Axioms ar Axioms ar A theorem is a statement that can be proved. An axiom, on the other hand, is a fundamental assumption — a statement that is accepted as true without having to be proved. You can use these axioms to justify solution steps when simplifying mathematical expressions, proving theorems, solving equations, and supporting mathematical arguments. You have to know all the axioms in this section, along with their names. Here’s a summary of them all together: Property Name Addition Multiplication Closure Property: a + b is a real number is a real number 3 × 5 = 15 Œ R Identity Property Inverse Property: a + (–a) = 0 = –a + a 3 + (–3) = 0 = (–3) + 3 a × a–1 = 1 = a–1 × a 3 × 3–1 = 1 = 3–1 × 3 Commutative Property Associative Property: (a + b) + c = a + (b + c) (3 + 5) + 6 = 3 + (5 + 6) (ab)c = a(bc) (3 × 5) × 6 = 3 × (5 × 6) Distributive Property of Multiplication over Addition: a(b + c) = ab + ac 3(5 + 6) = 3 × 5 + 3 × 6 and (b + c)a = ba + ca (5 + 6)3 = 5 × 3 + 6 × 3 3434343434 Section 1.2 Section 1.2 Section 1.2 — The Real Number System Section 1.2 Section 1.2 Check it out Any of the properties covered in the section as well as the axioms on the previous page can be used to justify steps. ps in Proofsoofsoofsoofsoofs ps in Pr ps in Pr ustify Ste ustify Ste Axioms to J Axioms to J Use the Use the ustify Steps in Pr Axioms to Justify Ste Use the Axioms to J ps in Pr ustify Ste Axioms to J Use the Use the These axioms can be used to justify steps in a mathematical proof. Example Example Example Example Example 11111 Show that (x + y) – x = y. Justify your steps. Solution Solution Solution Solution Solution (x + y) – x = (y + x) + (–x) dition, dition, ty of ad ad ad ad addition, ty of ty of oper oper CommCommCommCommCommutautautautautatititititivvvvve pre pre pre pre proper dition, operty of dition, ty of oper action action subtr subtr and the definition of and the definition of action subtraction and the definition of subtr action subtr and the definition of and the definition of = y + [x + (–x)] dition dition ty of ad ad ad ad addition ty of ty of oper oper e pre proper e pre pr Associatititititivvvvve pr Associa Associa dition operty of dition ty of oper Associa Associa = y + 0 = y an additiditiditiditiditivvvvve ine ine ine ine invvvvvererererersesesesese an ad an ad Definition of Definition of Definition of an ad an ad Definition of Definition of dition dition ty of ad ad ad ad addition ty of ty of oper oper Identity pr Identity pr dition operty of Identity proper dition ty of oper Identity pr Identity pr Example Example Example Example Example 22222 Check it out You don’t usually need to write down how to justify each step — but you should know how to. Show that (x + y)(x – y) = x2 – y2. Justify your steps.
Solution Solution Solution Solution Solution (x + y)(x – y) = (x + y)(x + (–y)) Definition of Definition of subtr subtr action action Definition of subtr subtraction action Definition of Definition of subtr action = (x + y)x + (x + y)(–y) Distrib Distrib Distributiutiutiutiutivvvvve lae lae lae lae lawwwww Distrib Distrib = x2 + yx + x(–y) + y(–y) Distrib Distrib Distributiutiutiutiutivvvvve lae lae lae lae lawwwww Distrib Distrib = x2 + xy + x(–1·y) + y(–1·y) CommCommCommCommCommutautautautautatititititivvvvve lae lae lae lae law ofw ofw ofw ofw of ×, ×, ×, ×, ×, and m and mand m ultiplica ultiplica tion pr tion pr oper oper opertytytytyty and multiplica ultiplication pr tion proper and m ultiplica tion pr oper of –1 of –1 of –1 of –1 of –1 = x2 + xy + [x(–1)]y + [y(–1)]y Associa Associa Associa Associatititititivvvvve lae lae lae lae law ofw ofw ofw ofw of × × × × × Associa = x2 + xy + (–xy) + (–1)(y2) CommCommCommCommCommutautautautautatititititivvvvve lae lae lae lae law ofw ofw ofw ofw of ×, ×, ×, ×, ×, opertytytytyty oper oper tion pr tion pr ultiplica ultiplica and m and mand m tion proper ultiplication pr and multiplica oper tion pr ultiplica and m of –1 of –1 of –1 of –1 of –1 = x2 + 0 + (–y2) ty of + + + + + ty of ty of oper oper se pr se pr InInInInInvvvvverererererse pr operty of se proper ty of oper se pr = x2 – y2 and and +, +, ty of ty of oper oper Identity pr Identity pr and +, and ty of +, operty of Identity proper and +, ty of oper Identity pr Identity pr action action subtr subtr definition of definition of action subtraction definition of subtr action subtr definition of definition of Section 1.2 Section 1.2 Section 1.2 — The Real Number System Section 1.2 Section 1.2 3535353535 Independent Practice State the real number property that justifies each statement below: 1. 3 + 5 is a real number. 2. 7 × 2 is a real number. 3. m + c = c + m for any real numbers m and c. 4. mc = cm for any real numbers m and c. 5. 12 × (7 × 4) = (12 × 7) × 4 6. 5(m – v) = 5m – 5v 7. 8 + (7 + 4) = (8 + 7) + 4 8. m–1 × m = m × m–1 = 1 9 10. 10 × 1 = 1 × 10 = 10 11. The following is a proof showing that 0c = 0, for any real c. Fill in the missing properties to support each step in the proof. 0c = 0c + 0 = 0c + [c + (–c)] = [0c + c] + (–c) = [0c + 1c] + (–c) = [0 + 1]c + (–c) = 1c + (–c) = c + (–c) = 0 Inverse property of + Identity property of × Identity property of × 12. Given real numbers m, c, and v, m(c – v) = mc – mv. Fill in the missing properties to support each step in the proof. m(c – v) = m[c + (–v)] = mc + m(–v) = mc + m(–1 × v) = mc + [m·(–1)] × v = mc + [(–1)·m] × v = mc + (–1)(mv) = mc + (–mv) = mc – mv Distributive property of × over + Associative property of × Associative property of × Definition of subtraction ound Up ound Up RRRRRound Up ound Up ound Up RRRRRound Up ound Up ound Up ound Up ound Up This Section is full of rules and properties. These aren’t just useless lists of abstract rules, though — you’ll be using them throughout Algebra I. You’ve been using most of these rules in previous grades without realizing it — but now you know all the proper names for them too. You’ll sometimes have to state which rules you’re using when you’re doing math problems. 3636363636 Section 1.2 Section 1.2 Section 1.2 — The Real Number System Section 1.2 Section 1.2 TTTTTopicopicopicopicopic 1.3.11.3.1 1.3.11.3.1 1.3.1 Section 1.3 Exponent Lawswswswsws Exponent La Exponent La Exponent La Exponent La Exponent Lawswswswsws Exponent La Exponent La Exponent La Exponent La California Standards: 2.0:2.0:2.0:2.0:2.0: Students under stand stand Students under Students under Students understand stand stand Students under tions as tions as h operaaaaations as h oper h oper and use suc and use suc tions as and use such oper tions as and use suc h oper and use suc finding finding taking the opposite, finding finding finding taking a root,oot,oot,oot,oot, taking a r taking a r ocal, ocal, ecipr ecipr the r the r ocal, taking a r eciprocal, the recipr ecipr the r taking a r ocal, the r actional actional aising to a fr aising to a fr and r and r actional aising to a fractional and raising to a fr actional aising to a fr and r and r popopopopowwwwwererererer..... TTTTThehehehehey under stand stand y under y under stand y understand stand y under and use the rules of and use the rules of and use the rules of and use the rules of and use the rules of xponents..... eeeeexponents xponents xponents xponents What it means for you: You’ll learn about the rules of exponents. Key words: exponent base power product quotient Exponents have a whole set of rules to make sure that all mathematicians deal with them in the same way. There are a lot of rules written out in this Topic, so take care. tions tions ted Multiplica s are Re Re Re Re Reeeeepeapeapeapeapeated Multiplica ted Multiplica s ars ar s ar PPPPPooooowwwwwererererers ar tions ted Multiplications tions ted Multiplica A power is a multiplication in which all the factors are the same. For example, m2 = m × m and m3 = m × m × m are both powers of m. In this kind of expression, “m” is called the base and the “2” or “3” is called the exponent. Example Example Example Example Example 11111 a) Find the volume of the cube shown. Write your answer as a power of e. b) If the edges of the cube are 4 cm long, what is the volume? e e e Solution Solution Solution Solution Solution a) V = e × e × e = e3 b) V = e3 = (4 cm)3 = 43 cm3 = 64 cm3 Guided Practice Expand each expression and evaluate. 1. 23 2. 32 3. 52 × 32 4. 24y3 5. Find the area, A, of the square shown. Write your answer as a power of s. 6. If the sides of the square are 7 inches long, what is the area? s s 7. Find the volume of a cube if the edges are 2 feet long. (Volume V = e3, where e is the edge length.) Section 1.3 Section 1.3 Section 1.3 — Exponents, Roots, and Fractions Section 1.3 Section 1.3 3737373737 Exponents Exponents ules of e Lots of R R R R Rules of ules of e Lots of TTTTTherherherherhere are are are are are Lots of e Lots of Exponents ules of Exponents Exponents ules of e Lots of 1) If you multiply m2 by m3, you get m5, since: m2 × m3 = (m × m) × (m × m × m = m5 The exponent of the product is the same as the exponents of the factors added together. This result always holds — to multiply powers with the same base, you simply add the exponents. ma × mb = ma+b 2) In a similar way, to divide powers, you subtract the exponents. ma ÷ mb = ma–b 3) When you raise a power to a power, you multiply the exponents — for example, (m3)2 = m3 × m3 = m6. 4) Raising a product or quotient to a power is the same as raising each of its elements to that power. For example: (mb)3 = mb × mb × mb = (m × b) × (m × b) × (m × b = m3b3. (ma)b = mab (mb)a = maba ⎛ ⎜⎜⎜ ⎝ m b ⎞ a ⎟⎟⎟ = ⎠ a m a b 5) Using rule 1 above: ma × m0 = ma + 0 = ma. So m0 equals 1. m0 = 1 6) It’s also possible to make sense of a negative exponent: ma × m–a = ma–a = m0 = 1 (using rules 1 and 5 above) So the reciprocal of ma is m–a. − 1 a )m ( −= m a = 1 a m 7) And taking a root can be written using a fractional power. 1 n= a n a These rules always work, unless the base is 0. The exponents and the bases can be positive, negative, whole numbers, or fractions. The only exception is you cannot raise zero to a negative exponent — zero does not have a reciprocal. 3838383838 Section 1.3 Section 1.3 Section 1.3 — Exponents, Roots, and Fractions Section 1.3 Section 1.3 Independent Practice In exercises 1–6, write each expression using exponents. 1. 2 × 2 × 2 × 2 2. a × a × a × 4 3. a × b × a × b 6. Show that 6 4 = k2. k k Simplify the expressions in exercises 8–25 using rules of exponents. 8. 170 11. 6 3 4 3 14. ( 2 2 )3 3 3 9. 2–3 12. (23)2 · 22 10. 22 · 23 13. ⋅ 4 3 2 3 7 3 15. (x4 ÷ x2) · x3 16. (x2)3 ÷ x4 17. ⋅ 3 x ( ax 5 x 2 ) 18. ( − 3 3 )x − 4 x ⋅ 5 x 19. (2x–2)3 · 4x2 20. 3x0y–2 21. (3x)0xy–2 22. 5x–1 × 6(xy)0 23. y ( 2x )4 x 2 24. y ( )2 3 2 x y− 2 ( 25. 32 26. An average baseball has a radius, r, of 1.45 inches. V Find the volume, V, of a baseball in cubic inches. ( = 4 3 = 1 2 , 27. The kinetic energy of a ball (in joules) is given by E 2 where m is the ball's mass (in kilograms) and v is its velocity (in meters per second). If a ball weighs 1 kilogram and is traveling at 10 meters per second, what is its kinetic energy in joules? 3π r mv ) at= 1 2 2 , where a is its acceleration (in meters per second 28. The speed of a ball (in meters per second) accelerating from rest is given by v squared) and t is its time of flight (in seconds). Calculate the speed of a ball in meters per second after 5 seconds of flight if it is accelerating at 5 meters per second squared. ound Up ound Up RRRRRound Up ound Up ound Up RRRRRound Up ound Up ound Up ound Up ound Up That’s a lot of rules, but don’t worry — you’ll get plenty of practice using them later in the program. Exponents often turn up when you’re dealing with area and volume. The next Topic will deal just with square roots, which is a special case of Rule 7 from the previous page. Section 1.3 Section 1.3 Section 1.3 — Exponents, Roots, and Fractions Section 1.3 Section 1.3 3939393939 TTTTTopicopicopicopicopic 1.3.21.3.2 1.3.21.3.2 1.3.2 California Standards: 2.0:2.0:2.0:2.0:2.0: Students under stand stand Students under Students under Students understand stand stand Students under tions as tions as h operaaaaations as h oper h oper and use suc and use suc tions as and use such oper and use suc tions as h oper and use suc taking the opposite, finding taking a root,oot,oot,oot,oot, taking a r taking a r the reciprocal, taking a r taking a r actional actional aising to a fr aising to a fr and r and r actional aising to a fractional and raising to a fr actional aising to a fr and r and r stand stand popopopopowwwwwererererer..... TTTTThehehehehey under y under y under stand y understand stand y under and use the rules of and use the rules of and use the rules of and use the rules of and use the rules of xponents..... eeeeexponents
xponents xponents xponents What it means for you: You’ll look more closely at the rules of square roots. Key words: square root radical radicand principal square root minor square root Check it out: Technically, the square root of a number p can be written as p2 , but you don’t usually write the 2. Square Re Re Re Re Rootsootsootsootsoots Squar Squar Squar Squar Square Re Re Re Re Rootsootsootsootsoots Squar Squar Squar Squar In the last Topic you learned about all the exponent rules — this Topic will look more closely at one rule in particular. Square roots are the type of root that you’ll come across most often in math problems — so it’s really important that you know how to deal with them. adical Sign adical Sign oot Sign is the R oot Sign is the R or the R or the R Another Name f Another Name f adical Sign oot Sign is the Radical Sign or the Root Sign is the R Another Name for the R adical Sign oot Sign is the R or the R Another Name f Another Name f The square root of p is written p . If you multiply p by itself, you get p — so p × p = p. Multiplying p by itself means you square it. The nth root of p is written pn . If you raise pn to the power n, you get p — so ( pn )n = p. The symbol is called the radical sign and shows the nonnegative root if more than one root exists. In the expression pn p is called the radicand. (the nth root of p), The square root of a number p is also written p 2 . 1 You can show this using the rules of exponents For any real number p > 0, the square root is written as p (or p p 1 2 ). If r = p , then r2 = p and (–r)2 = p. r is called the principal square root of p and –r is called the minor square root of p. Guided Practice Complete the following. 1. The radicand of 83 is ......... 2. The 6th root of t is written ........ in radical notation. 3. 9 × ........ = 9 4. b 2 = ........ in radical notation. 1 4040404040 Section 1.3 Section 1.3 Section 1.3 — Exponents, Roots, and Fractions Section 1.3 Section 1.3 o Square Re Re Re Re Rootsootsootsootsoots o Squar s Havvvvve e e e e TTTTTwwwwwo Squar o Squar s Ha s Ha e Number ositivvvvve Number e Number ositiositi ositi PPPPPositi e Numbers Ha o Squar s Ha e Number In practice, this means that every positive number has two square roots — a positive one (the principal square root) and a negative one (the minor square root). Example Example Example Example Example 11111 Find the square roots of the following numbers: a) 100 b) n2 Solution Solution Solution Solution Solution a) 100 = 10, so the principal square root is 10, and the minor square root is –10. b) n 2 = , so the principal square root is n , and the minor square n root is – n . The principal square root of n is written as n . The minor square root of n is written as – n . To indicate both square roots you can write ± n . Guided Practice Find the principal square root and minor square root of these numbers: 5. 4 6. 100 7. 81 Use the “±” symbol to give the principal and minor square root of the following numbers: 8. 9 11. 352 14. t2 9. 16 12. x2 10. 144 13. 81 15. 9 × 9 16. (st)2 Evaluate the following, giving the principal and minor roots: 1 2 17. 4 1 18. 121 2 Section 1.3 Section 1.3 Section 1.3 — Exponents, Roots, and Fractions Section 1.3 Section 1.3 4141414141 e Square Re Re Re Re Rootsootsootsootsoots e Squar Also Havvvvve Squar e Squar Also Ha Also Ha essions essions aic Expr AlgAlgAlgAlgAlgeeeeebrbrbrbrbraic Expr aic Expr essions Also Ha aic Expressions e Squar Also Ha essions aic Expr You can also take the square root of an algebraic expression. Example Example Example Example Example 22222 Find the square root of ( x +1 2 . ) Solution Solution Solution Solution Solution x +1 2 = \|x + 1\|, so the principal square root is x + 1 and the minor ) ( square root is – x + 1 . Guided Practice Give the principal and minor square root of each of the following expressions. 19. t t× 20. t 2 2× t 21. a 2× a2 22 ) 23 ) 24. ( a b+ 2 ) 25. ( t +1 2 ) t a 26. [ ( b+ 2 )] 2 27. [ ( a b+ 2 )] Independent Practice 1. Is this statement true or false? “The radicand of 325 is 5.” Evaluate the following. 2. 64 5. 25 1 2 3. ( )49 6. 122 4. a2 7. j× j Find the square roots of the following. ( 8. a2 ) 2 11. ( m 2 2 2+ n ) 9. ( k −1 2 ) ) ( 12. 2 pq 2 10. ( ⎡ +( 13. a ⎣ b m n+ 2 ) )× +( c d 2 ⎤ ) ⎦ ound Up ound Up RRRRRound Up ound Up ound Up ound Up ound Up RRRRRound Up ound Up ound Up Remember that when you take the square root of a positive number, you always have two possible answers — a positive one and a negative one. You can give both answers neatly using the ± sign. 4242424242 Section 1.3 Section 1.3 Section 1.3 — Exponents, Roots, and Fractions Section 1.3 Section 1.3 TTTTTopicopicopicopicopic 1.3.31.3.3 1.3.31.3.3 1.3.3 California Standards: 2.0:2.0:2.0:2.0:2.0: Students under stand stand Students under Students under Students understand stand stand Students under tions as tions as h operaaaaations as h oper h oper and use suc and use suc tions as and use such oper and use suc tions as h oper and use suc taking the opposite, finding taking a root,oot,oot,oot,oot, taking a r taking a r the reciprocal, taking a r taking a r actional actional aising to a fr aising to a fr and r and r actional aising to a fractional and raising to a fr actional aising to a fr and r and r stand stand popopopopowwwwwererererer..... TTTTThehehehehey under y under y under stand y understand stand y under and use the rules of and use the rules of and use the rules of and use the rules of and use the rules of xponents..... eeeeexponents xponents xponents xponents What it means for you: You’ll find out how to multiply and divide square roots. Key words: square root radical radicand numerator denominator Check it out: Remember, mc means the principal (nonnegative) square root of mc. ties of R R R R Rootsootsootsootsoots ties of ties of oper oper PrPrPrPrProper operties of ties of oper PrPrPrPrProper ties of R R R R Rootsootsootsootsoots ties of ties of oper oper operties of ties of oper In the last Topic you saw that positive numbers have two square roots. This Topic’s all about how to multiply and divide square roots, which is really important when you’re evaluating expressions involving more than one root sign. actorsssss actor actor s Can Be Split into F s Can Be Split into F Number Number s Can Be Split into Factor Numbers Can Be Split into F actor s Can Be Split into F Number Number A factor of a number is a number that divides into it without a remainder — for example, 1, 2, 5, and 10 are factors of 10. To factor a number or expression means to write it as a product of its factors — for example, 10 = 2 × 5. Factoring is a useful way of simplifying square roots — as you’ll see in the rest of this lesson. Square Re Re Re Re Rootsootsootsootsoots Squar Squar ty of ty of oper e Pre Proper oper e Pre Pr s a Multiplicatititititivvvvve Pr s a Multiplica TTTTTherherherherhere’e’e’e’e’s a Multiplica s a Multiplica ty of Squar operty of Squar ty of oper s a Multiplica mc = ⋅ m c This means that to make finding a square root easier, you can try to factor the radicand first. Example Example Example Example Example 11111 Find the following: a) 400 b) 8 c ) Solution Solution Solution Solution Solution a) 400 = × 4 100 b) 8 = × 4 2 = × 4 100 × 2 10 2 = 2 = × = 20 2 10 = × 4 2 2 = × 2 = 2 2 The same technique can work if you have an algebraic expression. You need to find factors that are squares of other expressions: c) +( x 2 ) 1 −( 3 x 2 ) = 1 +( x 2 ) 1 −( 3 x 2 ) = ( 1 x )( + 1 3 x ) 1– Section 1.3 Section 1.3 Section 1.3 — Exponents, Roots, and Fractions Section 1.3 Section 1.3 4343434343 Guided Practice Simplify the following square roots. 1. 900 2. 225 3. 20 4. 200 5. 32 6. 4 2x 7. 81 2t 8. 4 t +( 2 ) 2 9. 36 ( j − 3 2 ) 10. 64 ( k + 4 2 ) oots TTTTToooooooooo oots Square Re Re Re Re Roots oots Squar Squar ty of ty of oper oper vision Pr vision Pr s a Di TTTTTherherherherhere’e’e’e’e’s a Di s a Di ty of Squar operty of vision Proper s a Division Pr oots Squar ty of oper vision Pr s a Di = m c m c Again, the idea is to look for any factors in the numerator or denominator that are squares. Example Example Example Example Example 22222 Find the following: b) 3 16 c) 49 225 Solution Solution Solution Solution Solution a) 49 225 = = 49 225 2 7 2 15 = 7 15 b) = 3 16 = 3 16 3 42 = 3 4 Check it out: In Example 1 c), the numerator and the denominator are both squares. c) 2 +( 2 x −( 3 x ) 1 ) 1 2 = +( 2 x 2 ) 1 −( 4444444444 Section 1.3 Section 1.3 Section 1.3 — Exponents, Roots, and Fractions Section 1.3 Section 1.3 Guided Practice Find the following. 11. 12. 13. 14. 15. 16 4 25 9 125 16 50 4 x 2 36 Independent Practice Find the following. 1. 49 4× 2. 25 2m 3. 4. 5. 64 9 121 144 t 2 81 6. 48 2t 7. 72 2x 8. 8 2m m, ≠ 0 16. 200 2x , x π 0 17. 242 2a a, ≠ 0 18 19. 27 2y 10. 11. 12. 13. 14. 15. 300 2t , t π 0 2 x y2 81 36 2y 49 2 100 t 2 a 16 a, ≠ 0 2 x 2 16 y y, ≠ 0 ound Up ound Up RRRRRound Up ound Up ound Up ound Up ound Up RRRRRound Up ound Up ound Up Root signs are really tricky, so it’s a good idea to get rid of them whenever you can. When you multiply or divide roots, you’re left with simpler expressions, which makes math problems a lot easier. Section 1.3 Section 1.3 Section 1.3 — Exponents, Roots, and Fractions Section 1.3 Section 1.3 4545454545 TTTTTopicopicopicopicopic 1.3.41.3.4 1.3.41.3.4 1.3.4 California Standards: 2.0:2.0:2.0:2.0:2.0: Students under stand stand Students under Students under Students understand stand stand Students under tions as tions as h operaaaaations as h oper h oper and use suc and use suc tions as and use such oper tions as and use suc h oper and use suc finding finding taking the opposite, finding finding finding ocal, taking a root, ocal ocal ecipr ecipr the r the r eciprocal the recipr ocal ecipr the r the r and raising to a fractional power. They understand and use the rules of exponents. What it means for you: You’ll simplify fractions to their lowest terms. Key words: equivalent numerator d
enominator greatest common factor simplify actions actions alent Frrrrractions alent F alent F EquiEquiEquiEquiEquivvvvvalent F actions actions alent F alent Frrrrractions EquiEquiEquiEquiEquivvvvvalent F actions actions alent F alent F actions alent F actions Here’s another Algebra I Topic that you’ve seen in earlier grades. You’ve used fractions a lot before, but in Algebra I you’ll treat them more formally. This Topic goes over stuff on simplifying fractions that should feel quite familiar to you. tor and a Denominatortortortortor tor and a Denomina e a Numeraaaaator and a Denomina tor and a Denomina e a Numer actions Havvvvve a Numer e a Numer actions Ha FFFFFrrrrractions Ha actions Ha tor and a Denomina e a Numer actions Ha A fraction is any number expressed as one integer divided by another integer. Fractions are written in the form p q , where p and q are integers, and q π 0. The top number (p) is called the numerator and the bottom number (q) is the denominator. A fraction with a denominator of zero is undefined, because you can’t divide by zero. e the Same VVVVValuealuealuealuealue e the Same actions Havvvvve the Same e the Same actions Ha alent Frrrrractions Ha actions Ha alent F EquiEquiEquiEquiEquivvvvvalent F alent F e the Same actions Ha alent F Equivalent fractions are fractions of the same value — for example, 2 3 and 4 6 , 1 2 and . 4 8 To simplify 4 6 to 2 3 , you can rewrite the numerator and denominator as products of factors (factor them). You can then cancel any common factors by dividing both the numerator and denominator by those factors to produce an equivalent fraction. Example Example Example Example Example 11111 Convert 4 6 to 2 3 . Solution Solution Solution Solution Solution First factor the numerator and denominator . You can cancel factors that are common to both the numerator and denominator, so in this case, parts of the fractions cancel, leaving 4646464646 Section 1.3 Section 1.3 Section 1.3 — Exponents, Roots, and Fractions Section 1.3 Section 1.3 Guided Practice 1. Identify the numerator and the denominator in the fraction 9 10 . In Exercises 2-4, convert the fractions to 2. 12 16 3. 9 12 3 4 . In exercises 5-7, convert the fractions to 5 8 . 5. 25 40 6. 15 24 4. 21 28 7. 40 64 8. Show that 3 5 and 9 15 are equivalent fractions. 9. Show that 5 10 and 3 6 are equivalent fractions. actions TTTTToooooooooo actions alent Frrrrractions actions alent F te Equivvvvvalent F alent F te Equi ou Can Creaeaeaeaeate Equi te Equi ou Can Cr YYYYYou Can Cr ou Can Cr actions alent F te Equi ou Can Cr The greatest common factor of two numbers a and b is the largest possible number that will divide exactly into both a and b. To simplify a fraction (or reduce it to its lowest terms) is to convert it to an equivalent fraction in which the numerator and denominator have a greatest common factor of 1. Example Example Example Example Example 22222 Reduce 56 64 Solution Solution Solution Solution Solution = 56 64 ⋅ 7 8 ⋅ 8 8 to its lowest terms. FFFFFactor the n actor the n actor the n umer umer umeraaaaator and denomina tor and denomina tor and denomina tor and denominatortortortortor actor the numer actor the n umer tor and denomina actors frs frs frs frs fromomomomom actor actor Cancel common f Cancel common f Cancel common factor actor Cancel common f Cancel common f tor and denominatortortortortor tor and denomina tor and denomina umeraaaaator and denomina umer umer the n the n the numer tor and denomina umer the n the n The greatest common factor (GCF) of 7 and 8 is 1, so this is the simplest form of 56 64 . Similarly, you can produce an equivalent fraction by multiplying both the numerator and denominator by the same number. Section 1.3 Section 1.3 Section 1.3 — Exponents, Roots, and Fractions Section 1.3 Section 1.3 4747474747 Guided Practice Complete these statements. 10. The largest possible number that will divide exactly into two numbers a and b is called the ........................................... of a and b. 11. A fraction is expressed in its lowest terms if the numerator and denominator have a GCF of ........... Find the greatest common factor of each of these pairs. 12. 81 and 90 13. 56 and 77 14. 42 and 54 15. 13 and 19 Simplify these fractions. 12 14 4 12 16. 17. 18. 30 33 19. 9 24 Independent Practice 1. Identify the numerator in the fraction 11 13 . 2. Identify the denominator in the fraction 14 15 . In exercises 3–5, show how to simplify each fraction to 18 27 10 15 8 12 3. 4. 5. 2 3 . 6. Show that 10 16 and 5 8 are equivalent fractions. Find the greatest common factor of each of these pairs of numbers. 7. 15 and 20 8. 21 and 33 9. 26 and 39 Simplify these fractions. 12 20 11. 12 15 10. 13. 44 48 14. 81 90 12. 15. 21 39 56 77 ound Up ound Up RRRRRound Up ound Up ound Up RRRRRound Up ound Up ound Up ound Up ound Up Hopefully you recognized a lot of the stuff in this Topic from earlier grades. In the next couple of Topics you’ll go over multiplying, dividing, adding, and subtracting fractions — and you’ll need to be happy with simplifying fractions each time. 4848484848 Section 1.3 Section 1.3 Section 1.3 — Exponents, Roots, and Fractions Section 1.3 Section 1.3 TTTTTopicopicopicopicopic 1.3.51.3.5 1.3.51.3.5 1.3.5 California Standards: 2.0:2.0:2.0:2.0:2.0: Students under stand stand Students under Students under Students understand stand stand Students under tions as tions as h operaaaaations as h oper h oper and use suc and use suc tions as and use such oper tions as and use suc h oper and use suc finding finding taking the opposite, finding finding finding ocal, taking a root, ocal ocal ecipr ecipr the r the r eciprocal the recipr the r ocal ecipr the r and raising to a fractional power. They understand and use the rules of exponents. What it means for you: You’ll multiply and divide fractions, then simplify them. Key words: numerator denominator reciprocal Multiplying and Multiplying and Multiplying and Multiplying and Multiplying and Multiplying and Multiplying and Multiplying and Multiplying and Multiplying and actions actions viding Frrrrractions viding F viding F DiDiDiDiDividing F actions actions viding F viding Frrrrractions DiDiDiDiDividing F actions actions viding F viding F actions viding F actions You did lots of work on multiplying and dividing fractions in grade 7. This Topic is mainly a reminder of those techniques, because you’re going to be using them a lot in later math problems in Algebra I. ying the TTTTTop and Bottom op and Bottom op and Bottom ying the ying the y Multipl y Multipl actions b Multiply Fy Fy Fy Fy Frrrrractions b actions b Multipl Multipl op and Bottom y Multiplying the actions by Multipl op and Bottom ying the y Multipl actions b Multipl Multipl To multiply two fractions, find the product of the numerators and divide that by the product of the denominators. Given any numbers a, b, c, and d Œ R (b π 0, d π 0): ⋅ a c ⋅ b d c ⋅ = d ac bd a b = Example Example Example Example Example 11111 a) Multiply 2 5 and 7 3 . b) Multiply 3 4 and 9 2 . Solution Solution Solution Solution Solution a = 14 15 b = 27 8 Guided Practice Find each of these products. 1. ⋅ 2 3 4 5 4. 3 10 ⋅ 1 8 7. ⋅ 8 11 3 5 10. 3 8 ⋅ 15 2 2. ⋅ 3 5 1 2 5. ⋅ 7 8 3 4 8. 9 10 ⋅ 3 5 11. ⋅ 5 9 4 7 3. 6. 2 7 ⋅ 5 11 12. ⋅ 8 5 7 3 Section 1.3 Section 1.3 Section 1.3 — Exponents, Roots, and Fractions Section 1.3 Section 1.3 4949494949 Check it out: If there are two fractions multiplied together, you can cancel a factor from anywhere in the numerator with a matching factor in the denominator. They don’t have to have come from the same fraction. e Solutions in the Simplest Fororororormmmmm e Solutions in the Simplest F ys Givvvvve Solutions in the Simplest F e Solutions in the Simplest F ys Gi AlAlAlAlAlwwwwwaaaaays Gi ys Gi e Solutions in the Simplest F ys Gi You should always give your answer in its simplest form — so with more complicated examples, factor the numerators and denominators and cancel common factors. It’ll save time if you do this before you compute the products. Example Example Example Example Example 22222 Multiply and simplify ⋅ 56 64 8 28 . Solution Solution Solution Solution Solution 56 64 ⋅ ⋅ = 28 ⋅ and denominatortortortortorsssss s and denomina s and denomina umeraaaaatortortortortors and denomina umer umer actor the n actor the n F F F F Factor the n actor the numer s and denomina umer actor the n actorsssss actor actor Cancel all the common f Cancel all the common f Cancel all the common factor actor Cancel all the common f Cancel all the common f Guided Practice Multiply and simplify these expressions. 13. 16. 3 5 4 9 ⋅ 7 12 ⋅ 36 37 19. 40 15 ⋅ 24 64 14. 17. ⋅ 8 21 3 16 ⋅ 22 15 75 26 20. 81 90 ⋅ 10 90 15. 18. 14 25 36 55 ⋅ ⋅ 40 63 11 12 21. ⋅ 25 21 28 40 ocal ocal ecipr ecipr y the R y the R ying b ying b y Multipl y Multipl actions b vide Frrrrractions b actions b vide F DiDiDiDiDivide F vide F ocal eciprocal y the Recipr ying by the R y Multiplying b actions by Multipl ocal ecipr y the R ying b y Multipl actions b vide F The reciprocal of a fraction c d is d c , since c d ⋅ d c = 1. To divide by a fraction, you multiply by its reciprocal. Given any nonzero numbers a, b, c, and d Œ R: a b 5050505050 Section 1.3 Section 1.3 Section 1.3 — Exponents, Roots, and Fractions Section 1.3 Section 1. = ad bc Example Example Example Example Example 33333 Divide 72 96 by 9 144 . Solution Solution Solution Solution Solution 72 96 ÷ 9 144 72 = ⋅ 96 144 9 72 96 ⋅ 144 9 = ⋅ 8 9 ⋅ 12 8 ⋅ ⋅ 12 12 9 ocal ocal ecipr ecipr y the r y the r tion b tion b ultiplica ultiplica write as a m write as a m RRRRReeeeewrite as a m ocal eciprocal y the recipr tion by the r ultiplication b write as a multiplica ocal ecipr y the r tion b ultiplica write as a m s and denominatortortortortorsssss s and denomina s and denomina umeraaaaatortortortortors and denomina umer umer actor the n actor the n FFFFFactor the n actor the numer s and denomina umer actor the n 1 1 ⋅ 8 9 ⋅ 12 8 1 1 ⋅ 1 ⋅ 12 12 9
1 = 12 actorsssss actor actor Cancel all the common f Cancel all the common f Cancel all the common factor actor Cancel all the common f Cancel all the common f Guided Practice In Exercises 1–6, divide and simplify. 22. 4 9 ÷ 16 15 25. 28. 48 7 24 32 ÷ ÷ 16 35 9 56 23. 26. 29. 5 21 13 27 40 72 ÷ ÷ 25 14 2 63 ÷ 24 36 24. 27. 30. ÷ ÷ ÷ 34 39 15 56 15 70 42 13 3 32 36 56 Independent Practice Evaluate the following. Simplify your answer where appropriate. 1. 42 4 ⋅ 10 28 4. 24 6 ⋅ 30 8 7. 21 22 ÷ 21 14 10. ÷ 36 24 28 48 2. 18 20 ⋅ 16 24 5. 8. ÷ ÷ 30 50 21 32 45 55 24 36 11. 64 72 ⋅ 88 40 3. ⋅ 14 12 2 21 6. 9. 18 12 35 14 ÷ 8 6 ÷ 21 44 12. 12 50 ⋅ 88 44 ound Up ound Up RRRRRound Up ound Up ound Up ound Up ound Up RRRRRound Up ound Up ound Up Unless you’re told otherwise, you should always cancel fraction solutions down to the most simple form. In the next Topic you’ll look at adding and subtracting fractions, which is a bit tougher. Section 1.3 Section 1.3 Section 1.3 — Exponents, Roots, and Fractions Section 1.3 Section 1.3 5151515151 TTTTTopicopicopicopicopic 1.3.61.3.6 1.3.61.3.6 1.3.6 California Standards: 2.0:2.0:2.0:2.0:2.0: Students under stand stand Students under Students under Students understand stand stand Students under tions as tions as h operaaaaations as h oper h oper and use suc and use suc tions as and use such oper tions as and use suc h oper and use suc finding finding taking the opposite, finding finding finding ocal, taking a root, ocal ocal ecipr ecipr the r the r eciprocal the recipr the r ocal ecipr the r and raising to a fractional power. They understand and use the rules of exponents. What it means for you: You’ll add and subtract fractions with the same denominator, then fractions with different denominators. Key words: numerator denominator least common multiple prime factor ding and ding and AdAdAdAdAdding and ding and ding and ding and ding and AdAdAdAdAdding and ding and ding and actions actions acting Frrrrractions acting F acting F Subtr Subtr actions Subtracting F actions acting F Subtr Subtr acting Frrrrractions actions actions acting F acting F Subtr Subtr Subtracting F actions acting F Subtr actions Subtr You dealt with multiplying and dividing fractions in Topic 1.3.5. This Topic deals with adding and subtracting, which is a little bit harder if the denominators are different in each of the fractions. actions with the Same Denominatortortortortor actions with the Same Denomina + and – with Frrrrractions with the Same Denomina actions with the Same Denomina + and – with F + and – with F actions with the Same Denomina + and – with F + and – with F To find the sum (or the difference) of two fractions with the same denominator, just add (or subtract) the numerators, then divide by the common denominator. Example Example Example Example Example 11111 a) Calculate 1 7 5 + . 7 b) Calculate 5 7 1 − . 7 Solution Solution Solution Solution Solution a) Add the numerators and divide the answer by the common denominator, 7) Subtract the second numerator from the first and divide the answer by the common denominator, 7 Guided Practice Perform the indicated operations and simplify each expression in exercises 1–9. 1. + 5 6 7 6 4. 7. 15 32 53 32 + 21 32 − 15 32 2. 2 15 + 8 15 5. 8. 12 27 32 45 + 11 27 − 37 45 3. 13 18 + 11 18 6. − 24 49 11 49 9. − 21 16 15 16 5252525252 Section 1.3 Section 1.3 Section 1.3 — Exponents, Roots, and Fractions Section 1.3 Section 1.3 ent Denominatortortortortorsssss ent Denomina actions with Difffffferererererent Denomina ent Denomina actions with Dif ding Frrrrractions with Dif actions with Dif ding F AdAdAdAdAdding F ding F ent Denomina actions with Dif ding F You can find the sum (or difference) of two fractions with different denominators by first converting them into equivalent fractions with the same denominator. This common denominator should be the least common multiple (LCM) of the two denominators. The LCM of two numbers a and b is the smallest possible number that is divisible by both a and b. Check it out: See Topic 2.3.1 for more tions tions actoriza actoriza prime f prime f tions. about prime f actorizations prime factoriza tions actoriza prime f The prime factorizations of a and b can be used to calculate the LCM — the LCM is the product of the highest power of each prime factor that appears in either factorization. Example Example Example Example Example 22222 Calculate 2 3 7 + . 12 Solution Solution Solution Solution Solution To find the LCM of the denominators, write 3 and 12 as products of prime factors: 3 = 3, 12 = 22 × 3 So the LCM of 3 and 12 is 3 × 22 = 12 2 3 = 8 12 ConConConConConvvvvvererererert t t t t 2 3 to an equi to an equi to an equivvvvvalent fr alent fr alent fr action o action o action ovvvvver 12 er 12 er 12 alent fraction o er 12 to an equi to an equi alent fr action o er 12 8 12 7 + = 12 15 12 = 15 12 ⋅ 5 3 ⋅ 4 3 = 5 4 AdAdAdAdAdd frd frd frd frd fractions with the same denomina actions with the same denomina actions with the same denomina actions with the same denominatortortortortorsssss actions with the same denomina FFFFFactor the n actor the n actor the n umer umer umeraaaaator and denomina tor and denomina tor and denomina tor and denominatortortortortor actor the numer actor the n umer tor and denomina and cancel an and cancel an y common f y common f actor actor actorsssss and cancel any common f y common factor and cancel an and cancel an y common f actor So, to add or subtract fractions with different denominators: 1. Find the least common multiple of the denominators. 2. Convert each fraction into an equivalent fraction with the LCM as the denominator. 3. Add or subtract the fractions with the same denominators, then simplify the resultant fraction if possible. Section 1.3 Section 1.3 Section 1.3 — Exponents, Roots, and Fractions Section 1.3 Section 1.3 5353535353 Example Example Example Example Example 33333 Calculate 5 8 5 − . 12 Solution Solution Solution Solution Solution The LCM of 8 and 12 is 24. = 5 8 15 24 , 5 12 = 10 24 alent alent action into an equivvvvvalent action into an equi action into an equi t each frh frh frh frh fraction into an equi t eac t eac ConConConConConvvvvvererererert eac alent alent action into an equi t eac tor of 24 24 24 24 24 tor of tor of action with a denomina action with a denomina frfrfrfrfraction with a denomina action with a denominator of tor of action with a denomina 15 24 − = 5 10 24 24 Guided Practice actions with the same denominatortortortortorsssss actions with the same denomina actions with the same denomina act fr act fr Subtr Subtr act fractions with the same denomina Subtract fr actions with the same denomina act fr Subtr Subtr Find the least common multiple of each pair of numbers. 11. 5 and 6 10. 4 and 6 14. 21 and 49 13. 14 and 18 17. 16 and 40 16. 12 and 42 20. 24 and 32 19. 36 and 52 12. 9 and 12 15. 18 and 27 18. 15 and 36 21. 25 and 60 Work out and simplify the following. 22. 5 9 + 8 3 25. 28. 13 18 9 15 + 11 27 − 17 10 23. + 11 12 3 8 26. + 7 16 3 40 29. 9 8 − 19 20 24. 27. 30 12 Independent Practice Perform the indicated operations and simplify each expression. 1. 4. 7. 6 20 48 60 11 18 + 8 20 − + 18 60 4 12 2. 5. 8. 18 24 8 50 7 30 + + − 7 24 3 10 3 24 3. 6 16 − 4 16 6. 4 8 − 5 12 9. + 2 21 7 11 10. Steve and Joe are collecting money to donate to charity. They set a goal of collecting $100. Steve has collected $50, while Joe collected only 1/5 of the money needed to meet the goal. What fraction of their goal do they still need to collect? Simplify the following expressions as far as possible. 11. 3 7 7 + − 11 2 13 12 )( 1 + + 2 x 6 ) 1 1 ound Up ound Up RRRRRound Up ound Up ound Up ound Up ound Up RRRRRound Up ound Up ound Up Fractions will be popping up in all sorts of places throughout Algebra I, so it’s a good idea to make sure you really know the rules for them now. 5454545454 Section 1.3 Section 1.3 Section 1.3 — Exponents, Roots, and Fractions Section 1.3 Section 1.3 TTTTTopicopicopicopicopic 1.4.11.4.1 1.4.11.4.1 1.4.1 Section 1.4 tical Proofsoofsoofsoofsoofs tical Pr tical Pr thema thema MaMaMaMaMathema thematical Pr tical Pr thema MaMaMaMaMathema tical Proofsoofsoofsoofsoofs tical Pr tical Pr thema thema thematical Pr tical Pr thema California Standards: 24.2: Students identify the 24.2: Students identify the 24.2: Students identify the 24.2: Students identify the 24.2: Students identify the hhhhhypothesis and conc lusion lusion ypothesis and conc ypothesis and conc lusion ypothesis and conclusion lusion ypothesis and conc in logical deduction. in logical deduction. in logical deduction. in logical deduction. in logical deduction. 25.2: Students judge the 25.2: Students judge the 25.2: Students judge the 25.2: Students judge the 25.2: Students judge the gument gument an ar an ar alidity of alidity of vvvvvalidity of gument an argument alidity of an ar gument an ar alidity of hether the hether the ding to w ding to w accor accor ding to whether the according to w hether the ding to w accor hether the accor prprprprproper the realealealealeal the r the r ties of ties of oper oper ties of the r operties of oper the r ties of nnnnnumber system and the umber system and the umber system and the umber system and the umber system and the tions havvvvveeeee tions ha operaaaaations ha tions ha oper oper der of der of orororororder of der of oper tions ha oper der of pplied corrrrrrectlectlectlectlectly ay ay ay ay attttt pplied cor pplied cor been a been a been applied cor been a pplied cor been a h steppppp..... h ste eaceaceaceaceach ste h ste h ste What it means for you: You’ll justify each step of mathematical proofs, and you’ll learn about the hypothesis and conclusion of “if... then” statements. Key words: justify hypothesis conclusion proof Check it out: ties of ties of oper There are four prprprprproper oper ties of operties of ties of oper equality — adadadadaddition dition dition equality equality dition, dition equality equality ultiplicationtio
ntiontiontion, ultiplica ultiplica action, mmmmmultiplica action action subtr subtr subtraction action subtr ultiplica subtr and dididididivision vision vision vision. They basically vision say that you can do the same thing to both sides of an equation. For a formal definition of each, refer to Topic 2.2.1. A lot of Algebra I asks you to give formal proofs for stuff that you covered in earlier grades. You’re sometimes asked to state exactly which property you’re using for every step of a math problem. tical Proofoofoofoofoof tical Pr tical Pr thema thema a Ma h Step ofp ofp ofp ofp of a Ma a Ma h Ste h Ste ustify Eac ustify Eac ou Must J YYYYYou Must J ou Must J thematical Pr a Mathema ustify Each Ste ou Must Justify Eac tical Pr thema a Ma h Ste ustify Eac ou Must J A mathematical proof is a logical argument. When you write a mathematical proof, you have to justify each step in a logical way. In Algebra I, you do this using the axioms covered earlier in this chapter. You’ve seen lots of proofs already in this chapter — although some of them weren’t described as proofs at the time. Solving an equation to find the value of a variable is a form of mathematical proof. Look at the example below. It shows a mathematical proof written in two columns — with each step of the logical argument written on the left, and the justification for it written on the right. Example Example Example Example Example 11111 If 6x + 4 = 22, what is the value of x? Solution Solution Solution Solution Solution 6x + 4 = 22 (6x + 4) – 4 = 22 – 4 (6x + 4) + (–4) = 22 – 4 6x + (4 + (–4)) = 22 – 4 6x + (4 + (–4)) = 18 6x + 0 = 18 6x = 18 1 6 ⎛ ⎜⎜⎜ ⎝ × ( 6 )x 1 = × 6 18 × 6 ⎞ 1 ⎟⎟⎟⋅ = × ⎠ 6 x 18 1 6 ⎛ ⎜⎜⎜ ⎝ ⎛ ⎜⎜⎜ ⎝ 1 6 1 6 × 6 ⎞ ⎟⎟⎟⋅ =x ⎠ 18 6 × 6 ⎞ ⎟⎟⎟⋅ =x ⎠ 3 1·x = 3 x = 3 GiGiGiGiGivvvvven equa en equa en equa en equationtiontiontiontion en equa Subtr Subtr action pr action pr oper oper ty of ty of equality equality Subtraction pr action proper operty of ty of equality equality Subtr Subtr action pr oper ty of equality Definition of Definition of subtr subtr action action Definition of subtr subtraction action Definition of Definition of subtr action Associa Associa Associatititititivvvvve pr e pre proper e pre pr oper oper ty of ty of ty of ad ad ad ad addition dition dition operty of dition Associa Associa oper ty of dition Subtr Subtr acting acting Subtracting acting Subtr Subtr acting InInInInInvvvvverererererse pr se pr se pr oper oper ty of ty of ty of ad ad ad ad addition dition dition se proper operty of dition se pr oper ty of dition Identity pr Identity pr oper oper ty of ty of ty of ad ad ad ad addition dition dition Identity proper operty of dition Identity pr Identity pr oper ty of dition Multiplica Multiplica tion pr tion pr oper oper ty of ty of equality equality Multiplication pr tion proper operty of ty of equality equality Multiplica Multiplica tion pr oper ty of equality ultiplicationtiontiontiontion ultiplica ultiplica ty of m m m m multiplica ty of ty of oper oper e pre proper e pre pr Associatititititivvvvve pr Associa Associa operty of ultiplica ty of oper Associa Associa vision vision di di Definition of Definition of vision division Definition of di vision di Definition of Definition of viding viding DiDiDiDiDividing viding viding InInInInInvvvvverererererse pr oper oper se pr se pr ty of m m m m multiplica ty of ty of ultiplicationtiontiontiontion ultiplica ultiplica se proper operty of se pr ty of oper ultiplica Identity pr Identity pr oper oper ty of ty of ty of m m m m multiplica ultiplica ultiplica ultiplicationtiontiontiontion Identity proper operty of Identity pr Identity pr oper ty of ultiplica Section 1.4 Section 1.4 Section 1.4 — Mathematical Logic Section 1.4 Section 1.4 5555555555 Guided Practice Complete these statements: 1. A mathematical proof is called a ................. each step in a logical way using mathematical ................. 2. Mathematical proofs can be written in two columns, with the ..................... on the left and the ..................... on the right. ..................... because you have to y Combining Stepspspspsps y Combining Ste y Combining Ste tened b tened b oofs Can Often be Shor PrPrPrPrProofs Can Often be Shor oofs Can Often be Shor tened by Combining Ste oofs Can Often be Shortened b y Combining Ste tened b oofs Can Often be Shor Proofs can very often be written in the kind of two-column format used in the last example. The next statement in your argument goes on the left, and the justification for it goes on the right. Usually the justification will be something from earlier in this chapter. However, it’s not likely that you’d often need to include every single possible stage in a proof. Usually you’d solve an equation in a few lines, as shown below. Check it out: In Example 2 you’re doing several steps at once — but it’s the same thing as in Example 1. Example Example Example Example Example 22222 If 6x + 4 = 22, what is the value of x? Solution Solution Solution Solution Solution 6x + 4 = 22 6x = 18 x = 3 Usually it’s quicker (and a much better idea) to solve an equation the short way, like in Example 2. But you must be able to do it the long way if you need to, justifying each step using the real number axioms. lusion lusion es a Hypothesis and a Conc Givvvvves a Hypothesis and a Conc es a Hypothesis and a Conc Gi hen...””””” Gi Gi hen... ..., TTTTThen... hen... ..., “If“If“If“If“If..., ..., lusion es a Hypothesis and a Conclusion lusion es a Hypothesis and a Conc Gi hen... ..., Mathematical statements can often be written in the form: “If..., then...” For example, when you solve an equation like the one in Example 2, what you are really saying is: “If 6x + 4 = 22, then the value of x is 3.” A sentence like this can be broken down into two basic parts — a hypothesis and a conclusion. The hypothesis is the part of the sentence that follows “if” — here, it is 6x + 4 = 22. The conclusion is the part of the sentence that follows “then” — here, it is x = 3. IF hypothesis, THEN conclusion. 5656565656 Section 1.4 Section 1.4 Section 1.4 — Mathematical Logic Section 1.4 Section 1.4 Check it out: Watch out — a true conclusion doesn’t imply a true hypothesis, and a false hypothesis doesn’t imply that the conclusion is false. This doesn’t just apply to mathematical statements — it’s true for non-mathematical “If..., then...” sentences as well. For example: If an animal is an insect, then it has six legs. If you are in California, then you are in the United States. Now, both the hypothesis and the conclusion can be either true or false. For example, an animal may or may not be an insect, and it may or may not have six legs. However, the conclusion has to be a logical consequence of the hypothesis. Using the example above, this just means that if it is an insect, then it will have six legs. Once you’ve figured out a hypothesis and a conclusion, you can apply the following logical rules: If the hypothesis is true, then the conclusion will also be true. If the conclusion is false, then the hypothesis will also be false. So if an animal doesn’t have six legs, then it isn’t an insect. If you aren’t in the United States, then you’re not in California. And if x is not 3, then 6x + 4 π 22. 4x = 12 means that x = 3. x + y = 1 means that x = 1 – y. b + 4 = 17 – y means that b = 13 – y. Rewrite the following in “If..., then...” format. 1. 2. 3. Identify the hypothesis and conclusion in the following statements. 4. If 5y = 30, then y = 6. 5. If x2 + y2 = 16, then x2 = 16 – y2. 6. If d – 12 = 23z, then d = 23z + 12. 7. An animal has four legs if it is a dog. 8. Complete this proof by adding the missing justification steps. x – 7 = 17 (x – 7) + 7 = 17 + 7 [x + (–7)] + 7 = 17 + 7 [x + (–7)] + 7 = 24 x + [(–7) + 7] = 24 x + 0 = 24 x = 24 Given equation ................................... Definition of subtraction Adding ................................... Inverse property of + ................................... ound Up ound Up RRRRRound Up ound Up ound Up ound Up ound Up RRRRRound Up ound Up ound Up The important thing with mathematical proofs is to take each line of the math problem step by step. If you’re asked to justify your steps, make sure that you state exactly which property you’re using. Section 1.4 Section 1.4 Section 1.4 — Mathematical Logic Section 1.4 Section 1.4 5757575757 TTTTTopicopicopicopicopic 1.4.21.4.2 1.4.21.4.2 1.4.2 California Standards: xplain the xplain the Students e Students e 24.1: 24.1: 24.1: Students e xplain the Students explain the Students e 24.1: xplain the 24.1: difdifdifdifdifffffferererererence betw ence betweeneeneeneeneen ence betw ence betw ence betw e and deductivvvvveeeee e and deducti inductivvvvve and deducti e and deducti inducti inducti inducti e and deducti inducti rrrrreasoning and identify and easoning and identify and easoning and identify and easoning and identify and easoning and identify and eac eac xamples of xamples of vide e prprprprprooooovide e vide e each.h.h.h.h. xamples of eac vide examples of eac xamples of vide e 24.3: Students use 24.3: Students use 24.3: Students use 24.3: Students use 24.3: Students use countereeeeexamples to sho xamples to showwwww xamples to sho xamples to sho counter counter xamples to sho counter counter tion is falsealsealsealsealse thathathathathat an asser tion is f tion is f t an asser t an asser t an assertion is f tion is f t an asser t a singlelelelele t a sing t a sing e tha e tha and recoecoecoecoecognizgnizgnizgnizgnize tha and r and r e that a sing t a sing e tha and r and r countereeeeexample is xample is xample is counter counter xample is counter xample is counter sufsufsufsufsufficient to r efute an efute an ficient to r ficient to r efute an ficient to refute an ficient to r efute an tion. tion. asser asser tion. assertion. tion. asser asser What it means for you: You’ll identify inductive and deductive reasoning in math problems, and you’ll find how to use counterexamples to disprove a
rule. Key words: inductive reasoning deductive reasoning counterexample e and e and Inductivvvvve and Inducti Inducti e and e and Inducti Inducti e and e and Inductivvvvve and Inducti Inducti e and e and Inducti Inducti easoning easoning Deductivvvvve Re Re Re Re Reasoning Deducti Deducti easoning easoning Deducti Deducti Deductivvvvve Re Re Re Re Reasoning easoning easoning Deducti Deducti easoning Deducti easoning Deducti There are different types of mathematical reasoning. Two types mentioned in the California math standards are inductive reasoning and deductive reasoning. easoning Means Finding a General Ral Ral Ral Ral Ruleuleuleuleule easoning Means Finding a Gener Inductivvvvve Re Re Re Re Reasoning Means Finding a Gener easoning Means Finding a Gener Inducti Inducti easoning Means Finding a Gener Inducti Inducti Inductive reasoning means finding a general rule by considering a few specific cases. For example, look at this sequence of square numbers: 1, 4, 9, 16, 25, 36... If you look at the differences between successive terms, you find this: The difference between the first and second terms is 4 – 1 = 3. The difference between the second and third terms is 9 – 4 = 5. The difference between the third and fourth terms is 16 – 9 = 7. The difference between the fourth and fifth terms is 25 – 16 = 9. If you look at these differences, there’s a pattern — each difference is an odd number, and each one is 2 greater than the previous difference. So using inductive reasoning, you might conclude that: The difference between successive square numbers is always odd, and each difference is 2 greater than the previous one. Watch out though — this doesn’t actually prove the rule. This rule does look believable, but to prove it you’d have to use algebra. Guided Practice Use inductive reasoning to work out an expression for the nth term (xn) of these sequences. For example, the formula for the nth term of the sequence 1, 2, 3, 4,... is xn = n. 1. 2, 3, 4, 5,... 2. 11, 12, 13, 14,... 3. 2, 4, 6, 8,... 4. –1, –2, –3, –4,... In exercises 5-6, predict the next number in each pattern. 5. 1 = 12, 1 + 3 = 22, 1 + 3 + 5 = 32. 1, 1, 2, 3, 5, 8, 13, … 5858585858 Section 1.4 Section 1.4 Section 1.4 — Mathematical Logic Section 1.4 Section 1.4 ule Doesn’’’’’t t t t t WWWWWorororororkkkkk ule Doesn xample Prooooovvvvves es es es es TTTTThahahahahat a Rt a Rt a Rt a Rt a Rule Doesn ule Doesn xample Pr One Countereeeeexample Pr xample Pr One Counter One Counter ule Doesn xample Pr One Counter One Counter If you are testing a rule, you only need to find one counterexample (an example that does not work) to prove that the rule is not true. Once you have found one counterexample, you don’t need to look for any more — one is enough. Example Example Example Example Example 11111 Decide whether the following statement is always true: “2n + 1 is always a prime number, where n is a natural number.” Solution Solution Solution Solution Solution At first, the rule looks believable. If n = 1: 2n + 1 = 21 + 1 = 2 + 1 = 3. This is a prime number, so the rule holds for n = 1. If n = 2: 2n + 1 = 22 + 1 = 4 + 1 = 5. This is a prime number, so the rule holds for n = 2. If n = 3: 2n + 1 = 23 + 1 = 8 + 1 = 9. This is not a prime number, so the rule doesn’t hold for n = 3. So a counterexample is n = 3, and this proves that the rule is not always true. Once you’ve found one counterexample, you don’t need to find any more. Guided Practice Give a counterexample to disprove each of the following statements. 7. All odd numbers are of the form 4n + 1, where n is a natural number. 8. If a number is divisible by both 6 and 3, then it is divisible by 12. 9. \|x + 4\| ≥ 4 10. The difference between any two square numbers is always odd. 11. The difference between any two prime numbers is always even. 12. The difference between two consecutive cube numbers is always prime. 13. All quadrilaterals are squares. 14. All angles are right angles. 15. All prime numbers are odd. 16. A number is always greater than its multiplicative inverse. Section 1.4 Section 1.4 Section 1.4 — Mathematical Logic Section 1.4 Section 1.4 5959595959 ying a General Ral Ral Ral Ral Ruleuleuleuleule ying a Gener easoning Means AAAAApplpplpplpplpplying a Gener ying a Gener easoning Means Deductivvvvve Re Re Re Re Reasoning Means easoning Means Deducti Deducti ying a Gener easoning Means Deducti Deducti Deductive reasoning is almost the opposite of inductive reasoning. In deductive reasoning, you use a general rule to find out a specific fact. Example Example Example Example Example 22222 A number is a multiple of 3 if the sum of its digits is a multiple of 3. Use this information to decide whether 96 is a multiple of 3. Solution Solution Solution Solution Solution The sum of the digits is 9 + 6 = 15, which is divisible by 3. The statement says that a number is a multiple of 3 if the sum of its digits is a multiple of 3. Using deductive reasoning, that means that you can say that 96 is divisible by 3. Guided Practice Use deductive reasoning to work out the 10th term of these sequences: 18. xn = 6n 21. xn = 20n + 1 17. xn = n 20. xn = 3n – 1 Use deductive reasoning to reach a conclusion: 23. Ivy is older than Peter. Stephen is younger than Peter. 24. Lily lives in Maryland. Maryland is in the United States. 19. xn = 2n – 1 22. xn = n(n + 1) Independent Practice Use inductive reasoning in Exercises 1–3. 1. Give the next three numbers of the sequence: 25, 29, 34, 40, … 2. Write an expression for the nth term (xn) of the sequence: 24, 72, 216, 648, … 3. Audrey needs $650 to buy a digital camera. Her savings account shows the following balances: If the pattern continues, at the start of which month will she be able to buy the digital camera? Date Balance Jan 1 $100.00 Feb 1 $150.00 March 1 $250.00 April 1 $400.00 Use deductive reasoning in exercises 4-5. 4. Find the first five terms of the sequence xn = 3n. 5. Find the first five terms of the sequence xn = n(n – 1). ound Up ound Up RRRRRound Up ound Up ound Up ound Up ound Up RRRRRound Up ound Up ound Up Inductive reasoning means that you can make a general rule without having to check every single value — so it saves you a lot of work. 6060606060 Section 1.4 Section 1.4 Section 1.4 — Mathematical Logic Section 1.4 Section 1.4 TTTTTopicopicopicopicopic 1.4.31.4.3 1.4.31.4.3 1.4.3 California Standards: Givvvvven a specific en a specific en a specific Gi Gi 25.3: 25.3: en a specific 25.3: Gi 25.3: en a specific Gi 25.3: algalgalgalgalgeeeeebrbrbrbrbraic sta tement tement aic sta aic sta aic statement tement aic sta tement inininininvvvvvolving linear quadraaaaatictictictictic,,,,, quadr olving linear,,,,, quadr quadr olving linear olving linear quadr olving linear or aor aor aor aor absolute v bsolute valuealuealuealuealue bsolute v bsolute v bsolute v tions or tions or essions or equa essions or equa eeeeexprxprxprxprxpressions or equa tions or essions or equations or tions or essions or equa inequalities,,,,, students students students inequalities inequalities students inequalities students inequalities hether the hether the mine w mine w deter deter hether the mine whether the determine w mine w deter hether the deter stastastastastatement is tr tement is trueueueueue tement is tr tement is tr tement is tr alwwwwwaaaaaysysysysys,,,,, or al al sometimes,,,,, al sometimes sometimes or or or or al sometimes sometimes nenenenenevvvvvererererer..... What it means for you: You’ll learn how to recognise whether statements are true, and if they’re true sometimes or all the time. Key words: hypothesis conclusion Check it out: See Topic 1.4.1 for more on hypotheses and conclusions. Using the logic from that Topic, if x is not 3 or –3, then x2 is not equal to 9. tements tements aic Sta aic Sta AlgAlgAlgAlgAlgeeeeebrbrbrbrbraic Sta tements aic Statements tements aic Sta AlgAlgAlgAlgAlgeeeeebrbrbrbrbraic Sta tements tements aic Sta aic Sta aic Statements tements aic Sta tements When you’re trying to solve a mathematical problem, sometimes the solution is a single value. This Topic is about other situations, when things can be a bit more complicated. e than One Parararararttttt e than One P e than One P Mor Mor Solutions Can Consist of Solutions Can Consist of More than One P Solutions Can Consist of Mor e than One P Mor Solutions Can Consist of Solutions Can Consist of Example Example Example Example Example 11111 Find x given that x2 = 9. Solution Solution Solution Solution Solution You can think of the equation as a hypothesis — then you need to find a logical conclusion. One value that satisfies the equation is x = 3. However, the statement “If x2 = 9, then x = 3” is not true — because x = –3 also satisfies the equation. There are two values satisfying the equation, and you need to include both of them in your answer. So the solution is actually: “x = 3 or x = –3.” Or in “If..., then...” form: If x2 = 9, then x = 3 or x = –3. tions and Inequalities are e e e e AlAlAlAlAlwwwwwaaaaays ys ys ys ys TTTTTrrrrrueueueueue tions and Inequalities ar tions and Inequalities ar Some Equa Some Equa Some Equations and Inequalities ar tions and Inequalities ar Some Equa Some Equa If you’re given an algebraic statement (such as an equation or an inequality), there won’t always be a single value (or even two values) that satisfy the statement. Example Example Example Example Example 22222 What values of x satisfy x2 – 9 = (x + 3)(x – 3)? Solution Solution Solution Solution Solution If you look at the problem above, you might see that x = 0 satisfies the equation — because if you put x = 0, then both sides equal –9. But you might also realize that x = 3 satisfies the equation — since if x = 3, both sides equal 0. But that’s not all. If x = –3, both sides also equal 0. So x = –3 also satisfies the equation. In fact, the above equation is always true — no matter what value you pick for x. So to say “either x = 0, x = 3, or x = –3” is incorrect. You need to say that it is always
true. Section 1.4 Section 1.4 Section 1.4 — Mathematical Logic Section 1.4 Section 1.4 6161616161 Check it out: See Topic 1.2.3 for more about absolute value equations. Check it out: If \|a\| < \|b\|, then: 0 < a < b, or b < a < 0 or a < 0 < b, or b < 0 < a. tement is Nevvvvver er er er er TTTTTrrrrrueueueueue tement is Ne tement is Ne Sometimes a Sta Sometimes a Sta Sometimes a Statement is Ne tement is Ne Sometimes a Sta Sometimes a Sta There is another possibility. Look at the following absolute value equation: Example Example Example Example Example 33333 Find x given that \|x\| + 3 = 0. Solution Solution Solution Solution Solution You need to find values for x that satisfy the above equation. However, if you subtract 3 from both sides, you form the equivalent equation: \|x\| = –3 But the absolute value of a number is its distance from 0 on the number line — and a distance cannot be negative. That means that there are no values of x that satisfy the equation. So the equation is never true. Guided Practice 1. Find x given that x2 = 25. 2. What values of x satisfy x2 – 25 = (x + 5)(x – 5)? 3. Find x given that \|x\| + 5 = 0 4. Find x given that x2 = –16, x Œ R 5. What values of x satisfy x2 > 0? 6. Find x given that x2 + 1 = 17. tical Logicgicgicgicgic tical Lo tical Lo thema e Examples of Ma Ma Ma Ma Mathema thema e Examples of MorMorMorMorMore Examples of e Examples of thematical Lo tical Lo thema e Examples of You can combine the ideas from this section. Look at the following examples. Example Example Example Example Example 44444 Is the following statement true? If \|a\| < \|b\|, then a < b. Solution Solution Solution Solution Solution This is an example where you can find a counterexample to disprove the rule. For example, if a = –1 and b = –2, then \|a\| < \|b\|, but a > b. Here, the conclusion isn’t a logical consequence of the hypothesis — so you can’t apply the logical rules from earlier. That means that the statement is not true. 6262626262 Section 1.4 Section 1.4 Section 1.4 — Mathematical Logic Section 1.4 Section 1.4 Example Example Example Example Example 55555 If x is a real number, find the possible values for which the following is true: x2 + 1 < 2x Solution Solution Solution Solution Solution You can use the inequality as a hypothesis. Now you need to find a suitable conclusion using a series of logical steps. x2 + 1 < 2x ﬁ x2 – 2x + 1 < 0 ﬁ (x – 1)2 < 0 So your “If..., then...” statement is: If x2 + 1 < 2x, then (x – 1)2 < 0. Your hypothesis is: x2 + 1 < 2x Your conclusion is: (x – 1)2 < 0 When you square a real number, you can never get a negative number — so (x – 1)2 cannot be negative. This means that your conclusion is false. And using the logic from earlier, this means that your hypothesis is also false. So you’ve proved that there is no real number x for which x2 + 1 < 2x. The statement is never true. Example Example Example Example Example 66666 “If a number is odd, its square is odd.” The number 71774784 is a perfect square. Use the statement above to say whether the square roots of 71774784 are odd. Solution Solution Solution Solution Solution Hypothesis: a number is odd Conclusion: its square is odd The square of a number is 71774784 — which is an even number. So the conclusion is false. This means that the hypothesis is also false, and the square root of 71774784 is not odd. Section 1.4 Section 1.4 Section 1.4 — Mathematical Logic Section 1.4 Section 1.4 6363636363 Independent Practice 1. If x is a real number, find the possible values for which the following is true: x2 – 4x + 4 < 0 2. Is this statement true? “If \|x\| < 3, then x < 3.” 3. Find x given that x2 = 64 4. What values of x satisfy x2 – 100 = (x + 10)(x – 10)? 5. Find x given that \|x\| + 5 = 0. 6. If x is a real number, find the possible values for which the following is true: x2 + 9 < 6x In exercises 7-11, say whether the algebraic statements are true sometimes, always, or never. 7. 5 + x = 10 10. x2 < 0 8. x2 = –9 11. x2 > x 9. x2 = 49 12. Is this statement true? “If x > 0, then x3 > x2” 13. Find x given that \|x\| + 7 = 0. 14. What values of x satisfy x2 – 25 = (x + 5)(x – 5)? 15. The area of a rectangle is given by A = l × w where l and w are the length and width of the rectangle, respectively. Can the area of a rectangle ever be less than zero? 16. Find x given that x2 = 81 17. Is the following statement true? If x > 0 then x2 > x 18. Prove that the following statement is not true: “the square root of a number is always smaller than the number itself.” 19. Prove that the difference between successive square numbers is always odd, and each difference is 2 greater than the previous difference. ound Up ound Up RRRRRound Up ound Up ound Up RRRRRound Up ound Up ound Up ound Up ound Up There won’t always be a one-part solution to an algebraic statement such as an equation or inequality. If the statement is always true, it’s no good just giving one value the statement holds for. And if it’s never true, you have to state that too. 6464646464 Section 1.4 Section 1.4 Section 1.4 — Mathematical Logic Section 1.4 Section 1.4 Chapter 1 Investigation Counting Collections Counting Collections Counting Collections Counting Collections Counting Collections Counting Collections Counting Collections Counting Collections Counting Collections Counting Collections Sets and subsets aren’t only useful in Math class — they can be used to describe everyday situations. A cereal company is giving away baseball erasers free in their boxes of cereal. There are 7 erasers to collect. All of the children at an elementary school want to collect the whole set. At the moment, they all have different collections and none have more than one of any one eraser. What is the maximum number of children there could be at the school? Things to think about: How many children could there be if there were only two erasers to collect? How many children could there be if there were three erasers to collect? How many children could there be if there were four erasers to collect? Look at your answers — what do you notice? The collections can overlap — for example, one child could have erasers a, b, and d, while another could have erasers b, c, and g. Extension 1) If there were 8 erasers in the set, how many different collections could there be? What if there were 20 erasers to collect? Try to find a general rule for the number of different collections for sets with n items. 2) A set of trading cards consists of 78 numbered cards. • How many people could have different collections of cards? The cards come in sealed packs that cost $1.80 per pack. Each pack contains 8 randomly selected cards. Your friend says that it would only cost $18 to get the full set. Is your friend right? 3) Set G is the set of two-digit prime numbers. How many subsets of G are there? ound Up ound Up RRRRRound Up ound Up ound Up RRRRRound Up ound Up ound Up ound Up ound Up This Investigation shows that you can use sets and subsets to model real-life situations. In fact, you probably divide things into subsets without even realizing it — for example, sorting out your favourite types of candy from a mixed box. estigaaaaationtiontiontiontion — Counting Collections estigestig estig pter 1 Invvvvvestig pter 1 In ChaChaChaChaChapter 1 In pter 1 In pter 1 In 6565656565 Chapter 2 Single Variable Linear Equations Section 2.1 Algebra Basics .......................................................... 67 Section 2.2 Manipulating Equations ............................................. 76 Section 2.3 More Equations ......................................................... 81 Section 2.4 Using Equations ........................................................ 89 Section 2.5 Consecutive Integer Tasks, Time and Rate Tasks .... 95 Section 2.6 Investment and Mixture Tasks ................................. 108 Section 2.7 Work-Related Tasks ................................................ 121 Section 2.8 Absolute Value Equations ....................................... 128 Investigation Wildlife Park Trains .................................................. 135 6666666666 TTTTTopicopicopicopicopic 2.1.12.1.1 2.1.12.1.1 2.1.1 California Standards: 4.0: Students simplify 4.0: Students simplify 4.0: Students simplify 4.0: Students simplify 4.0: Students simplify eeeeexprxprxprxprxpressions essions essions essions before solving essions linear equations and inequalities in one variable, such as 3(2x – 5) + 4(x – 2) = 12..... What it means for you: You’ll combine like terms to simplify expressions. Key words: algebraic expression simplify like terms Don’t forget: See Topic 1.3.1 if you’re not sure what exponents are. Section 2.1 Simplifying AlgAlgAlgAlgAlgeeeeebrbrbrbrbraicaicaicaicaic Simplifying Simplifying Simplifying Simplifying Simplifying AlgAlgAlgAlgAlgeeeeebrbrbrbrbraicaicaicaicaic Simplifying Simplifying Simplifying Simplifying essions essions ExprExprExprExprExpressions essions essions essions essions ExprExprExprExprExpressions essions essions Unless you’re told otherwise, you always need to give algebraic solutions in the simplest form possible. essions Contain VVVVVariaariaariaariaariabbbbbleslesleslesles essions Contain essions Contain aic Expr AlgAlgAlgAlgAlgeeeeebrbrbrbrbraic Expr aic Expr aic Expressions Contain essions Contain aic Expr Algebraic expressions are made up of terms. A term can be a product of numbers and variables, like 4x2 or 5x, or just a number. For example, the algebraic expression 4x2 – 5x + 7 – 2x2 + 2x – 3 has six terms. The terms are separated by plus and minus signs. Each sign “belongs” to the term that it’s in front of. 2 8 + 2 x xy Invisible + sign Guided Practice – 6 + 7 – 2 y y 2 This minus sign belongs to the 6y term. Write the number of terms in the expressions in Exercises 1–4. 1. 3x2 + 4x – 2 3. 3x2 2. 8x4 + 7x3 + 2x2 – 8 4. 8 + 7xy – 2xy9 + 4x7 + 3y2 + 55y3 5. Which variable is multiplied by –4 in the algebraic expression 4x2 – 4y + 8 + 4xy? 6. Counting from left to right, which term is the
fourth term in the algebraic expression 8x2 + 2xy – 6y + 9xy3 – 4? LikLikLikLikLike e e e e TTTTTerererererms Can Be Combined ms Can Be Combined ms Can Be Combined ms Can Be Combined ms Can Be Combined Like terms are terms with identical variables that have identical exponents. The terms 4x2 and –2x2 are like terms because they have the same variable, x, with the same exponent, 2. Likewise, –5x and 2x are like terms, and 7 and –3 are like terms. Like terms can be combined using the distributive, commutative, and associative properties. Example Example Example Example Example 11111 Simplify 4x2 – 2x2. Solution Solution Solution Solution Solution 4x2 – 2x2 = (4 – 2)x2 = 2x2 Section 2.1 Section 2.1 Section 2.1 — Algebra Basics Section 2.1 Section 2.1 6767676767 Don’t forget: See Lesson 1.2.9 for more on number properties. Check it out: Subtracting a number is just the same as adding a negative number. So, 4x – 5 is the same as 4x + (–5). Don’t forget: When there are no more like terms to collect, the expression is simplified as much as possible. essions essions ms to Simplify Expr Combine Like e e e e TTTTTerererererms to Simplify Expr ms to Simplify Expr Combine Lik Combine Lik essions ms to Simplify Expressions essions ms to Simplify Expr Combine Lik Combine Lik To simplify an algebraic expression, use number properties to first group and then combine like terms. Example Example Example Example Example 22222 Simplify the following: a) (4x – 5) – 2x b) 5x2 – 3x + 7x2 – 4x + 9 Solution Solution Solution Solution Solution a) (4x – 5) – 2x = 4x + (–5 – 2x) = 4x + (–2x – 5) = (4x + –2x) – 5 = [(4 – 2)x] – 5 = 2x – 5 dition dition ty of ad ad ad ad addition ty of ty of oper oper e pre proper e pre pr Associatititititivvvvve pr Associa Associa dition operty of dition ty of oper Associa Associa dition dition ty of ad ad ad ad addition ty of ty of oper oper CommCommCommCommCommutautautautautatititititivvvvve pre pre pre pre proper dition operty of dition ty of oper dition dition ty of ad ad ad ad addition ty of ty of oper oper e pre proper e pre pr Associatititititivvvvve pr Associa Associa dition operty of dition ty of oper Associa Associa opertytytytyty oper oper e pre proper e pre pr Distributiutiutiutiutivvvvve pr Distrib Distrib oper Distrib Distrib b) 5x2 – 3x + 7x2 – 4x + 9 = 5x2 + 7x2 – 3x – 4x + 9 = (5x2 + 7x2) + (–3x – 4x) + 9 = (5 + 7)x2 + (–3 – 4)x + 9 = 12x2 + (–7)x + 9 = 12x2 – 7x + 9 Guided Practice Simplify the following expressions in Exercises 7–14. 7. 7a + 12a – 4 9. 9x + (20 – 5x) 11. 7x2 + 7 + 20x2 + 3x 13. 5a – 4 × 3a + 7 × 2 – 3 × 6 8. 7a +3b – 5a 10. 5 – 10x – 2x – 7 12. 3 – 8x2 + 4x2 + 6x2 – 10 14. 3 × 4a – 2 × 5a2 + 2 × 2a2 In Exercises 15–18, simplify the expressions and determine the number of terms in each simplified form. 15. (4a – 9) + (2a – 18) 16. 15n + 3n + 8 – 2 – 6 17. 6a2 + 3a – 9a2 + 2a + 7 + 6 18. 6a + 3 × 7b – 2 × 5c + 7 – 9 + 2 × 4c 6868686868 Section 2.1 Section 2.1 Section 2.1 — Algebra Basics Section 2.1 Section 2.1 Independent Practice In Exercises 1–5, determine the number of terms in each algebraic expression: 1. 7b + 14a – 4 3. (27x2 + 4x) – 13 5. 2x + 4xy + 4x2 – (10 + 12y + 19y2) 2. 2a 4. 5 +10x + 20x2 + 3a In Exercises 6–9, simplify each algebraic expression: 6. 12m – 7 + 3c – 7m – 8c 7. 4a + 3b + 11a – 8b 8. 2 × 6x – 3 × 5x – 3x × 4 + 5 × 2x + 12x 9. 5m × 3 + 2 × 7m – 4m × 4 + 7 × 2m – 17m In Exercises 10–13, simplify each algebraic expression: 10. 3x + 1 2 a – 3 11. 1 10 8 y – 1 x 2 b + 8 10 12. (17.8n + 13.08q) – 3.8q – 9.9n 13. 0.4x2 + 3 8 x2 – 0.14 – 5 8 b – 6 12 x – 1 2 a 14. Which expression below simplifies to 2x + 1? i. x + 4x + 4 – 3 – 3x + x ii. 7 + 5x – 6 + 5 – 4x + x – 5 15. Which expression below simplifies to 3x? i. 5 + 4x2 – 3x – 2 + 4x2 – 8x2 – 3 ii. –2x2 + 6 + 4x2 – 3 + 3x – 2x2 – 3 In Exercises 16–17, find a simplified expression for the perimeter of the figure. 16. 2x x 4 + 1 17. 3 2a 2a 2x 2x 7 18. Juan bought 3 baseball cards for b dollars each and 2 baseball cards for c dollars each. He has bought 4 comic books for $5.00 each. Write and simplify an algebraic expression showing the total money Juan spent on baseball cards and comic books. 19. Three friends Tom, Leo, and Maria have several pieces of candy to eat. Tom has (2x + 4) pieces of candy, Leo has (8 – 2x) pieces of candy, and Maria has 8 pieces of candy. Write and simplify an algebraic expression showing the total number of pieces of candy the three friends have to eat. ound Up ound Up RRRRRound Up ound Up ound Up ound Up ound Up RRRRRound Up ound Up ound Up You’ve combined like terms before, in earlier grades — so this Topic should feel like good practice. It’s always important to give your final answers to algebraic problems in the simplest form. Section 2.1 Section 2.1 Section 2.1 — Algebra Basics Section 2.1 Section 2.1 6969696969 TTTTTopicopicopicopicopic 2.1.22.1.2 2.1.22.1.2 2.1.2 California Standards: 4.0: Students simplify 4.0: Students simplify 4.0: Students simplify 4.0: Students simplify 4.0: Students simplify eeeeexprxprxprxprxpressions essions essions essions before solving essions linear equations and inequalities in one variable, such as 3(2x – 5) + 4(x – 2) = 12..... What it means for you: You’ll use the distributive property to simplify expressions. Key words: distributive property commutative property Don’t forget: After getting rid of the grouping symbols, collect the like terms together. Getting Rid of Getting Rid of Getting Rid of Getting Rid of Getting Rid of Getting Rid of Getting Rid of Getting Rid of Getting Rid of Getting Rid of ouping Symbols ouping Symbols GrGrGrGrGrouping Symbols ouping Symbols ouping Symbols GrGrGrGrGrouping Symbols ouping Symbols ouping Symbols ouping Symbols ouping Symbols You already saw the distributive property in Topic 1.2.7. In this Topic you’ll simplify expressions by using the distributive property to get rid of grouping symbols. ouping Symbols ouping Symbols es Gr operty Rty Rty Rty Rty Remoemoemoemoemovvvvves Gr es Gr oper oper e Pr he Distributiutiutiutiutivvvvve Pr e Pr he Distrib TTTTThe Distrib he Distrib ouping Symbols es Grouping Symbols e Proper ouping Symbols es Gr oper e Pr he Distrib The expression 5(3x + 2) + 2(2x – 1) can be simplified — both parts have an “x” term and a constant term. To simplify an expression like this, you first need to get rid of the grouping symbols. The way to do this is to use the distributive property of multiplication over addition: a(b + c) = ab + ac. Example Example Example Example Example 11111 Simplify 5(3x + 2) + 2(2x – 1). Solution Solution Solution Solution Solution 5(3x + 2) + 2(2x – 1) = 15x + 10 + 4x – 2 = 15x + 4x + 10 – 2 = 19x + 8 Guided Practice Distrib Distrib Distributiutiutiutiutivvvvve pr e pre proper e pre pr oper oper opertytytytyty Distrib Distrib oper CommCommCommCommCommutautautautautatititititivvvvve pre pre pre pre proper oper oper ty of ty of ty of ad ad ad ad addition dition dition operty of dition oper ty of dition In Exercises 1–7, simplify the following expressions: 1. 2(4x + 5) + 8 2. 12(5a – 8) + 4x + 3 3. 6(2j + 3c) + 8(5c + 4z) 4. 10(x + 2) + 7(3 – 4x) 5. 6(a – b) + 4(2b – 3) 6. 5(3x + 4) + 3(4x + 10) + 2(8x + 9) 7. 8(2n – 3) + 9(4n – 5) + 4(3n + 7) 7070707070 Section 2.1 Section 2.1 Section 2.1 — Algebra Basics Section 2.1 Section 2.1 y a Negggggaaaaatititititivvvvve Number e Number e Number y a Ne y a Ne ying b ying b hen Multipl e Care we we we we when Multipl hen Multipl e Car TTTTTakakakakake Car e Car e Number ying by a Ne hen Multiplying b e Number y a Ne ying b hen Multipl e Car If a number outside a grouping symbol is negative, like in –7(2x + 1), you have to remember to use the multiplicative property of –1. This means that the signs of the terms within the grouping symbols will change: “+” signs will change to “–” signs and vice versa. Example Example Example Example Example 22222 Simplify the following: a) –7(2x + 1) b) –6(–x – 3) c) –3(5x – 4) Solution Solution Solution Solution Solution a) The +2x and +1 become negative. –7(2x + 1) = –14x – 7 b) The two negative terms inside the grouping symbols are multiplied by the negative term outside. They both become positive. –6(–x – 3) = 6x + 18 c) –3(5x – 4) = –15x + 12 Example Example Example Example Example 33333 Simplify the expression 4(2x – 1) – 5(x – 2). Show your steps. Check it out 3x + 6 has no more like terms, so it can’t be simplified any further. Solution Solution Solution Solution Solution 4(2x – 1) – 5(x – 2) = 8x – 4 – 5x + 10 = 8x – 5x – 4 + 10 = 3x + 6 Guided Practice GiGiGiGiGivvvvven e en een exprxprxprxprxpression en een e ession ession ession ession Distrib Distrib Distributiutiutiutiutivvvvve pr e pre proper e pre pr oper oper opertytytytyty Distrib Distrib oper CommCommCommCommCommutautautautautatititititivvvvve pre pre pre pre proper oper oper ty of ty of ty of ad ad ad ad addition dition dition operty of dition oper ty of dition In Exercises 8–13, simplify each algebraic expression: 8. –2(5a – 3c) 9. –8(3c – 2) 10. –2(–3x – 4) + 4(6 – 2x) 11. 7(2a + 9) – 4(a + 11) 12. –8(2y + 4) – 5(y + 4) 13. − 2 ⎛ ⎜⎜⎜ ⎝ 1 2 + n 2 ⎞ ⎟⎟⎟− ⎠ ⎛ ⎜⎜⎜ 3 ⎝ 1 3 − n 4 ⎞ ⎟⎟⎟ ⎠ 14. Simplify 12(2n – 7) – 9(3 – 4n) + 6(4x – 9). 15. Simplify 5(x – 2) – 7(–4x + 3) – 3(–2x). Section 2.1 Section 2.1 Section 2.1 — Algebra Basics Section 2.1 Section 2.1 7171717171 Independent Practice In Exercises 1–6, simplify the algebraic expressions: 1. –4(a + 2b) 3. 5(3b – 2q) – (3q + 4) 5. –17(3a – 5b) – 4(x + 3) 2. 3(2n + 4) + n(–4) 4. –9(2 + 3b) + 3(3 – 2b) 6. –2(10n – 4x) + 2(n + 6x) – 3(7x – 2n) 7. Simplify –7(3p – 9q) – 4(25q – 3r) + 12(5p + 8). In Exercises 8–12, find and simplify an expression for the perimeter of the shape shown. 8 11. 9. 10. - 2 x- - – 1 4 – 3x 12. 3 n – 4 - - - 2 – x1 20 – 3n (3x + 4) – 1 2 In Exercises 13–15, simplify the algebraic expressions: 13. 3 8 14. –0.1(25n + 46) – 0.8(16n – 5) + 0.4(3 – 4n) 15. – 1 3 (2x – 4) – 0.2(–7x – 10) (5 – 9x) – 1 8 (4x – 8) 16. Mia has 4(x + 1) dolls, where x is Mia
's age. Madeline has 5(2x + 3) dolls. Write and simplify an algebraic expression showing the total number of dolls in Mia and Madeline's collection. 17. Ruby, Sara, and Keisha are counting stamps. If x represents Ruby's age, she has 4(x – 4) stamps, Sara has 2(8 – x) stamps, and Keisha has 8(7 + 2x) stamps. Write and simplify an algebraic expression showing the total number of stamps owned by the three friends. ound Up ound Up RRRRRound Up ound Up ound Up RRRRRound Up ound Up ound Up ound Up ound Up The distributive property is really useful — it’s always good to get rid of confusing grouping symbols whenever you can. The main thing you need to watch out for is if you’re multiplying the contents of parentheses by a negative number — it will change the sign of everything in the parentheses. 7272727272 Section 2.1 Section 2.1 Section 2.1 — Algebra Basics Section 2.1 Section 2.1 TTTTTopicopicopicopicopic 2.1.32.1.3 2.1.32.1.3 2.1.3 California Standards: 4.0: Students simplify 4.0: Students simplify 4.0: Students simplify 4.0: Students simplify 4.0: Students simplify eeeeexprxprxprxprxpressions essions essions essions before solving essions linear equations and inequalities in one variable, such as 3(2x – 5) + 4(x – 2) = 12..... What it means for you: You’ll use the rules of exponents to simplify expressions, and you’ll show that proposed solutions to equations are correct. Key words: exponents commutative property equation Don’t forget: After getting rid of the grouping symbols, collect any like terms together. Check it out: 2xy and yx are like terms. e Simplifying and e Simplifying and MorMorMorMorMore Simplifying and e Simplifying and e Simplifying and e Simplifying and e Simplifying and MorMorMorMorMore Simplifying and e Simplifying and e Simplifying and king AnsAnsAnsAnsAnswwwwwererererersssss king king ChecChecChecChecChecking king ChecChecChecChecChecking king AnsAnsAnsAnsAnswwwwwererererersssss king king king This Topic gives another method of simplifying expressions — using the rules of exponents. Exponents to Multiply y y y y VVVVVariaariaariaariaariabbbbbleslesleslesles Exponents to Multipl Exponents to Multipl ules of ules of Use R Use R ules of Exponents to Multipl Use Rules of Exponents to Multipl ules of Use R Use R To simplify expressions like 4x(x2 – 2x + 1), you need to apply the distributive property as well as rules of exponents, such as xa × xb = xa+b. Example Example Example Example Example 11111 Simplify –2x(x2 – 2y + 1) – x(–4xy + y). Solution Solution Solution Solution Solution –2x(x2 – 2y + 1) – x(–4xy + y) = –2x3 + 4xy – 2x + 4x2y – xy = –2x3 + 3xy – 2x + 4x2y Get rid of Get rid of Get rid of g g g g grrrrrouping symbols fir ouping symbols fir ouping symbols fir ouping symbols firststststst Get rid of Get rid of ouping symbols fir TTTTThen collect lik hen collect lik hen collect lik e ter e ter e termsmsmsmsms hen collect like ter hen collect lik e ter If you have two or more different variables multiplied together, it doesn’t matter what order they’re in. For example, xy is the same as yx, and ab is the same as ba. This is because of the commutative property of multiplication. Example Example Example Example Example 22222 Simplify 2x(y + 1) – y(x + 3). Solution Solution Solution Solution Solution 2x(y + 1) – y(x + 3) = 2xy + 2x – yx – 3y = 2xy – yx + 2x – 3y = xy + 2x – 3y Guided Practice In Exercises 1–9, simplify the algebraic expressions: 1. 2x(3x – 4) 2. –4x(x – 4) 3. –2y(3yx + 2) 4. 2x(x + 5y) + 3y(y + 3) 5. 2y(2x + 2) – 4(2x + 2) 6. 6y(yx – 4) + 5(yx – 4) 7. 7n(3a + b) – 4a(7n + 2b) 8. 2x(2x2 – x) + x(2x – 8) + 3x(x – 4) 9. 2x(k – 9) – k(x – 7) + xk(4 – 3x) Section 2.1 Section 2.1 Section 2.1 — Algebra Basics Section 2.1 Section 2.1 7373737373 Don’t forget: See Topic 2.2.1 for the difference between expressions and equations. Check it out: Continue simplifying until you end up with the same number on each side of the equation. ritten as an Equationtiontiontiontion ritten as an Equa essions Can Be WWWWWritten as an Equa ritten as an Equa essions Can Be essions Can Be o Equal Expr TTTTTwwwwwo Equal Expr o Equal Expr o Equal Expressions Can Be ritten as an Equa essions Can Be o Equal Expr For any equation: Left-hand side = Right-hand side When you think you know what numbers the variables in an equation represent, you should always put them into the equation to check that both sides of the equation are equal. Replace the variables with the numbers and make sure that the left side of the equation is equal to the right side of the equation. You do exactly the same thing if a question asks you to prove or show that an equation is true. Example Example Example Example Example 33333 Show that x = –1 is a solution of 4(x – 2) + 2 = 4x – 2(2 – x). Solution Solution Solution Solution Solution 4(x – 2) + 2 = 4x – 2(2 – x) You’re told that x = –1 is a solution, so replace each x with –1. ﬁ 4(–1 – 2) + 2 = 4(–1) – 2[2 – (–1)] ﬁ 4(–3) + 2 = 4(–1) – 2(3) ﬁ –12 + 2 = –4 – 6 ﬁ –10 = –10 This shows that 4(x – 2) + 2 = 4x – 2(2 – x) is a true statement for x = –1. Guided Practice 10. Given x = –1, show that 2(5x + 3) – 3(3x + 2) = 4x + 3. 11. Given b = –1, show that ( 12. Given x = – 5 3 , show that 4(2x – 1) – 5(x – 2) = 1. 13. Show whether x = –2 is or is not a solution of –6x – 15 = –17 – 9x. − − 14. If b = –10, show that . 15. Show whether x = – 2 3 is or is not a solution of –6x – 15 = –17 – 9x. 16. Verify that x = 0 is a solution of 3 8 (3x + 4) – 1 2 (4x – 8) = 5.5. 17. Verify that x = –1 is a solution of (4x – 9)2 = 169. 18. Verify that x = –7 is a solution of 0.2( 4 9 x + 3) – 3 29 45 . 7474747474 Section 2.1 Section 2.1 Section 2.1 — Algebra Basics Section 2.1 Section 2.1 o or More e e e e VVVVVariaariaariaariaariabbbbbleslesleslesles o or Mor y Contain TTTTTwwwwwo or Mor o or Mor y Contain y Contain tions ma EquaEquaEquaEquaEquations ma tions ma tions may Contain o or Mor y Contain tions ma Sometimes your equation will have two or more different variables. Example Example Example Example Example 44444 Given that a = 3, b = –2, and c = 0, show that –2a(4 + b) = b(5 – c) – 2. Solution Solution Solution Solution Solution Again, just substitute the numbers for the variables and simplify until you show that both sides of the equation are equal. –2a(4 + b) = b(5 – c) – 2 ﬁ –2(3)(4 – 2) = –2(5 – 0) – 2 ﬁ –6(2) = –2(5) – 2 ﬁ –12 = –10 – 2 ﬁ –12 = –12 \ –2a(4 + b) = b(5 – c) – 2 for the given values of a, b, and c. Guided Practice In Exercises 19–24, find the value of each algebraic expression when the given substitutions are made: 19. 2x(x + 5y) – 3y(y + 3) 20. 6y(yx – 4) – 5(yx – 4) if x = 2, y = 4 if x = 1, y = –1 if a = 0, b = 3, n = 1 7 if x = 4, y = –8 if a = 4, b = 0.2 21. 7n(30 + b) – 4a(7n + 2b) 22. –2y(3yx + 2) 23. –4a(b – 4) 24. 2x(k – a) – k(x – a) + xk(a – 3x) if a = –4.2, x = 0.1, k = 1 3 Independent Practice In Exercises 1–3, simplify the algebraic expressions: 1. –7a(8bc – 3a) 2. 6n(yn + 7) – 7n(n – yn) 3. 2ab(c – d) + 4cb(c – d) – 3ac(2b + d) 4. Find the value of 2a(4a – 3) + (8 – a), if a = 3. 5. Find the value of 2ab(c – d) + 4cb(c – d) – 3ac(2b + d), if a = 8, b = 7, c = 6, d = 0. 6. The formula P = 6.5h + 0.10x is used to find the weekly pay of a salesperson at a local electronic store, where P is the pay in dollars, h is the number of hours worked, and x is the total value of merchandise sold (in dollars) by the salesperson. If the salesperson worked 40 hours and sold $4,250 worth of merchandise, how much pay did she earn? ound Up ound Up RRRRRound Up ound Up ound Up ound Up ound Up RRRRRound Up ound Up ound Up The material on equations in this Topic leads neatly on to the next Section. The rest of the Sections in this Chapter are all about forming and manipulating equations. Section 2.1 Section 2.1 Section 2.1 — Algebra Basics Section 2.1 Section 2.1 7575757575 TTTTTopicopicopicopicopic 2.2.12.2.1 2.2.12.2.1 2.2.1 Section 2.2 Equality Equality ties of ties of oper oper PrPrPrPrProper Equality ties of Equality operties of Equality ties of oper PrPrPrPrProper Equality Equality ties of ties of oper oper ties of Equality operties of Equality ties of oper Equality California Standards: 4.0: Students simplify 4.0: Students simplify 4.0: Students simplify 4.0: Students simplify 4.0: Students simplify essions befororororore solving eeeeexprxprxprxprxpressions bef e solving e solving essions bef essions bef e solving e solving essions bef tions tions linear equa linear equa tions and linear equations tions linear equa linear equa in one variaariaariaariaariabbbbblelelelele,,,,, in one v in one v inequalities in one v in one v – 2) = 12. – 2) = 12. – 5) + 4( – 5) + 4( h as 3(23(23(23(23(2x – 5) + 4( h as sucsucsucsucsuch as h as – 2) = 12. – 5) + 4(x – 2) = 12. – 2) = 12. – 5) + 4( h as What it means for you: You’ll solve one-step equations using properties of equality. Key words: expression equation variable solve isolate equality Now it’s time to use the material on expressions you learned in Section 2.1. An equation contains two expressions, with an equals sign in the middle to show that they’re equal. In this Topic you’ll solve equations that involve addition, subtraction, multiplication, and division. e Equal e Equal essions ar essions ar o Expr tion Shows ws ws ws ws TTTTThahahahahat t t t t TTTTTwwwwwo Expr o Expr tion Sho tion Sho An Equa An Equa e Equal essions are Equal o Expressions ar An Equation Sho e Equal essions ar o Expr tion Sho An Equa An Equa An equation is a way of stating that two expressions have the same value. This equation contains only numbers — there are no unknowns: The expression on the left-hand side... 24 – 9 = 15 ...has the same value as the expression on the right-hand side Some equations contain unknown quantities, or variables. The left-hand side... 2x – 3 = 5 ...equals the right-hand side The value of x that satisfies the equation is called the solution (or root) of the equation. tions tions action in Equa action in Equa dition and Subtr AdAdAdAdAd
dition and Subtr dition and Subtr tions action in Equations dition and Subtraction in Equa tions action in Equa dition and Subtr Addition Property of Equality For any real numbers a, b, and c, if a = b, then a + c = b + c. Subtraction Property of Equality For any real numbers a, b, and c, if a = b, then a – c = b – c. These properties mean that adding or subtracting the same number on both sides of an equation will give you an equivalent equation. This may allow you to isolate the variable on one side of the equals sign. Finding the possible values of the variables in an equation is called solving the equation. 7676767676 Section 2.2 Section 2.2 Section 2.2 — Manipulating Equations Section 2.2 Section 2.2 Example Example Example Example Example 11111 Solve x + 9 = 16. Solution Solution Solution Solution Solution You want x on its own, but here x has 9 added to it. So subtract 9 from both sides to get x on its own. (x + 9) – 9 = 16 – 9 x + (9 – 9) = 16 – 9 x + 0 = 16 – 9 x = 16 – 9 x = 7 In Example 1, x = 7 is the root of the equation. If x takes the value 7, then the equation is satisfied. If x takes any other value, then the equation is not satisfied. For example, if x = 6, then the left-hand side has the value 6 + 9 = 15, which does not equal the right-hand side, 16. When you’re actually solving equations, you won’t need to go through all the stages each time — but it’s really important that you understand the theory of the properties of equality. If you have a “+ 9” that you don’t want, you can get rid of it by just subtracting 9 from both sides. If you have a “– 9” that you want to get rid of, you can just add 9 to both sides. In other words, you just need to use the inverse operations. Example Example Example Example Example 22222 Solve x + 10 = 12. Solution Solution Solution Solution Solution x + 10 = 12 x = 12 – 10 Subtr Subtr Subtr act 10 fr act 10 fr om both sides om both sides Subtract 10 fr act 10 from both sides om both sides Subtr act 10 fr om both sides x = 2 Example Example Example Example Example 33333 Solve x – 7 = 8. Solution Solution Solution Solution Solution = 15 d 7 to both sides d 7 to both sides AdAdAdAdAdd 7 to both sides d 7 to both sides d 7 to both sides Section 2.2 Section 2.2 Section 2.2 — Manipulating Equations Section 2.2 Section 2.2 7777777777 Guided Practice In Exercises 1–8, solve the equation for the unknown variable. 1. x + 7 = 15 2. x + 2 = –8 3. 4 3 + x = 2 3 5. –9 + x = 10 7. 4 9 = x – 1 3 4. x – (–9) = –17 6. x – 0.9 = 3.7 8. –0.5 = x – 0.125 tions tions vision in Equa vision in Equa tion and Di tion and Di Multiplica Multiplica tions vision in Equations tion and Division in Equa Multiplication and Di tions vision in Equa tion and Di Multiplica Multiplica Multiplication Property of Equality For any real numbers a, b, and c, if a = b, then a × c = b × c. Division Property of Equality For any real numbers a, b, and c, such that c π 0, if a = b, then a c = b c . These properties mean that multiplying or dividing by the same number on both sides of an equation will give you an equivalent equation. That can help you to isolate the variable and solve the equation. vide to Get the VVVVVariaariaariaariaariabbbbble on Its Own le on Its Own le on Its Own vide to Get the vide to Get the y or Di y or Di Multipl Multipl le on Its Own y or Divide to Get the Multiply or Di le on Its Own vide to Get the y or Di Multipl Multipl As with addition and subtraction, you can get the variable on its own by simply performing the inverse operation. If you have “× 3” on one side of the equation, you can get rid of that value by dividing both sides by 3. If you have a “÷ 3” that you want to get rid of, you can just multiply both sides by 3. Once again, you just need to use the inverse operations. 7878787878 Section 2.2 Section 2.2 Section 2.2 — Manipulating Equations Section 2.2 Section 2.2 Example Example Example Example Example 44444 Solve 2x = 18. Solution Solution Solution Solution Solution You want x on its own... but here you’ve got 2x. on its ownwnwnwnwn on its o on its o y 2 to get et et et et x on its o y 2 to g y 2 to g vide both sides b vide both sides b DiDiDiDiDivide both sides b vide both sides by 2 to g on its o y 2 to g vide both sides b x = 2 2 18 2 x = 9 2 2 1x = 9 x = 9 Example Example Example Example Example 55555 Solve m 3 = 7. Solution Solution Solution Solution Solution × 3 = 7 × 3 Multiply both sides by 3 to get Multiply both sides by 3 to get Multiply both sides by 3 to get on its own on its own Multiply both sides by 3 to get m on its own on its own Multiply both sides by 3 to get on its own m 3 m = 21 Some equations are a bit more complicated. Take them step by step. Example Example Example Example Example 66666 Solve 4x – 2(2x – 1) = 2 – x + 3(x – 4). Solution Solution Solution Solution Solution 4x – 2(2x – 1) = 2 – x + 3(x – 4) 4x – 4x + 2 = 2 – x + 3x – 12 2 = –10 + 2x 2x = 12 x = 6 Clear out an Clear out an Clear out any gy gy gy gy grrrrrouping symbols ouping symbols ouping symbols ouping symbols Clear out an Clear out an ouping symbols TTTTThen combine lik hen combine lik hen combine lik e ter e ter e termsmsmsmsms hen combine like ter hen combine lik e ter Guided Practice In Exercises 9–16, solve each equation for the unknown variable. 9. 4x = 144 11. x 4 = –7 13. –3x + 4 = 19 10. –7x = –7 12. 3 2x = 9 14. 4 – 2x = 18 15. 3x – 2(x – 1) = 2x – 3(x – 4) 16. 3x – 4(x – 1) = 2(x + 9) – 5x Section 2.2 Section 2.2 Section 2.2 — Manipulating Equations Section 2.2 Section 2.2 7979797979 Independent Practice Solve each of these equations: 1. –2(3x – 5) + 3(x – 1) = –5 2. 4(2a + 1) – 5(a – 2) = 8 3. 5(2x – 1) – 4(x – 2) = –15 4. 2(5m + 7) – 3(3m + 2) = 4m 5. 4(5x + 2) – 5(3x + 1) = 2(x – 1) 6. b – {3 – [b – (2 – b) + 4]} = –2(–b – 3) 7. 4[3x – 2(3x – 1) + 3(2x – 1)] = 2[–2x + 3(x – 1)] – (5x – 1) 8. 30 – 3(m + 7) = –3(2m + 27) 9. 8x – 3(2x – 3) = –4(2 – x) + 3(x – 4) – 1 10. –5x – [4 – (3 – x)] = –(4x + 6) In Exercises 11–17, solve the equations and check your solutions. You don’t need to show all your steps. 11. 4t = 60 13. 15. p 8 y 7 = 1 = 4 17. s −4 = –4 12. x + 21 = 19 14. 7 – y = –11 16. 40 – x = 6 18. Solve –(3m – 8) = 12 – m 19. Denzel takes a two-part math test. In the first part he gets 49 points and in the second part he gets 4 9 of the x points. If his overall grade for the test was 65, find the value of x. 20. Latoya takes 3 science tests. She scores 24%, 43%, and x% in the tests. Write an expression for her average percentage over the 3 tests. Her average percentage is 52. Calculate the value of x. Solve the following equations. Show all your steps and justify them by citing the relevant properties. 21. x + 8 = 13 23. y – 7 = 19 25. 4m = 16 22. 11 + y = 15 24. x 3 = 4 ound Up ound Up RRRRRound Up ound Up ound Up RRRRRound Up ound Up ound Up ound Up ound Up Solving an equation means isolating the variable. Anything you don’t want on one side of the equation can be “taken over to the other side” by using the inverse operation. You’re always aiming for an expression of the form: “x = ...” (or “y = ...,” etc.). 8080808080 Section 2.2 Section 2.2 Section 2.2 — Manipulating Equations Section 2.2 Section 2.2 TTTTTopicopicopicopicopic 2.3.12.3.1 2.3.12.3.1 2.3.1 Section 2.3 actions actions ving Frrrrractions ving F ving F RRRRRemoemoemoemoemoving F actions actions ving F ving Frrrrractions RRRRRemoemoemoemoemoving F actions actions ving F ving F actions ving F actions California Standards: 4.0: Students simplify 4.0: Students simplify 4.0: Students simplify 4.0: Students simplify 4.0: Students simplify essions befororororore solving eeeeexprxprxprxprxpressions bef e solving e solving essions bef essions bef e solving e solving essions bef tions tions linear equa linear equa tions and linear equations tions linear equa linear equa in one variaariaariaariaariabbbbblelelelele,,,,, in one v in one v inequalities in one v in one v – 2) = 12. – 2) = 12. – 5) + 4( – 5) + 4( h as 3(23(23(23(23(2x – 5) + 4( h as sucsucsucsucsuch as h as – 2) = 12. – 5) + 4(x – 2) = 12. – 2) = 12. – 5) + 4( h as What it means for you: You’ll use the least common multiple to remove fractional coefficients from equations. Key words: coefficient least common multiple Expressions often seem more complicated if they contain fractions. Getting rid of fractions isn’t too difficult — you just need to use the least common multiple (LCM) again. tions Easier tions Easier es Solving Equa es Solving Equa actions Mak ving Frrrrractions Mak actions Mak ving F RRRRRemoemoemoemoemoving F ving F tions Easier es Solving Equations Easier actions Makes Solving Equa tions Easier es Solving Equa actions Mak ving F To solve equations that contain fractional coefficients, you can get rid of all the fractional coefficients by multiplying both sides of the equation by any common multiple of the denominators of the fractions. You don’t have to remove fractions, but it can make solving the equation a lot easier. The most efficient thing to multiply by is the least common multiple (LCM). If you needed to solve 1 6 common multiple of 4 and 6. To find the LCM, list the prime factors: , you first need to multiply by the least − = Now write each prime factor the greatest number of times it appears in any of the factorizations. The prime factor 2 occurs twice in the factorization of 4, so count two of them. The prime factor 3 occurs only once in the factorization of 6, so just count one of them. Then multiply the factors together to get the LCM: 2 × 2 × 3 = 12 So the LCM of 4 and 6 is 12. Example Example Example Example Example 11111 Solve the equation − = 1 x 1 4 1 6 x . Solution Solution Solution Solution Solution The LCM of the denominators is 12 — so multiply both sides of the equation by 12: × − × = × 1 12 1 x 12 1 12 1 1 4 3x – 12 = 2x 3x – 2x = 12 x = 12 1 6 x Section 2.3 Section 2.3 Section 2.3 — More Equations Section 2.3 Section 2.3 8181818181 Guided Practice In Exercises 1–2, find the least common multiple of the den
ominators: 1. 1 9 x + 13 – In Exercises 3–6, solve the equations for the unknown variable. 3. 1 2 x – 1 = x 5. 1 10 . 1 10 x – 3 = 1 4 x 6 or More Fe Fe Fe Fe Frrrrractions actions actions o or Mor k Out the LCM fCM fCM fCM fCM for or or or or TTTTTwwwwwo or Mor o or Mor k Out the L ou Can WWWWWororororork Out the L k Out the L ou Can YYYYYou Can ou Can actions actions o or Mor k Out the L ou Can Example Example Example Example Example 22222 Solve and check the root of Solution Solution Solution Solution Solution LCM of 3, 6, and 2 is 6 × 2x – 1 × 5x – 6 × 3 = 3 × 1x – 6 × 5 4x – 5x – 18 = 3x – 30 –x – 18 = 3x – 30 –x – 3x = –30 + 18 –4x = –12 − − x 12 4 − − 4 4 x = 3 = Checking the solution 15 Check it out: Since the left side of the equation is equal to the right side of the equation, x = 3 is a correct solution of the equation. 5 3 2 − 3 10 2 −11 − = − − − 8282828282 Section 2.3 Section 2.3 Section 2.3 — More Equations Section 2.3 Section 2.3 Guided Practice In Exercises 7–14, solve your answer for the unknown variable: 7. 9. 2 3 1 8 11. 13. 14 + 22 – 2 3 x n – 1 8 n 10. 8 = – 3 5 (2x – 1 = 17 – 1 2 x – 3 8 x 12. 1 2 + 7 10 x – 1 5 x + 17 = 10 20 + 3 2 x (a – 16) + 3 5 = 7 – 1 15 (a + 4) m + 5 2 (m – 11 + m) Independent Practice In Exercises 1–2, find the least common multiple of the denominators: 1 In Exercises 3–8, solve the equation for the unknown variable: 3. 5. 7 10 . 6. 8 21 x – 1 3 x + 6 (m – 2) – 1 5 m = – 1 5 The sum of the measures of the angles of a triangle is 180°. In Exercises 9–11, find the value of x. 9. x 10. 11. 1 2( – 30) x 1 x 4 1 5 x x + 10 + 2 1 x + 5 10 x 2 5 x 2 5 x 12. Qiaofang bought a new car for $27,000. If the value of the car depreciates as 27,000 – 7100 3 n, where n is the number of years since purchase, when will it be worth $12,800? 13. Mary’s weekly allowance increases every year as 2 5 x + $2, where x is Mary’s age. How old is Mary if she gets $5.20 allowance? ound Up ound Up RRRRRound Up ound Up ound Up RRRRRound Up ound Up ound Up ound Up ound Up If you multiply one side of an equation by a value, then you need to multiply the other side by the same thing. Make sure that the least common multiple is a multiple of the denominator of every fraction in the equation — otherwise you’ll still end up with annoying fractions in the equation. Section 2.3 Section 2.3 Section 2.3 — More Equations Section 2.3 Section 2.3 8383838383 TTTTTopicopicopicopicopic 2.3.22.3.2 2.3.22.3.2 2.3.2 California Standards: 4.0: Students simplify 4.0: Students simplify 4.0: Students simplify 4.0: Students simplify 4.0: Students simplify essions befororororore solving eeeeexprxprxprxprxpressions bef e solving e solving essions bef essions bef e solving e solving essions bef tions tions linear equa linear equa tions and linear equations tions linear equa linear equa in one variaariaariaariaariabbbbblelelelele,,,,, in one v in one v inequalities in one v in one v – 2) = 12. – 2) = 12. – 5) + 4( – 5) + 4( h as 3(23(23(23(23(2x – 5) + 4( h as sucsucsucsucsuch as h as – 2) = 12. – 5) + 4(x – 2) = 12. – 2) = 12. – 5) + 4( h as What it means for you: You’ll use the least common multiple to remove fractional coefficients from equations. Key words: coefficient least common multiple ficients in ficients in actional Coef actional Coef FFFFFrrrrractional Coef ficients in actional Coefficients in ficients in actional Coef ficients in ficients in actional Coef actional Coef FFFFFrrrrractional Coef ficients in actional Coefficients in ficients in actional Coef essions essions aic Expr aic Expr AlgAlgAlgAlgAlgeeeeebrbrbrbrbraic Expr essions aic Expressions essions aic Expr AlgAlgAlgAlgAlgeeeeebrbrbrbrbraic Expr essions essions aic Expr aic Expr aic Expressions essions aic Expr essions In the last Topic you dealt with equations with fractional coefficients — but sometimes the coefficients apply to more complicated values. ficients Firststststst ficients Fir ficients Fir actional Coef Eliminate Fte Fte Fte Fte Frrrrractional Coef actional Coef Elimina Elimina actional Coefficients Fir ficients Fir actional Coef Elimina Elimina Some equations contain fractional coefficients — and you can’t isolate the variables until you sort out the fractions. (x – 2) is all divided by 5. You need to deal with the fraction before you can get x on its own 10 The denominators are all different, so multiply the equation by the LCM of all three denominators. If there are fractional coefficients in the equation, multiply both sides of the equation by the least common multiple of the denominators of the fractions to remove the fractional coefficients. You need to keep each numerator (the “x – 2,” “x – 3,” and “3”) as a group. Then you can apply the distributive property to eliminate the grouping symbols. Example Example Example Example Example 11111 Solve − 10 . Don’t forget: See Lesson 2.3.1 for more on the least common multiple. Solution Solution Solution Solution Solution There are 3 different denominators, so you need the LCM. List the prime factors, and use them to work out the LCM. Check it out: Use parentheses to group each numerator before you multiply by the LCM. 5 = 5 6 = 2 × 3 10 = 2 × 5 So LCM = 2 × 3 × 5 = 30 Multiply both sides of the equation by 30 to clear the fractional coefficients. 30 = × 1 3 10 − ) × 30 − × 1 30 1 6(x – 2) – 5(x – 3) = 3 × 3 6x – 12 – 5x + 15 = 9 6x – 5x – 12 + 15 = 9 x + 3 = 9 x = 6 8484848484 Section 2.3 Section 2.3 Section 2.3 — More Equations Section 2.3 Section 2.3 Check it out 12 = So LCM = 2 × 2 × 3 = 12 Example Example Example Example Example Solve x − 11 12 22222 − 1 6 = + 1 x 4 . Solution Solution Solution Solution Solution − 11 12 ( ) 11 12 × − × 12 12 1 1 11 – 2(x – 1) = 3(x + 1) 11 – 2x + 2 = 3x + 3 –5x = –10 x = 2 12 = × 1 ( x + 1 ) 4 Independent Practice Solve each of the following equations: 1. 2 − m = 3 10 . 2 5 y ( 5 2 5. 1 + = 10 − ) 1 6 x 6 11 + = 12 + . 8 10 ) 1 = − 7 10 12 ( 10 10. 11. 12 21 ) 5 + 2 ound Up ound Up RRRRRound Up ound Up ound Up ound Up ound Up RRRRRound Up ound Up ound Up This Topic is really important, because you can’t simplify an expression fully until you’ve got rid of any fractional coefficients. Luckily, you’ve got the LCM to help you. Section 2.3 Section 2.3 Section 2.3 — More Equations Section 2.3 Section 2.3 8585858585 TTTTTopicopicopicopicopic 2.3.32.3.3 2.3.32.3.3 2.3.3 California Standards: 4.0: Students simplify 4.0: Students simplify 4.0: Students simplify 4.0: Students simplify 4.0: Students simplify essions befororororore solving eeeeexprxprxprxprxpressions bef e solving e solving essions bef essions bef e solving e solving essions bef tions tions linear equa linear equa tions and linear equations tions linear equa linear equa in one variaariaariaariaariabbbbblelelelele,,,,, in one v in one v inequalities in one v in one v – 2) = 12. – 2) = 12. – 5) + 4( – 5) + 4( h as 3(23(23(23(23(2x – 5) + 4( h as sucsucsucsucsuch as h as – 2) = 12. – 5) + 4(x – 2) = 12. – 2) = 12. – 5) + 4( h as What it means for you: You’ll remove decimal coefficients from equations by multiplying by a suitable power of 10. Key words: coefficient least common multiple ting Decimal ting Decimal Elimina Elimina ting Decimal Eliminating Decimal ting Decimal Elimina Elimina ting Decimal ting Decimal Elimina Elimina Eliminating Decimal ting Decimal ting Decimal Elimina Elimina ficients ficients CoefCoefCoefCoefCoefficients ficients ficients CoefCoefCoefCoefCoefficients ficients ficients ficients ficients If the coefficients are decimal, you still just need to multiply the equation by a suitable number — in fact, this time it’s slightly easier. er of 10 10 10 10 10 er of y a Py a Pooooowwwwwer of er of y a Py a P ying b ying b y Multipl y Multipl Decimals b Decimals b Get Rid of Get Rid of ying by a P y Multiplying b Decimals by Multipl Get Rid of Decimals b er of ying b y Multipl Decimals b Get Rid of Get Rid of Solving an equation that contains decimals can be made easier if it is first converted to an equivalent equation with integer coefficients. The method is the same as with fractions — you just multiply both sides of the equation by a suitable number. The idea is to multiply both sides of the equation by a large enough power of 10 to convert all decimals to integer coefficients. For example, 0.35 means “35 hundredths,” so you can write it So multiplying by 100 gives you the integer 35. 35 100 . Example Example Example Example Example 11111 Solve the equation 0.35x – 12 = –0.15x. Solution Solution Solution Solution Solution Multiply both sides of the equation by 100: 100(0.35x – 12) = 100(–0.15x) 100 × 0.35x – 100 × 12 = 100 × –0.15x 35x – 1200 = –15x 50x = 1200 x = 24 Example Example Example Example Example 22222 Solve 0.75x – 0.65(13 – x) = 8.35. Solution Solution Solution Solution Solution Multiply both sides of the equation by 100: 100[0.75x – 0.65(13 – x)] = 100(8.35) 100(0.75x) – 100(0.65)(13 – x) = 100(8.35) 75x – 65(13 – x) = 835 75x – 845 + 65x = 835 75x + 65x = 835 + 845 140x = 1680 x = 12 8686868686 Section 2.3 Section 2.3 Section 2.3 — More Equations Section 2.3 Section 2.3 Guided Practice In Exercises 1–10, solve the equation for the unknown variable: 1. 0.04x – 0.12 = 0.01x 2. –0.01x + 0.3 = 0.03x + 0.18 3. 0.06x – 0.09 = 0.05x + 0.12 – 0.15 4. 0.11 – 0.03x – 0.03 = 0.02x – 0.02 5. 0.16m + 2 = 0.03(5m + 2) 6. 0.01a = 0.45 – (0.04a + 0.15) 7. 0.25(x + 4) + 0.10(x – 2) = 3.60 8. 0.10(x + 100) = 0.08(x – 6) 9. 0.25(x – 50) = 0.2(x – 10) + 5 10. 0.25(x + 8) + 0.10(8 – x) + 0.05x = 3.60 y Higher Numbersssss ou Need to Multiply by by by by by Higher Number y Higher Number y Higher Number ou Need to Multipl Sometimes YYYYYou Need to Multipl ou Need to Multipl Sometimes Sometimes y Higher Number ou Need to Multipl Sometimes Sometimes In Examples 1 and 2 you needed to multiply both sides of each equation by 100. You can’t always just multiply by 100, though. If any of the decimals have more than 2 decimal places, you’ll need a higher power of 10. Example Example Example Exa
mple Example 33333 Solve 0.015x = 0.2 – 0.025x. Solution Solution Solution Solution Solution Multiply both sides by 1000 this time. 1000(0.015x) = 1000(0.2 – 0.025x) 15x = 200 – 25x 40x = 200 x = 5 If your longest decimal has 3 decimal places, multiply by 103 = 1000. If the longest decimal has 4 decimal places, multiply by 104, and so on. Guided Practice In Exercises 11–20, solve each equation for the unknown variable. 11. 0.3(2x + 7) = 1.8 12. 2.8 = 0.2(4x + 2) 13. 0.016 = 0.002(a – 1) – 0.001a 14. 0.1x + 0.1(3x – 8) = – 230 15. 0.001x + 0.05(2x – 3) = 35.2 16. 0.012p – 0.065p – 0.7 = 0.5 17. 0.003(x + 7) = 0.01x 18. 0.006(x + 3) = 0.002(x + 31) 19. 0.072 – 0.006j = 0.032 + 0.08j 20. 0.003(2x – 0.3) + 0.001x = 0.1979 Section 2.3 Section 2.3 Section 2.3 — More Equations Section 2.3 Section 2.3 8787878787 Independent Practice Solve each of the equations in Exercises 1–17: 1. 0.11x – 2.3 + 0.4(2x – 1) = 0.25(3x + 8) 2. 0.5x – 1 = 0.7x + 0.2 3. 0.3v + 4.2v – 11 = 1.5v + 4 4. 1.8(y – 1) = 3.1y + 2.1 5. 0.25(3x – 2) + 0.10(4x + 1) + 0.75(x – 1) = 4.55 6. 0.125(3b – 8) – 0.25(b + 5) – 0.2 = 0.2(2b + 7) 7. 0.4(x + 7) – 0.15(2x – 5) = 0.7(3x – 1) – 0.75(4x – 3) 8. 0.21(x – 1) – 0.25(2x + 1) = 0.5(2x + 1) – 0.6(4x – 15) + 0.03 9. 0.9(2v – 5) + 0.20(–3v – 1) = 0.22(4v – 3) 10. 2.46 – 0.52(x – 10) = 0.35(4x + 8) – 2.82 11. –0.20x – 1.10(3x – 11) = 1.25(2x – 5) – 6.25 12. 16 + 0.50y = 0.60(y + 20) 13. 20 + 1.20m = 1.10(m + 25) 14. 0.03x + 0.30(900 – x) = 72 15. 0.06y + 0.3y – 0.1 = 0.26 16. 0.04k + 0.06(40,000 – k) = 2100 17. –0.02[0.4 – 0.1(2 + 3x)] = 0.004x + 0.005 18. The sides of an equilateral triangle measure 0.2(10x + 90) units each. If the perimeter of the triangle is 612.6 units, find x. 19. The area of a rectangle is 34 units2. If the width is 0.25 units and the length is (x + 10) units, find the value of x. 20. The cost per minute to make a call is $0.05. If Meimei talks for (x + 20) minutes and the call costs $4.85, what is the value of x? 21. A cell phone plan charges $25 per month plus $0.10 per minute. If your monthly bill is $39.80, write and solve an equation to find out the number of minutes on your bill. 22. A moving van rents for $40 a day plus $0.08 a mile. Ed’s bill is $58.24 and he had the van for one day. Write and solve an equation to find out how many miles he drove. 23. Eylora has x quarters with a value of $0.25x. Emily has dimes that value 0.10(x + 8). If they have a total of $5.00 in coins, how many coins does Eylora have? 24. Michael, William, and Daniel are playing a game. Michael has x points, William has 0.01(x + 18,000) points, and Daniel has 0.02(x – 800) points. If together they have earned 11,288 points, how many points does Michael have? ound Up ound Up RRRRRound Up ound Up ound Up RRRRRound Up ound Up ound Up ound Up ound Up Whether you’re dealing with fractional or decimal coefficients, the method’s essentially the same — you multiply everything by a number that will make the algebra easier and mistakes less likely. Then you can start isolating the variable. 8888888888 Section 2.3 Section 2.3 Section 2.3 — More Equations Section 2.3 Section 2.3 TTTTTopicopicopicopicopic 2.4.12.4.1 2.4.12.4.1 2.4.1 California Standards: 4.0: Students simplify 4.0: Students simplify 4.0: Students simplify 4.0: Students simplify 4.0: Students simplify essions befororororore solving eeeeexprxprxprxprxpressions bef e solving e solving essions bef essions bef e solving e solving essions bef tions tions linear equa linear equa tions and linear equations tions linear equa linear equa in one variaariaariaariaariabbbbblelelelele,,,,, in one v in one v inequalities in one v in one v – 2) = 12. – 2) = 12. – 5) + 4( – 5) + 4( h as 3(23(23(23(23(2x – 5) + 4( h as h as sucsucsucsucsuch as – 2) = 12. – 5) + 4(x – 2) = 12. – 2) = 12. – 5) + 4( h as 5.0:5.0:5.0:5.0:5.0: Students solv Students solv Students solveeeee Students solv Students solv ultiste mmmmmultiste lems,,,,, lems lems p prp proboboboboblems p prp pr ultiste ultistep pr lems ultiste d prd proboboboboblems incincincincincluding w lems,,,,, lems lems d prd pr luding wororororord pr luding w luding w luding w lems tions tions olving linear equa olving linear equa inininininvvvvvolving linear equa tions olving linear equations tions olving linear equa in one in one and linear inequalities in one in one in one vvvvvariaariaariaariaariabbbbble and pr le and prooooovidevidevidevidevide le and pr le and pr le and pr h steppppp..... h ste h ste or eac or eac tion f tion f justifica justifica or each ste tion for eac justification f h ste or eac tion f justifica justifica What it means for you: You’ll set up and solve equations that model real-life situations. Key words: application linear equation Check it out: Look out for words that give math clues — for example, “twice” means two times and the word “is” means equals. tions of tions of pplica pplica AAAAApplica tions of pplications of tions of pplica AAAAApplica tions of tions of pplica pplica pplications of tions of pplica tions of tions tions Linear Equa Linear Equa tions Linear Equations tions Linear Equa Linear Equa tions tions Linear Equa Linear Equa Linear Equations tions tions Linear Equa Linear Equa “Applications” are just “real-life” tasks. In this Topic, linear equations start to become really useful. asksasks eal-Life”e”e”e”e” TTTTTasks asksasks eal-Lif tions are “Re “Re “Re “Re “Real-Lif eal-Lif tions ar tions ar Equa Equa tions of tions of pplica AAAAApplica pplica Equations ar tions of Equa pplications of eal-Lif tions ar Equa tions of pplica Applications questions are word problems that require you to set up and solve an equation. First decide how you will label the variables... ...then write the task out as an equation... ...making sure you include all the information given... ...then you can solve your equation. Example Example Example Example Example 11111 The sum of twice a number c and 7 is 21. Set up and solve an equation to find c. Solution Solution Solution Solution Solution You’re given the label “c” in the question — so just write out the equation. 2c + 7 = 21 2c = 14 c = 7 Guided Practice 1. Twice a number c plus 17 is 31. Find the value of c. 2. Three times a number k minus 8 is 43. Find the number k. 3. The sum of four times a number m and 17 is the same as 7 less than six times the number m. Find the number m. 4. Seven minus five times the number x is equal to the sum of four times the number x and 25. Find the number x. Section 2.4 Section 2.4 Section 2.4 — Using Equations Section 2.4 Section 2.4 8989898989 Check it out: “4v + 15” is just the phrase “15 more than four times v” written in math-speak. en the VVVVVariaariaariaariaariabbbbbleslesleslesles en the ys Be Givvvvven the en the ys Be Gi YYYYYou ou ou ou ou WWWWWononononon’’’’’t t t t t AlAlAlAlAlwwwwwaaaaays Be Gi ys Be Gi en the ys Be Gi Sometimes you’ll have to work out for yourself what the variables are, and decide on suitable labels for them. Example Example Example Example Example 22222 Juanita’s age is 15 more than four times Vanessa’s age. The sum of their ages is 45. Set up and solve an equation to find their ages. Solution Solution Solution Solution Solution This time you have to decide for yourself how to label each term. Let v = Vanessa’s age 4v + 15 = Juanita’s age v + (4v + 15) = 45 v + 4v + 15 = 45 5v + 15 = 45 5v = 30 v = 6 es is 45 es is 45 their aggggges is 45 their a their a he sum of he sum of TTTTThe sum of es is 45 he sum of their a es is 45 their a he sum of Plug in the value for v to get Juanita’s age: 4v + 15 = 4 × 6 + 15 = 24 + 15 = 39 So Vanessa is 6 years old and Juanita is 39 years old. Guided Practice 5. The length of a rectangular garden is 3 meters more than seven times its width. Find the length and width of the garden if the perimeter of the garden is 70 meters. 6. A rectangle is 4 meters longer than it is wide. The perimeter is 44 meters. What are the dimensions and area of the rectangle? 7. Abraham’s age is 4 less than half of Dominique’s age. Dominique’s age is 6 more than three times Juan’s age. The sum of their ages is 104. Find the age of each person. 8. The sides of an isosceles triangle are each 2 inches longer than the base. If the perimeter of the triangle is 97 inches, what are the lengths of the base and sides of the triangle? 9. The sum of two consecutive integers is 117. What are the integers? 9090909090 Section 2.4 Section 2.4 Section 2.4 — Using Equations Section 2.4 Section 2.4 Independent Practice 1. Find the value of x, if the line segment shown on the right is 21 cm long. Also, find the length of each part of the line segment. (4x – 5) (3x – 2) cm 21 cm cm 2. Point M is the midpoint of the line segment shown. Find the value of x and the length of the entire line segment. 6x – 11 4x + 23 M 3. The perimeter of the rectangular plot shown below is 142 feet. Find the dimensions of the plot. (6 – 1) ft x (4 + 2) ft x 4. The sum of the interior angles of a triangle is 180°. Find the size of each angle in the triangle shown below. 4x 3x 5x 5. The sum of one exterior angle at each vertex of any convex polygon is 360°. Find the size of each exterior angle shown around the triangle below. 3 + 10 y 2 + 45 y 4 – 10 y 6. The interior angles of a triangle sum to 180°. Find the size of each angle in the triangle sketched on the right. 3x 3x + 5 x 7. A rectangular garden has a length that is five meters less than three times its width. If the length is reduced by three meters and the width is reduced by one meter, the perimeter will be 62 meters. Find the dimensions of the garden. ound Up ound Up RRRRRound Up ound Up ound Up ound Up ound Up RRRRRound Up ound Up ound Up The thing to do with any word problem is to write out a math equation that describes the same situation. Then you can use all the techniques you’ve learned to solve the equation. Section 2.4 Section 2.4 Section 2.4 — Using Equations Section 2.4 Section 2.4 9191919191 TTTTTopicopicopicopicopic 2.4.22.4.2 2.4.22.4.2 2.4.2
California Standards: 4.0: Students simplify 4.0: Students simplify 4.0: Students simplify 4.0: Students simplify 4.0: Students simplify essions befororororore solving eeeeexprxprxprxprxpressions bef e solving e solving essions bef essions bef e solving e solving essions bef tions tions linear equa linear equa tions and linear equations tions linear equa linear equa in one variaariaariaariaariabbbbblelelelele,,,,, in one v in one v inequalities in one v in one v – 2) = 12. – 2) = 12. – 5) + 4( – 5) + 4( h as 3(23(23(23(23(2x – 5) + 4( h as h as sucsucsucsucsuch as – 2) = 12. – 5) + 4(x – 2) = 12. – 2) = 12. – 5) + 4( h as 5.0:5.0:5.0:5.0:5.0: Students solv Students solv Students solveeeee Students solv Students solv ultiste mmmmmultiste lems,,,,, lems lems p prp proboboboboblems p prp pr ultiste ultistep pr lems ultiste d prd proboboboboblems incincincincincluding w lems,,,,, lems lems d prd pr luding wororororord pr luding w luding w luding w lems tions tions olving linear equa olving linear equa inininininvvvvvolving linear equa tions olving linear equations tions olving linear equa in one in one and linear inequalities in one in one in one vvvvvariaariaariaariaariabbbbble and pr le and prooooovidevidevidevidevide le and pr le and pr le and pr h steppppp..... h ste h ste or eac or eac tion f tion f justifica justifica or each ste tion for eac justification f h ste or eac tion f justifica justifica What it means for you: You’ll set up and solve equations that model real-life situations involving money. Key words: coin task Check it out: Convert everything in the question to the same units — in this example it’s all been converted into cents. asksasks asksasks Coin TTTTTasks Coin Coin Coin Coin Coin TTTTTasks asksasks asksasks Coin Coin Coin Coin Lots of math problems involve objects of value such as coins and stamps. They’re often grouped together under the title “coin tasks.” Solving coin tasks uses all the same methods that you used in the last Topic. About Moneyyyyy About Mone About Mone tions tions e Linear Equa asks Invvvvvolvolvolvolvolve Linear Equa e Linear Equa asks In Coin Coin TTTTTasks In asks In Coin Coin tions About Mone e Linear Equations About Mone tions e Linear Equa asks In Coin If you have a collection of coins, there are two quantities you can use to describe it — how many coins there are, and how much they are worth. In coin tasks you’ll have to use both quantities. Example Example Example Example Example 11111 Jazelle has a coin collection worth $3.50. She only has nickels, dimes, and quarters. If she has four more dimes than quarters and twice as many nickels as she has dimes, how many coins of each kind does she have in her collection? Solution Solution Solution Solution Solution The first thing to do is to write expressions for how many of each type of coin she has. Call the number of quarters q. Then the number of dimes is q + 4. And the number of nickels is 2(q + 4) = 2q + 8. Then write expressions for the total value (in cents) of each type of coin. There are q quarters, so the total value of the quarters is 25q. There are q + 4 dimes, so the total value of the dimes is 10(q + 4) = 10q + 40. There are 2q + 8 nickels, so the total value of the nickels is 5(2q + 8) = 10q + 40. Write an equation to represent the fact that the collection is worth $3.50. 25q + (10q + 40) + (10q + 40) = 100(3.50) Solving for q gives: 45q + 80 = 350 45q = 270 q = 6 So Jazelle has: q = 6 quarters, q + 4 = 6 + 4 = 10 dimes, 2(q + 4) = 2(6 + 4) = 20 nickels. 9292929292 Section 2.4 Section 2.4 Section 2.4 — Using Equations Section 2.4 Section 2.4 Example Example Example Example Example 22222 Tickets to a puppet show sell at $2.50 for children and $4.50 for adults. There are five times as many children at a performance as there are adults and the show raises $3009. How many adult and children’s tickets were sold for the show? Solution Solution Solution Solution Solution Let x = adult tickets sold. Then 5x = children’s tickets sold. The value is the number of tickets multiplied by the cost of each ticket. So 4.50x = amount raised from adults (in dollars) 5x(2.50) = amount raised from children Write an equation showing that the sum of these amounts is $3009: 4.50x + 5x(2.50) = 3009 17x = 3009 x = 177 So 177 adult tickets and 5 × 177 = 885 children’s tickets were sold. Independent Practice 1. The total cost of buying some music CDs and having them shipped to Charles was $211.50. If the CDs cost $11.50 each and the shipping for the box of CDs was $4.50, how many CDs did Charles receive? 2. Jerome bought 12 CDs. Some of the CDs cost $7.50 each and the rest cost $6.50 each. How many CDs were bought at each price if Jerome spent a total of $82? 3. Rajan’s coin collection is valued at $22.70. He has one fewer halfdollar coin than twice the number of dimes and four more quarters than three times the number of dimes. How many dimes, quarters, and half-dollars does Rajan have, assuming he has no other coins? 4. Liza has 60 coins in her collection. The coin collection consists of nickels, dimes, and quarters. She has five fewer quarters than nickels and ten more dimes than quarters. How many coins of each kind does Liza have? 5. A school cafeteria cashier has collected $243 in one-dollar, fivedollar, and ten-dollar bills. The number of one-dollar bills is eight more than 20 times the number of ten-dollar bills. The cashier also has seven more than twice the number of ten-dollar bills in five-dollar bills. How many bills of each value does the cashier have? Section 2.4 Section 2.4 Section 2.4 — Using Equations Section 2.4 Section 2.4 9393939393 6. Dan’s algebra class is planning a summer afternoon get-together. Dan is supposed to bring some melons at $1.25 each, juice boxes at $0.50 each, and granola bars for $0.75 each. If he buys nine more than ten times the number of melons in juice boxes and seven more than five times the number of melons in granola bars, how many items of each kind did he buy with $29.75? 7. Martha bought some baseball uniforms for $313 and had them shipped to her. If the baseball uniforms cost $23.75 each and the shipping was $4.25 for the whole order, how many baseball uniforms did Martha buy? 8. Fifteen children’s books cost $51.25. Some were priced at $2.25 each, and the rest of the books were sold at $4.75 each. How many books were purchased at each price? 9. Dwight purchased various stamps for $16.35. He purchased 12 more 25¢ stamps than 35¢ stamps. The number of 30¢ stamps was four times the number of 35¢ stamps. Finally, he bought five 15¢ stamps. How many of each kind of stamp did Dwight buy? 10. A waiter has collected 150 coins from the tips he receives from his customers. The coins consist of nickels, dimes, and quarters. He has five more than twice the nickels in dimes and five more than four times the nickels in quarters. i) How many coins of each kind does the waiter have? ii) How much money does the waiter have? 11. Jessica has two more nickels than dimes, and three more quarters than nickels, but no other coins. If she has a total of $5.35, how many coins of each kind does she have? 12. A grocer’s deposit box contains 150 coins worth $12.50. They are all nickels and dimes. Find the number of each coin in the box. 13. A store sells nineteen different video games. Several games are priced at $19.99, while half that number are priced at $39.99, and 4 are priced at $49.99. How many games are priced at $19.99? 14. Mark bought packets of popcorn, drinks, and bags of nuts for him and some friends at the movie theater. Everyone got one of each. Drinks cost $4.50 each, packets of popcorn cost $3.75 each, and bags of nuts cost $2.00 each. If he spent $41, how many friends did Mark have with him? 15. John has quarters, nickels, and dimes. He has 4 more nickels than quarters and twice as many dimes and nickels. If he has $6.00, how many quarters does he have? ound Up ound Up RRRRRound Up ound Up ound Up RRRRRound Up ound Up ound Up ound Up ound Up To answer these questions, you need to use (i) the number of items, and (ii) their value. Then, when you have set up your equation and solved it, be sure to give your final answer in the form asked for in the problem. 9494949494 Section 2.4 Section 2.4 Section 2.4 — Using Equations Section 2.4 Section 2.4 TTTTTopicopicopicopicopic 2.5.12.5.1 2.5.12.5.1 2.5.1 Section 2.5 asksasks asksasks e Integgggger er er er er TTTTTasks e Inte e Inte Consecutivvvvve Inte Consecuti Consecuti e Inte Consecuti Consecuti e Integgggger er er er er TTTTTasks Consecutivvvvve Inte asksasks asksasks e Inte e Inte Consecuti Consecuti e Inte Consecuti Consecuti California Standards: 4.0: Students simplify 4.0: Students simplify 4.0: Students simplify 4.0: Students simplify 4.0: Students simplify essions befororororore solving eeeeexprxprxprxprxpressions bef e solving e solving essions bef essions bef e solving e solving essions bef tions tions linear equa linear equa tions and linear equations tions linear equa linear equa in one variaariaariaariaariabbbbblelelelele,,,,, in one v in one v inequalities in one v in one v – 2) = 12. – 2) = 12. – 5) + 4( – 5) + 4( h as 3(23(23(23(23(2x – 5) + 4( h as h as sucsucsucsucsuch as – 2) = 12. – 5) + 4(x – 2) = 12. – 2) = 12. – 5) + 4( h as 5.0:5.0:5.0:5.0:5.0: Students solv Students solv Students solveeeee Students solv Students solv ultiste mmmmmultiste lems,,,,, lems lems p prp proboboboboblems p prp pr ultiste ultistep pr lems ultiste d prd proboboboboblems incincincincincluding w lems,,,,, lems lems d prd pr luding wororororord pr luding w luding w luding w lems tions tions olving linear equa olving linear equa inininininvvvvvolving linear equa tions olving linear equations tions olving linear equa in one in one and linear inequalities in one in one in one vvvvvariaariaariaariaariabbbbble and pr le and prooooovidevidevidevidevide le and pr le and pr le and pr h steppppp..... h ste h ste or eac or eac tion f tion f justifica justifica
or each ste tion for eac justification f h ste or eac tion f justifica justifica What it means for you: You’ll solve word problems that refer to patterns of integers. Key words: consecutive integer common difference When you’re solving word problems, the most important thing to do is to write down what you know. Then create an equation that represents the relationship between the known and unknown quantities. Finally, you have to solve that equation. Sequences with a Common Difffffferererererence ence ence Sequences with a Common Dif Sequences with a Common Dif ence ence Sequences with a Common Dif Sequences with a Common Dif A sequence of integers with a common difference is a set of integers that increase by a fixed amount as you move from term to term 10 Common difference = 2 1 2 3 4 5 6 87 9 10 3 , 3 , 6 3 , 3 9 , Common difference = 3 Consecutive EVEN integers (for example, 2, 4, and 6) form a sequence with common difference 2. Consecutive ODD integers (for example, 1, 3, and 5) also form a sequence with common difference 2. So if you are given an even integer (or an odd integer) and you are asked to find the next even (or odd) integer, just add two. Example Example Example Example Example 11111 Check it out: Add 2 to x to find the second even integer, and another 2 to find the third. Find the three consecutive even integers whose sum is 48. Solution Solution Solution Solution Solution Call the first (smallest) even integer x. Then you can write down an expression for the other even integers in terms of x. 1st even integer = x 2nd even integer = x + 2 3rd even integer = x + 4 But the sum of the three even integers is 48, so x + (x + 2) + (x + 4) = 48. Now you can rearrange and simplify this formula, and then solve for x. x + (x + 2) + (x + 4) = 48 (x + x + x) + (2 + 4) = 48 3x + 6 = 48 3x = 42 x = 14 Section 2.5 Section 2.5 Section 2.5 — Consecutive Integer Tasks, Time and Rate Tasks Section 2.5 Section 2.5 9595959595 Example 1 continueduedueduedued Example 1 contin Example 1 contin Example 1 contin Example 1 contin So the smallest of the three even integers is 14. Now just add 2 to find the second even integer, and add another 2 to find the third. So the three consecutive even integers are: x = 14, x + 2 = 16, and x + 4 = 18. wns Firststststst wns Fir wns Fir or the Unkno or the Unkno essions f essions f rite Expr WWWWWrite Expr rite Expr or the Unknowns Fir essions for the Unkno rite Expressions f wns Fir or the Unkno essions f rite Expr The most important thing with any word problem is to first write it out again in math-speak. Example Example Example Example Example 22222 Find a sequence of four integers with a common difference of 4 whose sum is 92. Solution Solution Solution Solution Solution Call the first (smallest) integer v, for example. Then you can write the other three in terms of v. 1st integer = v 2nd integer = v + 4 3rd integer = v + 8 4th integer = v + 12 The sum of the four integers is 92, so: v + (v + 4) + (v + 8) + (v + 12) = 92 Now you can solve for v: (v + v + v + v) + (4 + 8 + 12) = 92 4v + 24 = 92 4v = 68 v = 17 So the smallest of the integers is 17. Since they differ by 4, the others must be 21, 25, and 29. Once you’ve completed the problem, do a quick answer check — add the integers together and see if you get what you want: 1st integer = v = 17 2nd integer = v + 4 = 21 3rd integer = v + 8 = 25 4th integer = v + 12 = 29 92 9696969696 Section 2.5 Section 2.5 Section 2.5 — Consecutive Integer Tasks, Time and Rate Tasks Section 2.5 Section 2.5 ✓ Example Example Example Example Example 33333 Find a sequence of four integers with a common difference of 2 whose sum is 944. Decide whether these integers are even or odd. Solution Solution Solution Solution Solution Call the first integer x, then write expressions for the others: Check it out: The difference between the integers is 2, whether they are odd or even. 1st integer = x 2nd integer = x + 2 3rd integer = x + 4 4th integer = x + 6 The sum of the four integers is 944, so: x + (x + 2) + (x + 4) + (x + 6) = 944 Now solve for x: 4x + 12 = 944 4x = 932 x = 233 So the smallest of the integers is 233, which means the others must be 235, 237, and 239. These are consecutive odd integers. Now do a quick answer check — add the integers together and see if you get what you want: 1st integer = x = 233 2nd integer = x + 2 = 235 3rd integer = x + 4 = 237 4th integer = x + 6 = 239 944 Guided Practice Find the unknown numbers in each of the following cases. 1. The sum of two consecutive integers is 103. 2. The sum of three consecutive integers is –138. 3. The sum of two consecutive even integers is 194. 4. The sum of three consecutive odd integers is –105. 5. The sum of a number and its double is 117. 6. Two integers have a difference of three and their sum is 115. 7. Three integers have a common difference of 3 and their sum is –78. 8. Four rational numbers have a common difference of 5 and a sum of 0. Section 2.5 Section 2.5 Section 2.5 — Consecutive Integer Tasks, Time and Rate Tasks Section 2.5 Section 2.5 9797979797 ✓ Independent Practice 1. Three consecutive integers have a sum of 90. Find the numbers. 2. Find the four consecutive integers whose sum is 318. 3. Find three consecutive integers such that the difference between three times the largest and two times the smallest integer is 30. 4. Find three consecutive integers such that the sum of the first two integers is equal to three times the highest integer. 5. Two numbers have a sum of 65. Four times the smaller number is equal to 10 more than the larger number. Find the numbers. 6. Four consecutive even integers have a sum of 140. What are the integers? 7. Find three consecutive even integers such that six more than three times the smallest integer is 54. 8. Find three consecutive odd integers whose sum is 273. 9. Find four consecutive odd integers such that 12 more than four times the smallest integer is 144. 10. Find three consecutive even integers such that six more than twice the first number is 94. 11. Find three consecutive even integers such that the product of 16 and the third integer is the same as the product of 20 and the second integer. 12. A 36-foot pole is cut into two parts such that the longer part is 11 feet longer than 4 times the shorter part. How long is each piece of the pole? 13. Find three consecutive odd integers such that four times the largest is one more than nine times the smallest integer. 14. Ten thousand people attended a three-day outdoor music festival. If there were 800 more girls than boys, and 1999 fewer adults than boys, how many people of each group attended the festival? ound Up ound Up RRRRRound Up ound Up ound Up RRRRRound Up ound Up ound Up ound Up ound Up Consecutive integer tasks are a strange application of math equations — but they appear a lot in Algebra I. Always make sure you’ve answered the question — you’ve always got to remember that your solution isn’t complete until you’ve stated what the integers actually are. 9898989898 Section 2.5 Section 2.5 Section 2.5 — Consecutive Integer Tasks, Time and Rate Tasks Section 2.5 Section 2.5 TTTTTopicopicopicopicopic 2.5.22.5.2 2.5.22.5.2 2.5.2 California Standards: 4.0: Students simplify 4.0: Students simplify 4.0: Students simplify 4.0: Students simplify 4.0: Students simplify essions befororororore solving eeeeexprxprxprxprxpressions bef e solving e solving essions bef essions bef e solving e solving essions bef tions tions linear equa linear equa tions and linear equations tions linear equa linear equa in one variaariaariaariaariabbbbblelelelele,,,,, in one v in one v inequalities in one v in one v – 2) = 12. – 2) = 12. – 5) + 4( – 5) + 4( h as 3(23(23(23(23(2x – 5) + 4( h as h as sucsucsucsucsuch as – 2) = 12. – 5) + 4(x – 2) = 12. – 2) = 12. – 5) + 4( h as 5.0:5.0:5.0:5.0:5.0: Students solv Students solv Students solveeeee Students solv Students solv ultiste mmmmmultiste lems,,,,, lems lems p prp proboboboboblems p prp pr ultiste ultistep pr lems ultiste d prd proboboboboblems incincincincincluding w lems,,,,, lems lems d prd pr luding wororororord pr luding w luding w luding w lems tions tions olving linear equa olving linear equa inininininvvvvvolving linear equa tions olving linear equations tions olving linear equa in one in one and linear inequalities in one in one in one vvvvvariaariaariaariaariabbbbble and pr le and prooooovidevidevidevidevide le and pr le and pr le and pr h steppppp..... h ste h ste or eac or eac tion f tion f justifica justifica or each ste tion for eac justification f h ste or eac tion f justifica justifica What it means for you: You’ll solve equation problems that refer to people’s ages. Key words: age task asksasks asksasks ted TTTTTasks ted ted AgAgAgAgAge-Re-Re-Re-Re-Relaelaelaelaelated ted ted TTTTTasks AgAgAgAgAge-Re-Re-Re-Re-Relaelaelaelaelated asksasks asksasks ted ted ted Age tasks relate ages of different people at various time periods — in the past, the present, or the future. asksasks ted TTTTTasks asksasks ted AgAgAgAgAge-Re-Re-Re-Re-Relaelaelaelaelated ted ted Your age at any point in your life can always be written as your current age plus or minus a certain number of years. If your current age is x years... ...then 5 years ago, your age was 5 fewer than x... ...and in 5 years’ time, your age will be 5 more than x. More generally, your age c years ago was: And your age in c years will be In much the same way, anybody’s age can always be related to someone else’s by adding or subtracting a certain number of years. wn Quantities wn Quantities or the Unkno or the Unkno essions f essions f rite Expr WWWWWrite Expr rite Expr wn Quantities or the Unknown Quantities essions for the Unkno rite Expressions f wn Quantities or the Unkno essions f rite Expr Solving an age task is pretty similar to solving a consecutive integer task. You need to write down expressions for the unknown quantities in terms of one variable (like x). Then you can use the information
in the question to write an equation that you can go on to solve. Example Example Example Example Example 11111 Charles is 7 years older than Jorge. In 20 years’ time, the sum of their ages will be 81 years. How old is each one now? Solution Solution Solution Solution Solution The first thing to do is write expressions relating all the ages to one another. Present You’re not told Jorge’s age, so call it x. Charles is currently 7 years older — that is, 7 more than x. Jorge’s age = x Charles’s age = x + 7 Future (in 20 years) Jorge will be 20 years older than at present. Charles will be 20 years older than at present. Section 2.5 Section 2.5 Section 2.5 — Consecutive Integer Tasks, Time and Rate Tasks Section 2.5 Section 2.5 9999999999 Example 1 continueduedueduedued Example 1 contin Example 1 contin Example 1 contin Example 1 contin Jorge’s future age = x + 20 Charles’s future age = (x + 7) + 20 = x + 27 Now use the information from the question to combine these expressions into an equation. The sum of their ages in 20 years will be 81, so: (x + 20) + (x + 27) = 81 Now solve your equation for x: (x + 20) + (x + 27) = 81 2x + 47 = 81 2x = 34 x = 17 So Jorge is currently 17 years old. That means that Charles is 17 + 7 = 24 years old. Example Example Example Example Example 22222 Juanita is twice as old as Vanessa. If 5 years were subtracted from Vanessa’s age and 2 years added to Juanita’s age, then Juanita’s age would be five times Vanessa’s. How old are the girls now? Solution Solution Solution Solution Solution As before, start by writing down mathematical expressions for the ages mentioned in the question. Present You’re not given Vanessa’s age, so call it v. Juanita is twice as old — that is, twice v. Vanessa’s age = v Juanita’s age = 2v “Adjusted” ages “If 5 years is subtracted from Vanessa’s age...” = v – 5 “If 2 years is added to Juanita’s age...” = 2v + 2 Now you can write an equation. Remember that Juanita’s “adjusted” age is five times as big as Vanessa’s. 2v + 2 = 5(v – 5) Now solve for v to find Vanessa’s current age: 2v + 2 = 5(v – 5) 2v + 2 = 5v – 25 –3v = –27 v = 9 So Vanessa is 9 years old. And since Juanita is twice as old, Juanita is 2 × 9 = 18 years old. Check it out: Watch out — you haven’t answered the question unless you say what v = 9 means for the people’s ages. 100100100100100 Section 2.5 Section 2.5 Section 2.5 — Consecutive Integer Tasks, Time and Rate Tasks Section 2.5 Section 2.5 Example Example Example Example Example 33333 A father is three times as old as his daughter. In 15 years’ time, the father will be twice as old as his daughter. What are their current ages? Solution Solution Solution Solution Solution Present You are not given the daughter’s age, so call it x. The father is three times as old — three times x. Daughter’s age = x Father’s age = 3x Future (in 15 years) The daughter and father will both be 15 years older. Daughter’s future age = x + 15 Father’s future age = 3x + 15 Don’t forget: Remember to say what your value of x means in the context of the question. In 15 years, the father’s age will be twice as great as his daughter’s. Write this as an equation, and solve: 3x + 15 = 2(x + 15) 3x + 15 = 2x + 30 x = 15 So the daughter is 15 years old, and the father is 3 × 15 = 45 years old. Guided Practice 1. Eylora is 4 times as old as Leo. If the sum of their ages is 5, how old is Eylora? 2. Clarence is 8 years older than Maria. In 24 years, the sum of their ages will be 100. How old is Clarence? 3. Tyler, Nick, and Sid are brothers. The sum of their ages is 54. The oldest brother, Nick, is 2 years older than Sid, and Sid is 2 years older than Tyler. How old is Sid? 4. A father is 9 times older than his daughter and 2 years older than his wife. If the sum of their ages is 74, how old is the father? 5. Ruby and Emily are twins. Rebecca is 6 more than 2 times Ruby and Emily's age. Altogether the sum of their three ages is 50. How old are Emily and Ruby? 6. Santos is 30 years older than his daughter Julia. If their ages are increased by 10% and added together the sum is 77. How old is Santos? 7. James is 4 years less than 7 times the age of his daughter, who is 4 more than half of her brother's age. The sum of their ages is 38. How old is James? Section 2.5 Section 2.5 Section 2.5 — Consecutive Integer Tasks, Time and Rate Tasks Section 2.5 Section 2.5 101101101101101 Independent Practice 1. Keisha, Juan, and Jose are friends who are all different ages. There is a 2 year difference in age between the oldest and youngest. Juan is not as old as Jose, but he is older than Keisha. If the sum of their ages is 36, how old is the oldest child? 2. Sonita’s father is three times as old as she is now. In ten years, her father will be twice as old as Sonita will be then. How old are Sonita and her father now? 3. Kadeeja is four times as old as her niece. In three years, Kadeeja will be three times as old as her niece. How old is each of them now? 4. Sally is twice as old as Daniel. Ten years ago, the sum of their ages was 70 years. How old is each one of them now? 5. Andy is four times as old as Alejandro. Five years ago, Andy was nine times as old as Alejandro. How old is each one now? 6. Chris is 40 years younger than his uncle. In ten years’ time the sum of their ages will be 80 years. How old are they now? 7. Jorge is three times Martha’s age. If 30 years is added to Martha’s age and 30 years is subtracted from Jorge’s age, their ages will be equal. How old is each person now? 8. Mia’s age in 20 years will be the same as Simon’s age is now. Ten years from now, Simon’s age will be twice Mia’s age. How old is each one now? 9. Paula is three times as old as Duncan. If four is subtracted from Duncan’s age and six is added to Paula’s age, Paula will then be four times as old as Duncan. How old are they now? 10. If you decrease Marvin's age by 25%, you will find his age 4 years ago. How old is Marvin now? 11. If you increase Qin's age by 75% then you will find his age 6 years from now. How old will Qin be in 6 years? 12. Jaya is 10 years younger than Sid. If you increase both of their ages by 20%, the difference between their ages is 18 less than Jaya's current age. How old is Sid? ound Up ound Up RRRRRound Up ound Up ound Up RRRRRound Up ound Up ound Up ound Up ound Up Age tasks are just another example of real-life equations. As always, you have to set up an equation from the information you’re given, then solve the equation. Your answer is only complete when you include the actual ages of the people involved. 102102102102102 Section 2.5 Section 2.5 Section 2.5 — Consecutive Integer Tasks, Time and Rate Tasks Section 2.5 Section 2.5 TTTTTopicopicopicopicopic 2.5.32.5.3 2.5.32.5.3 2.5.3 California Standards: Students applpplpplpplpplyyyyy Students a Students a 15.0: 15.0: 15.0: Students a 15.0: Students a 15.0: algalgalgalgalgeeeeebrbrbrbrbraic tec hniques to hniques to aic tec aic tec aic techniques to hniques to hniques to aic tec solvsolvsolvsolvsolve re re re re raaaaate pr lems lems te proboboboboblems te pr te pr lems, work lems te pr problems, and percent mixture problems. What it means for you: You’ll learn the formulas for rate, distance, and time, and use them to set up equations to solve real-life problems involving rates. Key words: rate speed distance time RRRRRaaaaatetetetete,,,,, TimeTimeTimeTimeTime,,,,, RRRRRaaaaatetetetete,,,,, TimeTimeTimeTimeTime,,,,, asksasks asksasks and Distance TTTTTasks and Distance and Distance and Distance and Distance and Distance TTTTTasks asksasks asksasks and Distance and Distance and Distance and Distance The greater an object’s speed, the greater the distance it travels in a given amount of time. Rate, time, and distance tasks are a little different to the ones you’ve seen in this Section so far because they have particular formulas that you need to learn. y a Fy a Fororororormmmmmulaulaulaulaula y a Fy a F ted b and Distance are Re Re Re Re Relaelaelaelaelated b ted b and Distance ar Speed, TimeTimeTimeTimeTime,,,,, and Distance ar and Distance ar Speed, Speed, ted by a F ted b and Distance ar Speed, Speed, The quantities of distance, time, and speed are related by a formula. Speed is the distance traveled per unit of time — for instance, the distance traveled in one second, one hour, etc. speed = distance time The units of speed depend on the units used for the distance and time. For example, if the distance is in miles and the time is in hours, then the speed will be in miles per hour. Time Fororororormmmmmulaulaulaulaula Time F Distance,,,,, Time F Time F Distance Distance e the Speed, ou Can Rearearearearearrrrrrangangangangange the Speed, e the Speed, ou Can R YYYYYou Can R ou Can R e the Speed, Distance Time F Distance e the Speed, ou Can R The above formula can be rearranged to give these important formulas for distance and time: distance = speed × time time = distance speed Guided Practice Use the speed, distance, and time formulas to work out the following: 1. Find the average speed if distance = 116 miles and time = 2 hours. 2. Find the average speed if distance = 349 km and time = 5 h. 3. Find the distance if speed = 75 mph and time = 2.5 h. 4. A car completes a 125 mile journey traveling at an average speed of 50 miles per hour. Work out the time taken. 5. A train travels a distance of 412.5 km at an average speed of 120 km per hour. How long did the journey take? 6. A train travels at an average speed of 140 mph. If it takes 4 hours to reach its destination, how far did it travel? 7. A long-distance runner completes a half marathon (13.1 miles) in a time of 1 hour 45 minutes. Find the runner’s average speed. Section 2.5 Section 2.5 Section 2.5 — Consecutive Integer Tasks, Time and Rate Tasks Section 2.5 Section 2.5 103103103103103 e an Equationtiontiontiontion e an Equa e an Equa hen Solv ou Knowwwww,,,,, TTTTThen Solv hen Solv ou Kno rite Down wn wn wn wn WWWWWhahahahahat t t t t YYYYYou Kno ou Kno rite Do WWWWWrite Do ri
te Do hen Solve an Equa e an Equa hen Solv ou Kno rite Do Motion tasks normally involve two objects with different speeds. Example Example Example Example Example 11111 Tim drives along a road at 70 km/h. Josh leaves from the same point an hour later and follows exactly the same route. If Josh drives at 90 km/h, how long will it take for Josh to catch up with Tim? Solution Solution Solution Solution Solution As always, first write down what you know — and use sketches, arrows, and tables if they help you visualize what is happening. Drawing arrows helps remind you which way each person is traveling: Tim at 70 km/h Josh at 90 km/h Tim left first... ...but Josh is traveling faster (and in the same direction), so he will catch up. Suppose Josh catches Tim x hours after Josh left. At that time, Tim will have been traveling for (x + 1) hours. emiT )sruohni( deepS )h/mkni( ecnatsiD )mkni( hsoJ x miT x 1+ 09 07 09 x (07 x )1+ = speed × time Time is in hours and speed is in kilometers per hour — so all distances will be in kilometers. Josh catches Tim when they have both traveled the same distance. So you need to solve: 90x = 70(x + 1) 90x = 70x + 70 20x = 70 x = 3.5 So it will take Josh 3.5 hours to catch up with Tim. 104104104104104 Section 2.5 Section 2.5 Section 2.5 — Consecutive Integer Tasks, Time and Rate Tasks Section 2.5 Section 2.5 Example Example Example Example Example 22222 Lorraine is driving to a theme park at 45 miles per hour. Twenty minutes after Lorraine leaves, Rachel sets off along the same freeway. If Rachel is traveling at 55 miles per hour, how long does it take her to catch up with Lorraine? Solution Solution Solution Solution Solution Lorraine at 45 mph Rachel at 55 mph Both vehicles are moving in the same direction... ...but at different speeds. Lorraine left first, but she’s traveling slower. Rachel left afterwards but is traveling faster — so at some point Rachel will catch up. If you call Rachel’s travel time x (hours), then Lorraine’s travel time will be x +( (since Lorraine left 20 minutes (= of an hour) earlier). )1 1 3 3 emiT )sruohni( deepS )hpmni( ecnatsiD )selimni( = speed × time lehcaR eniarroL x x+1 3 55 54 55 x x+⎛ ⎜⎜⎜ 45 ⎝ ⎞ ⎟⎟⎟⎟ ⎠ 1 3 When Rachel catches up, Rachel and Lorraine have traveled equal x+⎛ ⎜⎜⎜ distances, so 55x = 45 ⎝ 1 3 ⎞ ⎟⎟⎟⎟. So solve this equation to find x: ⎠ 55 x = 45 x + 45 3 10 = x x = 15 3 2 x = 1 1 2 So, Rachel catches up with Lorraine 1 1 2 hours after Rachel left. Section 2.5 Section 2.5 Section 2.5 — Consecutive Integer Tasks, Time and Rate Tasks Section 2.5 Section 2.5 105105105105105 Guided Practice 8. Felipe sets off from Phoenix for Los Angeles, driving at 60 mph. Thirty minutes later an emergency vehicle takes off after Felipe, on the same route, at an average speed of 80 mph. How long will it take the emergency vehicle to overtake Felipe? 9. Lavasha and Keisha travel separately to their grandmother's house, 114 miles away. Keisha leaves from the same place 30 minutes after Lavasha. If Lavasha is traveling at 40 mph and Keisha travels at 55 mph, how long will it take Keisha to catch up with Lavasha? 10. An express train leaves Boston for Washington D.C., traveling at 110 mph. Two hours later, a plane leaves Boston for Washington D.C., traveling at a speed of 550 mph. If the plane flies above the train route, how long will it take the plane to pass the express train? ections ections ving in Opposite Dir ving in Opposite Dir hings Mo h Out for or or or or TTTTThings Mo hings Mo h Out f WWWWWaaaaatctctctctch Out f h Out f ections ving in Opposite Directions hings Moving in Opposite Dir ections ving in Opposite Dir hings Mo h Out f When you’re solving motion questions, you always need to work out which direction each of the objects is moving in. If things are traveling in opposite directions, you have to think carefully about the equation. Example Example Example Example Example 33333 Bus 1 leaves Bulawayo at 8 a.m. at a speed of 80 km/h. It is bound for Harare, 680 km away. Bus 2 leaves the Harare depot at 8 a.m., heading for Bulawayo along the same highway at a speed of 90 km/h. Calculate at what time the buses pass each other. Solution Solution Solution Solution Solution Bus 1 at 80 km/h Bus 2 at 90 km/h This time the buses are traveling in opposite directions. emiT )sruohni( deepS )h/mkni( ecnatsiD )mkni( 1suB 2suB x x 08 09 08 x 09 x Check it out: Careful — the question asked for the time the buses pass, not for how long they will have been traveling. The initial distance between the two buses is 680 km, and they pass when the distance between them is zero. This means they must pass when the distances they have traveled sum to 680 km. So solve 80x + 90x = 680. This gives 170x = 680, or x = 4. Now you have to say what this value of x means. Since they pass 4 hours after their departure time of 8 a.m., they must pass at noon. 106106106106106 Section 2.5 Section 2.5 Section 2.5 — Consecutive Integer Tasks, Time and Rate Tasks Section 2.5 Section 2.5 Guided Practice 11. An express train leaves Chicago for Atlanta at 100 mph. At the same time a freight train leaves Atlanta for Chicago at 80 mph. If the distance between the two cities is 720 miles, how long will it take the trains to pass each other? 12. Ernesto, a resident of Los Angeles, sets off to visit his uncle in San Francisco. At the same time, Chang, a resident of San Francisco, sets off to visit a friend in Los Angeles. The road between San Francisco and Los Angeles is 390 miles long. If Ernesto drives at 60 mph while Chang drives at 70 mph, how long will it take for the drivers to pass each other? 13. Mr. and Mrs. Ding leave their home traveling in opposite directions on a straight road. Mrs. Ding drives 10 mph faster than Mr. Ding. After 2 hours, they are 200 miles apart. Find the rates that Mr. and Mrs. Ding are traveling. Independent Practice 1. A plane leaves Miami for Seattle at 9:00 a.m., traveling at 450 mph. At the same time another plane leaves Seattle for Miami, flying at 550 mph. At what time will the planes pass if the distance between Miami and Seattle is 3300 miles? 2. A jet leaves the airport traveling at a speed of 560 km/h. Another jet, leaving the same airport and traveling in the same direction, leaves 45 minutes later traveling at 750 km/h. About how long will it take for the second jet to overtake the first jet? 3. Two planes depart Los Angeles at the same time. One plane flies due east at 500 mph while the other plane flies due west at 600 mph. How long will it be before the planes are 3300 miles apart? 4. Two planes leave from the same city at the same time. One plane flies west at 475 mph and the other plane flies east at 550 mph. How long will it be before the planes are 4100 miles apart? 5. Two boaters leave a boat ramp traveling in opposite directions. The first boat travels 15 mph faster than the second one. If after 4 hours, the boats are 220 miles apart, how fast are the boats traveling? 6. Bus A leaves Eastport at 7 a.m. at a speed of 40 km/h and is bound for Westport. Bus B leaves Eastport at 7.45 a.m. and drives along the same freeway at 50 km/h. Bus C leaves Westport at 7.45 a.m. traveling towards Eastport along the same freeway at 60 km/h. Westport and Eastport are 220 km apart. Will bus B or bus C pass bus A first? ound Up ound Up RRRRRound Up ound Up ound Up ound Up ound Up RRRRRound Up ound Up ound Up As long as you learn the formulas for speed, distance, and time, then these problems are just like solving the rest of the real-life problems in this Section. As always, first set up an equation, then solve. Section 2.5 Section 2.5 Section 2.5 — Consecutive Integer Tasks, Time and Rate Tasks Section 2.5 Section 2.5 107107107107107 TTTTTopicopicopicopicopic 2.6.12.6.1 2.6.12.6.1 2.6.1 Section 2.6 asksasks asksasks estment TTTTTasks estment estment InInInInInvvvvvestment estment estment TTTTTasks InInInInInvvvvvestment asksasks asksasks estment estment estment California Standards: Students applpplpplpplpplyyyyy Students a Students a 15.0: 15.0: 15.0: Students a 15.0: Students a 15.0: algalgalgalgalgeeeeebrbrbrbrbraic tec hniques to hniques to aic tec aic tec hniques to aic techniques to hniques to aic tec solvsolvsolvsolvsolve e e e e rate problems, work cent cent per per problems, and per cent percent cent per lems..... lems lems e pre proboboboboblems e pre pr mixtur mixtur mixture pr mixtur lems mixtur What it means for you: You’ll learn what annual interest is, and you’ll solve real-life problems involving interest on investment. Key words: annual interest rate investment Interest is the money you earn by investing — that’s why it’s also known as “return on investment.” Money can be invested in lots of ways (for example, savings accounts, stocks, property, etc.), and each can have a different rate of interest. After One YYYYYearearearearear After One est is the Returetureturetureturn n n n n After One After One est is the R est is the R ual Inter AnnAnnAnnAnnAnnual Inter ual Inter ual Interest is the R After One est is the R ual Inter If your money is invested in a plan that pays interest at a rate of r per year, then the compound interest formula tells you the value of your savings after t years. A = amount including interest A = p(1 + r)t t = length of investment in years p = initial amount invested r = annual interest rate Putting t = 1 shows that after one year, you will have A = p + pr. But p is the original amount you invest (called the principal), and so the amount of interest earned in a single year (I) is equal to pr. Annual interest: I = pr Example Example Example Example Example 11111 Maria invests $5000 in a savings account with an annual interest rate of 10%. What is the return on her investment at the end of one year? Solution Solution Solution Solution Solution The formula you need is I = pr, where p = $5000. Convert the percent to a decimal: r = 10 100 = 0.10 Substitute in your values for p and r: I = pr = 5000 × 0.10 = 500 So Maria mak
es $500 in interest on her investment. 108108108108108 Section 2.6 Section 2.6 Section 2.6 — Investment and Mixture Tasks Section 2.6 Section 2.6 Guided Practice 1. A banker invested $7000 at an annual interest rate of 8%. What would be the return on the investment at the end of the year? 2. A banker invested $5000 at an annual rate of 2.5%. What would be the return on the investment at the end of one year? 3. A banker invested $10,000 at an annual rate of 13.25%. What would be the return on the investment at the end of one year? 4. At the end of one year an investment at 6% earned $795.00 in interest. What amount was invested? 5. At the end of one year an investment at 3.5% earned $28.00 in interest. What amount was invested? 6. At the end of one year an investment of $1250 earned $31.25 in interest. What was the interest rate? 7. At the end of one year an investment of $6000 earned $630 in interest. What was the interest rate? t Once t Once hemes a est in TTTTTwwwwwo Sco Sco Sco Sco Schemes a hemes a est in ou Could Invvvvvest in est in ou Could In YYYYYou Could In ou Could In t Once hemes at Once t Once hemes a est in ou Could In If you invest your money in two different schemes, the total interest over one year is the sum of the interest earned from each of the different schemes. where p1 is the amount invested in Scheme 1 at an interest rate of r1, and p2 is the amount invested in Scheme 2 at an interest rate of r2. I = p1r1 + p2r2 Example Example Example Example Example 22222 Francis has $10,000 to invest for one year. He plans to invest $6000 in stocks, and put the rest in a savings account. If the stocks pay 10% annually and the savings account pays 8%, how much interest will he make over the year? Solution Solution Solution Solution Solution I = p1r1 + p2r2 where p1 = 6000, r1 = 0.1 (= 10%) p2 = 10,000 – 6000 = 4000, r2 = 0.08 (= 8%) I = p1r1 + p2r2 = (6000 × 0.10) + (4000 × 0.08) = 600 + 320 = 920 So Francis will earn $920 over the year. Section 2.6 Section 2.6 Section 2.6 — Investment and Mixture Tasks Section 2.6 Section 2.6 109109109109109 Guided Practice 8. Tyler invested $12,000 in two different savings accounts for one year. One account had $8000 and paid a 10% return annually and the other had $4000 and paid a 12.5% return. How much interest did he earn over the year? 9. Maria invested some money in two different stock accounts for one year. If she invested $5000 in an account with an annual return of 2.5%, how much would she have had to invest at 3% in order to receive $260 interest for the year? 10. Yang is investing the same amount of money into two different savings accounts. He invests in an account returning 10% and an account returning 12% for one year. If he earned $1226.50 in interest, how much did he invest in each account? k Backwkwkwkwkwararararardsdsdsdsds k Bac ou to WWWWWororororork Bac k Bac ou to equire e e e e YYYYYou to ou to equir equir lems R Some Proboboboboblems R lems R Some Pr Some Pr lems Requir k Bac ou to equir lems R Some Pr Some Pr Sometimes you’re given information about total return on investment — and you have to work backwards to figure out how money was invested in the first place. Example Example Example Example Example 33333 A banker invested $12,000 for one year. He invested some of this money in an account with an interest rate of 5%, and the rest of the money in stocks which paid 9% interest annually. How much money did he invest in each plan if the total return from the investments was $700? Solution Solution Solution Solution Solution Call p1 the amount invested at 5%, and p2 the amount invested at 9%. Let p1 = x, then p2 = 12,000 – x. Also, r1 = 0.05 (= 5%), and r2 = 0.09 (= 9%). Substitute these values into the annual interest formula: I = p1 r1 + p2r2 = 0.05x + 0.09(12,000 – x) = 700 Now you can solve for x: 0.05x + 0.09(12,000 – x) = 700 5x + 9(12,000 – x) = 70,000 Get rid of ficients ficients decimal coef decimal coef Get rid of Get rid of ficients decimal coefficients Get rid of decimal coef ficients decimal coef Get rid of 5x + 108,000 – 9x = 70,000 5x – 9x = 70,000 – 108,000 Gr e termsmsmsmsms e ter e ter oup lik oup lik Gr Gr oup like ter Group lik e ter oup lik Gr Simplify –4x = –38,000 Simplify Simplify Simplify Simplify x = 9500 So he invested $9500 in the account paying 5%, and 12,000 – 9500 = $2500 in stocks. 110110110110110 Section 2.6 Section 2.6 Section 2.6 — Investment and Mixture Tasks Section 2.6 Section 2.6 Guided Practice 11. Latisha invested a total of $60,000. She invested a part of her money at an annual interest rate of 6% and the rest at 10%. If the total return at the end of the year was $5200, how much was invested at each rate? 12. Robin invested $80,000. He invested some of his money at an annual interest rate at 10% and the rest at 12%. If the total interest earned at the end of one year was $8300, how much was invested at each rate? 13. Leon invests $15,000 in two different accounts. He invests some at a 3% interest rate and the remaining at 4.5%. If he earns $621.00 in interest for one year, how much did he invest at each rate? 14. Fedder invested some money in three different savings accounts. If he invested $7500 in an account with an annual return of 8% and $4000 in an account earning 8.9%, how much money, to the nearest dollar, did he invest at 8.25% if he earned $1166.30 total interest in a year? est are Re Re Re Re Relaelaelaelaelatedtedtedtedted est ar est ar Inter Inter Amounts of Sometimes TTTTTwwwwwo o o o o Amounts of Amounts of Sometimes Sometimes Interest ar Amounts of Inter est ar Inter Amounts of Sometimes Sometimes In Example 4, instead of relating the sum of the individual amounts of interest to a given total, you have to relate them to each other. Example Example Example Example Example 44444 A banker had $80,000 to invest. She put some of the money in a deposit account paying 10% a year, and invested the rest at 8%. The annual return on the money invested at 8% was $100 more than the return on the 10% investment. How much money did she invest at each rate? Solution Solution Solution Solution Solution Call p1 the amount invested in the deposit account, and p2 the amount invested at 8%. Let p1 = x, then p2 = 80,000 – x. Also r1 = 0.1 (= 10%), and r2 = 0.08 (= 8%). This time, you know the return from the 8% plan was $100 more than the return from the 10% plan. Writing this as an equation gives: p1r1 + 100 = p2r2 Substitute all the information into this formula, then solve for x: 0.1x + 100 = 0.08(80,000 – x) 10x + 10,000 = 8(80,000 – x) 10x + 10,000 = 640,000 – 8x 18x = 630,000 x = 35,000 So she invested $35,000 at a rate of 10%, and 80,000 – 35,000 = $45,000 at a rate of 8%. Section 2.6 Section 2.6 Section 2.6 — Investment and Mixture Tasks Section 2.6 Section 2.6 111111111111111 Guided Practice 15. Dorothy divided $14,500 among two accounts paying 11% and 8% interest annually. The interest earned after 1 year in the 8% account was $455 less than that earned in the 11% account. How much money was invested in each account? 16. Michael invested $18,000 in two different accounts. The account paying 14% annually earned $1357.10 more than the one earning 15% interest. How much was invested in each account? 17. Lavasha invested $16,000 among two different accounts paying 10% and 12% in one year. If she earned twice as much interest in the account paying 12%, how much did she invest in each account? Independent Practice 1. In one year, an investment at 8% interest earned $1060. How much money was invested? 2. Louise invested $25,000 in two different accounts. She invested some money at a 4.5% interest rate and the remaining amount at a 3.25% interest rate. If she earned $1062.50 in interest in a year, how much did she invest at each interest rate? 3. Mackey invested some money in an account paying 6% interest for one year. She invested $1000 more than this amount in an account paying 8.5%. How much did she invest in total if the total interest earned in the year was $737.50? 4. Maxwell invested a total of $100,000. He invested his money in three different accounts. He invested $35,000 at 10% annual interest rate, $42,000 at 8% annual interest rate, and the remainder at 14% interest rate. How much interest did Maxwell receive in one year? 5. Garrett invested $12,000 at an annual interest rate of 6%. How much money would Garrett have had to invest at 10% so that the combined interest rate for both investments was 7% over the year? 6. Two business partners decided to borrow $30,000 start-up money for their business. One borrowed a certain amount at a 16.5% interest rate and the other one borrowed the rest of the money at a 12.5% interest rate. How much money did each business partner borrow if the total interest amount at the end of the year was $4150? ound Up ound Up RRRRRound Up ound Up ound Up ound Up ound Up RRRRRound Up ound Up ound Up Any problems that involve adding together investments with different interest rates are actually a type of mixture problem. In the next Topic you’ll deal with mixture problems not involving money. 112112112112112 Section 2.6 Section 2.6 Section 2.6 — Investment and Mixture Tasks Section 2.6 Section 2.6 TTTTTopicopicopicopicopic 2.6.22.6.2 2.6.22.6.2 2.6.2 California Standards: Students applpplpplpplpplyyyyy Students a Students a 15.0: 15.0: 15.0: Students a 15.0: Students a 15.0: algalgalgalgalgeeeeebrbrbrbrbraic tec hniques to hniques to aic tec aic tec hniques to aic techniques to hniques to aic tec solvsolvsolvsolvsolve e e e e rate problems, work cent cent per per problems, and per cent percent cent per lems..... lems lems e pre proboboboboblems e pre pr mixtur mixtur mixture pr mixtur lems mixtur What it means for you: You’ll set up and solve equations involving mixtures. Key words: mixture task volume mass concentration ingredients asksasks asksasks Mixture e e e e TTTTTasks Mixtur Mixtur Mixtur Mixtur Mixture e e e e TTTTTasks asksasks asksasks Mixtur Mixtur Mixtur Mi
xtur The interest problems in Topic 2.6.1 were actually mixture problems because you had to add together returns from investments with different interest rates. In this Topic you’ll see some mixture problems that don’t involve money. olume” olume” Amount per Unit Mass/V Amount per Unit Mass/V asks Use “ Some TTTTTasks Use “ asks Use “ Some Some olume” Amount per Unit Mass/Volume” asks Use “Amount per Unit Mass/V olume” Amount per Unit Mass/V asks Use “ Some Some Mixtures are made by combining ingredients. In math, mixture problems involve using algebra to work out the precise amounts of each ingredient. To do this, you have to make use of an equation of the form: concentration = amount of substance total volume (or total mass) Check it out: Concentration means “amount per liter,” or “amount per kilogram,” etc. The “amount of substance” could be given to you as a volume or as a mass — but it doesn’t make any difference to the math. Just use whatever units they give you in the question. Concentration can also be described as a percent: percent of ingredient = amount of ingredient total weight or volume of mixture × 100 The following example doesn’t involve a mixture, but it shows how the above formula can be used. Example Example Example Example Example 11111 If a 1 kg bag of granola consists of 15% raisins (by mass), how many grams of raisins are there in the bag? Solution Solution Solution Solution Solution The formula tells you: Percent of raisins = total amount of raisins total amount of granola 0.15 = total amount of raisins 1000 So total mass of raisins = 1000 × 0.15 = 150 grams Section 2.6 Section 2.6 Section 2.6 — Investment and Mixture Tasks Section 2.6 Section 2.6 113113113113113 Guided Practice 1. A 19-ounce can of a disinfectant spray consists of 79% ethanol by weight. How many ounces of pure ethanol does the spray contain? 2. A 2 kg bag of mixed nuts contains 20% walnuts (by mass). How many grams of walnuts are in the bag? 3. What is the percent of juice if a 2 liter bottle of juice drink has 250 ml of juice? 4. A 142 g bottle of baby powder contains 15% zinc oxide. How many grams of zinc oxide are in the powder? 5. What is the concentration of a hydrogen peroxide solution if 500 ml of the solution contains 15 ml of hydrogen peroxide? tion as a Pererererercent cent cent tion as a P te the Concentraaaaation as a P tion as a P te the Concentr te the Concentr ou Can Calcula YYYYYou Can Calcula ou Can Calcula cent ou Can Calculate the Concentr cent tion as a P te the Concentr ou Can Calcula This example shows how you can apply the formula to mixtures of various concentrations. Example Example Example Example Example 22222 If you combine 5 liters of a 10% fruit juice drink and 15 liters of a 20% fruit juice drink, what percent of fruit juice do you have? Solution Solution Solution Solution Solution Here, you have to use your formula more than once, but the principle is the same. Concentration of fruit juice = volume of fruit juice volume of fruit drink For the 10% fruit drink, concentration = 0.1 = x 5 , where x is the volume of fruit juice. The volume of fruit juice in the 10% solution is x = 0.1 × 5 = 0.5 liters. And for the 20% fruit drink, concentration = 0.2 = y 15 , where y is the volume of fruit juice. The amount of fruit juice in the 20% solution is y = 0.2 × 15 = 3 liters. Also, the total volume of the combined fruit drink is 5 + 15 = 20 liters. The total amount of fruit juice in the mixture is x + y = 0.5 + 3 = 3.5 liters. Therefore, the concentration of the final mixture is: Concentration = volume of fruit juice volume of fruit drink = . 3 5 20 = 0.175 = 17.5% 114114114114114 Section 2.6 Section 2.6 Section 2.6 — Investment and Mixture Tasks Section 2.6 Section 2.6 Guided Practice 6. If you combine 10 liters of juice cocktail made with 100% fruit juice with 5 liters of another juice cocktail containing 10% fruit juice, what percent of the 15 liters of juice cocktail will be fruit juice? 7. There is 4% hydrogen peroxide in a 475 ml hydrogen peroxide solution. The 4% solution is mixed with 375 ml of a hydrogen peroxide solution containing 15% hydrogen peroxide. What percent of hydrogen peroxide is in the combined solution? 8. Tina mixes 4.75 mg of a lotion containing 1% vitamin E with 2 mg of a lotion containing 7% vitamin E. What is the percentage of vitamin E in the combined lotion? Independent Practice 1. A 12 fl. oz. bottle of lotion contains 2.5% lavender. How many fluid ounces of lavender are contained in the bottle of lotion? 2. A 2.5 fl. oz. bottle of nose spray contains 0.65% sodium chloride. What volume of sodium chloride does the spray contain? 3. A 5-gallon car radiator should contain a mixture of 40% antifreeze and 60% water. What volumes of water and antifreeze should the radiator contain? 4. A 1.75 ml bottle of medicine contains 80 mg of an active ingredient per 0.8 ml of medicine. How many milligrams of active ingredient does the bottle contain? 5. A 473 ml bottle of rubbing alcohol contains 331.1 ml of isopropyl alcohol. What percentage of the bottle is isopropyl alcohol? 6. Two liters of 100% pure pineapple juice is mixed with 2 liters of soda water and 1 liter of orange juice to make a party punch. What percentage of the party punch is pineapple juice? 7. A 2 kg bag of walnuts is mixed with 3 kg of pecans, 4 kg of hazelnuts, and 1 kg of brazil nuts to make a 10 kg bag of mixed nuts. What percentage of the 10 kg bag of nuts is pecans and hazelnuts? 8. 473 ml of 70% rubbing alcohol is combined with 473 ml of 90% rubbing alcohol and 473 ml of water. What is the percentage of rubbing alcohol in the mixture? 9. Two liters of juice cocktail are combined with 6 liters of a different juice cocktail to make 8 liters of juice cocktail containing 50% juice. If the 6 liters of juice cocktail contains 33 percentage of fruit juice is the 2 liters of juice cocktail? 1 3 % fruit juice, what ound Up ound Up RRRRRound Up ound Up ound Up ound Up ound Up RRRRRound Up ound Up ound Up It’s a good idea to memorize the two formulas at the start of this Topic — knowing them will make it a lot easier to deal with any mixture tasks involving liquids or ingredients. Section 2.6 Section 2.6 Section 2.6 — Investment and Mixture Tasks Section 2.6 Section 2.6 115115115115115 TTTTTopicopicopicopicopic 2.6.32.6.3 2.6.32.6.3 2.6.3 California Standards: Students applpplpplpplpplyyyyy Students a Students a 15.0: 15.0: 15.0: Students a 15.0: Students a 15.0: algalgalgalgalgeeeeebrbrbrbrbraic tec hniques to hniques to aic tec aic tec hniques to aic techniques to hniques to aic tec solvsolvsolvsolvsolve e e e e rate problems, work cent cent per per problems, and per cent percent cent per lems..... lems lems e pre proboboboboblems e pre pr mixtur mixtur mixture pr mixtur lems mixtur What it means for you: You’ll set up and solve equations involving mixtures. Key words: mixture task volume mass concentration ingredients asksasks asksasks e Mixture e e e e TTTTTasks e Mixtur e Mixtur MorMorMorMorMore Mixtur e Mixtur e Mixture e e e e TTTTTasks MorMorMorMorMore Mixtur asksasks asksasks e Mixtur e Mixtur e Mixtur Remember that a mixture is made up of only the original ingredients. Nothing can mysteriously appear in the final mixture that was not part of the original ingredients, and nothing can disappear either. t Changeeeee t Chang edient Doesn’’’’’t Chang t Chang edient Doesn h Ingrrrrredient Doesn edient Doesn h Ing h Ing Eac Eac Amount of Amount of otal TTTTThe he he he he TTTTTotal otal Each Ing Amount of Eac otal Amount of t Chang edient Doesn h Ing Eac Amount of otal total amount of each substance in the ingredients = amount of that substance in the final mixture So if you know the amount of a substance in each of the ingredients you are mixing, you can always calculate the amount of that substance in the final mixture. Similarly, if you know the amount of a substance in one ingredient and in the final mixture, you can find out how much was in the other ingredient(s). Example Example Example Example Example 11111 There are 1000 gallons of water in a wading pool. The water is 5% chlorine. What quantity of a 65% chlorine solution would need to be added to bring the pool’s chlorine concentration up to 15%? Solution Solution Solution Solution Solution Ingredient 1: You know the volume and concentration of the water in the wading pool. Volume = 1000 gallons. Concentration of chlorine = 0.05 (= 5%). So the amount of chlorine in the wading pool to begin with is given by: Original amount of chlorine = volume × concentration = 1000 × 0.05 = 50 Ingredient 2: You also know that the concentration of the chlorine solution to be added is 0.65 (or 65%). So if you call the volume of the added chlorine solution x, then: The amount of chlorine added is 0.65x (= volume × concentration) Mixture: This means the final volume of water in the wading pool is 1000 + x, and so the final amount of chlorine in the pool is 0.15(1000 + x). Therefore 50 + 0.65x = 0.15(1000 + x). Now solve for x: 50 + 0.65x = 150 + 0.15x 0.5x = 100 x = 200 Therefore 200 gallons of the 65% solution are required. 116116116116116 Section 2.6 Section 2.6 Section 2.6 — Investment and Mixture Tasks Section 2.6 Section 2.6 Guided Practice 1. A pest control company has 200 gallons of 60% pure insecticide. To create a 70% pure insecticide solution, the company must mix the 60% pure insecticide solution with a 90% pure insecticide solution. How many gallons of the 90% pure insecticide solution should the company mix with the existing solution? 2. A chemist needs to dilute a 60% citric acid solution to a 20% citric acid solution. She needs 30 liters of the 20% solution. How many liters of the 60% solution and water should be used? (Hint: water has 0% acid.) 3. An alloy containing 40% gold is mixed with an 80% gold alloy to get 1000 kilograms of an alloy that is 50% gold. How many kilograms of each alloy are used? h Substance h Substance Eac Eac Amount of Amount of or the o
r the ula f s a Fororororormmmmmula f ula f s a F TTTTTherherherherhere’e’e’e’e’s a F s a F h Substance Each Substance Amount of Eac or the Amount of ula for the h Substance Eac Amount of or the ula f s a F Multiplying the volume by the concentration gives you the amount of substance in each solution. And since the total amount of substance is the same before and after, you can use this handy formula: c1v1 + c2v2 = cv where c1 and v1 are the concentration and volume of the first ingredient, c2 and v2 are the concentration and volume of the second ingredient, c and v are the concentration and volume of the mixture of the two. and The next example is quite similar to Example 1, but it uses the formula shown above. Example Example Example Example Example 22222 A pharmacist has 500 cm3 of a 30% acid solution. He replaced x cm3 of the 30% solution with a 70% acid solution to get 500 cm3 of a new 40% acid solution. What volume of the 30% acid solution did he replace with the 70% acid solution? Solution Solution Solution Solution Solution Assume that the pharmacist poured away x cm3 of the 30% solution and replaced it with 70% solution. Then 0.3(500 – x) + 0.7x = 0.4(500) 150 + 0.4x = 200 x = 125 So the pharmacist replaced 125 cm3 of the 30% acid solution. Section 2.6 Section 2.6 Section 2.6 — Investment and Mixture Tasks Section 2.6 Section 2.6 117117117117117 Guided Practice 4. A doctor prescribes 20 grams of a 62% solution of a generic medicine. The pharmacist has bottles of 50% and 70% solutions in stock. How many grams of each solution can the pharmacist use to fill the prescription for the patient? 5. Fifty pounds of special nuts costing $4.50 per pound were mixed with 120 pounds of generic nuts that cost $2 per pound. What is the value of each pound of the nut mixture? 6. Cherry juice that costs $5.50 per liter is to be mixed with 50 liters of orange juice that costs $2.50 per liter. How much cherry juice should be used if the value of the mixture is to be $3.50 per liter? 7. Sixty liters of a syrup that costs $8.50 per liter were mixed with honey that costs $4.75 per liter. How many liters of honey were used if the value of the mixture is $6 per liter? After” TTTTTotals otals otals After” After” and “ te the “Befororororore”e”e”e”e” and “ and “ te the “Bef AlAlAlAlAlwwwwwaaaaays Rys Rys Rys Rys Relaelaelaelaelate the “Bef te the “Bef otals and “After” otals After” and “ te the “Bef Example Example Example Example Example 33333 Marie ordered 40 pounds of walnuts at $1.50 per pound. She mixed this with hazelnuts costing $1.00 per pound, and sold the mixed nuts at $1.25 per pound. How many pounds of hazelnuts did she use, if she broke even? Solution Solution Solution Solution Solution Here, your “amount per unit weight” formula is: Price per pound (p) = v total value ( ) ) weight ( w Check it out: Notice that p1w1 + p2w2 = pw. Compare this to the formula on the previous page — it’s essentially the same. First write down what you know about the original ingredients and the final mixture (calling the quantity you need to find x, for example). Then relate a “before” total to the “after” total — so the total value of the ingredients is the same as the total value of the mixture — v1 + v2 = v. $1.50-per-pound walnuts: weight = 40 lb. Total value = 1.50 × 40 = 60. $1.00-per-pound hazelnuts: weight = x lb. Total value = 1.00 × x = x. Mixed ($1.25) nuts: weight = (40 + x) lb. Total value = 1.25(40 + x). To break even, the value of the mixed nuts must equal the sum of the values of the original nuts. That is, x + 60 = 1.25(x + 40). Solve this to find x: x + 60 = 1.25(x + 40) x + 60 = 1.25x + 50 –0.25x = –10 x = 40 Therefore she used 40 pounds of hazelnuts. 118118118118118 Section 2.6 Section 2.6 Section 2.6 — Investment and Mixture Tasks Section 2.6 Section 2.6 Guided Practice 8. Mrs. Roberts owns a pet store and wishes to mix 4 pounds of cat food worth $2.00 per pound with another cat food costing $3.00 per pound. How much of the $3.00 per pound cat food should be used if the mixture is to have costed $2.75 per pound? 9. A local garden center makes a medium grade plant seed by mixing a low grade plant seed bought for $2.50 a pound with 14 pounds of superior quality seed bought for $5.00 a pound. How much of the low grade plant seed needs to be mixed with the superior quality seed if it is to be worth $3.50 per pound? 10. A tea blend is made by using 2 kg of $2.00 per kg tea leaves and another tea leaf costing $4.00 per kilogram. How many kilograms of the $4.00 tea is needed to make a tea blend worth $3.75 per kilogram? 11. A landscaper wants to make a blend of grass seed using 300 pounds of $0.40 per pound grass seed and another seed costing $0.75 per pound. How much of the $0.75 seed does the landscaper need to make a $0.60 per pound blend? ut the Same Maththththth ut the Same Ma ut the Same Ma xt — b xt — b ent Conte A Difffffferererererent Conte ent Conte A Dif A DifA Dif xt — but the Same Ma ent Context — b ut the Same Ma xt — b ent Conte A Dif Example Example Example Example Example 44444 A retailer mixed two fruit drinks. He mixed 30 liters of a fruit drink that cost him $1.90 per liter with an unknown amount of a fruit drink that cost him $2.50 per liter. If the mixture’s ingredients cost $2.40 per liter, what volume of the $2.50-per-liter drink did he use? Solution Solution Solution Solution Solution This time, use: price per liter (p) = c total cost ( ) v volume ( ) $1.90-per-liter drink: volume = 30 liters. Total cost = 1.90 × 30 = 57. $2.50-per-liter drink: volume = x liters. Total cost = 2.50x. Mixed ($2.40) drink: volume = (30 + x) liters. Total cost = 2.40(30 + x). But the cost of the mixture is the sum of the ingredients’ costs, so 57 + 2.50x = 2.40(30 + x). Solving this for x gives: 57 + 2.5x = 72 + 2.4x 0.1x = 15 x = 150 So he used 150 liters of the $2.50-per-liter fruit drink. Section 2.6 Section 2.6 Section 2.6 — Investment and Mixture Tasks Section 2.6 Section 2.6 119119119119119 Guided Practice 12. A pet store owner mixed two different bird seeds. She mixed 20 pounds of a seed that costs $2.25 per pound with 50 pounds of another seed. If the mixture of seeds cost $2.75 per pound, how much per pound did the 50 pounds of seed cost? 13. Apples cost $0.80 per pound and grapes cost $1.10 per pound. Michael wishes to make a fruit tray using only apples and grapes that costs $1.00 per pound. If Michael has 8 lbs of apples, how many pounds of grapes are needed? 14. A roast coffee blend costing $6.60 per pound is made by mixing a bean that costs $2.00 per pound with another one that costs $7.00 per pound. If 5 pounds of the $2.00 bean are used, how many pounds of the $6.60 beans would be produced? Independent Practice 1. Dave and his parents went to the movies. Adult tickets cost $7.00 and children’s tickets cost $3.00. For the screening of one movie, 500 tickets were sold for a total of $2000. Find the number of each kind of ticket that was sold for the movie. 2. A 70% salt solution was diluted with purified water to produce about 50 liters of 55% salt solution. Approximately how much purified water was used? 3. A chemist wants to dilute a 60% boric acid solution to a 15% solution. He needs 30 liters of the 15% solution. How many liters of the 60% solution and water must the chemist use? 4. A student wants to dilute 35 liters of a 35% salt solution to a 16% solution. How many liters of distilled water does the student need to add to the 35% salt solution to obtain the 16% salt solution? 5. A tea blend is made by mixing 2 kg of a $2.00 per kg tea with 5 kg of another tea. If the total cost of the ingredients in the blend is $24.00, what is the price per kg of the 5 kg of tea? 6. How many liters of a 100% alcohol solution must be mixed with 20 liters of a 50% solution to get a 70% solution? 7. A chemist has 8 liters of a 30% solution of a compound. How much of a 100% solution of the compound must be added to get a 50% solution? 8. Milk that contains 2% fat is mixed with milk containing 0.5% fat. How much 0.5% fat milk is needed to be added to 10 gallons of 2% fat milk to obtain a mixture of milk containing 1% fat? ound Up ound Up RRRRRound Up ound Up ound Up ound Up ound Up RRRRRound Up ound Up ound Up Over the last couple of Topics you’ve seen lots of examples of mixture tasks. The most difficult part of a mixture task is setting up the equation — once you’ve done that it’s all a lot easier. 120120120120120 Section 2.6 Section 2.6 Section 2.6 — Investment and Mixture Tasks Section 2.6 Section 2.6 TTTTTopicopicopicopicopic 2.7.12.7.1 2.7.12.7.1 2.7.1 Section 2.7 WWWWWororororork Rk Rk Rk Rk Raaaaatetetetete WWWWWororororork Rk Rk Rk Rk Raaaaatetetetete California Standards: Students applpplpplpplpplyyyyy Students a Students a 15.0: 15.0: 15.0: Students a 15.0: Students a 15.0: algalgalgalgalgeeeeebrbrbrbrbraic tec hniques to hniques to aic tec aic tec aic techniques to hniques to hniques to aic tec solvsolvsolvsolvsolve re re re re raaaaate pr lems,,,,, w w w w worororororkkkkk lems lems te proboboboboblems te pr te pr lems te pr prprprprproboboboboblems lems,,,,, and percent lems lems lems mixture problems..... What it means for you: You’ll solve combined work rate problems by calculating each work rate in turn. Key words: work rate combined work rate Check it out: Problems like this usually assume that both people are working all the time (rather than one of them stopping early). Work problems are similar to the mixture problems you saw in Section 2.6. Again, the only new thing is that there are a few formulas that you need to know in order to set up the equations. About Speeds of WWWWWorororororkingkingkingkingking About Speeds of asks are e e e e About Speeds of About Speeds of asks ar ted TTTTTasks ar asks ar ted WWWWWororororork-Rk-Rk-Rk-Rk-Relaelaelaelaelated ted About Speeds of asks ar ted This area of math is concerned with calculating how long certain jobs will take if the people doing the job are
working at different rates. Example Example Example Example Example 11111 John takes 1 hour to deliver 100 newspapers, and David takes 90 minutes to deliver 100 newspapers. How long would it take them to deliver 100 newspapers between them? Assume that they work independently, that they both start at the same time, and that they are both working the whole time. Solution Solution Solution Solution Solution It’s tempting just to work out how long it takes John and David to deliver half the papers each. But that does not take into account the fact that they are working as a team — and since John works faster than David, he will deliver more papers. Instead, you have to figure out how quickly they work as a team — not just as two individuals. The problem is that you don’t know how many papers each of them delivers — you only have the information given in the question, which is shown in this diagram: John David 0 30 100 newspapers 60 90 mins 100 newspapers You need to work out their rate of delivering the papers. Based on how long it takes each person to deliver 100 papers, you can calculate how many papers are delivered every minute in total. If John can deliver 100 newspapers in 1 hour (60 minutes), he can deliver 100 ÷ 60 = 5 3 newspapers in 1 minute. If David can deliver 100 newspapers in 90 minutes, 10 9 10 9 he can deliver 100 ÷ 90 = newspapers in 1 minute. So in total 5 3 15 + = + = 9 10 9 25 9 newspapers are delivered each minute. This means that 100 papers will take 100 ÷ 25 9 = 36 minutes to deliver. Section 2.7 Section 2.7 Section 2.7 — Work-Related Tasks Section 2.7 Section 2.7 121121121121121 Guided Practice 1. Akemi can weed the garden in 4 hours. Keira can weed the garden in 12 hours. How long would it take the two of them to weed the garden together? 2. Martha can clean a statue in 15 hours. Chogan can clean the same statue in 9 hours. If they clean the statue together, how long would it take them to finish? 3. A carpenter can build a cabinet in 10 hours. Her assistant can build the same cabinet in 15 hours. How long would it take them to build the cabinet together, assuming they can work independently? k Done ÷ TimeTimeTimeTimeTime k Done ÷ te = WWWWWororororork Done ÷ k Done ÷ te = WWWWWororororork Rk Rk Rk Rk Raaaaate = te = k Done ÷ te = Work rate is the amount of work carried out per unit time. The work completed can be given as a fraction. For example, if only half the task is completed, write 1 2 . If the whole task is completed, write 1. Work rate = work completed time When you are solving a problem like this, you need to start by identifying the “work completed” and the time that it took. The “work completed” part in the previous example was quite straightforward. The example below is not so simple — make sure you understand each step. Example Example Example Example Example 22222 An inlet pump can fill a water tank in 10 hours. However, an outlet pump can empty the tank in 12 hours. An engineer turns on the inlet pump but forgets to switch off the outlet pump. With both pumps running, how long does it take to fill the tank? Solution Solution Solution Solution Solution Inlet pump’s work rate = work completed time = 1 10 Outlet pump’s work rate = work completed time = 1 12 The two pumps are working in opposite directions. The combined work rate is the difference between the two rates. So the combined rate = 1 10 1 − = 12 ⋅ 6 1 60 − ⋅ 5 1 60 = − 6 5 60 = 1 60 So after 60 hours, the net amount of water that has gone in is 1 tank’s worth — which means the tank will be full. So the answer is 60 hours. Check it out: The inlet pump’s “work completed” is 1 tank of water ininininin, and the outlet pump’s “work completed” is 1 tank of water outoutoutoutout. Check it out: The fractions have been subtracted using the least common multiple (LCM) of 10 and 12, which is 60. See Topic 2.3.1 for more on LCMs. 122122122122122 Section 2.7 Section 2.7 Section 2.7 — Work-Related Tasks Section 2.7 Section 2.7 Guided Practice 4. Megan, Margarita, and James work in a fast-food restaurant after school. It takes Megan 6 hours, Margarita 3 hours, and James 4 hours to clean the utensils individually. How long would it take the three of them to clean the utensils if they worked together? 5. A bathtub can be filled from a faucet in 10 minutes. However, a pump can empty the bathtub in 15 minutes. If the faucet and the pump are on at the same time, how long will it take to fill the bathtub? 6. Isabel can fence the family three-acre lot in 8 days. If it would take José 6 days and Marvin 12 days to fence the same lot, how long would it take the three people to fence the lot together? Independent Practice 1. Bill can paint a house in 5 days and Samantha can paint the same house in 7 days. How long will it take them, working together, to paint the house? 2. There are two drains in a tub filled with water. One drain would empty the tub in 3 hours, if opened. The other drain would empty the tub in 4 hours. If both drains are opened at the same time, how long will it take to empty the tub? 3. A tub has 2 drains. One drain can empty the full tub in 20 minutes. The other drain can empty half the tub in 30 minutes. How long will it take to empty the full tub if both drains are opened together? 4. Jose can paint a house in 4 days. Leroy can paint the same house in 5 days. How long will it take Jose and Leroy, working together, to paint the house? 5. Sam, Doris, and Alice are weeding a yard. Working by himself, Sam could weed the yard in 1½ hours. Doris could do the same yard by herself in 1 hour and Alice could do it in only 45 minutes. How long will it take the three of them working together to weed the yard? 6. A barrel has two filling pipes and one draining faucet. One pipe could fill the barrel in 2 hours and the other could fill it in 2½ hours. The faucet could empty the barrel in 4 hours. How long will it take to fill the barrel if both pipes are filling and the draining faucet is opened? ound Up ound Up RRRRRound Up ound Up ound Up ound Up ound Up RRRRRound Up ound Up ound Up The very first thing you need to do with a work rate problem is look through the word problem to identify the work completed and the time taken. Section 2.7 Section 2.7 Section 2.7 — Work-Related Tasks Section 2.7 Section 2.7 123123123123123 TTTTTopicopicopicopicopic 2.7.22.7.2 2.7.22.7.2 2.7.2 California Standards: Students applpplpplpplpplyyyyy Students a Students a 15.0: 15.0: 15.0: Students a 15.0: Students a 15.0: algalgalgalgalgeeeeebrbrbrbrbraic tec hniques to hniques to aic tec aic tec aic techniques to hniques to hniques to aic tec solvsolvsolvsolvsolve re re re re raaaaate pr lems,,,,, w w w w worororororkkkkk lems lems te proboboboboblems te pr te pr lems te pr prprprprproboboboboblems lems,,,,, and percent lems lems lems mixture problems..... What it means for you: You’ll solve combined work rate problems by first calculating the combined work completed. Key words: work rate combined work rate Check it out: The fractions have been multiplied by the LCM of 10, 12, and x, to get rid of the fractional coefficients. Combined WWWWWororororork Rk Rk Rk Rk Raaaaatetetetete Combined Combined Combined Combined Combined WWWWWororororork Rk Rk Rk Rk Raaaaatetetetete Combined Combined Combined Combined In Topic 2.7.1 you calculated each work rate in turn, then added or subtracted them. This Topic contains a method for combining each piece of information directly into one equation — which can make the whole calculation a lot more straightforward. All the Infororororormamamamamationtiontiontiontion All the Inf All the Inf lude lude tes Inc Combined WWWWWororororork Rk Rk Rk Rk Raaaaates Inc tes Inc Combined Combined lude All the Inf tes Include All the Inf lude tes Inc Combined Combined Another way to approach work rate problems is to put all the information from the question directly into this equation, which can then be solved: Combined work rate = combined work completed combined work completed total time total time Here’s the same problem that you saw in Example 2 in Lesson 2.7.1 — but now using the new method. Example Example Example Example Example 11111 An inlet pump can fill a water tank in 10 hours. However, an outlet pump can empty the tank in 12 hours. An engineer turns on the inlet pump but forgets to switch off the outlet pump. With both pumps running, how long does it take to fill the tank? Solution Solution Solution Solution Solution Combined work rate = total water in the tank time taken to fill the tank Let x = number of hours to fill the tank, and then substitute everything you know into the formula: 1 10 1 − = 12 1 x 1 tank’s worth of water total time (left-hand side of equation = combined rate) Then rearrange to solve: 60 x ⎛ ⎜⎜⎜ ⎝ 1 10 x 60 10 − − ⎞ ⎟⎟⎟ = ⎠ = 1 12 x 60 12 x x 60 x 60 x 6x – 5x = 60 x = 60 So it takes 60 hours to fill the tank. 124124124124124 Section 2.7 Section 2.7 Section 2.7 — Work-Related Tasks Section 2.7 Section 2.7 Example Example Example Example Example 22222 Jesse can paint a wall in 6 hours. Melinda can paint the same wall in 4 hours. How long would it take the two of them to paint the wall together if they worked independently and started at the same time? Solution Solution Solution Solution Solution Jesse can complete the whole task in 6 hours, so her work rate = 1 6 . Melinda can complete the whole task in 4 hours, so her work rate = 1 4 . You need to find out how many hours they would take to do the work together. They are working on the same wall, so add the contribution from each person: Combined work rate = 1 6 + 1 4 Use x for the hours it takes in total, and write out the equation for combined work rate. 1 x 1 6 1 + = 4 ⎞ ⎟⎟⎟ = ⎠ ⎛ ⎜⎜⎜ ⎝ x + 12 1 4 1 6 12 x 2x + 3x = 12 5x = 12 x and x Multiply by by by by by 12y 12y 12y 12y 12x,,,,, the L Multipl Multipl the L the L CM of CM of 6, 6, 4, 4, and and the LCM of CM of 6, 6, 4, 4, and Multipl Multipl the L CM of 6, 4, and hen solve fe fe fe fe for or or
or or x TTTTThen solv hen solv hen solv hen solv x = 2 2 5 Therefore painting the wall would take Jesse and Melinda 2 2 5 hours = 2 hours and 24 minutes. Example Example Example Example Example 33333 Liza can dig a garden in 7 hours alone. If Marisa helps her, they finish all the digging in just 3 hours, working independently. How long would it take Marisa to dig the garden alone? Solution Solution Solution Solution Solution Liza can complete the whole task in 7 hours, so her work rate = 1 7 . Together they can complete the whole task in 3 hours, so their combined work rate = 1 3 . Section 2.7 Section 2.7 Section 2.7 — Work-Related Tasks Section 2.7 Section 2.7 125125125125125 Example 3 continueduedueduedued Example 3 contin Example 3 contin Example 3 contin Example 3 contin You need to find out how fast Marisa could do the work on her own. Use x for the hours it takes Marisa, so her work rate = 1 x . This time you know the combined work rate, but not Marisa’s work rate. The formula is still: combined work rate = combined work completed total time . 1 3 rite out the equationtiontiontiontion rite out the equa rite out the equa WWWWWrite out the equa rite out the equa 7, x,,,,, and 3 Multiply by by by by by 21y 21y 21y 21y 21x,,,,, the L 7, 7, CM of CM of and 3 and 3 the L the L Multipl Multipl CM of 7, and 3 the LCM of 7, CM of and 3 the L Multipl Multipl hen solve fe fe fe fe for or or or or x hen solv hen solv TTTTThen solv hen solv 1 7 + =1 x ⎞ ⎟⎟⎟ = ⎠ ⎛ ⎜⎜⎜ ⎝ + 1 x 1 7 x 21 x 21 3 3x + 21 = 7x 4x = 21 x = 5 1 4 So Marisa can dig the garden in 5 1 4 hours = 5 hours and 15 minutes. Guided Practice 1. A pump can fill a fuel tank in 30 minutes. A second pump can fill the same tank in 60 minutes. How long would it take to fill the fuel tank if both pumps were filling the tank together? 2. Machine A can pack 50 crates of canned dog food in 15 minutes. When Machine A and Machine B are working at the same time, they can pack 50 crates of canned dog food in 9 minutes. How long would it take Machine B to pack 50 crates of canned dog food by itself? 3. Three student-service workers are cataloging books in a school library. Joaquin can catalog the books in 8 hours, Caroline can do the same job in 4 hours, and Joshua can do it in 6 hours. If the three students work together, how long will it take them to finish cataloging the books? 4. A central heater can warm a house in 24 minutes. When the central heater and a floor heater are used together they can warm the house in 16 minutes. How long would the floor heater take to warm the house alone? 5. When Dionne’s cell phone has been fully charged, it can be used for 5 hours before its battery runs out. When its battery runs out, the phone can be fully charged in 3 hours if it is not in use. Dionne’s phone’s battery has run out. If she sets it to recharge, and uses the phone constantly while it is recharging, how long will it take to fully charge her phone? 126126126126126 Section 2.7 Section 2.7 Section 2.7 — Work-Related Tasks Section 2.7 Section 2.7 Independent Practice 1. A window cleaner can clean the windows of a house in 5 minutes and his trainee can clean the same windows in 20 minutes. How long would it take the window cleaner and his trainee to clean the windows together? 2. Susan and John are hired to stuff envelopes with parent notices about an upcoming school event. If the task would take Susan 120 minutes and John 90 minutes individually, how long would it take the two of them to stuff the envelopes together? 3. Lorraine can tile a room in 20 hours. Juan can tile the same room in 30 hours, and Oliver can tile it in 40 hours. How long would it take them to tile the room if they worked together? 4. A faucet can fill a barrel in 2 hours. However, there is a hole in the bottom of the barrel that can empty it in 6 hours. Without knowing of the hole, Leo tries to fill the barrel. How long will it take? 5. Tyrone, Jerry, and Dorothy are painting a fence surrounding a field. Tyrone could paint the fence in 4 days by himself. Jerry could paint the same fence, but it would take him 6 days if he did it by himself. Dorothy could paint the fence in 3½ days. How long should it take the three of them, working together, to paint the fence? 6. Michael can decorate the cafeteria for a dance in 4 hours. Emily can decorate the cafeteria in 3 hours. Eylora can decorate the cafeteria in 1½ hours. If the three of them have 1 hour, working together, to decorate for the party, will it be done in time? 7. When turned on, a faucet can fill a tub in 3 hours. The tub has 2 drains — the first can empty the tub in 8 hours and the second can empty the tub in 7 hours. If the faucet is turned on while both drains are open, how long will it take to fill the tub? 8. Manuel and Anita have extensive gardens. Manuel can mow all the lawns with his lawn mower in 5 hours. If Anita helps him with her lawn mower, they can mow all the lawns in 2 hours. How long would it take Anita to mow all the lawns alone with her lawn mower? 9. Albert and Po work in a factory that makes leotards. Albert cuts out the material, and Po stitches the side seams. Albert can cut the parts for 1 box of leotards in 9 hours, and Po can stitch the side seams for 1 box of leotards in 27 hours (including any breaks). If Albert and Po start work at the same time, how long will it be before there is 1 full box of leotards cut and waiting for their side seams to be stitched? ound Up ound Up RRRRRound Up ound Up ound Up ound Up ound Up RRRRRound Up ound Up ound Up These work rate problems often include a lot of information, and it’s easy to get the values mixed up. It’s a good idea to check that your solutions make sense in the original word problem. Section 2.7 Section 2.7 Section 2.7 — Work-Related Tasks Section 2.7 Section 2.7 127127127127127 TTTTTopicopicopicopicopic 2.8.12.8.1 2.8.12.8.1 2.8.1 Section 2.8 tions tions alue Equa alue Equa Absolute VVVVValue Equa Absolute Absolute tions alue Equations tions alue Equa Absolute Absolute Absolute VVVVValue Equa tions tions alue Equa alue Equa Absolute Absolute alue Equations tions alue Equa Absolute tions Absolute California Standards: 3.0:3.0:3.0:3.0:3.0: Students solv Students solveeeee Students solv Students solv Students solv tions tions equa equa tions and inequalities equations equa tions equa inininininvvvvvolving a alues..... alues alues bsolute v bsolute v olving a olving a bsolute values olving absolute v alues bsolute v olving a What it means for you: You’ll solve equations involving absolute values. Key words: absolute value Check it out: In other words, an absolute value is never negative. You already saw absolute values on the number line in Topic 1.2.3. In this Topic you’ll see that absolute values can turn up in equations too — and you’ll learn how to get rid of them to solve the equations. ositivvvvveeeee ositiositi ositi alue is AlAlAlAlAlwwwwwaaaaays Pys Pys Pys Pys Positi alue is Absolute VVVVValue is alue is Absolute TTTTThe he he he he Absolute Absolute alue is Absolute The absolute value of x is defined as: x if x is positive or zero, and –x if x is negative. The absolute value of any real number, x, is written as \|x\|. Another way to think about it is that the absolute value is the distance of a number from zero on the number line. For example, the distance between 0 and 9 is the same as the distance between 0 and –9. This can be written as \|9\| = \|–9\| = 9. –9 9 –10 –9 –8 –7 –6 –5 –4 –3 –2 – 10 –3 3 The distance between 0 and 3 is the same as the distance between 0 and –3. This can be written as \|3\| = \|–3\| = 3. Guided Practice Find the distance that each letter is from zero: B D –5 –4 –3 –2 –. A 3. C Simplify: 5. \|–9\| 7. –\|–2\| 2. B 4. D 6. \|–20\| 8. –\|–7\| 128128128128128 Section 2.8 Section 2.8 Section 2.8 — Absolute Value Equations Section 2.8 Section 2.8 tions Havvvvve e e e e TTTTTwwwwwo Solutions o Solutions o Solutions tions Ha tions Ha alue Equa Absolute VVVVValue Equa alue Equa Absolute Absolute o Solutions alue Equations Ha o Solutions tions Ha alue Equa Absolute Absolute Watch out when you’re solving an equation such as \|2x\| = 10. There are actually two numbers whose distance from zero (their absolute value) is 10: –10 and 10. So, the equation \|2x\| = 10 can be rewritten as two separate equations: 2x = 10 or 2x = –10 Here’s the same information about absolute values written in math-speak: Let c ≥≥≥≥≥ 0. If \|x\| = c, then x = c or x = –c . Example Example Example Example Example 11111 Solve \|3x\| = 12. Solution Solution Solution Solution Solution There are two separate equations to solve: 3x = 12 or 3x = –12 Solve each equation for x. x = 3 12 3 3 x = 4 12 3 x = − 3 3 x = –4 Check your answers by substituting back into the original equation. \|3(4)\| = 12 \|12\| = 12 12 = 12 ✓ \|3(–4)\| = 12 \|–12\| = 12 12 = 12 ✓ So x = 4 and x = –4 are the solutions of the equation. Independent Practice Solve: 1. \|4x\| = 16 4. \|3x\| = 9 2. \|–2x\| = 8 3. \|–8x\| = 24 5. \|4.8x\| = 144 6. \|0.02x\| = 9 7. \|–1.04x\| = 0.2392 8. \|2x + 1x\| = 171 9. \|10x – 5x\| = 100 + 5 Write these as absolute value equations and find the solutions: 10. The product of four and a number is a distance of 20 from 0. 11. A number has a distance of 8 from 0. 12. Twice a number has a distance of 0.6 from 0. 13. The product of 1.6 and a number has a distance of 0.304 from 0. ound Up ound Up RRRRRound Up ound Up ound Up ound Up ound Up RRRRRound Up ound Up ound Up The main thing to remember with absolute values in an equation is that they result in two possible solutions. In the next Topic you’ll work through more complicated equations involving absolute values. Section 2.8 Section 2.8 Section 2.8 — Absolute Value Equations Section 2.8 Section 2.8 129129129129129 TTTTTopicopicopicopicopic 2.8.22.8.2 2.8.22.8.2 2.8.2 California Standards: 3.0:3.0:3.0:3.0:3.0: Students solv Students solveeeee Students solv Students solv Students solv tions tions equa equa tions and inequalities equations equa tions equa inininininvvvvvolving
a alues..... alues alues bsolute v bsolute v olving a olving a bsolute values olving absolute v alues bsolute v olving a What it means for you: You’ll solve equations involving absolute values. Key words: absolute value Absolute VVVVValuealuealuealuealue Absolute Absolute e on e on MorMorMorMorMore on e on Absolute Absolute e on MorMorMorMorMore on Absolute VVVVValuealuealuealuealue Absolute Absolute e on e on e on Absolute Absolute e on This Topic is quite similar to the last one — but this time the absolute values have more than one term. That means there’s a little more solving to do once you’ve removed the absolute value signs. e than One TTTTTererererermmmmm e than One e than One lude Mor lude Mor y Inc y Inc alues Ma Absolute VVVVValues Ma alues Ma Absolute Absolute lude More than One y Include Mor alues May Inc e than One lude Mor y Inc alues Ma Absolute Absolute If there’s more than just a single term in the absolute value signs, you need to keep those terms together until the absolute value signs have been removed. Example Example Example Example Example 11111 Solve \|2x – 5\| = 7. Solution Solution Solution Solution Solution Rewrite this as two separate equations: 2x – 5 = 7 or 2x – 5 = –7 Solve both equations for x: 2x = 12 x = 6 2x = –2 x = –1 Check your answers by substituting back into the original equation. \|2(6) – 5\| = 7 \|12 – 5\| = 7 \|7\| = 7 7 = 7 ✓ \|2(–1) – 5\| = 7 \|–2 – 5\| = 7 \|–7\| = 7 7 = 7 ✓ So x = 6 and x = –1 are the correct solutions of the equation. Guided Practice Find all possible solutions to these absolute value equations: 1. \|12 – 4x\| = 18 3. \|5x – 3x – 1\| = 10 5. − 3 g = 15 2. \|2x – 8\| = 4 4. \|3j + 1\| = 10 6. − 8 x = 1.6 Find all possible solutions to these equations when a = –4: 7. \|2x + a\| = 10 9. \|3x + 8\| = a 8. \|ax + 8\| = 144 10. \|ax – a\| = –a 130130130130130 Section 2.8 Section 2.8 Section 2.8 — Absolute Value Equations Section 2.8 Section 2.8 tion Firststststst tion Fir tion Fir e the Equa e to Rearearearearearrrrrrangangangangange the Equa e the Equa e to R ou Might Havvvvve to R e to R ou Might Ha YYYYYou Might Ha ou Might Ha e the Equation Fir tion Fir e the Equa e to R ou Might Ha If the absolute value is not alone on one side of the equals sign, the equation must be rearranged to get the absolute value on its own before the equation can be split into two parts. To solve an equation of the form d\|ax + b\| + c = k, rewrite the equation in the form \|ax + b\| = k − c d , then solve for x. Example Example Example Example Example 22222 Solve 2\|3x – 1\| + 4 = 12. Solution Solution Solution Solution Solution Rearrange the equation to get the absolute value on its own: 2\|3x – 1\| + 4 = 12 2\|3x – 13x – 1\| = 4 Now split this into the two parts: 3x – 1 = 4 3x = 5 x = 5 3 Check your answers: 2\|3 ⎛ ⎜⎜⎜ ⎝ 5 3 ⎞ ⎟⎟⎟ – 1\| + 4 = 12 ⎠ 2\|5 – 1\| + 4 = 12 2\|4\| + 4 = 12 8 + 4 = 12 ✓ 3x – 1 = –4 3x = –3 x = –1 2\|3(–1) – 1\| + 4 = 12 2\|–4\| + 4 = 12 8 + 4 = 12 ✓ So x = 5 3 and x = –1 are the correct solutions of the equation. Section 2.8 Section 2.8 Section 2.8 — Absolute Value Equations Section 2.8 Section 2.8 131131131131131 Guided Practice Find all possible solutions to these equations involving absolute values: 11. \|2y – 7\| + 5 = 20 12. 11 + \|–7 – 2k\| = 21 13. \|x – 9\| – 5 = 1 14. –11 + \|5 – 4y\| = 2 15. 4\|3y – 4\| = 8 16. –2\|6 – 5y\| = –8 17. –3\|12 – 7x\| = –6 18. 7 – 2\|8 – 4x\| = –9 3 = 8 2 5 19. 21 20. 4 3 x − 5 22. 10 + = 7 5 2 x − 4 = 11 Absolute VVVVValues on Both Sides alues on Both Sides alues on Both Sides Absolute ou Might Havvvvve e e e e Absolute Absolute ou Might Ha YYYYYou Might Ha ou Might Ha alues on Both Sides alues on Both Sides Absolute ou Might Ha When it comes to solving equations with absolute values on both sides, you end up with four equations rather than two: Example Example Example Example Example 33333 Solve \|3x – 2\| = \|4 – x\|. Solution Solution Solution Solution Solution \|3x – 2\| = \|4 – x\| ±(3x – 2) = ±(4 – x) There are four possible solutions: (1) 3x – 2 = 4 – x (2) 3x – 2 = –(4 – x) (3) –(3x – 2) = 4 – x (4) –(3x – 2) = –(4 – x) But — this set of four equations only contains two different equations: if you take equation (4) and divide both sides by –1, you get equation (1). if you take equation (3) and divide both sides by –1, you get equation (2). So if there are absolute values on both sides of an equation, then you can treat one of them as though it is not an absolute value — so writing \|3x – 2\| = \|4 – x\| is equivalent to writing \|3x – 2\| = 4 – x or 3x – 2 = \|4 – x\|. More generally: \|ax + b\| = \|cx + d\| is equivalent to \|ax + b\| = cx + d or ax + b = \|cx + d\|. 132132132132132 Section 2.8 Section 2.8 Section 2.8 — Absolute Value Equations Section 2.8 Section 2.8 Example Example Example Example Example 44444 Solve \|5x – 2\| = \|10 – x\|. Solution Solution Solution Solution Solution Rewrite the equation with only one absolute value: \|5x – 2\| = 10 – x Write out two separate equations to solve. 5x – 2 = +(10 – x) 5x – 2 = –(10 – x) Solve the equations for x. 5x – 2 = 10 – x 5x + x – 2 = 10 – x + x 6x – 2 = 10 6x – 2 + 2 = 10 + 2 6x = 12 x = 2 5x – x – 2 = –10 + x – x 4x – 2 = –10 4x – 2 + 2 = –10 + 2 4x = –8 x = –2 Check your answers by substituting them into the original equation. \|5x – 2\| = \|10 – x\| \|5(2) – 2\| = \|10 – (2)\| \|10 – 2\| = \|8\| \|8\| = \|8\| 8 = 8 ✓ \|5x – 2\| = \|10 – x\| \|5(–2) – 2\| = \|10 – (–2)\| \|–10 – 2\| = \|10 + 2\| \|–12\| = \|12\| 12 = 12 ✓ So x = 2 and x = –2 are the correct solutions of the equation. Guided Practice Find all possible solutions to these absolute value equations: 23. \|2x – 8\| = \|3x – 12\| 24. \|5x – 7\| = \|3x + 15\| 25. \|9x + 18\| = \|4x – 2\| 26. \|6x – 13\| = \|15 – 8x\| 27. \|4x – 36\| = \|2x – 4\| 28. \|8x + 4\| = \|12x – 2\| 29. The distance of (–2x – 8) from 0 is 3. What are the possible values of x? 30. If (2x + 4) and (3x + 8) are the same distance from 0, what are the possible values of x? 31. If (x + 8) is the same distance from 0 as (4x – 8), what are the possible values of x? If c = 10, find all possible solutions to these equations: 32. \|2x – c\| = \|c – x\| 33. \|cx – 5\| = \|x – c\| 34. \|4x – c\| = \|2x + c\| 35. \|2cx + 4\| = \|x – 3c\| Section 2.8 Section 2.8 Section 2.8 — Absolute Value Equations Section 2.8 Section 2.8 133133133133133 Independent Practice Solve: 1. \|2x\| = 84 3. \|x + 8\| = 24 2. \|3z\| = –9 4. \|–x + 4\| = 3 5. \|2x + 8\| = \|4 – 3x\| 6. \|7x – 4\| = \|3x + 9\| 7. \|0.1x – 0.3\| = \|0.3x + 4.1\| 8. –2\|x – 20\| = –8 9. 1 3 x 1 + = 8 1 12 + x 1 4 11. 1 2 x + 1 = 5 13. − 8 x . 0 1 ( 10 = 4 ) − x 3 10. − x 4 = 12 12 14 14 In Exercises 15–18 you will need to form an absolute value equation and solve it to find the unknown. 15. If (x + 4) is 3x from 0, what are the possible values of x? 16. If (4x – 5) is (2x + 1) from 0, what are the possible values of x? 17. If (3w + 2) and (w – 4) are the same distance away from 0, what are the possible values of w? 18. If (4x – 5 + x) and (7 + 5x + 2) are the same distance from 0, what are the possible values of x? 19. Given that \|3x – 5\| = \|2x + 6\|, find the two possible values of b2 – 2bx + x2 if b = –3. If a = 2, b = 4, and c = 6 then solve each absolute value equation for x: 20. \|ax – b\| = \|x + c\| 21. \|ax + b\| + c = a – bx 22. 1 a cx + = − ab ax c ound Up ound Up RRRRRound Up ound Up ound Up ound Up ound Up RRRRRound Up ound Up ound Up Equations with absolute values on both sides look really difficult at first. Make sure you understand how Example 3 shows that you still only get two distinct equations when there are two absolute values. 134134134134134 Section 2.8 Section 2.8 Section 2.8 — Absolute Value Equations Section 2.8 Section 2.8 Chapter 2 Investigation ildlife Pe Pe Pe Pe Pararararark k k k k TTTTTrrrrrainsainsainsainsains ildlif ildlif WWWWWildlif ildlif WWWWWildlif ildlife Pe Pe Pe Pe Pararararark k k k k TTTTTrrrrrainsainsainsainsains ildlif ildlif ildlif This Investigation is all about writing and solving equations about rates. A large wildlife park is circular in shape and has a diameter of 7 miles. Around the outskirts of the park is a circular train track. Two automatic trains travel around the park, both in a clockwise direction. The trains are designed to both average 20 mph so that they never meet. However, one train has developed a fault and now travels at 18 mph. The trains set off from stations on opposite sides of the park at 9 a.m. At what time will the faster train catch up with the slower train? (Assume that the trains instantly reach their average speeds and that you can ignore the lengths of the trains.) North Station South Station Things to think about: • What is the distance around the track? Circumference = p × diameter. Use p = 22 7 . When the trains meet, the faster train will have traveled further than the slower train. How much further? If the slower train has traveled x miles, how far has the faster train traveled? When they get to the meeting point, they will have both been traveling for the same amount of time. How long will each train have been traveling for in terms of x? Speed = distance time Extension 1) Where will the trains be when one catches up with the other? How many times will each train pass its starting point before they meet? 2) When the faster train has caught up with the slower train, it changes direction. Both trains are now traveling in opposite directions. After how many minutes will the trains meet again? Open-ended extension The park manager wants the trains to run each day while the park is open without one train catching up with the other. Unfortunately, the speeds of the trains cannot be changed for technical reasons. The opening times of the park are shown on the right. OPENING TIMES Mon - Fri: 9 a.m. - 5 p.m. Saturday: 9 a.m. - 8.30 p.m. Sunday: 10 a.m. - 7 p.m. Write a report to the manager recommending how he can achieve this. You may wish to include diagrams of the track in your report. ound Up ound Up RRRRRound Up ound Up ound Up ound Up ound Up RRRRRound Up ound Up ound Up Parts of this Investigation are tough because the trains are traveling in a circle — but the main th
ing with any real-life problem is to write down all the math carefully before you start solving. estigaaaaationtiontiontiontion — Wildlife Park Trains estigestig estig pter 2 Invvvvvestig pter 2 In ChaChaChaChaChapter 2 In pter 2 In pter 2 In 135135135135135 Chapter 3 Single Variable Linear Inequalities Section 3.1 Inequalities ............................................ 137 Section 3.2 Applications of Inequalities ................... 147 Section 3.3 Compound Inequalities ......................... 155 Section 3.4 Absolute Value Inequalities ................... 158 Investigation Mailing Packages .................................. 162 136136136136136 TTTTTopicopicopicopicopic 3.1.13.1.1 3.1.13.1.1 3.1.1 Section 3.1 Inequalities Inequalities Inequalities Inequalities Inequalities Inequalities Inequalities Inequalities Inequalities Inequalities California Standards: 4.0: Students simplify 4.0: Students simplify 4.0: Students simplify 4.0: Students simplify 4.0: Students simplify eeeeexprxprxprxprxpressions bef essions befororororore solving e solving e solving essions bef essions bef e solving e solving essions bef linear linear linear equations and linear linear inequalities in one variaariaariaariaariabbbbblelelelele,,,,, inequalities in one v inequalities in one v inequalities in one v inequalities in one v such as 3(2x – 5) + 4(x – 2) = 12..... What it means for you: You’ll go over inequality notation from grade 7 and you’ll show inequalities on the number line. Key words: inequality endpoint interval Inequalities work like equations, but they tell you whether one expression is bigger or smaller than the expression on the other side. In grade 7 you solved linear inequalities — so the stuff in the next couple of Topics should be fairly familiar. Inequality Symbols Inequality Symbols Inequality Symbols Inequality Symbols Inequality Symbols Expressions such as 3x > 8, x < –5, x ≥ 10, and x £ 10 are inequalities. An inequality is a mathematical sentence that states that two expressions are not equal. You read the inequality symbols like this: < “is less than” > “is greater than” £££££ “is less than or equal to” ≥≥≥≥≥ “is greater than or equal to” For example, you read m < c as “m is less than c” and you read k ≥ 5 as “k is greater than or equal to 5.” w Inequalities on the Number Line w Inequalities on the Number Line ou Can Sho YYYYYou Can Sho ou Can Sho w Inequalities on the Number Line ou Can Show Inequalities on the Number Line w Inequalities on the Number Line ou Can Sho The inequality x > 4 represents the interval (part of the number line) where the numbers are greater than 4. Similarly k ≥≥≥≥≥ 5 represents all real numbers greater than or equal to 5 on the number line. The numbers 4 and 5 in these examples represent the endpoints of the intervals of the number line under consideration. However, k ≥≥≥≥≥ 5 includes the endpoint 5, while x > 4 excludes the endpoint 4. Example Example Example Example Example 11111 Show the inequality x > 4 on the number line. Solution Solution Solution Solution Solution You can show x > 4 on the number line like this: Check it out: The symbol • and the closing parenthesis indicate that the list of numbers continues indefinitely (which is also shown by the arrow on the graph). –6 –4 –2 0 2 4 6 Note that the circle at the endpoint is open — this shows that the endpoint 4 is not part of the interval of the number line defined by the inequality x > 4. This is called an open interval, denoted as (4, •••••). Section 3.1 Section 3.1 Section 3.1 — Inequalities Section 3.1 Section 3.1 137137137137137 Check it out: The “[” bracket indicates that the endpoint is part of the interval, which is also shown by the closed circle in the graph. Check it out: Note also that the interval is closed at one end, but open at the other — which is why it’s called a half-open interval. Example Example Example Example Example 22222 Show the inequality k ≥ 5 on the number line. Solution Solution Solution Solution Solution –6 –4 –2 0 2 654 Note that the closed circle at 5 indicates that the endpoint 5 is included in the interval. This is called a half-open interval, denoted as [5, •••••). Guided Practice Write each of the following inequalities in interval notation, and show each graphically. 1. k > 6 4. j > –5 7. 3 ≥ y 2. m < –2 5. n £ 3.5 8. j ≥ 10 3. x ≥ 1 6. l £ 3 9. 10 £ x 10. State the inequality represented on the number line opposite. 11. State the inequality represented on the number line opposite using interval notation. Independent Practice –4 –2 –4 –2 0 0 2 2 4 4 In Exercises 1–3 write each inequality in interval notation. 1. r > 8 2. t £ –9 3. 3 £ x In Exercises 4–6 write each interval as an inequality in x. 6. (–•, 0] 5. (–6, •) 4. (–•, 2) In Exercises 7–12 show each inequality or interval graphically. 7. k > –7 10. (–11, •) 9. 1.5 £ y 12. (–•, 5] 8. k £ 2 11. [0, •) 13. Anthony is shopping for a birthday gift for his cousin Robert. He has $25 in his wallet. Write an inequality that shows how many dollars he can spend on the gift. 14. Teresa is only allowed to swim outside if the temperature outside is at least 85 °F. Write an inequality that shows the temperatures in degrees Fahrenheit at which Teresa is allowed to swim. 15. In order to achieve an ‘A’ in math, Ivy needs to score more than 95% on her next test. Write an inequality that shows the test score Ivy needs to achieve in order to earn her ‘A’ in math. ound Up ound Up RRRRRound Up ound Up ound Up ound Up ound Up RRRRRound Up ound Up ound Up The most difficult thing is remembering that a “[“ bracket shows the endpoint is included in the interval, and a “(“ parenthesis means the endpoint isn’t included. 138138138138138 Section 3.1 Section 3.1 Section 3.1 — Inequalities Section 3.1 Section 3.1 TTTTTopicopicopicopicopic 3.1.23.1.2 3.1.23.1.2 3.1.2 California Standards: 4.0: Students simplify 4.0: Students simplify 4.0: Students simplify 4.0: Students simplify 4.0: Students simplify eeeeexprxprxprxprxpressions bef essions befororororore solving e solving e solving essions bef essions bef e solving e solving essions bef linear linear linear equations and linear linear inequalities in one variaariaariaariaariabbbbblelelelele,,,,, inequalities in one v inequalities in one v inequalities in one v inequalities in one v such as 3(2x – 5) + 4(x – 2) = 12..... Students solveeeee 5.0:5.0:5.0:5.0:5.0: Students solv Students solv Students solv Students solv lems,,,,, lems lems p prp proboboboboblems mmmmmultiste p prp pr ultiste ultiste ultistep pr lems ultiste including word problems,,,,, olving inininininvvvvvolving olving olving linear equations olving linear inequalities in linear inequalities in and linear inequalities in linear inequalities in linear inequalities in le and prooooovidevidevidevidevide one variaariaariaariaariabbbbble and pr le and pr le and pr one v one v one v le and pr one v h steppppp..... h ste h ste or eac or eac tion f tion f justifica justifica or each ste tion for eac justification f h ste or eac tion f justifica justifica What it means for you: You’ll solve inequalities that contain + and – signs. Key words: inequality endpoint interval action action dition and Subtr dition and Subtr AdAdAdAdAddition and Subtr action dition and Subtraction action dition and Subtr action action dition and Subtr dition and Subtr AdAdAdAdAddition and Subtr action dition and Subtraction action dition and Subtr Inequalities Inequalities ties of ties of oper oper PrPrPrPrProper Inequalities ties of Inequalities operties of Inequalities ties of oper PrPrPrPrProper Inequalities Inequalities ties of ties of oper oper ties of Inequalities operties of Inequalities ties of oper Inequalities To solve inequalities such as x + 1 > 4, 3x + 2 > 7, or x – 7 £ 12, you need to apply properties of inequalities — they’re just like the properties of equality you applied to equations. Inequalities Inequalities ty of ty of oper oper dition Pr AdAdAdAdAddition Pr dition Pr Inequalities ty of Inequalities operty of dition Proper Inequalities ty of oper dition Pr Addition Property of Inequalities Given real numbers a, b, and c, if a > b, then a + c > b + c. In other words, adding the same number to both sides of an inequality gives an equivalent inequality. Example Example Example Example Example 11111 Solve and graph the solution of x – 2 > 5 on a number line. Write the solution in interval notation. Solution Solution Solution Solution Solution Solve Using ad Using ad dition pr dition pr oper oper ty of ty of inequalities inequalities Using addition pr dition proper operty of ty of inequalities inequalities Using ad Using ad dition pr oper ty of inequalities Check it out: The circle is open at the endpoint 7, which shows that 7 isn’t part of the solution interval. Graph: –7 0 7 Solution in interval notation: (7, •••••) Guided Practice Solve and graph each inequality. Write each solution set in interval notation. 1. l – 1 ≥ 3 2. r – 3 < –5 3. x – 4 £ –1 5. –2m < 2 – 3m 7. –2j + 3 ≥ –j 4. m – 7 > –10 6. –k £ –2k + 1 8. –7(j + 1) < –6j Section 3.1 Section 3.1 Section 3.1 — Inequalities Section 3.1 Section 3.1 139139139139139 Inequalities Inequalities ty of ty of oper oper action Pr action Pr Subtr Subtr Inequalities ty of Inequalities operty of action Proper Subtraction Pr Inequalities ty of oper action Pr Subtr Subtr Subtraction Property of Inequalities Given real numbers a, b, and c, if a > b, then a – c > b – c. That is, subtracting the same number from both sides of an inequality gives an equivalent inequality. Example Example Example Example Example 22222 Solve and then graph the solution of 3x £ 6 + 2x on a number line. Write the solution in interval notation and then state the maximum integer value of x that satisfies the inequality. Solution Solution Solution Solution Solution Solve: 3x £ 6 + 2x 3x – 2x £ 6 + 2x – 2x Subtr action pr action pr Subtr Subtr ty of ty of oper oper inequalities inequalities Subtraction pr operty of action proper inequalities ty o
f inequalities Subtr oper action pr inequalities ty of x £ 6 Graph: –1 0 1 2 3 4 5 6 Solution in interval notation: (–•••••, 6] So the maximum integer value of x is 6. opertiestiestiestiesties oper oper ou to Use Both Pr lems Need YYYYYou to Use Both Pr ou to Use Both Pr lems Need Some Proboboboboblems Need lems Need Some Pr Some Pr ou to Use Both Proper oper ou to Use Both Pr lems Need Some Pr Some Pr Example Example Example Example Example 33333 Solve and graph the solution set of 5x – 2 £ 4x – 3 on a number line. State the maximum integer value of x that satisfies the inequality, and write the solution set in interval notation. Solution Solution Solution Solution Solution Solve: 5x – 2 £ 4x – 3 5x – 2 + 2 £ 4x – 3 + 2 AdAdAdAdAddition pr inequalities inequalities ty of ty of oper oper dition pr dition pr inequalities ty of inequalities operty of dition proper inequalities ty of oper dition pr 5x £ 4x – 1 5x – 4x £ 4x – 4x – 1 Subtr inequalities inequalities ty of ty of oper oper action pr action pr Subtr Subtr inequalities ty of inequalities operty of action proper Subtraction pr inequalities ty of oper action pr Subtr opertytytytyty oper oper and commutautautautautatititititivvvvve pre pre pre pre proper and comm and comm oper and comm and comm x £££££ –1 Graph: –4 –3 –2 –1 0 1 2 Solution in interval notation: (–•••••, –1] So the maximum integer value of x is –1. Check it out: In this example, the half-open interval closes with a bracket to show that the endpoint 6 is part of the solution interval. Check it out: The closed circle of the graph shows that the endpoint is included in the solution interval. 140140140140140 Section 3.1 Section 3.1 Section 3.1 — Inequalities Section 3.1 Section 3.1 Guided Practice Solve and graph each inequality, and write each solution set in interval notation. Find also the maximum or minimum integer value that satisfies each inequality. 9. x + 2 < 8 11. j + 2 ≥ –3 13. 3k + 2 ≥ 2k 15. 2t + 1 £ t – 10 10. t + 5 £ 6 12. y + 3 > –9 14. 4x – 3 < 3x 16. 6(x – 1) ≥ 5(x + 2) Independent Practice Solve the following inequalities. Justify each step and write each solution set in interval notation. 1. j – 3 > –1.5 3. x + 3 < –11 5. 5k – 1.25 < 4k – 4 2. k – 2 ≥ –5 4. 2v – 2.5 > v + 1 6. –7j – 5 < –6j Solve and graph each inequality in Exercises 7–18. Write each solution set in interval notation. 7. 6x – 1 ≥ 5x + 3 9. 4(x – 3) – 3x < –11 11. 7t + 6(1 – t) £ –4 13. 2(t + 5) £ 3t – 11 15. 5(–2 + 3x) ≥ –2(1 – 7x) 17. 13 + 3(x – 5) £ 2(–3 + x) 8. 3(k – 2) – 2k ≥ 9 10. 4n – 3(n + 1) ≥ 1 12. 5t – 4(t – 1) > –3 14. –3x + 2 + 7x < –3 + 3x – 4 16. –4(1 – 4x) ≥ –7.5(–1 – 2x) 18. 5x – (3x – 4) > –(2 – x) 19. Find the maximum integer value of x if 6x + 4 £ 5x – 8. 20. Find the maximum integer value of x if –3(3 – 2x) < 5(x – 5). 21. Find the least integer value of x if –2(1 – x) > x – 1. 22. Find the least integer value of x if 4(x – 3) £ 5x + 2 23. Stephen needs to buy a new uniform for soccer. He already has $25, but the uniform costs $55. He participates in car washes to help pay for the uniform. Write an inequality to represent the amount of money, x, that Stephen needs to earn from the car washes in order to be able to afford the new uniform. Use this inequality to find the minimum amount of money he needs to earn. 24. An art gallery sells Peter’s paintings for $x, and keeps $100 commission. This means Peter is paid $(x – 100) for each painting. If Peter wants to make at least $750 for a particular painting, write an inequality to represent the amount, x, that the gallery needs to sell that painting for. Use this inequality to find the minimum price of the painting. ound Up ound Up RRRRRound Up ound Up ound Up ound Up ound Up RRRRRound Up ound Up ound Up Adding and subtracting with inequalities is a lot like dealing with normal equations — so there’s nothing in this Topic that should cause you too much trouble. Next up is multiplying and dividing. Section 3.1 Section 3.1 Section 3.1 — Inequalities Section 3.1 Section 3.1 141141141141141 TTTTTopicopicopicopicopic 3.1.33.1.3 3.1.33.1.3 3.1.3 California Standards: 4.0: Students simplify 4.0: Students simplify 4.0: Students simplify 4.0: Students simplify 4.0: Students simplify eeeeexprxprxprxprxpressions bef essions befororororore solving e solving e solving essions bef essions bef e solving e solving essions bef linear linear linear equations and linear linear inequalities in one variaariaariaariaariabbbbblelelelele,,,,, inequalities in one v inequalities in one v inequalities in one v inequalities in one v such as 3(2x – 5) + 4(x – 2) = 12..... Students solveeeee 5.0:5.0:5.0:5.0:5.0: Students solv Students solv Students solv Students solv lems,,,,, lems lems p prp proboboboboblems mmmmmultiste p prp pr ultiste ultiste ultistep pr lems ultiste including word problems,,,,, olving inininininvvvvvolving olving olving linear equations olving linear inequalities in linear inequalities in and linear inequalities in linear inequalities in linear inequalities in le and prooooovidevidevidevidevide one variaariaariaariaariabbbbble and pr le and pr le and pr one v one v one v le and pr one v h steppppp..... h ste h ste or eac or eac tion f tion f justifica justifica or each ste tion for eac justification f h ste or eac tion f justifica justifica What it means for you: You’ll solve inequalities that contain × and ÷ signs. Key words: inequality reverse 142142142142142 Section 3.1 Section 3.1 Section 3.1 — Inequalities Section 3.1 Section 3.1 vision vision tion and Di tion and Di Multiplica Multiplica vision tion and Division Multiplication and Di vision tion and Di Multiplica Multiplica vision vision tion and Di tion and Di Multiplica Multiplica tion and Division vision Multiplication and Di tion and Di vision Multiplica Multiplica Inequalities Inequalities ties of ties of oper oper PrPrPrPrProper Inequalities ties of Inequalities operties of Inequalities ties of oper PrPrPrPrProper Inequalities Inequalities ties of ties of oper oper ties of Inequalities operties of Inequalities ties of oper Inequalities After a Topic on addition and subtraction with inequalities, you know what to expect: multiplication and division of inequalities. Inequalities Inequalities ty of ty of oper oper tion Pr tion Pr Multiplica Multiplica Inequalities ty of Inequalities operty of tion Proper Multiplication Pr Inequalities ty of oper tion Pr Multiplica Multiplica Multiplication Property of Inequalities Given real numbers a, b, and c, if a > b and c > 0 then ac > bc. That is, multiplying both sides of an inequality by a positive number gives an equivalent inequality. For example: Start with an inequality: Multiplying by 2 gives: 2 × 4 < 2 × 10 4 < 10 ...is true. 8 < 20 ...which is also true. Example Example Example Example Example 11111 Solve 1 5 x > 2. Solution Solution Solution Solution Solution Multiplica Multiplica Multiplica tion pr tion pr oper oper ty of ty of inequalities inequalities Multiplication pr tion proper operty of ty of inequalities inequalities Multiplica tion pr oper ty of inequalities x > 10 Guided Practice Solve each inequality in Exercises 1–12. 1. 1 2 x £ 3 4. 6 7. j 9 10. 1 > x 3 12≤ ≤ y 20 − 1 5 2. 5. 1 4 x 4 p > 12 5> − 8. 1 3 1 < a 9 11. 2 5 p < 4 3. 6 15 9. ≥ k 12 1 4 12. − <6 3 4 a Inequalities Inequalities ty of ty of oper oper vision Pr DiDiDiDiDivision Pr vision Pr Inequalities ty of Inequalities operty of vision Proper Inequalities ty of oper vision Pr Division Property of Inequalities Given real numbers a, b, and c, if a > b and c > 0 then a c b > . c In other words, dividing both sides of an inequality by a positive number gives an equivalent inequality. For example: Start with an inequality: Dividing by 2 gives: 4 < 10 4 ÷ 2 < 10 ÷ 2 2 < 5 ...is true. ...which is also true. Example Example Example Example Example 22222 Solve 3x < 11. Solution Solution Solution Solution Solution 3x < 11 x 3 3 < x < 11 3 11 3 DiDiDiDiDivision pr vision pr vision pr oper oper ty of ty of inequalities inequalities vision proper operty of ty of inequalities inequalities vision pr oper ty of inequalities Guided Practice Solve each inequality in Exercises 13–26. 13. 10x < 5 15. 32 ≥ 8n 17. 40 < 8y 19. 72 ≥ 9j 21. 20k < 5 23. 4 > 28y 25. 6(6a) ≥ 4 14. 4y ≥ 20 16. 7z £ 28 18. 56 > 7j 20. 6k £ 48 22. 7 < 77x 24. 40a < 5 26. 9 £ 3b(20 + 7) Section 3.1 Section 3.1 Section 3.1 — Inequalities Section 3.1 Section 3.1 143143143143143 e Numbersssss ty with Negggggaaaaatititititivvvvve Number e Number e Number ty with Ne ty with Ne oper oper tion Pr tion Pr Multiplica Multiplica operty with Ne tion Proper Multiplication Pr e Number ty with Ne oper tion Pr Multiplica Multiplica This is really important — so make sure you read this carefully. You’ve just seen that multiplying both sides of an inequality like 4 > 3 by, say, 3, gives a true new inequality, 12 > 9. However, if the original inequality is multiplied by a negative number like –3, the resulting inequality is –12 > –9, which is false. To make the resulting inequality true, you have to reverse the inequality sign. That gives –12 < –9, which is true. Multiplication Property of Inequalities — Negative Numbers Given real numbers a, b, and c, if a > b and c < 0 then ac < bc. In other words, if you multiply both sides of an inequality by a negative number, you have to reverse the inequality symbol — otherwise the statement will be false. For example: Check it out: Watch out — the difference between this rule and the normal multiplication property is the “c < 0 then ac < bc” part. Start with an inequality: Multiplying by –2 and reversing the inequality sign gives: ...is true. –3 < 8 –2 × –3 > –2 × 8 6 > –16 ...which is also true. Example Example Example Example Example 33333 Solve – 1 2 x > 2. Solution Solution Solution Solution Solution x > 2 – 1 2 –2 × – 1 2 x < –4 x < –2 × 2 Multiplica Multiplica Multiplica tion pr tion pr oper oper ty of ty of inequalities inequalities Multiplication pr tion proper operty of ty of inequalitie
s inequalities Multiplica tion pr oper ty of inequalities Guided Practice Solve each inequality in Exercises 27–37. 27. 30. 33. ≤1 x 7 − 10 − 4 > −d 3 − ≥j 8 7 28. 31. 34. − <1 c 9 − 1 4 − < −k 11 a 5 29. 11 < < − 12 32. − ≤ 4 1 35 36. − 1 9 1 ⋅ > 4 x( 4 − 13 3 ) 37. 1 + ⋅ 5 10 1 ≥ − g 6 4 144144144144144 Section 3.1 Section 3.1 Section 3.1 — Inequalities Section 3.1 Section 3.1 e Numbersssss ty with Negggggaaaaatititititivvvvve Number e Number e Number ty with Ne ty with Ne oper oper vision Pr DiDiDiDiDivision Pr vision Pr operty with Ne vision Proper e Number ty with Ne oper vision Pr Division Property of Inequalities — Negative Numbers Given real numbers a, b, and c, if a > b and c < 0 then a c b < . c In other words, if you divide both sides of an inequality by a negative number, you have to reverse the inequality symbol — otherwise the statement will be false. Start with an inequality: Dividing by –3 and reversing the inequality sign gives: ...is true. –3 ...which is also true. Example Example Example Example Example 44444 Solve –13y < 39. Solution Solution Solution Solution Solution –13y < 39 − 39 − − 13 y > –3 13 13 > y DiDiDiDiDivision pr vision pr vision pr oper oper ty of ty of inequalities inequalities vision proper operty of ty of inequalities inequalities vision pr oper ty of inequalities Guided Practice Solve each inequality in Exercises 38–47. 39. 36 < –9x 38. –5x £ 5 41. –x > –1 40. –6j > 48 43. –77c < –11 42. 45 £ –9x 45. 49 > –7a 44. –72y ≥ –8 − ≥ −1 46. − ≥4 y 47. 6c 1 4 10 Section 3.1 Section 3.1 Section 3.1 — Inequalities Section 3.1 Section 3.1 145145145145145 Independent Practice Solve each inequality in Exercises 1–6. 1. 32 < 4h − >1 g 7 5. 2 £ 3. 6. 7 2. 9a £ –45 4. –8c £ 48 w 8 <5 5 7 (1 – 2y) < 5, and state the largest possible integer − h 6 5 7. Solve 3 2 value of y. y – 2 3 Solve each inequality in Exercises 8–21. 8. 2(x – 3) – 3(2 – x) > 8 10. 3(x – 1) < 7 – 2x 12. 3y – (5y + 4) > 7y + 2(y – 5) 13. 5(2x – 3) – 3(x – 7) ≥ 4(3x + 2) + 2x – 9 14. 5 – 4(x + 2) £ 7 + 5(2x – 1) 9. –4(3 – 2x) > 5x + 9 11. 5(y + 3) – 7y £ 3(2y + 3) – 5y 15. 7 – 2(m – 4) £ 2m + 11 − + x 2 3 1 5 3 − ≥ 16. 17. x 4 11 12 5 9 18. 0.5(x – 1) – 0.75(1 – x) < 0.65(2x – 1) 19. 7 – 3(x – 7) £ 4(x + 5) + 1 20. 0.35(x – 2) – 0.45(x + 1) ≥ 8 + 0.15(x – 10) 21. 3(2x + 6) – 5(x + 8) £ 2x – 22 22. Laura has $5.30 to spend on her lunch. She wants to buy a chicken salad costing $4.20 and decides to spend the rest on fruit. Each piece of fruit costs 45¢. Write an inequality to represent this situation, and then solve it to find how many pieces of fruit Laura can buy. 23. Audrey is selling magazine subscriptions to raise money for the school library. The library will get $2.50 for every magazine subscription she sells. Audrey wants to raise at least $250 for the library. Write and solve an inequality to represent the number of magazine subscriptions, x, Audrey needs to sell to reach her goal. 24. The total cost of food and supplies for a cat is x dollars per year, and medical expenses can be 1.5 times the cost of food and supplies per year. If Maddie can spend no more than $500 a year on the cat, what is the most that she can spend on food and supplies? Write and solve an inequality to represent the situation. ound Up ound Up RRRRRound Up ound Up ound Up RRRRRound Up ound Up ound Up ound Up ound Up Multiplication and division with negative numbers can sometimes be difficult. If you multiply by a negative number and forget to reverse the direction of the inequality sign, then your solution will be wrong — so watch out. 146146146146146 Section 3.1 Section 3.1 Section 3.1 — Inequalities Section 3.1 Section 3.1 TTTTTopicopicopicopicopic 3.2.13.2.1 3.2.13.2.1 3.2.1 Section 3.2 Multistep Inequalities Multistep Inequalities Multistep Inequalities Multistep Inequalities Multistep Inequalities Multistep Inequalities Multistep Inequalities Multistep Inequalities Multistep Inequalities Multistep Inequalities California Standards: 4.0: Students simplify 4.0: Students simplify 4.0: Students simplify 4.0: Students simplify 4.0: Students simplify eeeeexprxprxprxprxpressions bef essions befororororore solving e solving e solving essions bef essions bef e solving e solving essions bef linear linear linear equations and linear linear inequalities in one variaariaariaariaariabbbbblelelelele,,,,, inequalities in one v inequalities in one v inequalities in one v inequalities in one v such as 3(2x – 5) + 4(x – 2) = 12..... Students solveeeee 5.0:5.0:5.0:5.0:5.0: Students solv Students solv Students solv Students solv lems,,,,, lems lems p prp proboboboboblems mmmmmultiste p prp pr ultiste ultiste ultistep pr lems ultiste including word problems,,,,, olving inininininvvvvvolving olving olving linear equations olving linear inequalities in linear inequalities in and linear inequalities in linear inequalities in linear inequalities in le and prooooovidevidevidevidevide one variaariaariaariaariabbbbble and pr le and pr le and pr one v one v one v le and pr one v h steppppp..... h ste h ste or eac or eac tion f tion f justifica justifica or each ste tion for eac justification f h ste or eac tion f justifica justifica What it means for you: You’ll solve inequalities involving several steps. Key words: inequality isolate Check it out: This example uses the addition, subtraction and division properties of inequalities, as well as the associative property. Inequality problems often involve using more than one of the properties of inequalities that you saw in Topics 3.1.2 and 3.1.3. The multistep inequalities in this Topic are a little harder than the ones you saw in Section 3.1 — but you still solve them using the same methods. p Inequalities Combine Lots of TTTTTececececechniques hniques hniques p Inequalities Combine Lots of p Inequalities Combine Lots of Multiste Multiste hniques Multistep Inequalities Combine Lots of hniques p Inequalities Combine Lots of Multiste Multiste To simplify and therefore solve an inequality in one variable such as x, you need to isolate the terms in x on one side and isolate the numbers on the other. It’s often easiest to keep the x-terms on the side of the inequality sign where they have a positive value. Example Example Example Example Example 11111 Solve 4x – 7 > 2x. Solution Solution Solution Solution Solution 4x – 7 > 2x 4x – 7 + 7 > 2x + 7 4x > 2x + 7 Then get rid of the 2x on the right 4x – 2x > 2x + 7 – 2x Then get rid of the 2 on the right on the right Then get rid of the 2 Then get rid of the 2 on the right on the right Then get rid of the 2 2x > 7 Elimina Elimina Elimina te the –7 fr te the –7 fr om the left fir om the left fir om the left firststststst Eliminate the –7 fr te the –7 from the left fir Elimina te the –7 fr om the left fir Guided Practice 1. 5x – 2 £ 3 3. –3a – 3 < –9 5. 7 to get et et et et x on its o DiDiDiDiDivide both sides b on its o on its o on its ownwnwnwnwn vide both sides b vide both sides b y 2 to g y 2 to g vide both sides by 2 to g on its o vide both sides b y 2 to g 2. 4x – 1 > 2 4. 5x – 7 £ 8 6. 8. x 2 j 3 + 9 £ –2 – 8 ≥ 7 9. 8g – 10 £ 9g + 4 11. 4c – 9 ≥ 5c + 16 10. 4a + 5 £ 6a + 9 12. k – 4 £ 2k + 20 Section 3.2 Section 3.2 Section 3.2 — Applications of Inequalities Section 3.2 Section 3.2 147147147147147 Check it out: Instead of “x,” the variable might be “y,” “z,” or any other letter. ted Inequalities into Smaller Stepspspspsps ted Inequalities into Smaller Ste ted Inequalities into Smaller Ste eak Complica BrBrBrBrBreak Complica eak Complica eak Complicated Inequalities into Smaller Ste ted Inequalities into Smaller Ste eak Complica Here’s a useful checklist for tackling more complicated inequality questions by breaking them down into easier steps: Solving Inequalities 1. Multiply out any parentheses. 2. Simplify each side of the inequality. 3. Remove number terms from one side. 4. Remove x-terms from the other side. 5. Multiply or divide to get an x-coefficient of 1. Example Example Example Example Example 22222 Solve the inequality: 7(x – 2) – 3(x – 4) > 2(x – 5) Solution Solution Solution Solution Solution 7x – 14 – 3x + 12 > 2x – 10 4x – 2 > 2x – 10 4x – 2 + 2 > 2x – 10 + 2 4x > 2x – 8 4x – 2x > 2x – 8 – 2x 2x > –8 − 8 2 x > –4 x > 2 2 Multipl Multipl y out the par y out the par entheses entheses Multiply out the par y out the parentheses entheses Multipl Multipl y out the par entheses Simplify Simplify Simplify Simplify Simplify Elimina Elimina te the –2 fr te the –2 fr om the left om the left Eliminate the –2 fr te the –2 from the left om the left Elimina Elimina te the –2 fr om the left Get rid of the 2x on the right on the right on the right Get rid of the 2 Get rid of the 2 on the right on the right Get rid of the 2 Get rid of the 2 vide to get et et et et x on its o on its o on its o DiDiDiDiDivide to g on its ownwnwnwnwn vide to g vide to g on its o vide to g Guided Practice 13. 4x – 2 £ 3(x + 5) 15. 4x + 1 > 2(x + 2) 17. 8(x + 3) £ 7(x + 3) 19. 21 > 3x £ 3(x – 2) 14. 6x – 8 > 5(x + 2) 16. 5x – 4 < 3(x + 6) 18. 2(x + 1) ≥ 4(x – 2) 20. 22 < 3x ≥ 10(x + 1) 23. 2(x – 1) ≥ 4(x – 2) – 8 24. 12(b + 1) – 10b > 7(b + 3) + 6 148148148148148 Section 3.2 Section 3.2 Section 3.2 — Applications of Inequalities Section 3.2 Section 3.2 Independent Practice In Exercises 1–7, solve each inequality. 1. 9x – 7 £ 11 3. 6x – 12 > 5x + 8 2. 8c – 10 ≥ 7c + 6 4. 11(x + 8) ≥ 12x – 12 5. 9(p + 5) > 10p – 5 7. a 6 – 3 ≥ 5 6. a 2 < a + 4 In Exercises 8–28, solve each inequality and write the solution set in interval notation. 8. 6x – 2 £ 4(x + 5) 9. 3(x + 1) < 5x + 5 10. 5x + 1 > 3(x + 3) 11. 8(x – 1) ≥ 4x – 4 12. 6(j – 2) > 7(j + 4) 13. 3(x + 2) ≥ 5(x – 2) ( 7 4 ( ( 7 14. 15. 16. ( 8 17 £ 4a > 2x < 2(x + 4) £ 3(x – 2) 18. 3(k – 1) + 2(k + 1) < 4 19. 7(d + 3) + 2(d – 4) > –5 20. 6(x – 4) – 5(x + 1) £ 9 21. 4(t + 0.25) – 8(t – 7) ≥ –3 22. 4(x – 1) ≥ 8(x – 2) – 6x 23. –5(t – 2) + 4 > –(t + 2) 24. 4(t + 0.5) £ 0.5(4t – 12) + 6 25. 5(a – 1) – 2a < 4(a + 4) 26. 4(a + 3) + 4 ≥ –(a – 1) 27. 6(t + 1) –
10 > 7(t + 3) + 4t 28. 5(a – 5) – 3 > 4(a + 8) + 7a ound Up ound Up RRRRRound Up ound Up ound Up ound Up ound Up RRRRRound Up ound Up ound Up If you ever get stuck when you’re solving inequalities with more than one step, refer to the checklist on the previous page. Just take it one step at a time, as if you were dealing with an equation. Section 3.2 Section 3.2 Section 3.2 — Applications of Inequalities Section 3.2 Section 3.2 149149149149149 TTTTTopicopicopicopicopic 3.2.23.2.2 3.2.23.2.2 3.2.2 California Standards: 4.0: Students simplify 4.0: Students simplify 4.0: Students simplify 4.0: Students simplify 4.0: Students simplify eeeeexprxprxprxprxpressions bef essions befororororore solving e solving e solving essions bef essions bef e solving e solving essions bef linear linear linear equations and linear linear inequalities in one variaariaariaariaariabbbbblelelelele,,,,, inequalities in one v inequalities in one v inequalities in one v inequalities in one v such as 3(2x – 5) + 4(x – 2) = 12..... Students solveeeee 5.0:5.0:5.0:5.0:5.0: Students solv Students solv Students solv Students solv lems,,,,, lems lems p prp proboboboboblems ultiste mmmmmultiste p prp pr ultiste ultistep pr lems ultiste incincincincincluding w lems,,,,, d prd proboboboboblems lems lems d prd pr luding wororororord pr luding w luding w luding w lems olving olving inininininvvvvvolving olving linear equations olving linear inequalities in linear inequalities in and linear inequalities in linear inequalities in linear inequalities in le and prooooovidevidevidevidevide one variaariaariaariaariabbbbble and pr le and pr le and pr one v one v one v le and pr one v h steppppp..... h ste h ste or eac or eac tion f tion f justifica justifica or each ste tion for eac justification f h ste or eac tion f justifica justifica What it means for you: You’ll solve real-life problems involving inequalities. Key words: inequality isolate Check it out: These rules are pretty much the same as when you’re dealing with real-life equation problems. Inequalities Inequalities tions of tions of pplica pplica AAAAApplica Inequalities tions of Inequalities pplications of Inequalities tions of pplica AAAAApplica Inequalities Inequalities tions of tions of pplica pplica tions of Inequalities pplications of Inequalities tions of pplica Inequalities Some real-life problems include phrases like “at least” or “at most,” or deal with maximums or minimums. If you come across these phrases, chances are you’ll need to model the situation as an inequality. Proboboboboblems lems lems Pr eal-Life”e”e”e”e” Pr Pr eal-Lif Inequalities are “Re “Re “Re “Re “Real-Lif eal-Lif Inequalities ar Inequalities ar tions of tions of pplica AAAAApplica pplica lems tions of Inequalities ar pplications of lems Pr eal-Lif Inequalities ar tions of pplica In the same way that applications of equations are real-life problems, applications of inequalities are real-life inequalities problems. Real-life problems involving inequalities could be about pretty much anything — from finding the area of a field to figuring out how many CDs you can buy with a certain amount of money. What they all have in common is that they’ll all be word problems — and you’ll always have to set up and solve an inequality. Solving Real-Life Inequality Problems 1. First decide how you will label the variables... 2. ...then write the task out as an inequality... 3. ...make sure you include all the information given... 4. ...then solve the inequality. Example Example Example Example Example 11111 Find the three smallest consecutive even integers whose sum is more than 60. Solution Solution Solution Solution Solution First you need to label the variables: Let x = first (smallest) even integer x + 2 = next (second) even integer x + 4 = next (third) even integer 150150150150150 Section 3.2 Section 3.2 Section 3.2 — Applications of Inequalities Section 3.2 Section 3.2 Check it out: In this example, you know you need a “>” and not a “≥” because the problem says “more than,” not “more than or equal to.” Example 1 continueduedueduedued Example 1 contin Example 1 contin Example 1 contin Example 1 contin Then you need to write it out as an inequality: In English: The sum of three consecutive even integers is more than 60. In math: x + (x + 2) + (x + 4) > 60 Then simplify > 60 3x + 6 > 60 3x + 6 – 6 > 60 – 6 3x > 54 54 3 x > 18 x > 3 3 Answer in math: x > 18 Answer in English: The smallest even integer is more than 18. So the three smallest consecutive even integers whose sum is greater than 60 are 20, 22, and 24. Guided Practice 1. Find the largest three consecutive odd integers whose sum is at most 147. 2. Find the three smallest consecutive odd integers whose sum is more than 45. 3. Find the smallest three consecutive odd integers such that the sum of the first two integers is greater than the sum of the third integer and 11. 4. Find the smallest three consecutive even integers whose sum is greater than 198. 5. The difference between a number and twice that number is at least 7. Find the smallest possible integer that satisfies this criterion. 6. Three times a number is added to 11, and the result is less than 7 plus twice the number. Find the highest possible integer the number could be. 7. José scored 75 and 85 on his first and second algebra quizzes. If he wants an average of at least 83 after his third quiz, what is the least score that José must get on the third quiz? 8. Lorraine’s test scores for the semester so far are 60%, 70%, 75%, 80%, and 85%. If the cutoff score for a letter grade of B is 78%, what is the least score Lorraine must get on the final test to earn a B? Section 3.2 Section 3.2 Section 3.2 — Applications of Inequalities Section 3.2 Section 3.2 151151151151151 Example Example Example Example Example 22222 The perimeter of a rectangular field, F, is given by the formula p = 2l + 2w. As shown in the diagram, the length l = (10x – 6) m and the width w = (5x – 3) m. Find the possible values of x for which the given rectangular field would have a perimeter of at least 1182 meters. (5 – 3) m x F x (10 – 6) m Solution Solution Solution Solution Solution The variables are already labeled, so you just need to write it out as an inequality: Substituting the variables into the expression for the perimeter gives: p = 2(10x – 6) + 2(5x – 3) You’re also told that the perimeter is at least 1182 meters — in math, you write that as p ≥ 1182. So the inequality is: 2(10x – 6) + 2(5x – 3) ≥ 1182 20x – 12 + 10x – 6 ≥ 1182 30x – 18 ≥ 1182 30x – 18 + 18 ≥ 1182 + 18 30x ≥ 1200 x ≥ 1200 30 30 30 x ≥≥≥≥≥ 40 Answer in math: x ≥ 40 Answer in English: x would have to be greater than or equal to 40. Guided Practice 9. Lily wants to build a fence along the perimeter of her rectangular garden. She cannot afford to buy more than 14 meters of fencing. The length of the garden, l, is (2x – 3) m and the width, w, is (3x – 10) m. Write an inequality to represent the situation and solve it. What would the dimensions of the garden fence have to be to keep the fencing no longer than 14 meters. 10. A car travels (17x – 5) miles in (x – 7) hours. The car travels at a constant speed not exceeding the speed limit of 55 miles per hour. If speed = distance ÷ time, write an inequality to represent this situation and find the minimum possible number of miles traveled. 11. The formula for calculating the speed of an accelerating car is v = u + at, where v is the final speed, u is the original speed, a is the acceleration, and t is the time taken. Car A starts at 5 m/s and accelerates at 4 m/s2. At the same time, car B starts at 10 m/s and accelerates at 2 m/s2. Write and solve an inequality to find out how long it will be before Car A is traveling faster than Car B. 152152152152152 Section 3.2 Section 3.2 Section 3.2 — Applications of Inequalities Section 3.2 Section 3.2 Example Example Example Example Example 33333 A long-distance telephone call from Los Angeles, California, to Harare, Zimbabwe, costs $9.50 for the first three minutes, plus $0.80 for each additional minute (or fraction of a minute). Colleen has $18.30 to spend on a call. What is the maximum number of additional minutes she can spend on the phone? Solution Solution Solution Solution Solution First you need to label the variables: Let x = the additional number of minutes after the first 3 minutes. Then you need to write it out as an inequality: In English: The total cost is $9.50 plus $0.80 per additional minute. The total cost must be less than or equal to $18.30. In math: 9.50 + 0.80x = Total cost for the call Total cost for the call £ 18.30 \ 9.50 + 0.80x £ 18.30 Then simplify: 950 + 80x £ 1830 950 – 950 + 80x £ 1830 – 950 80x £ 880 x ≤ 880 80 80 80 x £££££ 11 Answer in math: x £ 11 Answer in English: The number of additional minutes must be no more than 11. So Colleen can spend up to 11 additional minutes on the phone. Guided Practice 12. Marisa is buying a new car. Car A costs $20,000, and has an average annual fuel cost of $1000. Car B costs $22,500, and has an average annual fuel cost of $500. After how many years will Car A have cost more than Car B? Assume that all other maintenance costs are equal for both cars. 13. A cell phone company offers its customers either Plan A or Plan B. Plan A costs $90 per month with unlimited air time. Plan B costs $60 per month, plus 50¢ for each minute of cell phone time. How many minutes can a customer who chooses Plan B use the cell phone before the cost of the calls exceeds the amount it would have cost under Plan A? Section 3.2 Section 3.2 Section 3.2 — Applications of Inequalities Section 3.2 Section 3.2 153153153153153 Independent Practice 1. Bank T’s checking account has monthly charges of an $8 service fee plus 6¢ per check written. Bank S’s checking account has monthly charges of a $10 service fee plus 4¢ per check written. A company has 150 employees, and pays them monthly by check. The company’s financial adviser suggests that Bank S woul
d be cheaper to use. Set up and solve an inequality that supports this recommendation. 2. Jack is doing a sponsored swim to raise money for charity. His mom sponsors him $10, plus $1 for every length of the pool he completes. His uncle sponsors him just $1.50 for every length he completes. How many lengths will Jack have to complete for his uncle to pay more than his mom? 3. On average Sendi uses 350 minutes of air time per month. Company M offers a cell phone plan of $70 per month plus 85¢ for each minute of air time. Company V offers a cell phone plan of $130 per month plus 65¢ for each minute of air time. Sendi chooses Company V. Use an inequality to show that this plan is cheaper for her. 4. A group of friends want to drive to a beach resort and spend 5 days there. A car rental firm offers them two rental plans; $15 a day plus 30¢ per mile traveled, or $20 a day plus 10¢ per mile. Which rental plan would be better if the beach resort is 150 miles from home, and why? 5. A bank charges a $10 monthly service fee plus 5¢ handling fee per check processed through its Gold checking account. The bank also offers a Platinum checking account and charges a $15 monthly service fee plus 3¢ handling fee per check drawn from this account. What is the highest number of checks per month for which the Gold account is cheaper than the Platinum account? 6. A group of executives is traveling to a meeting, so they decide to hire a car and travel together. The car rental agency rents luxury cars at $65 per day plus 65¢ per mile traveled, or $55 per day plus 85¢ per mile traveled. What is the maximum number of miles that they can drive before the $55 per day plan becomes more expensive than the $65 per day plan? ound Up ound Up RRRRRound Up ound Up ound Up ound Up ound Up RRRRRound Up ound Up ound Up This Topic is very similar to real-life applications of equations, which is covered in Sections 2.4–2.7. Always remember to give your solution as a sentence that answers the original problem. 154154154154154 Section 3.2 Section 3.2 Section 3.2 — Applications of Inequalities Section 3.2 Section 3.2 TTTTTopicopicopicopicopic 3.3.13.3.1 3.3.13.3.1 3.3.1 Section 3.3 Compound Inequalities Compound Inequalities Compound Inequalities Compound Inequalities Compound Inequalities Compound Inequalities Compound Inequalities Compound Inequalities Compound Inequalities Compound Inequalities California Standards: 4.0: Students simplify 4.0: Students simplify 4.0: Students simplify 4.0: Students simplify 4.0: Students simplify eeeeexprxprxprxprxpressions bef essions befororororore solving e solving e solving essions bef essions bef e solving e solving essions bef linear linear linear equations and linear linear inequalities in one variaariaariaariaariabbbbblelelelele,,,,, inequalities in one v inequalities in one v inequalities in one v inequalities in one v such as 3(2x – 5) + 4(x – 2) = 12..... Students solveeeee 5.0:5.0:5.0:5.0:5.0: Students solv Students solv Students solv Students solv lems,,,,, lems lems p prp proboboboboblems mmmmmultiste p prp pr ultiste ultiste ultistep pr lems ultiste including word problems, olving inininininvvvvvolving olving olving linear equations olving linear inequalities in linear inequalities in and linear inequalities in linear inequalities in linear inequalities in le and prooooovidevidevidevidevide one variaariaariaariaariabbbbble and pr le and pr le and pr one v one v one v le and pr one v h steppppp..... h ste h ste or eac or eac tion f tion f justifica justifica or each ste tion for eac justification f h ste or eac tion f justifica justifica What it means for you: You’ll learn about problems involving two inequalities, and how to solve them. Key words: conjunction disjunction inequality Check it out: The solution set of the conjunction is the numbers for both both both inequalities are which both both true. Sometimes a math problem gives you two different restrictions on a solution, using inequality signs. A compound inequality is two inequalities together — for example, 2x + 1 < 5 and 2x + 1 > –1. lude the WWWWWororororord “d “d “d “d “And”And”And”And”And” lude the lude the lems Inc Conjunction Proboboboboblems Inc lems Inc Conjunction Pr Conjunction Pr lems Include the lude the lems Inc Conjunction Pr Conjunction Pr The word “and” means the compound inequality below is a “conjunction.” You can rewrite a conjunction as a single mathematical statement, usually involving two inequality signs, like this: 2x + 1 < 5 and 2x + 1 > –1 can be rewritten as –1 < 2x + 1 < 5. Guided Practice In Exercises 1–9, express each conjunction as a single mathematical statement. 1. 3y – 1 > 5 and 3y – 1 < 11 2. 4a + 7 > –10 and 4a + 7 < –2 3. –8c + 2 £ 16 and –8c + 2 ≥ –3 4. 7x – 2 < 14 and 7x – 2 > 4 5. 10a – 7 < 2 and 10a – 7 > –5 6. 9t + 4 £ 4 and 9t + 4 ≥ 3 7. –4g – 5 < –5 and –4g – 5 > –10 8. 7c – 9 < 7 and 7c – 9 > –4 9. 8y + 9 £ 2 and 8y + 9 > –6 Solving Conjunctions Solving Conjunctions Solving Conjunctions Solving Conjunctions Solving Conjunctions The solution to a conjunction must satisfy both inequalities — both inequalities must be true. Example Example Example Example Example 11111 Solve and graph the inequality –1 < 2x + 1 < 5. Solution Solution Solution Solution Solution The aim is to get x by itself. –1 – 1 < 2x + 1 – 1 < 5 – 1 Subtr act 1 act 1 Subtr Subtr act 1 Subtract 1 act 1 Subtr –2 < 2x < 4 –1 < x < 2 y 2 to get et et et et x in the mid y 2 to g y 2 to g in the middledledledledle vide b vide b in the mid in the mid DiDiDiDiDivide b vide by 2 to g y 2 to g vide b in the mid So the solution is any number greater than –1 but less than 2. This is graphed as: –1 0 1 2 Section 3.3 Section 3.3 Section 3.3 — Compound Inequalities Section 3.3 Section 3.3 155155155155155 Guided Practice In Exercises 10–13, solve and graph each inequality. 10. 5 < 3y – 1 < 11 11. –10 < 4a + 6 < –2 12. 7 £ 7x – 7 £ 14 13. –5 £ 9t + 4 £ 22 In Exercises 14–19, solve each inequality. 14. –5 < a 10 16. –11 < –4g + 5 < –3 < 3 18. –15 < 8y + 9 £ 9 15. –11 < c − 9 7 < 5 17. –9 < 6y – 9 < 3 a + − 6 19. 7 < £ 9 6 Check it out: In this example, the lines head in opposite directions — they have no points in common. Any number in the solution set only satisfies one of the inequalities. lude the WWWWWororororord “Or” d “Or” d “Or” lude the lude the lems Inc Disjunction Proboboboboblems Inc lems Inc Disjunction Pr Disjunction Pr d “Or” lems Include the d “Or” lude the lems Inc Disjunction Pr Disjunction Pr Here’s an example of a disjunction: 3x – 4 < –4 or 3x –4 > 4 The solution to a disjunction is all the numbers that satisfy either one inequality or the other. Example Example Example Example Example 22222 Solve and graph the solution set of 3x – 4 < –4 or 3x – 4 > 4. Solution Solution Solution Solution Solution 3x – 4 + 4 < –4 + 4 3x < 0 x < 0 –1 0 or or or 1 Guided Practice 3x – 4 + 4 > 4 + 4 AdAdAdAdAdd 4d 4d 4d 4d 4 3x > 8 x > 8 3 2 8 3 DiDiDiDiDivide b vide b vide b vide by 3y 3y 3y 3y 3 vide b In Exercises 20–23, solve the inequality and graph each solution set. 20. 7a – 7 < –7 or 7a – 7 > 21 21. 5x – 4 £ 6 or 5x – 4 ≥ 26 22. c − 9 5 < 3 or c − 9 5 ≥ 9 23. t − 7 3 £ –9 or t − 7 3 > 6 In Exercises 24–27, solve each disjunction. 24. 8c – 4 > 92 or 8c – 4 < –12 25. –9g – 7 £ 2 or –9g – 7 > 20 26. 6c + 5 > 8 or 6c + 5 £ 5 27. j − 13 6 £ –10 or j − 13 6 ≥ 5 156156156156156 Section 3.3 Section 3.3 Section 3.3 — Compound Inequalities Section 3.3 Section 3.3 Independent Practice Solve each conjunction or disjunction in Exercises 1–19. 1. –7 < 2x + 3 < 9 3. 9 < 4x + 5 < 17 5. 3 £ 3(2x – 5) £ 9 2. 8 < 3x – 4 < 14 4. –1 £ x – 3 £ 5 6. –6 £ 9 – 5x £ 19 7. 5 £ 7 – 2(x – 3) £ 21 8. –. –3 £ 11. – 10. 3 £ 4x – 9 and 4x – 9 £ 15 2 ) and 2 ) ( 3 x − 4 < 10 12. 2y + 2 < 4y – 4 or 4y – 4 > 5y + 2 13. 5y + 7 < –13 or 7 – 3y < –5 14. 11 + 2y < 4y – 3 or 4y – 3 > 6y + 7 15. –2x £ –3x + 4 and –3x + 4 £ 4x + 18 16. 3 8 x + 3 < –4 or 3 – 3 7 x < –6 17 18. –15 £ 4x – 6 £ –10 19. 4x – 9 < 27 and 10x – 16 > 2x + 8 20. The sum of three consecutive even integers is between 82 and 85. Find the numbers. 21. The sum of three consecutive odd integers is between 155 and 160. Find the consecutive odd integers. The formula C = degrees Celsius. Use this fact to answer Exercises 22–23. (F – 32) is used to convert degrees Fahrenheit to 5 9 22. The temperature inside a greenhouse falls to a minimum of 65 °F at night and rises to a maximum of 120 °F during the day. Find the corresponding temperature range in degrees Celsius. 23. The usual temperature range of liquid water is 0 degrees Celsius (freezing point) to 100 degrees Celsius (boiling point). Find the corresponding temperature range of water in degrees Fahrenheit. ound Up ound Up RRRRRound Up ound Up ound Up RRRRRound Up ound Up ound Up ound Up ound Up This Topic can be a little hard to understand at first. You can write the solution to a conjunction in one statement, but disjunction solutions usually have to be written in two parts because they cover two different parts of the number line. Section 3.3 Section 3.3 Section 3.3 — Compound Inequalities Section 3.3 Section 3.3 157157157157157 TTTTTopicopicopicopicopic 3.4.13.4.1 3.4.13.4.1 3.4.1 Section 3.4 alue Inequalities alue Inequalities Absolute VVVVValue Inequalities Absolute Absolute alue Inequalities alue Inequalities Absolute Absolute Absolute VVVVValue Inequalities alue Inequalities alue Inequalities Absolute Absolute alue Inequalities Absolute alue Inequalities Absolute California Standards: 3.0:3.0:3.0:3.0:3.0: Students solv Students solveeeee Students solv Students solv Students solv inequalities inequalities equations and inequalities inequalities inequalities alues..... alues alues bsolute v bsolute v olving a inininininvvvvvolving a olving a bsolute values olving absolute v alues bsolute v olving a What it means for you: You’ll solve absolute value inequalities of the form Ωmx ± cΩ < v or Ωmx ± cΩ > v. Key words: absolute value conjunction disjunction inequality e
ndpoint interval Check it out: The solution interval is the open interval (–4, 18). On the graph, the endpoints –4 and 18 aren’t included. You last saw absolute value equations in Section 2.8 — now you’re going to see inequalities involving absolute values. As with normal inequalities, you can have conjunctions and disjunctions with absolute value inequalities. e Compound Inequalities e Compound Inequalities alue Inequalities ar Absolute VVVVValue Inequalities ar alue Inequalities ar Absolute Absolute e Compound Inequalities alue Inequalities are Compound Inequalities e Compound Inequalities alue Inequalities ar Absolute Absolute An absolute value inequality can have a form such as \|x\| < m. \|x\| < m means that x is restricted to points on the number line less than m units from 0 — in either the positive or the negative direction, as shown on this number line: \|x\| < m is equivalent to –m < x < m. –m unitsm 0 m units m Guided Practice In Exercises 1–4, write the equivalent compound inequality and graph the inequality on a number line. 1. \|x\| < 3 3. \|g\| < 2 2. \|a\| < 7 4. \|x – 1\| < 10 nequalities CCCCCan an an an an BBBBBe e e e e CCCCConjunctions onjunctions onjunctions nequalities alue IIIIInequalities nequalities alue Absolute VVVVValue alue Absolute Absolute onjunctions onjunctions nequalities alue Absolute Absolute You can write an absolute value inequality of the form \|mx + c\| < v as the conjunction mx + c < v and mx + c > –v (or you could write it mx + c < v and –(mx + c) < v). So, to solve an inequality like this, you can solve the compound inequality –v < mx + c < v. Example Example Example Example Example 11111 Solve \|x – 7\| < 11. Write the solution in interval notation and graph its solution interval on a number line. Solution Solution Solution Solution Solution x – 7 < 11 and x – 7 > –11 –11 < x – 7 < 11 –11+ 7 < x – 7 + 7 < 11 + 7 –4 < x < 18 Solution interval: (–4, 18) rite the inequality as a conjunction rite the inequality as a conjunction WWWWWrite the inequality as a conjunction rite the inequality as a conjunction rite the inequality as a conjunction e the conjunction to get et et et et x b b b b by itself y itself y itself e the conjunction to g e the conjunction to g SolvSolvSolvSolvSolve the conjunction to g y itself y itself e the conjunction to g Graph: –4 0 18 158158158158158 Section 3.4 Section 3.4 Section 3.4 — Absolute Value Inequalities Section 3.4 Section 3.4 ndpoints are e e e e IIIIIncncncncncluded luded luded ndpoints ar Sign Means the he he he he EEEEEndpoints ar ndpoints ar Sign Means t TTTTThe he he he he £££££ Sign Means t Sign Means t luded luded ndpoints ar Sign Means t Example Example Example Example Example 22222 Solve \|4x – 9\| £ 11. Write the solution in interval notation and graph its solution interval on a number line. Solution Solution Solution Solution Solution –11 £ 4x – 9 £ 11 –11 + 9 £ 4x – 9 + 9 £ 11 + 9 –2 £ 4x £ 20 rite the inequality as a conjunction rite the inequality as a conjunction WWWWWrite the inequality as a conjunction rite the inequality as a conjunction rite the inequality as a conjunction e the conjunction to get et et et et x b b b b by itself y itself y itself e the conjunction to g e the conjunction to g SolvSolvSolvSolvSolve the conjunction to g y itself y itself e the conjunction to g Check it out: In this example, the solution interval is closed — so the endpoints are included in the graph. – 1 2 £££££ x £££££ 5 Solution interval: [– 1 2 , 5] Graph: 0 –0.5 5 Guided Practice Solve each conjunction and write each solution set in interval notation. 5. \|t – 10\| < 1 6. \|a + 2\| £ 4 7. \|3a\| £ 21 8. \|15j\| < 5 9. r 8 £ 2 10. h 12 < 1 11. \|x + 8\| < 10 12. \|x – 11\| < 4 14. \|3x – 7\| < 8 15. \|5y + 7\| £ 22 13. \|2x – 5\| < 11 y + 2 16. 4 5 £ 13 1 4 17. y − 3 20. \|3(2 – x)\| £ 5 £ 5 1 5 18. 2 x − 5 21. \|5x – 11\| £ 19 £ 4 19. 3 2 x − 2 £ 4 Check it out: The graphs head in opposite directions — they have no points in common. junctions junctions nequalities CCCCCan an an an an BBBBBe e e e e DisDisDisDisDisjunctions nequalities nequalities alue IIIIInequalities alue alue Absolute VVVVValue Absolute Absolute junctions junctions nequalities alue Absolute Absolute An absolute value inequality can have a form such as \|x\| > m. \|x\| > m means that x is restricted to points on the number line more than m units from 0 — in either the positive or the negative direction, as shown on this number line: unitsm \|x\| > m is equivalent to x < –m or x > m. –m 0 m units m Section 3.4 Section 3.4 Section 3.4 — Absolute Value Inequalities Section 3.4 Section 3.4 159159159159159 Check it out: In this example, the solution set is in two parts. The endpoints –2 and 5 are not included. Don’t forget: The “» ” sign is the “union” sign — here it means that the solution set is (–•, –2) andandandandand (5, •). See Section 1.1 for a reminder about unions. Guided Practice In Exercises 22–25, write the equivalent compound inequality, and graph the inequality on a number line. 22. \|x\| > 14 24. \|a\| > 10 23. \|t\| > 8 25. \|t + 8\| > 23 junctions junctions mal Dis mal Dis s into Nor Absolute VVVVValuealuealuealuealues into Nor s into Nor Absolute ConConConConConvvvvvererererert t t t t Absolute Absolute junctions mal Disjunctions s into Normal Dis junctions mal Dis s into Nor Absolute An absolute value inequality of the form \|mx + c\| > v is the same as the disjunction mx + c < –v or mx + c > v. To solve this type of absolute value inequality, you solve the disjunction. Example Example Example Example Example 33333 Solve the compound inequality \|2x – 3\| > 7. Write the solution set in interval notation and graph the solution set. Solution Solution Solution Solution Solution 2x – 3 < –7 or 2x < –4 x < –2 2x – 3 > 7 WWWWWrite the inequality as a disjunction rite the inequality as a disjunction rite the inequality as a disjunction rite the inequality as a disjunction rite the inequality as a disjunction 2x > 10 e the disjunction to get et et et et x b b b b by itself x > 5 Solv Solv Solv y itself y itself e the disjunction to g e the disjunction to g Solve the disjunction to g y itself Solv y itself e the disjunction to g Solution set: (–•••••, –2) »»»»» (5, •••••) Graph: –2 0 5 ndpoints are e e e e IIIIIncncncncncluded luded luded ndpoints ar Sign Means the he he he he EEEEEndpoints ar ndpoints ar Sign Means t TTTTThe he he he he ≥≥≥≥≥ Sign Means t Sign Means t luded luded ndpoints ar Sign Means t Remember that the solution interval is closed if there is a “greater than or equal to” sign. Example Example Example Example Example 44444 Solve and graph \|5x + 11\| ≥ 6. Write the solution set in interval notation. Solution Solution Solution Solution Solution 5x + 11 £ –6 or 5x £ –17 x £££££ – 17 5 5x + 11 ≥ 6 5x ≥ –5 x ≥≥≥≥≥ –1 e to get et et et et x b b b b by itself SolvSolvSolvSolvSolve to g y itself y itself e to g e to g y itself y itself e to g Check it out: Here, the solution intervals are half-open. The endpoints –3.4 and –1 are included. Solution set: (–•••••, –3.4] »»»»» [–1, •••••) Graph: –3.4 –1 0 160160160160160 Section 3.4 Section 3.4 Section 3.4 — Absolute Value Inequalities Section 3.4 Section 3.4 Guided Practice Solve each disjunction and write each solution set in interval notation. 26. \|a – 8\| > 1 27. \|t + 2\| > 8 28. \|7a\| ≥ 14 29. \|4j\| ≥ 16 30. c 3 > 6 31. c 12 ≥ 1 3 32. \|x + 9\| > 7 33. \|x – 11\| > 12 34. \|3x – 7\| > 13 35. \|5y + 11\| > 21 36. \|4m + 9\| ≥ 11 37. \|7b – 8\| ≥ 13 38. 3 4 x − 3 ≥ 5 39. 2 1 x − 7 ≥ 3 40. \|3(x – 2) + 7\| ≥ 8 41. \|4(3 + x)\| ≥ 13 42. \|2(3x – 7) + 15\| ≥ 11 Independent Practice 1. Write \|a\| > 5 as a compound inequality. 2. Write \|w\| £ c as a compound inequality. In Exercises 3–4 write the equivalent compound inequality in interval notation. 3. \|a\| < 4 4. \|b\| ≥ 2 In Exercises 5–12, solve each inequality. 5. g 2 ≥ 3 7. \|w – 17\| £ 8 9. \|5t + 10\| ≥ 15 11. 8 t − 7 3 ≥ 2 6. \|c + 2\| > 24 8. \|4a – 3\| £ 9 x + 2 10. 3 1 < 1 12. 17 2 c + 5 > 4 13. Solve and graph the solution set of 7\|2(x – 7) + 5\| ≥ 14. 14. Given m > 0, graph the solution set of \|x\| £ m on a number line. 15. Given m > 0, solve and graph the solution set of \|x\| ≥ m. 16. Solve \|3m – 5\| > m + 7. 17. Solve \|5m + 3\| > 2m – 1. 18. A floor tile must measure 20 cm along its length, to within 2 mm. Write and solve an absolute value inequality to find the maximum and minimum possible lengths for the tile. ound Up ound Up RRRRRound Up ound Up ound Up ound Up ound Up RRRRRound Up ound Up ound Up Like in Section 3.3, you’ve got to watch out for the difference between conjunctions and disjunctions. Also, look back at Section 1.1 if you’ve forgotten what the “»“ sign means. Section 3.4 Section 3.4 Section 3.4 — Absolute Value Inequalities Section 3.4 Section 3.4 161161161161161 Chapter 3 Investigation Mailing Pacacacacackakakakakagggggeseseseses Mailing P Mailing P Mailing P Mailing P Mailing Pacacacacackakakakakagggggeseseseses Mailing P Mailing P Mailing P Mailing P Inequalities turn up all the time in everyday life. In this investigation you’ll see that using inequalities makes it much easier to model some real-life problems. A shipping company will deliver packages of any weight, as long as the conditions on the right are satisfied. Part 1: Design a set of six boxes of different shapes and sizes that meet these requirements. Part 2: What is the maximum volume a box can have while still satisfying the shipping company’s requirements? Packages must satisfy the following: Length + Width + Height £ 120 cm Length £ 80 cm Width £ 80 cm Height £ 80 cm Volume = length × width × height Things to think about: Try to include a wide range of different sizes and shapes in your set of boxes. Aim to make your set as useful as possible for mailing different types of item. Extension For these extensions, only boxes in the shapes of rectangular prisms should be considered. 1) What is the maximum surface area a box can have while still satisfying the shipping company’s requirements? What are the dimensions of this box? 2) You need to ship a picture th
at has a height of 83 cm. How wide could the picture be so that it fits in a rectangular prism shaped box that satisfies the shipping company’s requirements? Write an inequality to represent the possible width of the picture. width 83 c m Open-ended Extension 1) You work for an organization that produces pocket dictionaries. The dimensions of the dictionaries are shown on the right. What size box will hold the greatest number of dictionaries, while still satisfying the shipping company’s requirements? The dictionaries don’t all have to be placed in the box the same way. Experiment with placing some dictionaries vertically and some horizontally. 10 cm 6cm c m 3 2) Your company is concerned about the environment and wants to use the smallest area of cardboard possible to package the books in. Design the box that you think will be most efficient. (Assume the box does not need any tabs to stick it together.) ound Up ound Up RRRRRound Up ound Up ound Up ound Up ound Up RRRRRound Up ound Up ound Up In general, problems that use the words “maximum,” “minimum,” “limits,” or “tolerance” often mean that you need to use inequalities to model the situations. 162162162162162 estigaaaaationtiontiontiontion — Mailing Packages estigestig estig pter 3 Invvvvvestig pter 3 In ChaChaChaChaChapter 3 In pter 3 In pter 3 In Chapter 4 Linear Equations and Their Graphs Section 4.1 The Coordinate Plane ........................... 164 Section 4.2 Lines...................................................... 175 Section 4.3 Slope ..................................................... 189 Section 4.4 More Lines ............................................ 197 Section 4.5 Inequalities ............................................ 212 Investigation Tree Growth........................................... 227 163163163163163 TTTTTopicopicopicopicopic 4.1.14.1.1 4.1.14.1.1 4.1.1 Section 4.1 te Plane te Plane he Coordinadinadinadinadinate Plane he Coor he Coor TTTTThe Coor te Plane te Plane he Coor he Coordinadinadinadinadinate Plane TTTTThe Coor te Plane te Plane he Coor he Coor te Plane he Coor te Plane California Standards: Students grrrrraaaaaph a linear 6.0:6.0:6.0:6.0:6.0: Students g ph a linear ph a linear Students g Students g ph a linear ph a linear Students g tion tion equa equa tion and compute the xequation equa tion equa and y-intercepts (e.g., graph 2x + 6y = 4). They are also able to sketch the region defined by linear inequality (e.g., they sketch the region defined by 2x + 6y < 4). What it means for you: You’ll refresh your knowledge of plotting coordinates. Key words: coordinate horizontal vertical point of intersection origin Check it out: The x-coordinate is sometimes called the abscissa, and the y-coordinate is sometimes called the ordinate. Check it out: The coordinates of the origin are (0, 0). Plotting coordinates isn’t anything new to you — you’ve had lots of practice in earlier grades. This Topic starts right at the beginning though, to remind you of the earlier work. e Used to Locate Pte Pte Pte Pte Points on a Plane oints on a Plane oints on a Plane e Used to Loca e Used to Loca tes ar CoorCoorCoorCoorCoordinadinadinadinadinates ar tes ar oints on a Plane tes are Used to Loca oints on a Plane e Used to Loca tes ar A plane is a flat surface, kind of like a blackboard, a tabletop, or a sheet of paper. However, a plane extends indefinitely in all directions — it goes on and on forever. Planes are made up of an infinite number of points. To locate one of these points, two perpendicular number lines are drawn in the plane. The horizontal number line is called the x-axis. y-axis The vertical number line is called the y-axis. The point of intersection of the two number lines is called the origin. origin x-axis You can locate each point in the plane using an ordered pair of numbers (x, y), where x represents the horizontal distance and y represents the vertical distance of the point from the origin. The numbers in an ordered pair are often called coordinates. Example Example Example Example Example 11111 Plot the coordinates (3, 4) on a graph. Solution Solution Solution Solution Solution (3, 4) x-coordinate y-coordinate 3 units across (3, 4) 4 units up 164164164164164 Section 4.1 Section 4.1 Section 4.1 — The Coordinate Plane Section 4.1 Section 4.1 Guided Practice In Exercises 1–4, plot the coordinates on a coordinate plane. 1. (4, 0) 3. (8, 2) 2. (–2, 5) 4. (5, –2) Use the graph opposite to answer Exercises 5–6. 5. What are the coordinates of point A? 6. What are the coordinates of point B? B –4 –2 A 2 4 4 2 0 –2 –4 han One Pointointointointoint han One P y Need to Plot More e e e e TTTTThan One P han One P y Need to Plot Mor y Need to Plot Mor ou Ma YYYYYou Ma ou Ma ou May Need to Plot Mor han One P y Need to Plot Mor ou Ma You can often join up several plotted points to create the outline of a shape. Example Example Example Example Example 22222 Draw a coordinate plane and plot and label the points M (3, 4), A (–3, 4), C (–3, –4), and V (3, –4). Connect each pair of consecutive points and find the perimeter of the resulting quadrilateral. Solution Solution Solution Solution Solution A negative x-coordinate means the point is left of the y-axis. A -5-6 -4 -3 -2 C 6 5 4 3 2 1 -1 0 -1 -2 -3 -4 -5 - negative y-coordinate means the point is below the x-axis. The length of each square on the coordinate grid represents one unit of measure. So, the quadrilateral is 8 units long and 6 units wide. Perimeter (distance around the edge) = 2(6) + 2(8) = 12 + 16 = 28 units Section 4.1 Section 4.1 Section 4.1 — The Coordinate Plane Section 4.1 Section 4.1 165165165165165 Guided Practice Work out Exercises 7–9 by plotting and labeling the points on copies of the coordinate plane used in Example 2. 7. A (0, 5), B (–5, 2), C (–5, –4), D (5, –4), and E (5, 2). Name the figure formed when the points are connected in order. 8. A (3, 3), B (–3, 3), C (–3, –3), and D (3, –3). Connect the points in order and then name and find the area of the figure formed. 9. A (2, 5), B (2, –2), C (5, –2), D (5, –4), E (–5, –4), F (–5, –2), G (–2, –2), and H (–2, 5). Connect the points in order and find the perimeter of the figure formed. Independent Practice 1. What is the difference between the x-axis and the y-axis? 2. Explain in words how to graph the coordinates (5, 3). Use the graph opposite to answer Exercises 3–7. 3. What are the coordinates of point A? 4. What are the coordinates of point B? 5. What are the coordinates of point C? 6. What are the coordinates of point D? 7. What are the coordinates of point E? D –4 –2 4 2 0 –2 –4 C A 2 B 4 E In Exercises 8–11, plot the coordinates on a coordinate plane. 8. The origin 9. (2, 1) 10. (–4, 1) 11. (3, –2) 12. Plot and label the following points: A (–1, –2), B (2, 4), C(5, –2), D (–2, 2), and E (6, 2). Connect the points in order, then connect E to A, and name the figure formed. ound Up ound Up RRRRRound Up ound Up ound Up RRRRRound Up ound Up ound Up ound Up ound Up That Topic should have felt quite familiar. Remember that coordinate pairs always list the x-coordinate first, then the y-coordinate — and watch out for any negative numbers. In the next Topic you’ll look at each part of the coordinate plane in more detail. Section 4.1 166166166166166 Section 4.1 Section 4.1 Section 4.1 — The Coordinate Plane Section 4.1 TTTTTopicopicopicopicopic 4.1.24.1.2 4.1.24.1.2 4.1.2 California Standards: Students grrrrraaaaaph a linear 6.0:6.0:6.0:6.0:6.0: Students g ph a linear ph a linear Students g Students g ph a linear ph a linear Students g tion tion equa equa tion and compute the xequation equa tion equa and y-intercepts (e.g., graph 2x + 6y = 4). They are also able to sketch the region defined by linear inequality (e.g., they sketch the region defined by 2x + 6y < 4). What it means for you: You’ll look in more detail at the four main regions of the coordinate plane, and the axes. Key words: coordinate quadrant axes the Plane the Plane ants of ants of Quadr Quadr the Plane ants of the Plane Quadrants of the Plane ants of Quadr Quadr the Plane the Plane ants of ants of Quadr Quadr ants of the Plane Quadrants of the Plane ants of Quadr the Plane Quadr There are four main regions of the coordinate plane — they’re divided up by the x- and y-axes. This Topic is about spotting where on the coordinate plane points lie. our Quadrantsantsantsantsants our Quadr he Plane Consists of F F F F Four Quadr our Quadr he Plane Consists of TTTTThe Plane Consists of he Plane Consists of our Quadr he Plane Consists of The coordinate axes divide the plane into four regions called quadrants. The quadrants are numbered counterclockwise using Roman numerals. The signs of the coordinates differ from quadrant to quadrant, as shown in the diagram below. If the x-coordinate is negative and the y-coordinate is positive, the point is in quadrant II. If the x- and y-coordinates are both positive, the point is in quadrant I. If the x- and y-coordinates are both negative, the point is in quadrant III. If the x-coordinate is positive and the y-coordinate is negative, the point is in quadrant IV. So, you can easily tell which quadrant a particular point is in by simply looking at the signs of its coordinates. Section 4.1 Section 4.1 Section 4.1 — The Coordinate Plane Section 4.1 Section 4.1 167167167167167 Example Example Example Example Example 11111 State the quadrant where each point is located. Justify your answer. a) (–2, 3) c) (a, –a) where a > 0 b) (111, 2111) d) (–m, –m) where m > 0 Solution Solution Solution Solution Solution a) (–2, 3) is in quadrant II, since the x-coordinate is negative and the y-coordinate is positive. b) (111, 2111) is in quadrant I, since the x- and y-coordinates are both positive. c) (a, –a) is in quadrant IV, since the x-coordinate is positive and the y-coordinate is negative (the problem states that a > 0). d) (–m, –m) is in quadrant III, since the x and y-coordinates are both negative (the problem states that m > 0).
Guided Practice In Exercises 1–6, name the quadrant or axis where each point is located. Justify your answers. 1. (1, 0) 4. (0, –5) In Exercises 7–12, a > 0, k > 0, m < 0, v < 0, and p Œ R. Name the quadrant or axis where each point is located. Justify your answers. 7. (a, k) 10. (p, 0) 3. (–4, –1) 6. (–p, p) 8. (m, k) 11. (0, p) 2. (2, 3) 5. (–5, 8) 9. (a, v) 12. (v, v) In Exercises 13–18, a < –2. Name the quadrant where each point is located. Justify your answers. 13. (3, a) 16. (2a, 2) 15. (–3, –a) 18. (4, a – 3) 14. (a, –1) 17. (a – 2, 3) 168168168168168 Section 4.1 Section 4.1 Section 4.1 — The Coordinate Plane Section 4.1 Section 4.1 - or y-Ax-Ax-Ax-Ax-Axeseseseses e on the x- or - or - or e on the e on the oints ar oints ar Some P Some P oints are on the Some Points ar - or e on the oints ar Some P Some P Points in the plane are located in one of the four quadrants or on the axes. All points whose coordinates are in the form (x, 0) are on the x-axis — for example, (3, 0), (1, 0), (–2, 0), (–3.5, 0). All points whose coordinates are in the form (0, y) are on the y-axis — for example, (0, 2), (0, 4), (0, –3), (0, –2). (0, 4) (0, 2) (–3.5, 0) (–2, 0) (1, 0) (3, 0) (0, –2) (0, –3) Example Example Example Example Example 22222 State which axis, if either, these points lie on. Justify your answer. a) (0, 3) b) (12, 2) c) (45, 0) Solution Solution Solution Solution Solution a) (0, 3) is on the y-axis, since x = 0. b) (12, 2) isn’t on an axis, since neither x nor y is 0. c) (45, 0) is on the x-axis, since y = 0. Guided Practice In Exercises 19–24, let t > 0. State which axis, if either, each point lies on. 19. (0, 1) 22. (t, 0) 20. (0, –3) 23. (–4, 0) 21. (t, 1) 24. (t + 1, 0) Independent Practice In Exercises 1–12, let a > 0 and b < 0. Name the quadrant or axis where each point is located, and justify your answers. 1. (5, a) 4. (0, –3) 7. (b, 1) 10. (2b, a) 2. (a, 4) 5. (–2, 0) 8. (b, 3) 11. (–b, a) 3. (a + 2, 4) 6. (0, 0) 9. (2a, b) 12. (–a, –2b) ound Up ound Up RRRRRound Up ound Up ound Up RRRRRound Up ound Up ound Up ound Up ound Up Another straightforward Topic, really. Working out which quadrant a coordinate pair appears in is all about checking the signs of each of the numbers — and a coordinate pair can only be on one of the axes if one of the numbers is 0. In the next Topic you’ll look at lines plotted on the coordinate plane. Section 4.1 Section 4.1 Section 4.1 — The Coordinate Plane Section 4.1 Section 4.1 169169169169169 TTTTTopicopicopicopicopic 4.1.34.1.3 4.1.34.1.3 4.1.3 Lines Lines Lines Lines Lines Lines Lines Lines Lines Lines California Standards: Students grrrrraaaaaph a linear 6.0:6.0:6.0:6.0:6.0: Students g ph a linear ph a linear Students g Students g ph a linear ph a linear Students g tion tion equa equa tion and compute the xequation equa tion equa and y-intercepts (e.g., graph 2x + 6y = 4). They are also able to sketch the region defined by linear inequality (e.g., they sketch the region defined by 2x + 6y < 4). What it means for you: You’ll learn a formal definition of a line, and you’ll refresh your knowledge of line equations. Key words: coordinate quadrant axes That’s enough of learning about the coordinate plane itself — now it’s time to plot lines. If you join two points in a coordinate plane you’ll form part of a line. Lines Havvvvve No Endpoints e No Endpoints e No Endpoints Lines Ha Lines Ha e No Endpoints e No Endpoints Lines Ha Lines Ha A straight line extends indefinitely in opposite directions. Lines have: no endpoints (they have no beginning or end) infinite length no thickness To identify a straight line you just need two points on it. So, to draw a straight line in a plane, simply plot two points and connect them using a straightedge. Example Example Example Example Example 11111 Draw the straight line defined by the points (–2, 3) and (5, –1). Solution Solution Solution Solution Solution -5-6 -4 -3 -2 6 5 4 3 2 1 -1 0 -1 -2 -3 -4 -5 -6 1 2 3 4 5 6 The arrowheads show that the line continues indefinitely (without ending). Guided Practice Complete the sentence in Exercise 1. 1. A straight line extends ___________ in opposite directions. In Exercises 2–8, draw the straight line defined by the pairs of coordinates. 2. (3, 1) and (–4, 2) 3. (0, 0) and (–3, –3) 4. (–2, 1) and (3, 1) 5. (3, –2) and (–1, –2) 6. (1, 3) and (1, –4) 7. (5, 1) and (–3, 2) 8. (3, 3) and (–1, –1) 170170170170170 Section 4.1 Section 4.1 Section 4.1 — The Coordinate Plane Section 4.1 Section 4.1 y an Equationtiontiontiontion y an Equa y an Equa A Line Can Be Described b A Line Can Be Described b A Line Can Be Described by an Equa y an Equa A Line Can Be Described b A Line Can Be Described b Check it out: At the point (3, 5), x = 3. You can check that (3, 5) is of the form (x, 2x – 1) by substituting 3 for x. In other words: (x, 2x – 1) ﬁ (3, 2(3) – 1) ﬁ (3, 5). All the points that lie on the line shown have coordinates (x, y), where y = 2x – 1. This means that the coordinates all have the form (x, 2x – 1) — for example, (–1, –3), (1, 1), (2, 3), (3, 5). y = 2x – 1 is called the equation of the line. (3, 5) (2, 3) (1, 11 -2 -3 -4 -5 -6 -6 -5 -4 -3 -2 -1 (–1, –3) y x= 2 – 1 Example Example Example Example Example 22222 The points on the line y = 7x + 3 are defined by (x, 7x + 3). Find the coordinates of the points where x Œ {0, 1, 2}. Solution Solution Solution Solution Solution Just substitute each value of x into (x, 7x + 3) to find the coordinates. x = 0 means (x, 7x + 3) = (0, 7(0) + 3) = (0, 3) x = 1 means (x, 7x + 3) = (1, 7(1) + 3) = (1, 10) x = 2 means (x, 7x + 3) = (2, 7(2) + 3) = (2, 17) So the coordinates of the points are (0, 3), (1, 10), and (2, 17). Guided Practice Draw out a coordinate grid spanning –6 to 6 on the x-axis and –6 to 6 on the y-axis. Draw and label the following lines on the grid. 9. y = 3 12. x = 2 10. y = –4 13. x = –5 11. y = 0 14. x = 0 Draw the graphs for Exercises 15–16 on coordinate grids spanning –6 to 6 on the x-axis and –6 to 6 on the y-axis. 15. Draw the graph of the set of all points (x, y) such that x = y. 16. Draw the graph of the set of points (x, y) such that x = 5 and y Œ R. Describe the line you have drawn. 17. If x Œ {–1, 1, 3}, find the set M of points defined by (x, –2x + 1). Exercises 18 and 19 are about the set of ordered pairs (x, 5x – 6). 18. Name the five members of the set if x is a natural number less than 6. 19. Which members of this set are also members of the set of ordered pairs (x, x2)? Section 4.1 Section 4.1 Section 4.1 — The Coordinate Plane Section 4.1 Section 4.1 171171171171171 Independent Practice In Exercises 1–2, x is an integer greater than –4 and less than 0. 1. Find the set of points defined by (x, 3x – 2). 2. Which member of this set is also a member of the set of ordered pairs (x, x – 4)? Work out Exercises 3–4 by plotting and labeling the points on a grid spanning –6 to 6 on the x-axis and –3 to 9 on the y-axis. 3. If x Œ {–3, –2, –1, 0, 1, 2, 3}, plot the set of points defined by (x, x2). 4. If x Œ {–5, –4, –3, –2, –1, 0, 1, 2, 3}, plot the set of points defined by (x, \|x + 1\|). 5. What shape would you expect the graph to be for the set of points defined by (x, \|x\|)? Work out Exercises 6–11 by plotting and labeling the points on a grid spanning –6 to 6 on the x-axis and –6 to 6 on the y-axis. 6. Plot two lines l1 and l2 whose points are defined by l1 = (x, 2x + 1) and l2 = (x, 2x + 3). 7. Describe the two lines defined in Exercise 6 and the relationship between them. 8. Plot the two lines whose points are defined by (x, –2x + 1) and (x, –2x – 2). 9. Describe the two lines defined in Exercise 8 and the relationship between them. 10. Draw two lines whose points are defined by (x, 2x + 1) and (x, –0.5x + 1). 11. Give the coordinates of the point where the two lines defined in Exercise 10 intersect. Work out Exercises 12–14 by plotting and labelling the points on a grid spanning –6 to 6 on the x-axis and –9 to 3 on the y-axis. 12. If x Œ {–5, –4, –3, –2, –1, 0, 1, 2, 3}, plot all points defined by (x, –\|x + 1\|). 13. Compare the graph of the set of points defined by (x, \|x + 1\|) (in Exercise 4) with the graph of the set of points defined by (x, –\|x + 1\|) above. Describe the relationship between the two graphs. 14. What would you expect the relationship to be between the graph of the set of points defined by (x, \|x\|) and the graph of the set of points defined by (x, –\|x\|)? ound Up ound Up RRRRRound Up ound Up ound Up RRRRRound Up ound Up ound Up ound Up ound Up You know what a line is from everyday life — but notice that the Math definition is a little more precise. In Math, a line is infinitely long and doesn’t actually have any thickness. It’s hard to imagine, but it’s OK to just carry on without worrying too much about how strange that seems. 172172172172172 Section 4.1 Section 4.1 Section 4.1 — The Coordinate Plane Section 4.1 Section 4.1 TTTTTopicopicopicopicopic 4.1.44.1.4 4.1.44.1.4 4.1.4 California Standards: Students grrrrraaaaaph a linear 6.0:6.0:6.0:6.0:6.0: Students g ph a linear ph a linear Students g Students g ph a linear ph a linear Students g tion tion equa equa tion and compute the xequation equa tion equa and y-intercepts (e.g., graph 2x + 6y = 4). They are also able to sketch the region defined by linear inequality (e.g., they sketch the region defined by 2x + 6y < 4). What it means for you: You’ll check whether lines are horizontal or vertical. Key words: horizontal vertical constant tical Lines tical Lines ontal and VVVVVererererertical Lines ontal and ontal and HorizHorizHorizHorizHorizontal and tical Lines tical Lines ontal and ontal and VVVVVererererertical Lines HorizHorizHorizHorizHorizontal and tical Lines tical Lines ontal and ontal and tical Lines ontal and tical Lines You can tell a lot about a line just by looking at the points it goes through. One of the simplest things to spot without plotting the graph is whether the line is horizontal or vertical. e the Same x-Coor -Coordinadinadinadinadinatetete
tete -Coor -Coor e the Same tical Line Havvvvve the Same e the Same tical Line Ha oints on a VVVVVererererertical Line Ha tical Line Ha oints on a PPPPPoints on a oints on a -Coor e the Same tical Line Ha oints on a The x-coordinate tells you how far to the left or right of the y-axis a point is. Points with the same x-coordinate are all the same horizontal distance from the y-axis. So, if a set of points all have the same x-coordinate, that set will fall on a vertical line. The equation of a vertical line is x = c, where c is a constant (fixed) number. For example, x = 3, x = –1. Example Example Example Example Example 11111 Draw and label the lines x = –3 and x = 2. Solution Solution Solution Solution Solution (–3, 4) -3 -4 (–3, –2) -2 4 3 2 1 -1 0 -1 -2 -3 -4 (2, 0) 1 2 3 4 (2, –4) The values of the y-coordinates are different for each point on the line, but the values of x are the same (–3 for the line x = –3, 2 for the line x = 2). PPPPPoints on a Horiz oints on a Horiz oints on a Horiz ontal Line ha ontal Line ha ontal Line havvvvve the Same e the Same e the Same -Coor -Coor -Coordinadinadinadinadinatetetetete oints on a Horizontal Line ha e the Same y-Coor oints on a Horiz ontal Line ha e the Same -Coor The y-coordinate tells you how far above or below the x-axis a point is. Points with the same y-coordinate are all the same vertical distance from the x-axis. So, if a set of points all have the same y-coordinate, that set will fall on a horizontal line. Section 4.1 Section 4.1 Section 4.1 — The Coordinate Plane Section 4.1 Section 4.1 173173173173173 The equation of a horizontal line is y = c, where c is a constant (fixed) number. For example, y = 6, y = –3. Example Example Example Example Example 22222 Draw and label the lines y = 4 and y = –1. Solution Solution Solution Solution Solution y = 4 (–3, 4) y = –1 -3 -2 -4 (–3, –1) (2, 4) 1 2 3 4 (4, –1) 4 3 2 1 -1 0 -1 -2 -3 -4 Guided Practice The values of the x-coordinates are different for each point on the line, but the values of y are the same (–1 for the line y = –1, 4 for the line y = 4). In Exercises 1–10, draw and label each set of lines on a coordinate plane. 1. x = 3 and x = –1 3. x = 2 and x = –3 5. y = 4 and y = 1 7. y = 3 and x = 2 9. y = –7 and x = –5 2. x = 4 and x = 1 4. y = 3 and y = –1 6. y = 2 and y = –3 8. y = 6 and x = –4 10. x = 6 and y = 4 Independent Practice In Exercises 1–6, draw and label each set of lines on a coordinate plane. 1. x = 8 and x = 0 3. x = 4 and x = –6 5. y = –4 and y = –6 2. x = –3 and x = 1 4. y = 0 and y = 3 6. y = 2 and y = –3 In Exercises 7–12, write the equation for each line on the graph. y 4 3 2 1 –6 –5 –4 –3 –2 –1 0 –1 –2 –3 –4 –5 –6 –7 –8 7. 654321 x 8. 9. 10. y 20 18 16 14 12 10 8 6 4 2 –12–10–8–6 –4 –2 0 –2 –4 8642 10 x 12 11. 8642 10 x 12 y 8 6 4 2 –12–10–8–6 –4 –2 0 –2 –4 –6 –8 –10 –12 –14 –16 12. ound Up ound Up RRRRRound Up ound Up ound Up RRRRRound Up ound Up ound Up ound Up ound Up One thing that can look confusing at first is that the line x = 0 is actually the y-axis, while the line y = 0 is the x-axis. And a line of the form x = c (for any constant c) is a vertical line (that is, it is parallel to the y-axis) — while a line y = c is a horizontal line (that is, it’s parallel to the x-axis). 174174174174174 Section 4.1 Section 4.1 Section 4.1 — The Coordinate Plane Section 4.1 Section 4.1 Topic 4.2.1 Section 4.2 Points on a Line Points on a Line California Standards: 6.0: Students graph a linear equation and compute the xand y-intercepts (e.g., graph 2x + 6y = 4). They are also able to sketch the region defined by linear inequality (e.g., they sketch the region defined by 2x + 6y < 4). 7.0: Students verify that a point lies on a line, given an equation of the line. Students are able to derive linear equations by using the point-slope formula. What it means for you: You’ll learn how to show mathematically that points lie on a line. Key words: linear equation variable solution set verify You already dealt with lines in Topics 4.1.3 and 4.1.4. In this Topic you’ll see a formal definition relating ordered pairs to a line — and you’ll also learn how to show that points lie on a particular line. Graphs of Linear Equations are Straight Lines An equation is linear if the variables have an exponent of one and there are no variables multiplied together. For example, linear: 3x + y = 4, 2x = 6, y = 5 – x nonlinear: xy = 12, x2 + 3y = 1, 8y3 = 20 Linear equations in two variables, x and y, can be written in the form Ax + By = C. The solution set to the equation Ax + By = C consists of all ordered pairs (x, y) that satisfy the equation. All the points in this solution set lie on a straight line. This straight line is the graph of the equation. If the ordered pair (x, y) satisfies the equation Ax + By = C, then the point (x, y) lies on the graph of the equation. The ordered pair (–2, 5) satisfies the equation 2x + 2y = 6. So the point (–2, 5) lies on the line 2x + 2y = 6. y-axis The point (2, 1) lies on the line 2x + 2y = 6. So the ordered pair (2, 1) satisfies the equation 2x + 2y = 6. (2, 1) 1 2 3 4 5 6 x-axis (–2, 5) –4 –3 –2 6 5 4 3 2 1 0 –1 –1 –2 –3 Verifying That Points Lie on a Line To determine whether a point (x, y) lies on the line of a given equation, you need to find out whether the ordered pair (x, y) satisfies the equation. If it does, the point is on the line. You do this by substituting x and y into the equation. Section 4.2 — Lines 175 Check it out: This is the method for showing that (2, –3) is a solution of x – 3y = 11. Example 1 a) Show that the point (2, –3) lies on the graph of x – 3y = 11. b) Determine whether the point (–1, 1) lies on the graph of 2x + 3y = 4. Solution a) 2 – 3(–3) = 11 2 + 9 = 11 11 = 11 Substitute 2 for x and –3 for y A true statement So the point (2, –3) lies on the graph of x – 3y = 11, since (2, –3) satisfies the equation x – 3y = 11. b) If (–1, 1) lies on the line, 2(–1) + 3(1) = 4. But 2(–1) + 3(1) = –2 + 3 = 1 Since 1 π 4, (–1, 1) does not lie on the graph of 2x + 3y = 4. Guided Practice Determine whether or not each point lies on the line of the given equation. 1. (–1, 2); 2x – y = –4 3. (–3, –1); –5x + 3y = 11 5. (–2, –2); y = 3x + 4 7. (–2, –1); 8x – 15y = 3 2. (3, –4); –2x – 3y = 6 4. (–7, –3); 2y – 3x = 15 6. (–5, –3); –y + 2x = –7 8. (1, 4); 4y – 12x = 3 ⎞ ⎟⎟⎟; –3x – 10y = 2 ⎠ 10. ,− ⎛ ⎜⎜⎜ ⎝ 2 3 2 5 ⎛ ⎜⎜⎜ ⎝ 1 3 ,− 1 4 ⎞ ⎟⎟⎟; 6x – 16y = 7 ⎠ 9. Independent Practice In Exercises 1–4, determine whether or not each point lies on the graph of 5x – 4y = 20. 1. (0, 4) 3. (2, –3) 2. (4, 0) 4. (8, 5) In Exercises 5–8, determine whether or not each point lies on the graph of 6x + 3y = 15. 5. (2, 1) 8. (3, –1) 7. (–1, 6) 6. (0, 5) In Exercises 9–12, determine whether or not each point lies on the graph of 6x – 6y = 24. 9. (4, 0) 11. (100, 96) 10. (1, –3) 12. (–400, –404) 13. Explain in words why (2, 31) is a point on the line x = 2 but not a point on the line y = 2. 14. Determine whether the point (3, 4) lies on the lines 4x + 6y = 36 and 8x – 7y = 30. Round Up Round Up You can always substitute x and y into the equation to prove whether a coordinate pair lies on a line. That’s because if the coordinate pair lies on the line then it’s actually a solution of the equation. 176 Section 4.2 — Lines Topic 4.2.2 California Standards: 6.0: Students graph a linear equation and compute the xand y-intercepts (e.g., graph 2x + 6y = 4). They are also able to sketch the region defined by linear inequality (e.g., they sketch the region defined by 2x + 6y < 4). 7.0: Students verify that a point lies on a line, given an equation of the line. Students are able to derive linear equations by using the point-slope formula. What it means for you: You’ll learn how to graph a straight line by joining two points. Key words linear equation Graphing Ax + By = C Graphing Ax + By = C Every point on a line is a solution to the equation of the line. If you know any two solutions (any two coordinate pairs), then you can join the points with a straight line. Graphing the Line Ax + By = C Using Two Points The graph of the equation Ax + By = C consists of all points (x, y) whose coordinates satisfy Ax + By = C. To graph the line, you just need to plot two points on it and join them together with a straight line. Rearrange the equation so it is in the form y = Px + Q. Choose two values of x and substitute them into your equation to find the corresponding values of y. Plot the two points and draw a straight line through them. Plot a third point to check that the line is correct — the point should lie on the line. Example 1 Plot and label the graph of the equation x – y = –3. Solution Rearrange this first to get y = x + 3. Choose two values of x, then draw a table to help you find the y-values. x 2– 4 y (x, y) y = x 3+ 3+2–= 1= )1,2–( y = x 3+ 3+4= 7= )7,4( When you plot the graph, the line should be straight. y-axis –3 (4, 7) y = x – (1, 4) 1 2 3 4 5 6 x-axis 1 –1 –2 –3 (–2, 1) –6 –5 –4 –3 –2 Checkx, y) = (1, 4) (1, 4) lies on the line — which means the line is correct. Section 4.2 — Lines 177 Example 2 Graph and label the line of the equation y = –2x – 4. Solution x 2– 2 y y 2–= x 4– 4–)2–(2–= 0= y 2–= x 4– 4–)2(2–= 8–= (x, y) )0,2–( )8–,2( (–2, 0) –3 –5 –4 –2 –6 y-axis 2 1 1 2 3 4 5 6 (0, –4) y = – 2 x – 4 (2, –8) 0 –1 –1 –2 –3 –4 –5 –6 –7 –8 –9 –10 x-axis Check: x = 0 ﬁ y = –2x – 4 = –2(0) – 4 = 0 – 4 = –4 ﬁ (x, y) = (0, –4) (0, –4) lies on the line — which means that the line is correct. Guided Practice Graph the line through the two points in each of Exercises 1–2. 1. (–1, –3) and (3, 5) 2. (–3, 4) and (4, –3) Graph and label the lines of the equations in Exercises 3–6. 3. –x – 2y = 4 5. 5y – 3x = 15 4. 2x – 3y = 6 6. 7y – 2x = –14 Independent Practice In Exercises 1–4, graph the line through each set of two points. 1. (–1, –2) and (2, 4) 3. (0, 0) and (2, 6) 2. (–1, –1) and (1, 3) 4. (0, –2) and (1, 1) Graph and label the lines of the equations in Exercises 5–16. 5. x + y = 8 7. 2x + y = –3 9. –
3x + y = –6 11. 2x – y = –14 13. 8x + 4y = 24 15. 3x – 9y = –27 6. y – x = 10 8. 5x + y = –12 10. –10x + y = 21 12. 6x + 2y = 18 14. 12x – 4y = 8 16. 2x – 8y = 16 Round Up Round Up It’s easy to make a mistake when working out y-values, so choose x-values that will make the algebra easy (for example, 0 and 1). And it’s always a good idea to check your line by plotting a third point. 178 Section 4.2 — Lines Topic 4.2.3 California Standards: 7.0: Students verify that a point lies on a line, given an equation of the line. Students are able to derive linear equations by using the point-slope formula. What it means for you: You’ll rearrange line equations to find unknown values. Key words: linear equation substitute Check it out: Because (2k, 3) lies on the line you can replace x and y with 2k and 3, and the equation will still hold. Using Line Equations Using Line Equations This Topic carries straight on from Topic 4.2.2. If there’s an unknown in a pair of coordinates then you need to substitute the x and y values into the equation, and rearrange to find the unknown value. Points on a Line Satisfy the Equation of the Line In Lesson 4.2.1 you saw that if the ordered pair (x, y) satisfies the equation Ax + By = C, then the point (x, y) lies on the corresponding graph. Similarly, if a point (x, y) lies on the line Ax + By = C, then its coordinates satisfy the equation. So if you substitute the coordinates for x and y into the equation, it will make a true statement. Here are some examples showing how you can use this property to calculate unknown values: Example 1 (2k, 3) is a point on the line x – 3y = 7. Find k. Justify your answer. Solution x – 3y = 7 2k – 3(3) = 7 2k – 9 = 7 2k – 9 + 9 = 7 + 9 2k = 16 k = 2 2 k = 8 16 2 Substituting 2k for x and 3 for y Addition Property of Equality Division Property of Equality So the point on the line x – 3y = 7 with y-coordinate 3 is (16, 3). Section 4.2 — Lines 179 Example 2 ⎛ ⎜⎜⎜ ⎝ 2 , m 4 3 ⎞ ⎟⎟⎟ is a point on the line 2x – 9y = –10. Find the value of m. ⎠ Solution 2x – 9y = –10 ⎞ ⎟⎟⎟ = –10 ⎠ 2(4) – 9 ⎛ ⎜⎜⎜ ⎝ m 2 3 8 – 6m = –10 –6m = –18 − − m 18 6 − − 6 6 m = 3 = So the point on the line 2x – 9y = –10 with x-coordinate 4 is (4, 2). Guided Practice ⎛ ⎜⎜⎜ ⎝ 2 3 1. 4p, ⎞ ⎟⎟⎟ lies on the graph of 6x – 5y = –32. Find the value of p. ⎠ ⎛ ⎜⎜⎜ 2. The point − ⎝ 4 5 Find the value of h. 2, ⎞ ⎟⎟⎟ h lies on the graph of –x – 3y = 3. ⎠ 3. The point (2a, –3a) lies on the graph of –x – 2y = –12. Find the value of a. 4. Find the coordinates of the point in Exercise 5. ⎛ ⎜⎜⎜ ⎝ 1 4 5. k , − 3 4 ⎞ ⎟⎟⎟ is a point on the line x – 3y = 5. Find the value of k. ⎠ k 6. Find the coordinates of the point in Exercise 7. 180 Section 4.2 — Lines Example 3 The point (1, 3) lies on the line bx + y = 6. Find the value of b. Solution Here you have to use the coordinates to identify the equation. The question is different, but the method is the same. bx + y = 6 b(1 Guided Practice 7. The point (–1, 2) lies on the line –bx + 2y = –4. Find the value of b. 8. The point (–3, –5) lies on the graph of 2x – 3ky = 24. Find the value of k. 9. The point (8, 7) lies on the line bx + y = 11. Find b. 10. The point (6, 3) lies on the line bx – y = 9. Find b. 11. The point (14, 3) lies on the line x + by = –10. Find b. 12. The point (23, –4) lies on the line x – ky = –21. Find k. 13. The point (4, 4) lies on the line 4x – 2ky = 0. Find k. 14. The point (5, –1) lies on the line 9x + 3ky = 30. Find k. 15. The point (–2, –2) lies on the line 7kx – 3y = 13. Find k. 16. The point (–1, –4) lies on the line 7kx – 2ky = 0.25. Find k. Independent Practice In Exercises 1–5, the given point lies on the line 3x – 4y = 24. Find the value of each k and find the coordinates of each point. 1. (2k, 0) 4. (4k, k) 2. (4k, –6) 5. (–4k, –6k) 3. (4k, –18) In Exercises 6–10, the point (2, –4) lies on the given line. Find the value of b in each case. 6. bx + 3y = –6 9. 7x + by = –18 7. –bx + 5y = 10 10. 9bx – 2by = 4 8. 6x – by = 32 11. The point (1, 1) lies on the line bx – 2by = 4. Find b. 12. The point (–3, 6) lies on the line 4x + 6ky = 24. Find k. 13. The point (4k, 2k) lies on the line 2x – 6y = 12. Find k and the coordinates of the point. Round Up Round Up This Topic’s really just an application of the method you learned in Topic 4.2.1. Once you’ve substituted the x and y values into the equation then you just solve it as normal. Section 4.2 — Lines 181 The x- and y-Intercepts The x- and y-Intercepts The intercepts of a graph are the points where the graph crosses the axes. This Topic is all about how to calculate them. The x-Intercept is Where the Graph Crosses the x-Axis The x-axis on a graph is the horizontal line through the origin. Every point on it has a y-coordinate of 0. That means that all points on the x-axis are of the form (x, 0). The x-intercept of the graph of Ax + By = C is the point at which the graph of Ax + By = C crosses the x-axis. The x-intercept here is (–1, 0). y-axis 3 2 1 1 2 3 4 x-axis –4 –3 –2 0 –1 –1 –2 –3 Computing the x-Intercept Using “y = 0” Since you know that the x-intercept has a y-coordinate of 0, you can find the x-coordinate by letting y = 0 in the equation of the line. Example 1 Find the x-intercept of the line 3x – 4y = 18. Solution Let y = 0, then solve for x: 3x – 4y = 18 3x – 4(0) = 18 3x – 0 = 18 3x = 18 x = 6 So (6, 0) is the x-intercept of 3x – 4y = 18. Topic 4.2.4 California Standards: 6.0: Students graph a linear equation and compute the xand y-intercepts (e.g., graph 2x + 6y = 4). They are also able to sketch the region defined by linear inequality (e.g., they sketch the region defined by 2x + 6y < 4). What it means for you: You’ll learn about x- and y-intercepts and how to compute them from the equation of a line. Key words: intercept linear equation 182 Section 4.2 — Lines Check it out: Always write the x-intercept as a point, not just as the value of x where the graph crosses the x-axis. For example, (6, 0), not 6. Example 2 Find the x-intercept of the line 2x + y = 6. Solution Let y = 0, then solve for x: 2x + y = 6 2x + 0 = 6 2x = 6 x = 3 So (3, 0) is the x-intercept of 2x + y = 6. Guided Practice In Exercises 1–8, find the x-intercept. 1. x + y = 5 2. 3x + y = 18 3. 5x – 2y = –10 4. 3x – 8y = –21 5. 4x – 9y = 16 6.15x – 8y = 5 7. 6x – 10y = –8 8. 14x – 6y = 0 The y-Intercept is Where the Graph Crosses the y-Axis The y-axis on a graph is the vertical line through the origin. Every point on it has an x-coordinate of 0. That means that all points on the y-axis are of the form (0, y). The y-intercept of the graph of Ax + By = C is the point at which the graph of Ax + By = C crosses the y-axis. The y-intercept here is (0, 3). y-axis 6 5 4 3 2 1 –4 –3 –2 –1 0 1 2 3 x-axis Section 4.2 — Lines 183 Computing the y-Intercept Using “x = 0” Since the y-intercept has an x-coordinate of 0, find the y-coordinate by letting x = 0 in the equation of the line. Example 3 Find the y-intercept of the line –2x – 3y = –9. Solution Let x = 0, then solve for y: –2x – 3y = –9 –2(0) – 3y = –9 0 – 3y = –9 –3y = –9 y = 3 So (0, 3) is the y-intercept of –2x – 3y = –9. Example 4 Find the y-intercept of the line 3x + 4y = 24. Solution Let x = 0, then solve for y: 3x + 4y = 24 3(0) + 4y = 24 0 + 4y = 24 4y = 24 y = 6 So (0, 6) is the y-intercept of 3x + 4y = 24. Guided Practice In Exercises 9–16, find the y-intercept. 9. 4x – 6y = 24 10. 5x + 8y = 24 11. 8x + 11y = –22 12. 9x + 4y = 48 13. 6x – 7y = –28 14. 10x – 12y = 6 15. 3x + 15y = –3 16. 14x – 5y = 0 Check it out: Always write the y-intercept as a point, not just as the value of y where the graph crosses the y-axis. For example, (0, 3), not 3. 184 Section 4.2 — Lines Independent Practice 1. Define the x-intercept. 2. Define the y-intercept. Find the x- and y-intercepts of the following lines: 3. x + y = 9 5. –x – 2y = 4 7. 3x – 4y = 24 9. –5x – 4y = 20 4. x – y = 7 6. x – 3y = 9 8. –2x + 3y = 12 10. –0.2x + 0.3y = 1 11. 0.25x – 0.2y = 2 1 12 13. ( 3 5 ) is the x-intercept of the line –10x – 3y = 12. 0g, Find the value of g. )is the y-intercept of the line 2x – 15y = –3. ( 14. 0 1 , k 5 Find the value of k. 15. The point (–3, b) lies on the line 2y – x = 8. Find the value of b. 16. Find the x-intercept of the line in Exercise 15. 17. Another line has x-intercept (4, 0) and equation 2y + kx = 20. Find the value of k. In Exercises 18-22, use the graph below to help you reach your answer6 –5 –4 –3 –2 –1 0 –1 1 2 3 4 5 x 6 –2 –3 –4 –5 –6 18. Find the x- and y-intercepts of line n. 19. Find the x-intercept of line p. 20. Find the y-intercept of line r. 21. Explain why line p does not have a y-intercept. 22. Explain why line r does not have an x-intercept. Round Up Round Up Make sure you get the method the right way around — to find the x-intercept, put y = 0 and solve for x, and to find the y-intercept, put x = 0 and solve for y. In the next Topic you’ll see that the intercepts are really useful when you’re graphing lines from the line equation. Section 4.2 — Lines 185 Topic 4.2.5 California Standards: 6.0: Students graph a linear equation and compute the xand y-intercepts (e.g., graph 2x + 6y = 4). They are also able to sketch the region defined by linear inequality (e.g., they sketch the region defined by 2x + 6y < 4). 7.0: Students verify that a point lies on a line, given an equation of the line. Students are able to derive linear equations by using the point-slope formula. What it means for you: You’ll graph lines by first calculating the x- and y-intercepts. Key words: linear equation intercept 186 Section 4.2 — Lines Graphing Lines Graphing Lines In Topic 4.2.2 you learned how to graph a straight line by plotting two points. If you’re not given points on the line, it’s easiest to use the x- and y-intercepts. Graphing Lines by Computing the Intercepts The method below for plotting a straight-line graph is the same as in Topic 4.2.2. To graph the line, you plot two points — except this time you use the x-intercept and the y-intercept, then draw a straight line th
rough them. Graphing a Line Find the x-intercept — let y = 0, then solve the equation for x. Find the y-intercept — let x = 0, then solve the equation for y. Draw a set of axes and plot the two intercepts. Draw a straight line through the points. Check your line by plotting a third point. Example 1 Draw the graph of 5x + 3y = 15 by computing the intercepts. Solution x-intercept: 5x + 3(0) = 15 5x + 0 = 15 5x = 15 x = 3 y-intercept: 5(0) + 3y = 15 0 + 3y = 15 3y = 15 y = 5 Therefore (3, 0) is the x-intercept and (0, 5) is the y-intercept. Check: x = 1 5x + 3y = 15 3 –2 0 –1 –1 –2 –3 y-axis 1 2 3 4 5 6 x-axis − 15 5 3 ) lies on the line — which means the line is correct. 10 Finding the intercepts is the quickest way of finding two points. When you substitute 0 for y to solve for x, the y-term disappears, and vice versa — making the equations easier to solve. Example 2 Draw the graph of y = –x + 3 by computing the intercepts. Solution x-intercept: 0 = –x + 3 x = 3 y-intercept: y = –(0) + 3 y = 3 Therefore (3, 0) is the x-intercept and (0, 3) is the y-intercept. –3 –2 y-axis 6 5 4 3 2 1 Check: x = 1 y = –x + 3 ﬁ y = –1 + 3 = 2 (1, 2) lies on the line — which means the line is correct. 1 2 3 4 5 6 x-axis 0 –1 –1 –2 –3 Guided Practice Draw the graphs of the following equations by computing the intercepts. 1. 5x + 2y = 10 3. 6x – y = 3 5. 2x + y = 3 7. x + y = 10 9. 2x – y = 4 11. 3x + y = 9 13. y = 2x + 4 2. –3x – 3y = 12 4. –4x + 5y = 20 6. –x – 8y = 2 8. x – y = 4 10. x + 5y = 10 12. x – 4y = 8 14. y = 5x – 10 15. y = 3x + 9 17. y = – 2 5 x – 2 16. y = 18 Section 4.2 — Lines 187 Independent Practice Draw graphs of the lines using the x- and y-intercepts in Exercises 1–6. 1. x-intercept: (–3, 0) 2. x-intercept: (1, 0) 3. x-intercept: (4, 0) 4. x-intercept: (–6, 0) 5. x-intercept: (–1, 0) 6. x-intercept: (2, 0) y-intercept: (0, 2) y-intercept: (0, 6) y-intercept: (0, –3) y-intercept: (0, –4) y-intercept: (0, 7) y-intercept: (0, –5) Draw the graphs of the equations in Exercises 7–18 by computing the intercepts. 7. x + y = 6 9. x – y = –5 11. 3x + y = 6 13. 2x – y = –4 15. 4x + 3y = –12 17. 6x – 3y = 24 8. x + y = –4 10. x – y = 7 12. 2x + y = 8 14. 3x – y = –3 16. 5x – 2y = 10 18. 10x – 12y = 60 19. Show that the graphs of x + y = 6 and –6x – 6y = –36 are the same. 20. Explain why the graph of 5x + 8y = 0 cannot be drawn using the intercepts. Round Up Round Up This Topic follows on neatly from Topic 4.2.2, where you graphed lines by plotting two points and joining them with a straight line. You can use any two points — the main reason for using the intercepts is that they’re usually easier to calculate. 188 Section 4.2 — Lines TTTTTopicopicopicopicopic 4.3.14.3.1 4.3.14.3.1 4.3.1 Section 4.3 Slope of a Line Slope of a Line Slope of a Line Slope of a Line Slope of a Line Slope of a Line Slope of a Line Slope of a Line Slope of a Line Slope of a Line California Standards: 7.0:7.0:7.0:7.0:7.0: Students verify that a point lies on a line, given an equation of the line. le to derivvvvveeeee le to deri Students are ae ae ae ae abbbbble to deri le to deri Students ar Students ar le to deri Students ar Students ar tions tions linear equa linear equa tions by using the linear equations tions linear equa linear equa point-slope formula. What it means for you: You’ll find the slope of a line given any two points on the line. Key words: slope steepness horizontal vertical rise over run Check it out: Dx is pronounced “delta x” and just means “change in x.” By now you’ve had plenty of practice in plotting lines. Any line can be described by its slope — which is what this Topic is about. The Slope of a Line is Its Steepness The Slope of a Line is Its Steepness The Slope of a Line is Its Steepness The Slope of a Line is Its Steepness The Slope of a Line is Its Steepness The slope (or gradient) of a line is a measure of its steepness. The slope of a straight line is the ratio of the vertical change to the horizontal change between any two points lying on the line. The vertical change is usually written Dy, and it’s often called the rise. In the same way, the horizontal change is usually written Dx, and it’s often called the run. D = rise y D = run x O Slope = vertical change horizontal change = rise run = Δ Δ y x , provided DDDDDx πππππ 0 If you know the coordinates of any two points on a line, you can find the slope. The slope, m, of a line passing through points P1 (x1, y1) and P2 (x2, y2) is given by this formula , provided x2 – x1 πππππ 0 There is an important difference between positive and negative slopes — a positive slope means the line goes “uphill” ( ), whereas a line with a negative slope goes “downhill” ( ). Section 4.3 Section 4.3 Section 4.3 — Slope Section 4.3 Section 4.3 189189189189189 a Line a Line ula to Find the Slope of Use the Fororororormmmmmula to Find the Slope of ula to Find the Slope of Use the F Use the F a Line ula to Find the Slope of a Line a Line ula to Find the Slope of Use the F Use the F Example Example Example Example Example 11111 Find the slope of the line that passes through the points (2, 1) and (7, 4) and draw the graph. Check it out: In this example, (x1, y1) = (2, 1) and (x2, y2) = (7, 4). Solution Solution Solution Solution Solution So the slope is 3 5 . You know that the line passes through (2, 1) and (7, 4), so just join those two points up to draw the graph. y (7, 4) (2, 1) In the graph above, notice how the line has a positive slope, meaning it goes “uphill” from left to right. In fact, since the slope is 3 5 , the line goes 3 units up for every 5 units across. Guided Practice In Exercises 1–4, find the slope of the line on the graph opposite. (4) 5. Find the slope of the line that passes through the points (1, 5) and (3, 2), and draw the graph. 6. Find the slope of the line that passes through the points (3, 1) and (2, 4), and draw the graph. –6 –5 –4 –3 –2 –1 (3) (2) (11 –2 –3 –4 –5 –6 190190190190190 Section 4.3 Section 4.3 Section 4.3 — Slope Section 4.3 Section 4.3 e Coordinadinadinadinadinatestestestestes e Coor e Nee Negggggaaaaatititititivvvvve Coor e Coor e Nee Ne eful If TTTTTherherherherhere are are are are are Ne eful If eful If Be Car Be Car Be Careful If e Coor eful If Be Car Be Car Example Example Example Example Example 22222 Find the slope of the line that passes through the points (3, 4) and (6, –2). Solution Solution Solution Solution Solution 2 So the slope is –2. Check it out: It doesn’t matter which point you call (x1, y1) and which you call (x2, y2) — choose whichever makes the math easier. This time the line has a negative slope, meaning it goes “downhill” from left to right. Here the slope is –2, which means that the line goes 2 units down for every 1 unit across. w a Graaaaaphphphphph w a Gr e to Draaaaaw a Gr w a Gr e to Dr t Hat Havvvvve to Dr e to Dr t Hat Ha ou Don’’’’’t Ha ou Don YYYYYou Don ou Don w a Gr e to Dr ou Don Example Example Example Example Example 33333 Find the slope of the lines through: a) (2, 5) and (–4, 2) b) (1, –6) and (3, 3) Solution Solution Solution Solution Solution a) m = b Guided Practice Find the slope m of the line through each pair of points below. 7. (–1, 2) and (3, 2) 9. (5, –7) and (–3, –7) 11. (–1, –3) and (1, –4) 13. (–2, –2) and (–3, –17) 15. (0, –1) and (1, 0) 8. (0, –5) and (–6, 1) 10. (4, –1) and (–3, 5) 12. (5, 7) and (–11, –12) 14. (18, 2) and (–32, 7) 16. (0, 0) and (–14, –1) Section 4.3 Section 4.3 Section 4.3 — Slope Section 4.3 Section 4.3 191191191191191 lems Invvvvvolvolvolvolvolve e e e e VVVVVariaariaariaariaariabbbbbleslesleslesles lems In Some Proboboboboblems In lems In Some Pr Some Pr lems In Some Pr Some Pr Example Example Example Example Example 44444 If the slope of the line that passes through the points (4, –1) and (6, 2k) is 3, find the value of k. Solution Solution Solution Solution Solution Even though one pair of coordinates contains a variable, k, you still use the slope formula in exactly the same way as before. m = y x 2 2 − − , which means that But the slope is 3, so 1 2 k + 2 = 3 ﬁ 2k + 1 = 6 ﬁ 2k = 5 ﬁ k = 5 2 Guided Practice Find the slope m of the lines through the points below. 17. (7, –2c) and (10, –c) 19. (2, 2k) and (–5, –5k) 21. (3d, 7d) and (5d, 9d) 23. (9c, 12v) and (12c, 15v) 25. (10, 14d) and (d, –7) 18. (b, 1) and (3b, –3) 20. (3q, 1) and (2q, 7) 22. (4a, 5k) and (2a, 7k) 24. (p, q) and (q, p) 26. (2t, –3s) and (18s, 14t) In Exercises 27–32 you’re given two points on a line and the line’s slope, m. Find the value of the unknown constant in each Exercise. 27. (–2, 3) and (3k, –4), m = 28. (4, –5t) and (7, –8t), m = 2 5 2 7 29. (4b, –6) and (7b, –10), m = 3 4 30. (–8, –6) and (12, 4), m = – 31. (7k, –3) and (k, –1), m = 4 2 5 v 32. (1, –17) and (40, 41), m = – 174 78 t 192192192192192 Section 4.3 Section 4.3 Section 4.3 — Slope Section 4.3 Section 4.3 Independent Practice 1. Find the slope m of the lines shown below. (iii) (v) -5 -4 y -4 -5 (i) (ii) (iv) In Exercises 2–5, find the slope of the line that passes through the given points, and draw the graph. 2. (–2, 1) and (0, 2) 4. (–5, 2) and (–1, 3) 3. (4, 4) and (1, 0) 5. (3, –3) and (7, 3) In Exercises 6–10, find the slope of the line through each of the points. 6. (–3, 5) and (2, 1) 8. (2, 3) and (4, 3) 10. (2s, 2t) and (s, 3t) 7. (0, 4) and (–4, 0) 9. (6d, 2) and (4d, –1) In Exercises 11–15, you’re given two points on a line and the line’s slope, m. Find the value of the unknown constant in each Exercise. 11. (3t, 7) and (5t, 9), m = – 12. (3k, 1) and (2k, 7), m = 1 2 1 3 13. (0, 14d) and (10, –6d), m = –1 14. (2t, –3) and (–3t, 5), m = 5 4 15. (0, 8d) and (–1, 4d), m = – 1 3 ound Up ound Up RRRRRound Up ound Up ound Up RRRRRound Up ound Up ound Up ound Up ound Up Slope is a measure of how steep a line is — it’s how many units up or down you go for each unit across. If you go up or down a lot of units for each unit across, the line will be steep and the slope will be large (either large and positive if it goes up from left to
right, or large and negative if it goes down from left to right). Section 4.3 Section 4.3 Section 4.3 — Slope Section 4.3 Section 4.3 193193193193193 oint-Slope Fororororormmmmmulaulaulaulaula oint-Slope F oint-Slope F PPPPPoint-Slope F oint-Slope F PPPPPoint-Slope F oint-Slope Fororororormmmmmulaulaulaulaula oint-Slope F oint-Slope F oint-Slope F The point-slope formula is a really useful way of calculating the equation of a straight line. aight Line aight Line a Str a Str tion of tion of ula to Find the Equa Use the Fororororormmmmmula to Find the Equa ula to Find the Equa Use the F Use the F aight Line a Straight Line tion of a Str ula to Find the Equation of aight Line a Str tion of ula to Find the Equa Use the F Use the F If you know the slope of the line and a point on the line, you can use the point-slope formula to find the equation of the line. The point-slope formula for finding the equation of a line is: y – y1 = m(x – x1) where m is the slope and (x1, y1) is a point on the line. You substitute the x-coordinate of a point on the line for x1 and the y-coordinate of the same point for y1. Watch out though — x and y are variables and they stay as letters in the equation of the line. Example Example Example Example Example 11111 Find the equation of the line through (–4, 6) that has a slope of –3. Solution Solution Solution Solution Solution (x1, y1) = (–4, 6) and m = –3 y – y1 = m(x – x1) ﬁ y – 6 = –3[x – (–4)] ﬁ y – 6 = –3(x + 4) ﬁ y – 6 = –3x – 12 ﬁ y + 3x = –6 Guided Practice Write the equation of the line that passes through the given point and has the given slope. 1. Point (–2, –3), slope = –1 2. Point (3, –5), slope = 2 3. Point (–7, –2), slope = –5 4. Point (4, –3), slope = 2 3 5. Point (2, 6), slope = – 3 4 6. Point (–2, –3), slope = 5 8 7. Point (–5, –3), slope = – 6 7 ⎛ ⎜⎜⎜ 8. Point − ⎝ 2 3 , ⎞ ⎟⎟⎟ ⎠ 1 4 , slope = 2 5 TTTTTopicopicopicopicopic 4.3.24.3.2 4.3.24.3.2 4.3.2 California Standards: 7.0:7.0:7.0:7.0:7.0: Students verify that a point lies on a line, given an equation of the line. Students are ae ae ae ae abbbbble to deri le to derivvvvveeeee le to deri le to deri Students ar Students ar Students ar le to deri Students ar y using y using tions b tions b linear equa linear equa y using tions by using linear equations b y using tions b linear equa linear equa the point-slope fororororormmmmmula.ula.ula.ula.ula. the point-slope f the point-slope f the point-slope f the point-slope f What it means for you: You’ll learn about the pointslope formula and use it to find the equation of a line. Key words: slope point-slope formula Check it out: The point-slope formula is just the one from Topic 4.3.1 rearranged. (x, y) is any point on the line, so (x1, y1) and (x, y) are the two points on the line. 194194194194194 Section 4.3 Section 4.3 Section 4.3 — Slope Section 4.3 Section 4.3 Check it out You can use either point as (x1, y1) here — you’ll still get the same equation. Find the Slope Firststststst Find the Slope Fir e Givvvvven,en,en,en,en, Find the Slope Fir Find the Slope Fir e Gi e Gi oints ar IfIfIfIfIf TTTTTwwwwwo Po Po Po Po Points ar oints ar oints are Gi Find the Slope Fir e Gi oints ar If you know the coordinates of two points on a straight line, you can still find the equation using the point-slope formula — but you have to find the slope first. Example Example Example Example Example 22222 Write the equation of the straight line that contains the points (3, –2) and (–1, 5). Solution Solution Solution Solution Solution Step 1: Find the slope using the given points Slope Step 2: Write the equation y – y1 = m(x – x1) (x – 3) y – (–2 4y + 8 = –7(x – 3) ﬁ 4y + 8 = –7x + 21 ﬁ 4y + 7x = 13 (x – 3) Guided Practice Write the equation of the line that passes through the given pair of points. 9. (–1, 0) and (3, –4) 11. (–5, 7) and (3, 9) 13. (8, 7) and (–7, –5) 15. (3, 1) and (5, 4) 17. (6, 2) and (4, 1) 19. (1, 0) and (2, 0) 21. (4, –1) and (–1, –3) 23. (4, 5) and (2, 6) 25. (3, –5) and (0, 8) 10. (–1, 1) and (–3, –3) 12. (6, –8) and (–2, –10) 14. (–10, 11) and (5, –12) 16. (2, –5) and (3, –1) 18. (–3, 5) and (4, 3) 20. (–2, –2) and (7, 5) 22. (7, 2) and (3, 3) 24. (–2, –3) and (–3, –2) 26. (1, 1) and (4, –6) Section 4.3 Section 4.3 Section 4.3 — Slope Section 4.3 Section 4.3 195195195195195 Independent Practice In Exercises 1–6, write the equation of the line that passes through the given point and has the given slope. 1. Point (1, 5), slope = –3 2. Point (2, 0), slope = 1 2 3. Point (3, 1), slope = 1 4 5. Point (–8, 6), slope = – 4 5 4. Point (–3, 4), slope = – 6. Point (–3, –4), slope = 2 3 3 8 In Exercises 7–16, write the equation of the line that passes through the given pair of points. 7. (0, 3) and (4, –1) 9. (3, 8) and (4, 4) 11. (–6, 9) and (–4, –6) 13. (–4, –8) and (–5, 4) 15. (10, 5) and (4, 6) 8. (–1, 6) and (7, 5) 10. (4, –7) and (–3, 5) 12. (4, –9) and (–3, –9) 14. (–8, 3) and (8, 4) 16. (0, 0) and (–4, –6) 17. The points (5, 6) and (8, 7) lie on a line. Find the equation of this line. 18. The line in Exercise 17 forms one side of a triangle that has one vertex at the point (5, –4). If the slope of one of the edges of the triangle is –3, find the equation of this edge. 19. The point (8, k) lies on the third edge of the triangle in Exercise 18. Given that the triangle is isosceles, find, by graphing, the value of k. 20. Joshua is an architect who must build a wheelchair-accessible office building. To make a ramp that is easy to maneuver in a wheelchair, Joshua designs a ramp that is 27 inches high and 540 inches long. What is the slope of the ramp? 21. If the student population at a high school changes from 1372 in 1996 to 1768 in 2006, what is the average rate of change of the student population? (Hint: Use the pairs of coordinates (1996, 1372) and (2006, 1768) to reach your answer.) ound Up ound Up RRRRRound Up ound Up ound Up ound Up ound Up RRRRRound Up ound Up ound Up You should look over the point-slope formula until you can write it down from memory. It’s a really useful formula and makes finding the equation of a line much easier — but only if you remember it. 196196196196196 Section 4.3 Section 4.3 Section 4.3 — Slope Section 4.3 Section 4.3 TTTTTopicopicopicopicopic 4.4.14.4.1 4.4.14.4.1 4.4.1 Section 4.4 allel Lines allel Lines PPPPParararararallel Lines allel Lines allel Lines PPPPParararararallel Lines allel Lines allel Lines allel Lines allel Lines California Standards: 8.0:8.0:8.0:8.0:8.0: Students under stand stand Students under Students under stand Students understand stand Students under allel allelallel par par pts of pts of the conce the conce allel parallel pts of par the concepts of the conce par pts of the conce lines lines lines and perpendicular lines lines lines and how their slopes are related. Students are able to find the equation of a line perpendicular to a given line that passes through a given point..... What it means for you: You’ll work out the slopes of parallel lines and you’ll test if two lines are parallel. Key words: parallel intersect Now that you’ve practiced finding the slope of a line, you can use the method on a special case — parallel lines. allel Lines Nevvvvver Meet er Meet er Meet allel Lines Ne PPPPParararararallel Lines Ne allel Lines Ne er Meet er Meet allel Lines Ne Parallel lines are two or more lines in a plane that never intersect (cross). These lines are all parallel. No matter how long you draw them, they’ll never meet. The symbol \|\| is used to indicate parallel lines — you read this symbol as “is parallel to.” So, if l1 and l2 are lines, then l1 \|\| l2 means “line l1 is parallel to line l2.” allel Lines Havvvvve Identical Slopes e Identical Slopes e Identical Slopes allel Lines Ha PPPPParararararallel Lines Ha allel Lines Ha e Identical Slopes e Identical Slopes allel Lines Ha You can determine whether lines are parallel by looking at their slopes. Two lines are parallel if their slopes are equal. Example Example Example Example Example 11111 Prove that the three lines A, B, and C shown on the graph are parallel. Solution Solution Solution Solution Solution Using the rise over run formula (see Topic 4.3.1), you can see that they all have a slope of Section 4.4 — More Lines Section 4.4 Section 4.4 Section 4.4 Section 4.4 197197197197197 Guided Practice 1. Two lines on the same plane that never intersect are called _________________ lines. 2. To determine if two lines are parallel you can look at their _________________. 3. Prove that the line f defined by y – 3 = 2 3 (x – 4) is parallel to line g defined by y – 6 = 2 3 (x + 1). t Havvvvve Defined Slopes e Defined Slopes e Defined Slopes t Hat Ha t Ha tical Lines Don’’’’’t Ha tical Lines Don VVVVVererererertical Lines Don tical Lines Don e Defined Slopes e Defined Slopes tical Lines Don Vertical lines are parallel, but you can’t include them in the definition on page 197 because their slopes are undefined. Points on a vertical line all have the same x-coordinate, so they are of the form (c, y1) and (c, y2). The slope of a vertical line is undefined because is not defined. y Finding Slopes y Finding Slopes allel b Lines are Pe Pe Pe Pe Parararararallel b allel b Lines ar Lines ar est if TTTTTest if est if y Finding Slopes allel by Finding Slopes est if Lines ar y Finding Slopes allel b Lines ar est if To check if a pair of lines are parallel, just find the slope of each line. If the slopes are equal, the lines are parallel. Example Example Example Example Example 22222 Show that the straight line through (2, –3) and (–5, 1) is parallel to the straight line joining (7, –1) and (0, 3). Solution Solution Solution Solution Solution Step 1: Find the slope of each line using the formula m1 = m2 = Step 2: Compare the slopes and draw a conclusion. − = −4 7 4 7 , so m1 = m2. So the straight line through (2, –3) and (–5, 1) is parallel to the straight line through (7, –1) and (0, 3). Check it out: The slope of a vertical line is undefined because division by 0 is undefined. Don’t f
orget: See Topic 4.3.1 for more on calculating the slope of a line. Check it out: It’s possible that all the points lie on the same line — meaning there is actually only one line. To be absolutely accurate, you should check that this isn’t the case before you say the lines are parallel. You can do this by finding the equations of the lines and comparing them. 198198198198198 Section 4.4 — More Lines Section 4.4 Section 4.4 Section 4.4 Section 4.4 Guided Practice 4. Show that line a, which goes through points (7, 2) and (3, 3), is parallel to line b joining points (–8, –4) and (–4, –5). 5. Show that the line through points (4, 3) and (–1, 3) is parallel to the line though points (–6, –1) and (–8, –1). 6. Determine if line f joining points (1, 4) and (6, 2) is parallel to line g joining points (0, 8) and (10, 4). 7. Determine if the line through points (–5, 2) and (3, 7) is parallel to the line through points (–5, 1) and (–3, 6). 8. Determine if the line through points (–8, 4) and (–8, 3) is parallel to the line through points (6, 3) and (–4, 3). lems are e e e e TTTTTougher ougher ougher lems ar allel Line Proboboboboblems ar lems ar allel Line Pr Some Parararararallel Line Pr allel Line Pr Some P Some P ougher ougher lems ar allel Line Pr Some P Some P Example Example Example Example Example 33333 Find the equation of a line through (–1, 4) that is parallel to the straight line joining (5, 7) and (–6, –8). Solution Solution Solution Solution Solution Step 1: Find the slope m1 of the line through (5, 7) and (–6, –8). m1 = = − − 15 11 = 15 11 Step 2: The slope m2 of the line through (–1, 4) must be equal to since the lines are parallel. 15 11 So, m1 = m2 = 15 11 Step 3: Now use the point-slope formula to find the equation of the line through point (–1, 4) with slope 15 11 . y – y1 = m(x – x1) Check it out: This is the equation of the line through (–1, 4) that is parallel to the straight line joining (5, 7) and (–6, –8). [x – (–1)] ﬁ y – 4 = 15 11 ﬁ 11y – 44 = 15(x + 1) ﬁ 11y – 44 = 15x + 15 Equation: 11y – 15x = 59 Section 4.4 — More Lines Section 4.4 Section 4.4 Section 4.4 Section 4.4 199199199199199 Guided Practice 9. Find the equation of the line through (–3, 7) that is parallel to the line joining points (4, 5) and (–2, –8). 10. Find the equation of the line through (6, –4) that is parallel to the line joining points (–1, 6) and (7, 3). 11. Find the equation of the line through (–1, 7) that is parallel to the line joining points (4, –3) and (8, 6). 12. Write the equation of the line through (–3, 5) that is parallel to the line joining points (–1, 2.5) and (0.5, 1). 13. Write the equation of the line through (–2, –1) that is parallel to the line x + 3y = 6. Independent Practice 1. Line l1 has slope about l1 and l2? 1 2 and line l2 has slope 1 2 . What can you conclude 2. Line l1 has a slope of – 1 3 . If l1 \|\| l2, then what is the slope of l2? 3. Show that all horizontal lines are parallel. 4. Show that the line through the points (5, –3) and (–8, 1) is parallel to the line through (13, –7) and (–13, 1). 5. Determine if the line through the points (5, 4) and (0, 9) is parallel to the line through (–1, 8) and (4, 0). 6. Determine if the line through the points (–2, 5) and (6, 0) is parallel to the line through (8, –1) and (0, 4). 7. Determine if the line through the points (4, –7) and (4, –4) is parallel to the line through (–5, 1) and (–5, 5). 8. Determine if the line through the points (–2, 3) and (–2, –2) is parallel to the line through (1, 7) and (–6, 7). 9. Find the equation of the line through (1, –2) that is parallel to the line joining the points (–3, –1) and (8, 7). 10. Find the equation of the line through (–5, 3) that is parallel to the line joining the points (–2, 6) and (8, –1). 11. Write the equation of the line through (0, 6) that is parallel to the line 3x + 2y = 6. ound Up ound Up RRRRRound Up ound Up ound Up RRRRRound Up ound Up ound Up ound Up ound Up When you draw lines with different slopes on a set of axes, you might not see where they cross. But remember, you are only looking at a tiny bit of the lines — they go on indefinitely in both directions. If they don’t have identical slopes, they’ll cross sooner or later. 200200200200200 Section 4.4 — More Lines Section 4.4 Section 4.4 Section 4.4 Section 4.4 TTTTTopicopicopicopicopic 4.4.24.4.2 4.4.24.4.2 4.4.2 California Standards: 8.0:8.0:8.0:8.0:8.0: Students under stand stand Students under Students under stand Students understand stand Students under the concepts pts pts pts pts of parallel lines the conce the conce the conce the conce and perperperperperpendicular lines pendicular lines pendicular lines pendicular lines and pendicular lines how their slopes are related. le to find le to find Students are ae ae ae ae abbbbble to find Students ar Students ar le to find le to find Students ar Students ar a line a line tion of tion of the equa the equa a line tion of a line the equation of the equa a line tion of the equa pendicular to a givvvvvenenenenen perperperperperpendicular to a gi pendicular to a gi pendicular to a gi pendicular to a gi t passes thr line tha line tha ough a ough a t passes thr line that passes thr ough a t passes through a line tha t passes thr line tha ough a gigigigigivvvvven point. en point. en point. en point. en point. What it means for you: You’ll work out the slopes of perpendicular lines and you’ll test if two lines are perpendicular. Key words: perpendicular reciprocal Check it out: This definition doesn’t work for horizontal or vertical lines, since the slope of a vertical line can’t be defined in the same way as for other lines. But remember that the lines x = c and y = k (for constants c and k) are perpendicular. pendicular Lines pendicular Lines PPPPPerererererpendicular Lines pendicular Lines pendicular Lines PPPPPerererererpendicular Lines pendicular Lines pendicular Lines pendicular Lines pendicular Lines Math problems about parallel lines often deal with perpendicular lines too. “Perpendicular” might sound like a difficult term, but it’s actually a really simple idea. t Right AngAngAngAngAngleslesleslesles t Right t Right pendicular Lines Meet a PPPPPerererererpendicular Lines Meet a pendicular Lines Meet a pendicular Lines Meet at Right t Right pendicular Lines Meet a Two lines are perpendicular if they intersect at 90° angles, like in the graphs on the right. ocals ocals ecipr e Negggggaaaaatititititivvvvve Re Re Re Re Recipr ecipr e Nee Ne e Ne pendicular Lines ar Slopes of P P P P Perererererpendicular Lines ar pendicular Lines ar Slopes of Slopes of ocals eciprocals pendicular Lines are Ne ocals ecipr pendicular Lines ar Slopes of Slopes of Two lines are perpendicular if the slope of one is the negative reciprocal of the slope of the other. To get the reciprocal of a number you divide 1 by it. For example, the reciprocal of x is 1 x and so the negative reciprocal is – 1 x . The reciprocal of x y is 1 x y = and so the negative reciprocal is – y x y x . Example Example Example Example Example 11111 Prove that lines A and B, shown on the graph, are perpendicular to each other. Solution Solution Solution Solution Solution Using the rise over run formula: Don’t forget: See Topic 4.3.1 for the “rise over run” formula. Slope of A = m1 = Slope of B = m2 = 1 2 = 2 4 −4 2 = –2 4 2 2 4 B A 1 2 is the negative reciprocal of –2, so A and B must be perpendicular. Section 4.4 — More Lines Section 4.4 Section 4.4 Section 4.4 Section 4.4 201201201201201 Guided Practice 1. Perpendicular lines meet at ________________ angles. 2. Find the negative reciprocal of 3. 3. Find the negative reciprocal of – 4. Find the negative reciprocal of – 1 4 4 5 . . 5. Use the graph to prove that A and B are perpendicular. A –6 –4 –2 y 6 4 2 0 –2 –4 –6 2 4 6 X B = –1 = –1 × ines: m11111 × × ines: pendicular LLLLLines: ines: pendicular PPPPPerererererpendicular pendicular = –1 × m22222 = –1 = –1 × ines: pendicular When you multiply a number by its reciprocal, you always get 1. For example, 1 5 5 × = = and So because two perpendicular slopes are negative reciprocals of each other, their product is always –1. Here’s the same thing written in math-speak: Check it out: ^ is the symbol for “is perpendicular to.” If two lines l1 and l2 have slopes m1 and m2, l1 ^ l2 if and only if m1 × m2 = –1. Example Example Example Example Example 22222 P and Q are two straight lines and P ^ Q. P has a slope of –4. What is the slope of Q? Solution Solution Solution Solution Solution mP × mQ = –1 ﬁ – 4 × mQ = –1 1 4 − =1 − 4 ﬁ mQ = . So the slope of Q is 1 4 . 202202202202202 Section 4.4 — More Lines Section 4.4 Section 4.4 Section 4.4 Section 4.4 Guided Practice 6. Lines l1 and l2 are perpendicular. If the slope of l1 is 1 5 , find the slope of l2. 7. Lines A and B are perpendicular. If the slope of A is – 5 8 , find the slope of B. 8. Lines R and T are perpendicular. If R has slope – 7 11 , what is the slope of T? 9. The slope of l1 is –0.8. The slope of l2 is 1.25. Determine whether l1 and l2 are perpendicular. inding SSSSSlopes lopes lopes inding pendicular by y y y y FFFFFinding inding pendicular b ines are e e e e PPPPPerererererpendicular b pendicular b ines ar ShoShoShoShoShow w w w w TTTTThahahahahat t t t t LLLLLines ar ines ar lopes lopes inding pendicular b ines ar Example Example Example Example Example 33333 Determine the equation of the line passing through (3, 1) that is perpendicular to the straight line through (2, –1) and (4, 2). Solution Solution Solution Solution Solution Step 1: Find slope m1 of the line through (2, –1) and (4, 2): Step 2: Find the slope m2 of a line perpendicular to that line: m1 × m2 = –1 3 2 × m2 = –1 ﬁ m2 = – 2 3 Step 3: Now use the point-slope formula to find the equation of the line through (3, 1) with slope – 2 3 . y – y1 = m(x – x1x – 3) ﬁ 3y – 3 = –2(x – 3) ﬁ 3y – 3 = –2x + 6 Equation: 3y + 2x = 9 Section 4.4 — More Lines Section 4.4 Section 4.4 Section 4.4 Section 4.4 203203203203203 Guided Practice 10. S
how that the line through the points (5, –3) and (–8, 1) is perpendicular to the line through (4, 6) and (8, 19). 11. Show that the line through (0, 6) and (5, 1) is perpendicular to the line through (4, 8) and (–1, 3). 12. Show that the line through (4, 3) and (2, 2) is perpendicular to the line through (1, 3) and (3, –1). 13. Determine the equation of the line through (3, –4) that is perpendicular to the line through the points (–7, –3) and (–3, 8). 14. Determine the equation of the line through (6, –7) that is perpendicular to the line through the points (8, 2) and (–1, 8). 15. Find the equation of the line through (4, 5) that is perpendicular to the line –3y + 4x = 6. Independent Practice In Exercises 1–8, J and K are perpendicular lines. The slope of J is given. Find the slope of K. 1. mJ = –3 5 2 2. mJ = –14 4. mJ = 3. mJ = 6 7 5. mJ = – 8 3 6. mJ = – 14 15 8. mJ = 0.45 7. mJ = –0.18 9. Show that the line through (2, 7) and (–2, 8) is perpendicular to the line through (–3, –3) and (–2, 1). 10. Show that the line through (–4, 3) and (3, –2) is perpendicular to the line through (–7, –1) and (–2, 6). 11. Determine the equation of the line through (5, 9) that is perpendicular to a line with slope 1 3 . 12. Determine the equation of the line through (3, –5) that is perpendicular to the line through the points (–3, 2) and (–6, –4). ound Up ound Up RRRRRound Up ound Up ound Up RRRRRound Up ound Up ound Up ound Up ound Up “Perpendicular” is just a special math word to describe lines that are at right angles to each other. Remember that the best way to show that two lines are at right angles is to calculate their slopes — if they multiply together to make –1, then the lines are perpendicular. 204204204204204 Section 4.4 — More Lines Section 4.4 Section 4.4 Section 4.4 Section 4.4 TTTTTopicopicopicopicopic 4.4.34.4.3 4.4.34.4.3 4.4.3 California Standards: 8.0:8.0:8.0:8.0:8.0: Students under stand stand Students under Students under stand Students understand stand Students under allel allelallel par par pts of pts of the conce the conce allel parallel pts of par the concepts of par the conce pts of the conce pendicular pendicular lines and per lines and per pendicular lines and perpendicular lines and per pendicular lines and per lines and how their slopes lines and how their slopes lines and how their slopes lines and how their slopes lines and how their slopes ted. Students are ted. ararararare re re re re relaelaelaelaelated. ted. ted. able to find the equation of a line perpendicular to a given line that passes through a given point..... What it means for you: You’ll plot graphs and solve equations using a method called the slope-intercept form of an equation. he Slope-Intercececececeptptptptpt he Slope-Inter he Slope-Inter TTTTThe Slope-Inter he Slope-Inter he Slope-Intercececececeptptptptpt he Slope-Inter he Slope-Inter TTTTThe Slope-Inter he Slope-Inter a Line a Line FFFFForororororm ofm ofm ofm ofm of a Line a Line a Line FFFFForororororm ofm ofm ofm ofm of a Line a Line a Line a Line a Line Now that you’ve practiced calculating the slope and intercept of a line, you can use the two things together to make plotting graphs easier. a Line: y = mx + b a Line: he Slope-Intercececececept Fpt Fpt Fpt Fpt Forororororm ofm ofm ofm ofm of a Line: a Line: he Slope-Inter TTTTThe Slope-Inter he Slope-Inter a Line: he Slope-Inter In the slope-intercept form the y is alone on one side of the equation. The slope-intercept form of the equation is: y = mx + b Key words: slope-intercept form intercept Here are a few examples of equations in the slope-intercept form: y = 3x + Don’t forget: In the slope-intercept form of the line equation, m and b are numbers, and x and y are variables. Guided Practice In Exercises 1–6, decide whether each equation is in slope-intercept form or not. 1. y = 3x + 7 3. y – 3 = 2(x – 4) 2. 3x + 4y = 7 4. y – 8 = 3(x – 4) 5. y = 3 2 x + 18 6. y = –4x – 1 t Easy to Plot Graaaaaphsphsphsphsphs t Easy to Plot Gr m Makes es es es es IIIIIt Easy to Plot Gr t Easy to Plot Gr m Mak he Slope-Intercececececept Fpt Fpt Fpt Fpt Forororororm Mak m Mak he Slope-Inter TTTTThe Slope-Inter he Slope-Inter t Easy to Plot Gr m Mak he Slope-Inter The slope-intercept form of an equation is really useful for plotting graphs because m is the slope of the line and b is the y-coordinate of the y-intercept. y = mx + b slope y-coordinate of y-intercept Section 4.4 — More Lines Section 4.4 Section 4.4 Section 4.4 Section 4.4 205205205205205 Example Example Example Example Example 11111 Plot the graph of y = 1 2 x + 2. Solution Solution Solution Solution Solution y = mx + b = 1 2 x + 2 Slope = m = 1 2 — for a slope of 1 2 , go up 1 unit for every 2 units across. y-coordinate of the y-intercept = b = 2 — so the y-intercept is (0, 2). Example Example Example Example Example 22222 Plot the graph of y = –3x – 4. Solution Solution Solution Solution Solution y = mx + b = –3x – 4 Slope = m = –3 — for a slope of –3, go down 3 units for every unit across. y-coordinate of the y-intercept = b = –4 — so the y-intercept is (0, –4). y 1 x6 -5 -4 -3 -2 6 5 4 3 2 1 -1 0 -1 -2 -3 -4 -5 -6 y x = –3 – 4 -6 -5 -4 -3 -2 6 5 4 3 2 1 -1 0 -1 -2 -3 -4 -5 -6 1 2 3 4 5 6 Guided Practice 7. In the equation y = 8x + 5, find the slope. 8. In the equation y = –x + 10, find the slope. 9. In the equation y = 2x + 5, find the y-intercept. 10. In the equation y = 7b – 3, find the y-intercept. In Exercises 11–18, plot each equation on a graph. 11. y = 2x + 3 12. y = x – 6 13. y = –7x – 8 15. y = – 1 2 x + 6 17. y = 3 4 x 14. y = – 1 3 x – 4 16. y = 1 5 x – 3 18. y = 6 Don’t forget: You can write y = –3x – 4 as y = –3x + (–4), so the y-intercept is –4. 206206206206206 Section 4.4 — More Lines Section 4.4 Section 4.4 Section 4.4 Section 4.4 lope-IIIIInternternternterntercececececept pt pt pt pt FFFFFororororormmmmm lopeet the SSSSSlopelopeet the to GGGGGet the et the to SolvSolvSolvSolvSolve fe fe fe fe for or or or or y to to lopeet the to OK, so you know the slope-intercept form of an equation makes drawing graphs a lot easier. The trouble is, you’ll often be given an equation which isn’t in slope-intercept form. To get the equation into slope-intercept form, solve for y. Example Example Example Example Example 33333 A line has the equation Ax + By = C, where B π 0. Solve this equation for y, justifying each step. Solution Solution Solution Solution Solution Ax + By = C Ax – Ax + By = –Ax + C By = –Ax + C By B Ax = − + en equationtiontiontiontion en equa en equa GiGiGiGiGivvvvven equa en equa equality equality ty of ty of oper oper action pr action pr Subtr Subtr equality ty of equality operty of action proper Subtraction pr equality ty of oper action pr Subtr Subtr equality equality ty of ty of oper oper vision pr vision pr DiDiDiDiDivision pr equality ty of equality operty of vision proper equality ty of oper vision pr The equation is now in slope-intercept form. The slope, m = – the y-coordinate of the y-intercept, b = – C B . A B and Example Example Example Example Example 44444 Determine the slope and y-intercept of the line 2x – 3y = 9. Solution Solution Solution Solution Solution Step 1: Solve the given equation for y. 2x – 3y = 9 –3y = –2x + Now you’ve got the equation in slope-intercept form, y = mx + b. Step 2: Get the slope and y-intercept from the equation. The slope, m = 2 3 . The y-coordinate of the y-intercept, b = –3. So, the y-intercept = (0, –3). Section 4.4 — More Lines Section 4.4 Section 4.4 Section 4.4 Section 4.4 207207207207207 Guided Practice In Exercises 19-24, find the slope and y-intercept of the line. 19. 3x + 3y = 9 20. 2y – 6x = 10 21. 2x – 2y = 5 22. –7y + 5x = 14 23. –2y + 3x = 8 24. –5x + 4y = 12 Write down the equations of the following lines in slope-intercept form. 25. The line with slope 4 that passes through the point (0, 2). 26. The line with slope 2 that passes through the point (0, –6). 27. The line with slope –3 that passes through the point (0, 1). 28. The line with slope – 6 7 that passes through the point (0, –3). Independent Practice In Exercises 1–10, find the slope and y-intercept of each equation that’s given. x – 5 1. y = 1 2 3. y = 3x + 6 5. y = 2(x + 2) 7. 7 – y = 5(x + 4) 9. 3x + 4y = 8 x + 1 2. y = 2 3 4. –6y = 3x + 12 6. y – 4 = 2(x + 1) 8. y – 3 = 2(x – 9) 10. 2x + 3y = 9 In Exercises 11–15, plot the graph of the given equation. 11. y = 1 3 x + 5 13. y = x + 2 15. y = 2x 12. y = – 1 3 x – 6 14. y = –x + 2 In Exercises 16–20, write the equations of the lines in slope-intercept form. 4 3 1 2 16. A line with slope 17. A line with slope 18. 4x + 2y = 8 20. 3x – 4y = –16 that passes through the point (0, 4) that passes through the point (0, –2) 19. 6x – 3y = 15 In Exercises 21–22, write the equations of the lines in slope-intercept form. 21. The line with slope 0 that passes through the point (2, 6) 22. The line with slope 2 that passes through the point (6, 3) 208208208208208 Section 4.4 — More Lines Section 4.4 Section 4.4 Section 4.4 Section 4.4 TTTTTopicopicopicopicopic 4.4.44.4.4 4.4.44.4.4 4.4.4 California Standards: 8.0:8.0:8.0:8.0:8.0: Students under stand stand Students under Students under stand Students understand stand Students under allel allelallel par par pts of pts of the conce the conce allel parallel pts of par the concepts of par the conce pts of the conce pendicular pendicular lines and per lines and per pendicular lines and perpendicular lines and per pendicular lines and per lines and how their slopes lines and how their slopes lines and how their slopes lines and how their slopes lines and how their slopes Students areeeee Students ar Students ar ted. ted. ararararare re re re re relaelaelaelaelated. ted. Students ar Students ar ted. aaaaabbbbble to find the equa tion of tion of le to find the equa le to find the equa tion of le to find the equation of tion of le to find the equa pendicular to a pendicular to a a line per a line per a line perpendicular to a pendicular to a pendicular t
o a a line per a line per gigigigigivvvvven line tha t passes t passes en line tha en line tha en line that passes t passes en line tha t passes en point. en point. ough a givvvvven point. ough a gi ough a gi thrthrough a gi thrthr en point. en point. ough a gi thr What it means for you: You’ll learn how to tell whether lines are parallel or perpendicular by looking at the slope-intercept form. Key words: parallel perpendicular reciprocal Don’t forget: See Topic 4.3.1 for more on the slope of a line. Don’t forget: Use the point-slope formula once you have all the information you need. About Slopes About Slopes MorMorMorMorMore e e e e About Slopes About Slopes About Slopes MorMorMorMorMore e e e e About Slopes About Slopes About Slopes About Slopes About Slopes This Topic carries on from the material on parallel and perpendicular lines that you learned earlier in this Section. Lines are Pe Pe Pe Pe Parararararallel allel allelallel Lines ar Lines ar ou if alues of m TTTTTell ell ell ell ell YYYYYou if ou if alues of VVVVValues of alues of allel ou if Lines ar Lines ar ou if alues of Parallel lines all have the same slope, so the slope-intercept forms of their equations all have the same value of m. For example, the lines y = 3x + 2, y = 3x – 1 and y = 3x – 6 are all parallel. Example Example Example Example Example 11111 Find the equation of the line through (4, –4) that is parallel to the line 2x – 3y = 6. Solution Solution Solution Solution Solution Step 1: Write 2x – 3y = 6 in the slope-intercept form — that is, solve the equation for y. 2x – 3y = 6 ﬁ –3y = –2x + 6 ﬁ y = 2 3 x – 2 Step 2: Get the slope from the equation. The slope of the line y = 2 3 x – 2 is 2 3 . Since the required line through (4, –4) is parallel to the line y = 2 3 x – 2, its slope is also 2 3 . Step 3: Now write the equation of the line through (4, –4) with a slope of 2 3 . y – y1 = m(x – x1) ﬁ y – (–4) = 2 3 (x – 4) ﬁ 3(y + 4) = 2(x – 4) ﬁ 3y + 12 = 2x – 8 ﬁ 33333y – 2x = –20 Guided Practice 1. Give an example of a line that is parallel to y = 1 2 x + 1. 2. Is the line y = 4 5 x – 2 parallel to the line y = 4 5 x + 6? Explain. 3. Find the equation of the line through (–4, 3) that is parallel to the line y = 3x + 9. 4. Find the equation of the line through (3, 8) that is parallel to the line 3x + y = 1. Section 4.4 — More Lines Section 4.4 Section 4.4 Section 4.4 Section 4.4 209209209209209 Don’t forget: See Topic 4.4.2 for a reminder on perpendicular lines. Lines are Pe Pe Pe Pe Perererererpendicular pendicular pendicular Lines ar Lines ar ou if Also TTTTTell ell ell ell ell YYYYYou if ou if Also alues of m Also Also alues of VVVVValues of alues of pendicular ou if Lines ar pendicular Lines ar ou if Also alues of The slope-intercept forms of equations of perpendicular lines have values of m that are negative reciprocals of each other. For example, the lines y = 3x + 2 and y = – because the negative reciprocal of 3 is – 1 3 . 1 3 x – 1 must be perpendicular, Example Example Example Example Example 22222 Find the equation of the line through (2, –4) that is perpendicular to the line –3y – x = 5. Solution Solution Solution Solution Solution Step 1: Write –3y – x = 5 in the slope-intercept form (that is, solve the equation for y). –3y – x = 5 ﬁ –3y = Step 2: Get the slope (m1) of the line y = – 3 slope (m2) of the required line through (2, –4). x – 5 3 and determine the Since m1 = – 1 3 , and m2 is the negative reciprocal of – 1 3 , m2 must be 3. Step 3: Write the equation of the line through (2, –4) with a slope of 3. Use the point-slope formula here: y – y1 = m(x – x1) ﬁ y – (–4) = 3(x – 2) ﬁ y + 4 = 3x – 6 ﬁ y – 3x = –10 Guided Practice 5. Give an example of a line that’s perpendicular to the line y = 6x. 6. Is the line y = 4x + 2 perpendicular to the line y = – 1 4 x + 4? Explain your answer. 7. Find the equation of the line through (–2, 0) that is perpendicular to the line y = –2x – 4. 8. Find the equation of the line through (–4, 6) that is perpendicular to the line 3x – 4y = 24. 210210210210210 Section 4.4 — More Lines Section 4.4 Section 4.4 Section 4.4 Section 4.4 Independent Practice In Exercises 1–8, determine whether the pairs of lines are parallel, perpendicular, or collinear. Check it out: “Collinear” means “on the same straight line.” If two lines have the same slope, and pass through the same point, they must be collinear. 1. y = 2x + 1 and y = – 1 2 x – 6 2. y = 1 3 x + 5 and y = –3x – 4 3. y = 4x – 8 and y = – 1 4 x + 2 4. y = 6 and y = 3 5. x = 2 and y = –4 6. 5x – 2y = –10 and 10x – 4y = –20 7. 3x + y = 6 and 6x + 2y = –4 8. 2x – y = –4 and 6x – 3y = –12 In Exercises 9–20, find the equations of the lines. 9. The line through (5, 2) that’s parallel to a line with slope 1 2 . 10. The line through (3, –3) that’s parallel to a line with slope 2 5 . 11. The line through (2, –9) that’s perpendicular to a line with slope –3. 12. The line through (–3, 1) that’s perpendicular to a line with slope 5 8 . 13. The line through (0, 0) that’s parallel to 3x + y = 18. 14. The line through (3, 5) that’s parallel to 3x – 7y = –21. 15. The line through (4, –3) that’s parallel to 3x – 4y = 16. 16. The line through (–2, 6) that’s parallel to 6x – 10y = –20. 17. The line through (0, 6) that’s perpendicular to 2x + y = 18. 18. The line through (–3, –5) that’s perpendicular to 3x – 6y = –24. 19. The line through (6, –2) that’s perpendicular to 3x – 5y = –10. 20. The line through (8, 2) that’s perpendicular to the line joining the points (–6, 3) and (–2, 6). 21. Use slopes to decide whether the points (–3, –8), (3, –2), and (8, 3) are collinear (on the same line) or noncollinear. 22. Use slopes to decide whether the points (4, 5), (3, –2), and (8, 3) are collinear or noncollinear. 23. Use slopes to decide whether the points B (2, 10), K (–1, 3), and J (5, –3) are vertices of a right triangle. 24. Show that the points M (3, 11), A (–4, 4), and T (3, –3) are vertices of a right triangle. ound Up ound Up RRRRRound Up ound Up ound Up ound Up ound Up RRRRRound Up ound Up ound Up Hopefully you’ll see now why the slope-intercept form of a line is so useful — you can just glance at the equations to see whether lines are parallel or perpendicular, without having to plot the graphs. Section 4.4 — More Lines Section 4.4 Section 4.4 Section 4.4 Section 4.4 211211211211211 Section 4.5 gions Defined gions Defined RRRRReeeeegions Defined gions Defined gions Defined RRRRReeeeegions Defined gions Defined gions Defined gions Defined gions Defined by Inequalities by Inequalities by Inequalities by Inequalities by Inequalities by Inequalities by Inequalities by Inequalities by Inequalities by Inequalities Just like with equations, you can graph inequalities on the coordinate plane. The only tricky bit is showing whether the solution set is above or below the line. This Topic will show you how. vides the Plane into TTTTThrhrhrhrhree Ree Ree Ree Ree Reeeeegions gions gions vides the Plane into vides the Plane into A Line Di A Line Di gions A Line Divides the Plane into gions vides the Plane into A Line Di A Line Di The graph of a linear equation divides the plane into three regions: The set of points that lie on the line. The set of points that lie above the line. The set of points that lie below the line. The regions above and below the line of a linear equation are each represented by a linear inequality. Example Example Example Example Example 11111 Graph y = –x + 3 and show that it divides the plane into three regions. Solution Solution Solution Solution Solution Set of points bbbbbeloeloeloeloelowwwww the line — all these points satisfy the inequality . –6 –5 –4 –3 –2 6 5 4 3 2 1 0 –1 –1 –2 –3 –4 –5 –6 y-axis Set of points abababababooooovvvvveeeee the line — all these points satisfy the inequality . 1 2 3 4 5 6 x-axis Set of points on on on on on the line — all these points satisfy the equation y = –x + 3 y = –x + 3 y = –x + 3. The line y = –x + 3 y = –x + 3 forms a bbbbborororororder der der der der between the other two regions. The points on the line don’t satisfy either of the inequalities — that’s why the line is dashed. TTTTTopicopicopicopicopic 4.5.14.5.1 4.5.14.5.1 4.5.1 California Standards: 6.0:6.0:6.0:6.0:6.0: Students graph a linear equation and compute the xand y-intercepts (e.g., graph 2x + 6y = 4)..... TTTTThehehehehey ary ary ary ary are also e also e also e also e also h the reeeeegiongiongiongiongion h the r le to sketcetcetcetcetch the r h the r le to sk aaaaabbbbble to sk le to sk h the r le to sk defined by linear inequality defined by linear inequality defined by linear inequality defined by linear inequality defined by linear inequality (e(e(e(e(e.g.g.g.g.g.,.,.,.,., the the the the they sky sky sky sky sketcetcetcetcetch the r h the reeeeegiongiongiongiongion h the r h the r h the r < 4). < 4). defined by 2x + 6 + 6 + 6 + 6 + 6y < 4). defined by 2 defined by 2 < 4). < 4). defined by 2 defined by 2 What it means for you: You’ll learn to sketch the region defined by a linear inequality. Key words: inequality plane region Check it out: You’ll find more about the dashed line in Topic 4.5.2. 212212212212212 Section 4.5 Section 4.5 Section 4.5 — Inequalities Section 4.5 Section 4.5 Identifying the Reeeeegiongiongiongiongion Identifying the R Identifying the R Identifying the R Identifying the R To identify the inequality defining a region, choose a point from the region and substitute its coordinates into the equation of the line that borders the region. Since the point doesn’t lie on the line, the equation won’t be a true statement. To make the statement true, you need to replace the “=” with a “<” or “>” sign. The resulting inequality defines the region containing the point. Example Example Example Example Example 22222 State the inequality that defines the shaded region. y-axis Solution Solution Solution Solution Solution Choose a point in the shaded region, for example, (0, 2). Test this point in the equation of the line: y = 2x + 1 –6
–5 –4 –3 –2 So at (0, 2), you get 2 = 2(0) + 1, that is, 2 = 1, which is a false statement. 0 –1 –1 –2 –3 –4 –5 –6 1 2 3 4 5 6 x-axis Since 2 > 1, a “>” sign is needed to make it a true statement. So (0, 2) satisfies the inequality y > 2x + 1. Therefore the inequality that defines the shaded region is y > 2x + 1. Guided Practice In Exercises 1–2, state the inequality that defines the shaded region on each of the graphs. 1. –6 6 5 4 3 2 1 0 –1 –1 –2 –3 –4 –5 –6 –5 –4 –2 –3 2 + x 2 = y 26 –5 –4 –3 –2 1 2 3 4 5 6 0 –1 –1 –2 –3 –4 –5 –6 Section 4.5 Section 4.5 Section 4.5 — Inequalities Section 4.5 Section 4.5 213213213213213 y a Linear Inequality y a Linear Inequality gion Defined b hing the Reeeeegion Defined b gion Defined b hing the R SkSkSkSkSketcetcetcetcetching the R hing the R y a Linear Inequality gion Defined by a Linear Inequality y a Linear Inequality gion Defined b hing the R An ordered pair (x, y) is a solution of a linear inequality if its x and y values satisfy the inequality. The graph of a linear inequality is the region consisting of all the solutions of the inequality (the solution set). To sketch the region defined by a linear inequality, you need to plot the graph of the corresponding equation (the border line), then shade the correct region. To decide which is the correct region, just test a point. Example Example Example Example Example 33333 Sketch the region defined by 6x – 3y < 9. Solution Solution Solution Solution Solution First plot the graph of the corresponding equation. This is the border line. Rearrange the equation into the form y = mx + b: 6x – 3y = 9 ﬁ 3y = 6x – 9 ﬁ y = 2x – 3 Table of values for sketching the line: x 0 3 y ( y,x ) y 2= x 3– 3–)0(2= 3–= )3–,0( y 2= x 3– 3–)3(2= 3= )3,3( So the border line goes through the points (0, –3) and (3, 3), as shown in the graph below. Now test whether the point (0, 0) satisfies the inequality. Substitute x = 0 and y = 0 into the inequality. 6x – 3y < 9 0 – 0 < 9 0 < 9 — This is a true statement. Therefore (0, 0) lies in the region 6x – 3y < 9 — so shade the region containing (0, 0). y 6x – 6 –5 –4 –3 –2 0 –1 –1 –2 –3 –4 –5 –6 y-axis 1 2 3 4 5 6 x-axis Check it out: Use whichever method you find easiest for sketching the line — see Section 4.2. For example, you could plot the intercepts if you prefer. Check it out: (0, 0) has been used as a test point to make the algebra easy, but you can use any point — as long as it doesn’t lie on the line. 214214214214214 Section 4.5 Section 4.5 Section 4.5 — Inequalities Section 4.5 Section 4.5 Example Example Example Example Example 44444 Graph the solution set of y + 2x > 4. Solution Solution Solution Solution Solution First plot the graph of the corresponding equation (the border line). Table of values for sketching the line: x 0 3 y ( y,x ) y 2–= x 4+ 4+)0(2–= 4= )4,0( y 2–= x 4+ 4+)3(2–= 2–= )2–,3( So the border line goes through the points (0, 4) and (3, –2), as shown below. Test whether the point (0, 0) satisfies the inequality. Substitute x = 0 and y = 0 into the inequality. y + 2x > 4 0 + 0 > 4 0 > 4 — This is a false statement. Therefore (0, 0) does not lie in the region y + 2x > 4 — so shade the region that does not contain (0, 0). y-axis y x-axis 6 5 4 3 2 1 0 –1 –1 –2 –3 –4 –5 –6 –6 –5 –4 –3 –2 Guided Practice In each of Exercises 3–8, use a set of axes spanning from –6 to 6 on the x- and y-axes. For each Exercise, shade the region defined by the inequality. 3. y > 0.5x + 2 5. y + x > –2 7. –2y + 3x > 6 4. y + 2x < 0 6. 4x + 3y < 12 8. y < –x + 3 In Exercises 9–14, show whether the given point is a solution of –5x + 2y > –8. 9. (0, 0) 12. (2, 1) 11. (–3, 9) 14. (–15, 13) 10. (6, –3) 13. (39, –36) Section 4.5 Section 4.5 Section 4.5 — Inequalities Section 4.5 Section 4.5 215215215215215 Independent Practice In Exercises 1–4, use the graph opposite to determine if the given point is in the solution set. 1. (0, 0) 2. (1, 2) 3. (3, –2) 4. (–2, –2) y-axis 6 –5 –4 –3 –2 0 –1 –1 –2 –3 –4 –5 –6 x-axis In Exercises 5–8, state the inequality that defines the shaded region on each of the graphs. 2x + y = 4 y-axis 6 –5 –4 –3 –2 0 –1 –1 –2 –3 –4 –5 –6 x-axis 5. 7. y = 1 –6 –5 –4 –3 –2 y-axis 1 –1 –2 –3 –4 –5 –6 2 + 1 x 2 = y –6 –5 –4 –3 –2 y-axis 1 –1 –2 –3 –4 –5 –6 x-axis 6. 8. y-axis 2 0 –1 –1 –2 –3 –4 –5 –6 x-axis x-axis –6 –5 –4 –3 –2 9. Show whether (–2, –1) is a solution of 2x – 5y £ 10. 10. Show whether (–3, 5) is a solution of 4x + y < 5. 11. Show whether (2, 4) is a solution of x + 6y < 0. 12. Show whether (0, –3) is a solution of y ≥ –6x + 5. Graph the solution set in Exercises 13–22. 13. y < 3 4 x + 6 15. y < – 2 5 x – 2 17. x > 0 19. x + 2y > 8 21. 4x – 6y > 24 14. y < 4 5 x + 4 16. y < 1 18. x – 4y > 8 20. 4x + 3y < –12 22. 5x + 8y < 24 ound Up ound Up RRRRRound Up ound Up ound Up RRRRRound Up ound Up ound Up ound Up ound Up Well, that was quite a long Topic, with lots of graphs. Inequality graphs aren’t easy, so it’s always a good idea to check whether you’ve shaded the correct part by seeing whether a test point satisfies the original inequality. 216216216216216 Section 4.5 Section 4.5 Section 4.5 — Inequalities Section 4.5 Section 4.5 TTTTTopicopicopicopicopic 4.5.24.5.2 4.5.24.5.2 4.5.2 California Standards: 6.0:6.0:6.0:6.0:6.0: Students graph a linear equation and compute the xand y-intercepts (e.g., graph 2x + 6y = 4)..... TTTTThehehehehey ary ary ary ary are also e also e also e also e also h the reeeeegiongiongiongiongion h the r le to sketcetcetcetcetch the r h the r le to sk aaaaabbbbble to sk le to sk h the r le to sk defined by linear inequality defined by linear inequality defined by linear inequality defined by linear inequality defined by linear inequality (e(e(e(e(e.g.g.g.g.g.,.,.,.,., the the the the they sky sky sky sky sketcetcetcetcetch the r h the reeeeegiongiongiongiongion h the r h the r h the r < 4). < 4). defined by 2x + 6 + 6 + 6 + 6 + 6y < 4). defined by 2 defined by 2 < 4). < 4). defined by 2 defined by 2 What it means for you: You’ll learn how to show the different types of inequality on a graph. Key words: inequality strict inequality border region Don’t forget: Remember, £ means “less than or equal to,” ≥ means “greater than or equal to.” gions gions s of R R R R Reeeeegions s of s of BorBorBorBorBorderderderderders of gions gions s of s of R R R R Reeeeegions BorBorBorBorBorderderderderders of gions gions s of s of gions s of gions In Topic 4.5.1 you were dealing with regions defined by strict inequalities — the ones involving a < or > sign. This Topic shows you how to graph inequalities involving £ and ≥ signs too. Borderderderderdersssss Bor Bor ypes of e Difffffferererererent ent ent ent ent TTTTTypes of ypes of e Dif gions Can Havvvvve Dif e Dif gions Can Ha RRRRReeeeegions Can Ha gions Can Ha ypes of Bor Bor ypes of e Dif gions Can Ha The region defined by a strict inequality doesn’t include points on the border line, and you draw the border line as a dashed line. For example, the region defined by y > –x + 3 doesn’t include any points on the line y = –x + 3. Regions defined by inequalities involving a £££££ or ≥≥≥≥≥ sign do include points on the border line. In this case, you draw the border line as a solid line. For example, the region defined by y ≥ –x + 3 includes all the points on the line y = –x + 3. Example Example Example Example Example 11111 Graph 2x – y = –2 and show the three regions of the plane that include all the points on this line. Solution Solution Solution Solution Solution e and e and Set of points abababababooooovvvvve and e and e and on on on on on the line — all these points satisfy the inequality 22222x – – – – – y £££££ –2 –2 –2 –2 –2. Set of points on on on on on the line — all these points satisfy the equation = –2 22222x – – – – – y = –2 = –2 = –2 and both = –2 the inequalities. 22222x – – – – – y £££££ –2 –2 –2 –2 –2 –6 –5 –4 –3 –2 y-axis 22222x – – – – – y ≥≥≥≥≥ –2 –2 –2 –2 –2 0 –1 –1 –2 –3 –4 –5 –6 Set of points bbbbbeloeloeloeloelowwwww and on and on and on the line — all and on and on these points satisfy the inequality 22222x – – – – – y ≥≥≥≥≥ –2 –2 –2 –2 –2. x-axis Section 4.5 Section 4.5 Section 4.5 — Inequalities Section 4.5 Section 4.5 217217217217217 der Line der Line the Bor lusivvvvve ofe ofe ofe ofe of the Bor the Bor lusi lusi gions Inc hing Reeeeegions Inc gions Inc hing R SkSkSkSkSketcetcetcetcetching R hing R der Line the Border Line gions Inclusi der Line the Bor lusi gions Inc hing R The method for sketching the region is the same as the method in Topic 4.5.1, except now there’s an extra step for showing the border type — using a dashed line, or a solid line. Example Example Example Example Example 22222 Sketch the region of the coordinate plane defined by y £ –x – 5. Solution Solution Solution Solution Solution First plot the border line. The border line equation is y = –x – 5. Table of values for sketching the line: x 0 y ( y,x ) y –= x 5– 5–0–= 5–= )5–,0( 3– y –= x 5– 5–3= 2–= )2–,3–( So the border line goes through the points (0, –5) and (–3, –2), as shown in the graph below. Identify the border type: the border line is solid, since the sign is £££££. Test whether the point (0, 0) satisfies the inequality — substitute x = 0 and y = 0 into the inequality. y £ –x – 5 ﬁ 0 £ –0 – 5 0 £ –5 — this is a false statement. y-axis Therefore (0, 0) doesn’t lie in the region y £ –x – 5 — so shade the region that doesn’t contain (0, 0). 1 2 3 4 5 6 x-axis –6 –5 –4 –3 –2 y £ x– – 5 4 3 2 1 0 –1 –1 –2 –3 –4 –5 –6 –7 –8 Guided Practice In Exercises 1–4, show whether the given point is in the solution set of –2x + 3y £ –15. 1. (0, 0) 4. (–3, –5) 3. (4, –7) 2. (6, –1) In each of Exercises 5–8 use a set of axes spanning from –6 to 6 on the x- and y-axes, and shade the region defined by the inequality. 5. 4x + 3y £ 9 7. –2y ≥ x 6. y £ 2x + 3 8. 2x + y ≥ 4 218218218218218 Section 4.5 Section 4.5 Section 4.5 — Inequalities Section 4.5 Section 4.5 phing Reeeeegions gions gions phing R Graaaaaphing R phing R Gr Gr e Examp
les of MorMorMorMorMore Examples of e Examples of gions e Examples of Gr gions phing R Gr e Examples of Graphing regions isn’t always straightforward, so here are a couple more examples and some more practice exercises. Example Example Example Example Example 33333 Graph the solution set of 4y – 3x ≥ 12. Solution Solution Solution Solution Solution First form the border-line equation: 4y – 3x = 12 4y = 3x + 12 y = 3 4 x + 3 Table of values for sketching the line+ x 3,x ) 3+)0( 3= )3,0( 3+)4( 6= )6,4( So the border line goes through the points (0, 3) and (4, 6). The border line is solid, since the sign is ≥≥≥≥≥. Test whether the point (0, 0) satisfies the inequality. Substitute x = 0 and y = 0 into the inequality. 4y – 3x ≥ 12 4(0) – 3(0) ≥ 12 0 ≥ 12 — This is a false statement. y-axis Therefore (0, 0) doesn’t lie in the region 4y – 3x ≥ 12 — so shade the region that doesn’t contain (0, 0). 4 – 3 y x ≥ 12 8 7 6 5 4 3 2 1 –6 –5 –4 –3 –2 0 –1 –1 –2 –3 1 2 3 4 5 6 x-axis Section 4.5 Section 4.5 Section 4.5 — Inequalities Section 4.5 Section 4.5 219219219219219 Check it out: After you’ve sketched a region, it can be a good idea to test another point to make sure you’ve got it right. Pick a point in your shaded region and make sure the inequality is satisfied there as well. Example Example Example Example Example 44444 Sketch the region of the coordinate plane defined by 2y < 2x + 6. Solution Solution Solution Solution Solution First form the border-line equation: 2y = 2x + 6 y = x + 3 Table of values for sketching the line: x 0 3 y y = x 3+ 3+0= 3= y = x 3+ 3+3= 6= y,x ) ( )3,0( )6,3( So the border line goes through the points (0, 3) and (3, 6). The border line is dashed, since the sign is <. Test whether the point (0, 0) satisfies the inequality. Substitute x = 0 and y = 0 into the inequality. 2y < 2x + 6 2(0) < 2(0) + 6 0 < 6 — This is a true statement. Therefore (0, 0) lies in the region 2y < 2x + 6 — so shade the region containing (0, 0). –6 –5 –4 –3 –2 y-axis -axis 6 5 4 3 2 1 0 –1 –1 –2 –3 –4 –5 –6 Guided Practice In each of Exercises 9–18, use a set of axes spanning from –6 to 6 on the x- and y-axes. For each exercise, shade the region defined by the inequality. 9. y £ x 11. x + 2 > 0 13. y + 3 £ 0 15. y £ 2x – 5 17. y > 2x + 6 10. y – 2 < 0 12. x – 4 £ 0 14. y ≥ –3x 16. x + 4y < 4 18. 4x – 3y ≥ 8 19. Show whether (5, 4) is in the solution set of 4x – 3y £ 8. 20. Show whether (–4, 2) is in the solution set of 2x + y > –6. 220220220220220 Section 4.5 Section 4.5 Section 4.5 — Inequalities Section 4.5 Section 4.5 Independent Practice In Exercises 1–4, show whether the given point is in the solution set of the given line. 1. (–1, 3); 2x + y < 1 3. (0.5, 0.25); x + 2y £ 1 2. (–2, –8); x – y ≥ 6 4. (1, –5); x + 2y > –9 In Exercises 5–6, determine if the given point is in the solution set shown by the shaded region of the graph. 5. (4, 6) 6. (–1, 1) y-axis 6 5 4 3 2 1 y-axis 6 –5 –4 –3 –2 x-axis 1 2 3 4 5 6 0 –1 –1 –2 –3 –4 –5 –6 In Exercises 7–8, determine if the given point is in the solution set shown by the shaded region of the graph. 7. (3, –2) 8. (–2, 3) x-axis 1 2 3 4 5 6 y = –2 –6 –5 –4 –3 –2 0 –1 –1 –2 –3 –4 –5 –6 Graph the solution set in Exercises 9–20. 9. x > 4 11. x – 3y £ 9 13. x + y ≥ –1 15. x + 2y < 2 17. y ≥ 3x 19. x + y ≥ 3 10. y £ –2 12. x + y < 5 14. y ≥ 2x – 7 16. y £ 2x + 2 18. y > –2x + 1 20. 2x + y £ –3 ound Up ound Up RRRRRound Up ound Up ound Up ound Up ound Up RRRRRound Up ound Up ound Up Remember — graphs of inequalities including < and > signs will always have a dashed line, and graphs including £ and ≥ signs will always have a solid line. Section 4.5 Section 4.5 Section 4.5 — Inequalities Section 4.5 Section 4.5 221221221221221 TTTTTopicopicopicopicopic 4.5.34.5.3 4.5.34.5.3 4.5.3 California Standards: 9.0:9.0:9.0:9.0:9.0: Students solve a system of two linear equations in two variables algebraically and are able to interpret the Students Students Students answer graphically. Students Students ararararare ae ae ae ae abbbbble to solv e a system e a system le to solv le to solv le to solve a system e a system e a system le to solv ofofofofof tw tw tw tw two linear inequalities in o linear inequalities in o linear inequalities in o linear inequalities in o linear inequalities in twtwtwtwtwo vo vo vo vo variaariaariaariaariabbbbbles and to sk les and to sketcetcetcetcetchhhhh les and to sk les and to sk les and to sk the solution sets..... the solution sets the solution sets the solution sets the solution sets What it means for you: You’ll graph two inequalities to show the solution set that satisfies both inequalities. Key words: system of linear inequalities region point-slope formula gions Defined gions Defined RRRRReeeeegions Defined gions Defined gions Defined gions Defined gions Defined RRRRReeeeegions Defined gions Defined gions Defined o Inequalities o Inequalities bbbbby y y y y TTTTTwwwwwo Inequalities o Inequalities o Inequalities bbbbby y y y y TTTTTwwwwwo Inequalities o Inequalities o Inequalities o Inequalities o Inequalities The idea of graphing two different inequalities on one graph is really not as hard as it sounds. When you’ve finished, your graph will show the region where all the points satisfy both inequalities. y More e e e e TTTTThan One Linear Inequality han One Linear Inequality han One Linear Inequality y Mor y Mor gions Defined b R R R R Reeeeegions Defined b gions Defined b han One Linear Inequality gions Defined by Mor han One Linear Inequality y Mor gions Defined b A system of linear inequalities is made up of two or more linear inequalities that contain the same variables. For example, 3x + 2y > 6 and 4x – y < 5 are linear inequalities both containing the variables x and y. An ordered pair (x, y) is a solution of a system of linear inequalities if it is a solution of each of the inequalities in the system. For example, (1, –2) is a solution of the system of inequalities y < –x + 2 and 2y < 2x + 6. The graph of two linear inequalities is the region consisting of all the points satisfying both inequalities. Example Example Example Example Example 11111 Sketch the region satisfying both y < –x + 2 and 2y < 2x + 6. y-axis Line 2y = 2x + 6 1 2 3 4 5 6 x-axis x< – + 2 = region defined by y = region defined by 2 < 2 + 6 x y = region defined by < – + 2 x y and 2 < 2 + 6 x y Line y = –x + 2 Solution Solution Solution Solution Solution 6 5 4 3 2 1 0 –1 –1 –2 –3 –4 –5 –6 –6 –5 –4 –3 –2 Set of points satisfying bbbbbothothothothoth y < –x + 2 andandandandand 2y < 2x + 6. 222222222222222 Section 4.5 Section 4.5 Section 4.5 — Inequalities Section 4.5 Section 4.5 Check it out: The problem in Example 2 is just another way of asking you to sketch the region defined by 5y + 3x ≥ –25 and y – x £ –5. gions Defined by y y y y TTTTTwwwwwo Linear Inequalities o Linear Inequalities o Linear Inequalities gions Defined b hing Reeeeegions Defined b gions Defined b hing R SkSkSkSkSketcetcetcetcetching R hing R o Linear Inequalities o Linear Inequalities gions Defined b hing R To sketch the region defined by two linear inequalities, shade the regions defined by each inequality. The region defined by both inequalities is the area where your shading overlaps. Example Example Example Example Example 22222 Graph the solution set satisfying 5y + 3x ≥ –25 and y – x £ –5. Solution Solution Solution Solution Solution First line: 5y + 3x = –25 ﬁ 5y = –3x – 25 ,x ) ( y x 0 5 y –= y –= 3 5 3 5 x 5– x 5– 5–)0(–= 3 5 5–)5(–= 3 5 5–= )5–,0( 8–= )8–,5( The border line y = – 3 5 and is a solid line. x – 5 goes through the points (0, –5) and (5, –8), Second line: y – x = –,x ) y = x 5– 5–0= 5–= )5–,0( y = x 5– 5–3= 2–= )2–,3( The border line y = x – 5 goes through the points (0, –5) and (3, –2), and is a solid line. Test whether the point (0, 0) satisfies each inequality: 5y + 3x ≥ –25 5(0) + 3(0) ≥ –25 0 ≥ –25 This is a true statement, so (0, 0) lies in the region 5y + 3x ≥ –25. Shade the region above y = – 3 5 x – 5. y-axis Don’t forget: You need to graph both lines on the same axes. y – x £ –5 0 – 0 £ –5 0 £ –5 This is a false statement, so (0, 0) doesn’t lie in the region y – x £ –5. Shade the region below y = x – 5. The required region is the area where the shading overlaps. –3 –2 3 2 1 0 –1 –1 –2 –3 –4 –5 –6 –7 –8 –-axis 5 + 3 y –25 x ≥ and x £ – –5 y = –3 5x – 5 Section 4.5 Section 4.5 Section 4.5 — Inequalities Section 4.5 Section 4.5 223223223223223 Guided Practice In each of Exercises 1–2, use a set of axes spanning from –5 to 7 on the x-axis and –6 to 6 on the y-axis. For each exercise, shade the region containing all solution points for both inequalities. 1. 2x + 3y < 6 and y – 2x < 2 2. y – x ≥ 4 and 2x + y £ 5 In each of Exercises 3–6, use a set of axes spanning from –6 to 6 on the x- and y-axes. For each exercise, shade the region containing all solution points for both inequalities. 3. y < –x + 4 and y < x 5. y < x and y < –x – 2 4. y < x + 2 and y > –2x + 5 6. y £ 0.5x + 3 and y £ 2x + 2 Identifying the Inequalities Defining a Reeeeegiongiongiongiongion Identifying the Inequalities Defining a R Identifying the Inequalities Defining a R Identifying the Inequalities Defining a R Identifying the Inequalities Defining a R To identify the inequalities defining a region, you first need to establish the border line equations. Then examine a point in the region to identify the inequalities. Here’s the method: Find the equations — use the point-slope formula to find the equations of each of the border lines. Identify the signs — choose a point in the region (but not on a line) and substitute its coordinates into each equation. Since the point does not lie on either of the lines, you will have two false statements. Replace the “=” in each equation with a “<” or “>” sign to make the statements true. If the line is solid, use “£££££” or “≥≥≥≥≥.” Write the inequalities which define the region. Example Example Example Example Example 33333 Find the inequalities whose simultaneous solution set is
the shaded region shown on the right. Solution Solution Solution Solution Solution First line: Two points on this line are (0, 1) and (3, 2). –4 –3 –2 m1 = − 2 1 − 3 0 = 1 3 y-axis Line 2 Line 1 x-axis 1 –1 –2 –3 –4 –5 y – y1 = m(x – x1) ﬁ y – 2 = 1 3 (x – 3 224224224224224 Section 4.5 Section 4.5 Section 4.5 — Inequalities Section 4.5 Section 4.5 Example 3 continueduedueduedued Example 3 contin Example 3 contin Example 3 contin Example 3 contin Second line: Two points on this line are (3, 2) and (1, –2). m1 = − − – y1 = m(x – x1) ﬁ y – 2 = 2(x – 3) ﬁ y – 2 = 2x – 6 ﬁ y = 2x – 4 So the equations of the two border lines are y = 1 3 x + 1 and y = 2x – 4. Choose a point in the shaded region, for example, (5, 4). Substitute this point into the two equations. Equation for line 1 — this is a false statement. , so a > sign is needed to make it true. However, the line is solid so the sign should be ≥. So the first inequality is y ≥ 1 3 x + 1. Equation for line 2: y = 2x – 4 ﬁ 4 = 10 – 4 ﬁ 4 = 6 — this is a false statement. 4 < 6, so a < sign is needed to make it true. The line is dashed, so the < sign is correct. So the second inequality is y < 2x – 4. Therefore the inequalities defining the shaded region are y ≥≥≥≥≥ and y < 2x – 4. 1 3 x + 1 Guided Practice In Exercises 7–10, find inequalities whose simultaneous solution defines each of the shaded regions. 7. 6 5 4 3 2 1 0 –1 –1 –2 –3 –4 –5 –6 –2 2 –3 + –5 –0.5 x + 1 x y+3 =9 1 2 3 4 5 6 7 8. –5 –4 x + y – –2 –3 = –2 6 5 4 3 2 1 0 –1 –1 –2 –3 –4 –5 –6 Section 4.5 Section 4.5 Section 4.5 — Inequalities Section 4.5 Section 4.5 225225225225225 y-axis 10. 1 2 3 4 5 6 7 x-axis –5 –4 –3 –2 9. –5 –4 –3 –2 6 5 4 3 2 1 0 –1 –1 –2 –3 –4 –5 –6 y-axis 1 2 3 4 5 6 7 x-axis 6 5 4 3 2 1 0 –1 –1 –2 –3 –4 –5 –6 Independent Practice In Exercises 1–4, use the graph opposite to determine if the given point is included in the solution set. 1. (0, 0) 2. (2, 6) 3. (6, 2) 4. (4, –2) –6 –5 –4 –3 –2 5. Determine whether the point (–4, –3) lies in the solution region of both 3x – 4y £ 2 and x – 2y ≥ 1. y-axis 6 5 4 3 2 1 0 –1 –1 –2 –3 –4 –5 –-axis 6. Determine whether the point (0, 0) lies in the solution region of both 3x – 4y £ 2 and x – 2y ≥ 1. 7. Determine whether the point (–3, 1) lies in the solution region of both 3x – 4y £ 2 and x – 2y ≥ 1. 8. On axes spanning from –3 to 9 on the x-axis and –6 to 6 on the y-axis, graph the solution set that satisfies both the inequalities y + 2x ≥ 4 and y ≥ 1.5x – 2. 9. On axes spanning from –3 to 9 on the x-axis and –8 to 4 on the y-axis, graph the solution set that satisfies both the inequalities y > x – 6 and y > –x + 2. In each of Exercises 10–11, show on axes spanning from –5 to 7 on the x-axis and –6 to 6 on the y-axis the region defined by the set of inequalities: 10. 4y – 3x < 12, 2x + y < 0, and y ≥ 1 11. 3x + 5y > 10, x – y < 2, and y £ 2 12. The points (2, 6.6) and (7, 8.1) lie on a line bounding a region. Find the equation of the line. 13. The region in Exercise 12 is also bounded by the lines y = 0.3x and y = \|1.5x – 6\|. If the points (2, 3), (5, 1.5), and (4, 4) all lie within the region, draw and shade the region on axes spanning from –2 to 12 on the x- and y-axes. ound Up ound Up RRRRRound Up ound Up ound Up ound Up ound Up RRRRRound Up ound Up ound Up When you’re graphing a system of linear inequalities, don’t forget that you still have to pay attention to whether the lines should be solid or dashed. 226226226226226 Section 4.5 Section 4.5 Section 4.5 — Inequalities Section 4.5 Section 4.5 Chapter 4 Investigation ee Grooooowthwthwthwthwth ee Gr ee Gr TTTTTrrrrree Gr ee Gr TTTTTrrrrree Gr ee Grooooowthwthwthwthwth ee Gr ee Gr ee Gr This Chapter was all about turning raw data into graphs that show you patterns in the data. The loss of forest areas (deforestation) is a huge environmental problem. Many countries are cutting down trees at a faster rate than they can be replaced. If you look at a cross section of a tree, you’ll notice a pattern of rings. Each ring is a layer of wood that took one year to grow. 26 pine trees have been felled. The number of rings on each stump and the diameter are recorded below. diameter )ega(sgnirforebmuN 32 96 51 4 64 )mc(retemaiD 6.03 1.26 5.52 2.11 4.14 91 2.9 02 35 86 23 72 23 23 7.02 0.24 1.74 1.11 4.02 0.72 8.13 )ega(sgnirforebmuN 05 73 11 84 8 34 24 36 )mc(retemaiD 3.94 0.04 7.31 3.41 8.4 9.03 4.33 5.35 61 8.4 35 46 75 85 7.33 0.53 5.54 5.25 Draw a scatter diagram of the data. Put “Age (years)” on the x-axis and “Diameter (cm)” on the y-axis. Describe the correlation between the age of the tree and the diameter of its trunk. Why do you think that the correlation is not perfect? Draw a best-fit line for the data. A best-fit line is one that passes as close to as many of the points as possible. About half the points should be on each side of the line. Would you expect this line to pass through the origin? Why? What is the slope of the best-fit line? What does the slope of your graph represent? Find the equation of the best-fit line. a) How old would you expect a tree to be if its diameter is 23 cm? b) What would you expect the diameter of a 6-year-old tree to be? Extension It’s difficult to measure the diameter of a living tree. It’s much easier to measure the circumference of the tree instead. 1) Write an equation to link a tree’s circumference with its likely age. How old is a tree likely to be if its circumference is 157 cm? 2) What is the average annual increase in a tree’s circumference? Open-ended Extension 1) Record the number of pages and the thickness of each book (excluding the cover). (For this investigation you will need a selection of books of different types.) Produce a scatter diagram of the data and draw a best-fit line. Find the slope and equation of your line. What does the slope mean? Measure the thickness of some different books. Use your equation to predict the number of pages. How accurate are your predictions? Why do you think this is? 2) Think of two sets of data that you suspect may have a linear relationship. You should choose data sets that you can easily collect. Investigate the relationship using a scatter diagram. ound Up ound Up RRRRRound Up ound Up ound Up RRRRRound Up ound Up ound Up ound Up ound Up You can work out patterns in any collection of data, as long as the data is all about the same thing. estigaaaaationtiontiontiontion — Tree Growth estigestig estig pter 4 Invvvvvestig pter 4 In ChaChaChaChaChapter 4 In pter 4 In pter 4 In 227227227227227 Chapter 5 Systems of Equations Section 5.1 Systems of Equations ........................... 229 Section 5.2 The Elimination Method ........................ 244 Section 5.3 Applications of Systems of Equations .. 249 Investigation Breaking Even in an Egg Business ....... 261 228228228228228 TTTTTopicopicopicopicopic 5.1.15.1.1 5.1.15.1.1 5.1.1 Section 5.1 phing Method phing Method he Graaaaaphing Method he Gr he Gr TTTTThe Gr phing Method phing Method he Gr he Graaaaaphing Method TTTTThe Gr phing Method phing Method he Gr he Gr phing Method he Gr phing Method California Standards: Students solve ae ae ae ae a 9.0:9.0:9.0:9.0:9.0: Students solv Students solv Students solv Students solv system of tw tw tw tw two linear o linear o linear system of system of o linear system of o linear system of tions in two vo vo vo vo variaariaariaariaariabbbbbleslesleslesles tions in tw tions in tw equa equa equations in tw equa tions in tw equa le to le to and are ae ae ae ae abbbbble to and ar and ar algebraically and ar le to le to and ar et the answwwwwererererer et the ans interprprprprpret the ans et the ans inter inter et the ans inter inter gggggrrrrraaaaaphicall phicallyyyyy..... Students are phicall phicall phicall able to solve a system of two linear inequalities in two variables and to sketch the solution sets. What it means for you: You’ll solve systems of linear equations by graphing the lines and working out where they intersect. Key words: system of linear equations simultaneous equations In Section 4.5 you graphed two inequalities to find the region of points that satisfied both inequalities. Plotting two linear equations on a graph involves fewer steps, and it means you can show the solution to both equations graphically. tions tions Linear Equa Linear Equa Systems of Systems of tions Linear Equations Systems of Linear Equa tions Linear Equa Systems of Systems of A system of linear equations consists of two or more linear equations in the same variables. For example, 3x + 2y = 7 and x – 3y = –5 form a system of linear equations in two variables — x and y. The solution of a system of linear equations in two variables is a pair of values like x and y, or (x, y), that satisfies each of the equations in the system. For example, x = 1, y = 2 or (1, 2) is the solution of the system of equations 3x + 2y = 7 and x – 3y = –5, since it satisfies both equations. Equations in a system are often called simultaneous equations because any solution has to satisfy the equations simultaneously (at the same time). The equations can’t be solved independently of one another. y Graaaaaphing phing phing y Gr y Gr tions b tions b Equa Equa Solving Systems of Solving Systems of phing tions by Gr Equations b Solving Systems of Equa phing y Gr tions b Equa Solving Systems of Solving Systems of A system of two linear equations can be solved graphically, by graphing both equations in the same coordinate plane. Every point on the line of an equation is a solution of that equation. The point at which the two lines cross lies on both lines and so is the solution of both equations. Check it out: The point of intersection is the point where the lines cross. The solution of a system of linear equations in two variables is the point of intersection (x, y) of their graphs. Section 5.1 Section 5.1 Section 5.1 — Systems of Equations Section 5.1 Section 5.1 229229229229229 Check it out: Use whichever method you find easiest to plot the graphs. The tables of values
method’s been used in Example 1. Example Example Example Example Example 11111 Solve this system of equations by graphing: 2x – 3y = 7 –2x + y = –1 Solution Solution Solution Solution Solution Step 1: Graph both equations in the same coordinate plane. Line of first equation: 2x – 3y = 7 3y = 2x – – y 1– 3– The line goes through the points (2, –1) and (–1, –3). Line of second equation: –2x + y = –1 y = 2x – 1 x 0 1 y 1– 1 The line goes through the points (0, –1) and (1, 1). Now you can draw the graph: 6 5 4 3 2 1 –6 –5 –4 –3 –2 (–1, –3) 0 –1 –1 –2 –3 –4 –5 –6 y-axis – -axis Step 2: Read off the coordinates of the point of intersection. The point of intersection is (–1, –3). Step 3: Check whether your coordinates give true statements when they are substituted into each equation. 2x – 3y = 7 ﬁ 2(–1) – 3(–3) = 7 ﬁ 7 = 7 — True statement –2x + y = –1 ﬁ –2(–1) + (–3) = –1 ﬁ –1 = –1 — True statement Therefore x = –1, y = –3 is the solution of the system of equations. 230230230230230 Section 5.1 Section 5.1 Section 5.1 — Systems of Equations Section 5.1 Section 5.1 Guided Practice Solve each system of equations in Exercises 1–6 by graphing on x- and y-axes spanning from –6 to 6. 1. y + x = 2 and y = 2 3 x + 2 3. y = x – 3 and y + 2x = 3 2. y + x = 3 and 3y – x = 5 4. y – 3 2 x = 1 and y + 1 2 x = –3 5. y – x = 3 and y + x = –1 6. 2y – x = –6 and y + 1 2 x = –3 Independent Practice Solve each system of equations in Exercises 1–6 by graphing on x- and y-axes spanning from –6 to 6. 1. 2x + y = 7 and y = x + 1 3. y = –3 and x – y = 2 5. 2y + 4x = 4 and y = –x + 3 2. x + y = 0 and y = –2x 4. x – y = 4 and x + 4y = –1 6. y = –x and y = 4x Determine the solution to the systems of equations graphed in Exercises 7 and 8. 7. y = – x –6 –5 –4 –3 –1 –1 –2 –3 –4 –5 –6 y-axis 8-axis –6 –5 –4 –3 –2 6 5 4 3 2 1 0 –1 –1 –2 –3 –4 –5 –6 y-axis – -axis Solve each system of equations in Exercises 9–16 by graphing on x- and y-axes spanning from –6 to 6. 9. x – y = 6 and x + y = 0 11. 4x – 3y = 0 and 4x + y = 16 10. y = 2x – 1 and x + y = 8 12. x – y = 0 and x + y = 8 13. y = –x + 6 and x – y = –4 14. x – y = 1 and x + y = –3 15. x + y = 1 and x – 2y = 1 16. 2x + y = –8 and 3x + y = –13 ound Up ound Up RRRRRound Up ound Up ound Up RRRRRound Up ound Up ound Up ound Up ound Up There’s something very satisfying about taking two long linear equations and coming up with just a one-coordinate-pair solution. You should always substitute your solution back into the original equations, to check that you’ve got the correct answer. Section 5.1 Section 5.1 Section 5.1 — Systems of Equations Section 5.1 Section 5.1 231231231231231 TTTTTopicopicopicopicopic 5.1.25.1.2 5.1.25.1.2 5.1.2 California Standards: Students solve ae ae ae ae a 9.0:9.0:9.0:9.0:9.0: Students solv Students solv Students solv Students solv system of tw tw tw tw two linear o linear o linear system of system of o linear system of o linear system of tions in two vo vo vo vo variaariaariaariaariabbbbbleslesleslesles tions in tw tions in tw equa equa equations in tw equa tions in tw equa algalgalgalgalgeeeeebrbrbrbrbraicall aicallyyyyy and are able to aicall aicall aicall interpret the answer graphically..... Students are able to solve a system of two linear inequalities in two variables and to sketch the solution sets. What it means for you: You’ll learn about the substitution method and use it to solve systems of linear equations. Key words: substitution system of linear equations simultaneous equations Check it out: This is the same problem that you saw in Example 1 in the previous Topic — but this time it’s been solved with the substitution method. The Substitution Method The Substitution Method The Substitution Method The Substitution Method The Substitution Method The Substitution Method The Substitution Method The Substitution Method The Substitution Method The Substitution Method Graphing can work well if you have integer solutions, but it can be difficult to read off fractional solutions, and even integer solutions if the scale of your graph is small. e Graaaaaphsphsphsphsphs e Gr he Substitution Method Doesn’’’’’t Int Int Int Int Invvvvvolvolvolvolvolve Gr e Gr he Substitution Method Doesn TTTTThe Substitution Method Doesn he Substitution Method Doesn e Gr he Substitution Method Doesn The substitution method involves a bit more algebra than graphing, but will generally give you more accurate solutions. Substitution Method Take one of the equations and solve for one of the variables (x or y). Substitute the expression for the variable into the other equation. This gives an equation with just one variable. Solve this equation to find the value of the variable. Substitute this value into one of the earlier equations and solve it to find the value of the second variable. Example Example Example Example Example 11111 Solve this system of equations using substitution: 2x – 3y = 7 (Equation 1) –2x + y = –1 (Equation 2) Solution Solution Solution Solution Solution Step 1: Rearrange one equation so that one of the variables is expressed in terms of the other. In this case, it’s easiest to solve for y in Equation 2 because it has a coefficient of 1: –2x + y = –1 ﬁ y = 2x – 1 (Equation 3) Now you have y expressed in terms of x. Step 2: Substitute 2x – 1 for y in Equation 1. Then solve to find the value of x. 2x – 3y = 7 2x – 3(2x – 1) = 7 2x – 6x + 3 = 7 –4x + 3 = 7 –4x = 4 x = –1 232232232232232 Section 5.1 Section 5.1 Section 5.1 — Systems of Equations Section 5.1 Section 5.1 Example 1 continueduedueduedued Example 1 contin Example 1 contin Example 1 contin Example 1 contin Step 3: Substitute –1 for x into an equation to find y. Equation 3 is the best one to use here as y is already isolated — so you don’t have to do any rearranging. y = 2x – 1 ﬁ y = 2(–1) – 1 ﬁ y = –3 Therefore x = –1, y = –3 is the solution of the system of equations. It’s a good idea to check that the solution is correct by substituting it into the original equations. 2x – 3y = 7 ﬁ 2(–1) – 3(–3) = 7 ﬁ –2 + 9 = 7 ﬁ 7 = 7 — True statement –2x + y = –1 ﬁ –2(–1) + (–3) = –1 ﬁ 2 – 3 = –1 ﬁ –1 = –1 — True statement The solution makes both of the original equations true statements, so it must be correct. Guided Practice Solve each system of equations in Exercises 1–6 by the substitution method. 1. y = 2x – 3 and –3y + 2x = –15 2. x + 2y = 7 and 3x – 2y = 5 4. w + z = 13 and w – 2z = 4 3. a + b = 2 and 5a – 2b = –4 6. m + 6n = 25 and n – 3m = –18 5. y + 2x = 5 and 2x – 3y = 1 Independent Practice Solve by the substitution method: 1. y = 3x and x + 21 = –2y 3. 7x – 4y = 27 and –x + 4y = 3 2. x + y = 5 and 5x + 2y = 16 4. 9y – 6x = 0 and 13x + 9 = 24y Solve each system of three equations by the substitution method: 5. 3x + y = 10, 4x – z = 7, 7y + 2z = 17 6 = , 2a + b – 2c = –11, –4a – 4b + c = –32 7 ound Up ound Up RRRRRound Up ound Up ound Up RRRRRound Up ound Up ound Up ound Up ound Up With the substitution method, it doesn’t matter which equation you choose to rearrange, or which variable you choose to solve for — you will get the same answer as long as you follow the steps correctly and don’t make any mistakes along the way. The important thing is to keep the algebra as simple as you possibly can. Section 5.1 Section 5.1 Section 5.1 — Systems of Equations Section 5.1 Section 5.1 233233233233233 TTTTTopicopicopicopicopic 5.1.35.1.3 5.1.35.1.3 5.1.3 California Standards: Students solve ae ae ae ae a 9.0:9.0:9.0:9.0:9.0: Students solv Students solv Students solv Students solv system of tw tw tw tw two linear o linear o linear system of system of o linear system of o linear system of tions in two vo vo vo vo variaariaariaariaariabbbbbleslesleslesles tions in tw tions in tw equa equa equations in tw equa tions in tw equa algalgalgalgalgeeeeebrbrbrbrbraicall y and are ae ae ae ae abbbbblelelelele y and ar y and ar aicall aicall aically and ar y and ar aicall et the answwwwwererererer et the ans et the ans to interprprprprpret the ans to inter to inter et the ans to inter to inter gggggrrrrraaaaaphicall phicallyyyyy..... Students are phicall phicall phicall able to solve a system of two linear inequalities in two variables and to sketch the solution sets. What it means for you: You’ll show that the graphing and substitution methods both give the same solution to a system of linear equations. Key words: substitution system of linear equations simultaneous equations phing phing Comparing the Graaaaaphing Comparing the Gr Comparing the Gr phing phing Comparing the Gr Comparing the Gr Comparing the Graaaaaphing phing phing Comparing the Gr Comparing the Gr phing phing Comparing the Gr Comparing the Gr and Substitution Methods and Substitution Methods and Substitution Methods and Substitution Methods and Substitution Methods and Substitution Methods and Substitution Methods and Substitution Methods and Substitution Methods and Substitution Methods There’s nothing new in this Topic — you’ll just see that the graphing and substitution methods always give the same answer. An Example Using the Graaaaaphing Method phing Method phing Method An Example Using the Gr An Example Using the Gr phing Method phing Method An Example Using the Gr An Example Using the Gr Example Example Example Example Example 11111 Solve this system of equations by graphing: 2y + x = 8 2y + 2x = 10 Solution Solution Solution Solution Solution Step 1: Graph both equations in the same coordinate plane. Line of first equation: 2y + x = 8 ﬁ 2y = – The line goes through the points (0, 4) and (2, 3). Line of second equation: 2y + 2x = 10 ﬁ 2y = –2x + 10 ﬁ y = –x + 5 x 0 3 y 5 2 The line goes through the points (0, 5) and (3, 2). Step 2: Read off the coordinates of the point of intersection. The point of intersection is (2, 3). y = –0.6 –5 –4 –3 –2 0 –1 –1 –2 y-axis -ax s i Step 3: Check whether your coordinates give true statements when substituted into each of the equations. 2y + x = 8 ﬁ 2(3) + 2 = 8 ﬁ 8 = 8 — True statement 2y + 2x = 10 ﬁ 2(3) + 2(2) = 10 ﬁ 10 = 10 — Tr
ue statement Therefore x = 2, y = 3 is the solution of the system of equations. 234234234234234 Section 5.1 Section 5.1 Section 5.1 — Systems of Equations Section 5.1 Section 5.1 The Same Example Using the Substitution Method The Same Example Using the Substitution Method The Same Example Using the Substitution Method The Same Example Using the Substitution Method The Same Example Using the Substitution Method Example Example Example Example Example 22222 Solve this system of equations using substitution: 2y + x = 8 (Equation 1) 2y + 2x = 10 (Equation 2) Solution Solution Solution Solution Solution Step 1: Rearrange one equation so that one of the variables is expressed in terms of the other. In this case, it’s easiest to solve for x in Equation 1 because it has a coefficient of 1: 2y + x = 8 ﬁ x = –2y + 8 (Equation 3) Now you have x expressed in terms of y. Step 2: Substitute –2y + 8 for x in Equation 2. Then solve to find the value of y. 2y + 2x = 10 2y + 2(–2y + 8) = 10 2y – 4y + 16 = 10 –2y = –6 y = 3 Step 3: Substitute 3 for y in an equation to find x. Equation 3 is the best one to use here because x is already isolated — so you don’t have to do any rearranging. x = –2y + 8 x = –2(3) + 8 x = 2 So x = 2, y = 3, or (2, 3), is the solution of the system of equations. Check by substituting the solution in the original equations. 2y + x = 8 2(3) + 2 = 8 8 = 8 — True statement 2y + 2x = 10 2(3) + 2(2) = 10 10 = 10 — True statement The solution makes both of the original equations true statements, so it must be correct. Section 5.1 Section 5.1 Section 5.1 — Systems of Equations Section 5.1 Section 5.1 235235235235235 Check it out: The substitution method gives the same answer as the graphing method used in Example 1. Guided Practice Solve each system of equations in Exercises 1–4 by graphing on xand y-axes spanning from –6 to 6. 1. y = 5 and x + y = 9 2. x = 4 and x + y = 6 3. 4x + 4y = 12 and 2x + y = 5 4. 5x + y = 15 and y = 1 3 x – 1 5. Solve the systems of equations in Exercises 1–4 using the substitution method. Independent Practice In Exercises 1–2, say whether it is possible to find the exact solutions of each system of equations using their graphs. 1. –6 –5 –4 –1 –1 –2 –3 –4 –5 –6 –2 1 26 –5 –4 –3 –2 0 –1 –1 –2 –3 –4 –5 –6 1 2 3 4 5 6 –2 x – 3 y = 7 Solve Exercises 3–6 by graphing. 3. y = 2 and 3x – y = 10 5. x – y = 1 and 2x + y = –1 4. x – y = 0 and x + y = 6 6. x – 2y = 4 and y = –x + 4 7. Solve the systems of equations in Exercises 3–6 using the substitution method. Solve Exercises 8–13 by substitution. 8. 2x – y = 8 and y = 4 10. y = 2x – 1 and x + y = 5 12. 6x + y = – 2 and 4x – 3y = 17 13. 4x – 5y = 0 and 4x – 3y = 8 9. x + y = 0 and y = –3x 11. y = 3x and 2x + 3y = 44 ound Up ound Up RRRRRound Up ound Up ound Up RRRRRound Up ound Up ound Up ound Up ound Up The graphing method works well with simple equations with integer solutions, but you need to draw the graphs very carefully — if your graphs aren’t accurate enough, you’ll get the wrong solution. If the equations look complicated, the substitution method is likely to give more accurate results. 236236236236236 Section 5.1 Section 5.1 Section 5.1 — Systems of Equations Section 5.1 Section 5.1 TTTTTopicopicopicopicopic 5.1.45.1.4 5.1.45.1.4 5.1.4 California Standards: Students solve ae ae ae ae a 9.0:9.0:9.0:9.0:9.0: Students solv Students solv Students solv Students solv system of tw tw tw tw two linear o linear o linear system of system of o linear system of o linear system of tions in two vo vo vo vo variaariaariaariaariabbbbbleslesleslesles tions in tw tions in tw equa equa equations in tw equa tions in tw equa algalgalgalgalgeeeeebrbrbrbrbraicall y and are ae ae ae ae abbbbblelelelele y and ar y and ar aicall aicall aically and ar y and ar aicall et the answwwwwererererer et the ans et the ans to interprprprprpret the ans to inter to inter et the ans to inter to inter gggggrrrrraaaaaphicall phicallyyyyy..... Students are phicall phicall phicall able to solve a system of two linear inequalities in two variables and to sketch the solution sets. What it means for you: You’ll learn how to spot systems of equations that have no solutions. Key words: inconsistent point of intersection parallel substitution system of linear equations Check it out: Equations with identical slopes, but different intercepts, result in parallel lines. So you can tell if the system is inconsistent without even drawing the graphs, by rearranging the equations into the slope-intercept form. Inconsistent Systems Inconsistent Systems Inconsistent Systems Inconsistent Systems Inconsistent Systems Inconsistent Systems Inconsistent Systems Inconsistent Systems Inconsistent Systems Inconsistent Systems So far, there’s been one solution to each system of linear equations — but it doesn’t always work out like that. tions tions Linear Equa Linear Equa Systems of Systems of tions Linear Equations Systems of Linear Equa tions Linear Equa Systems of Systems of The systems so far in this Section have all had one solution. These are called independent systems. Some systems of two linear equations have no solutions — in other words, there’s no ordered pair (x, y) that satisfies both of the equations in the system. A system of equations with no solutions is called an inconsistent system. sect sect t Inter Inconsistent System Lines Don’’’’’t Inter t Inter Inconsistent System Lines Don Inconsistent System Lines Don sect t Intersect sect t Inter Inconsistent System Lines Don Inconsistent System Lines Don Example Example Example Example Example 11111 Solve this system of equations by graphing Solution Solution Solution Solution Solution Step 1: Graph both equations in the same coordinate plane. Line of first equation– 1– The line goes through (0, 1) and (–2, –1). Line of second equation– 0 The line goes through (0, –3) and (3, 0). y-axis -axis 6 5 4 3 2 1 0 –1 –1 –2 –3 –4 –5 –6 –2 –3 1 –6 –5 –4 y = x + Step 2: Read off the coordinates of the point of intersection. The lines are parallel, so there’s no point of intersection. So the system of equations has no solutions — the system is inconsistent. The lines of equations in an inconsistent system are parallel. Section 5.1 Section 5.1 Section 5.1 — Systems of Equations Section 5.1 Section 5.1 237237237237237 tement tement alse Sta alse Sta es a F he Substitution Method Givvvvves a F es a F he Substitution Method Gi TTTTThe Substitution Method Gi he Substitution Method Gi tement alse Statement es a False Sta tement alse Sta es a F he Substitution Method Gi Here’s the same problem you saw in Example 1, but this time using the substitution method. Example Example Example Example Example 22222 Solve this system of equations by substitutionEquation 1) (Equation 2) Solution Solution Solution Solution Solution Step 1: Rearrange one equation so that one of the variables is expressed in terms of the other. Equation 2 already expresses y in terms of x: y = x – 3 Step 2: Substitute x – 3 for y in Equation 1. Then solve to find x. Check it out You can’t substitute any value of x into Equation 3 and make a true statement — so there are no solutions. y – x = 1 (x – 3) – x = 1 (Equation 3) x – 3 – x = 1 –3 = 1, which is a false statement. Therefore Equation 3 doesn’t hold for any value of x. So the system of equations has no solutions — the system is inconsistent. Guided Practice 1. How many solutions does an independent system have? 2. How many solutions does an inconsistent system have? 3. Convert y = 3x – 4 and –9x + 3y = 6 to slope-intercept form. 4. Say why you can now tell that the equations from Exercise 3 must be inconsistent. 5. Use the substitution method to say whether y = 3x – 4 and –9x + 3y = 7 are inconsistent. Independent Practice In Exercises 1–4, graph each pair of equations and say whether they are independent or inconsistent. 1. y = x – 2 and y – x = –6 3. 2x – y = –1 and y = 2x + 2 2. 2y + x = 5 and y = –x + 1 4. 3x + 4y = 3 and y = – x – 4 3 4 Use the substitution method to show that each of the pairs of equations in Exercises 5–8 are inconsistent. 5. x + y = –5 and x + y = 3 7. 2x + y = –3 and 2x + y = 1 6. x + y = –6 and y = 4 – x 8. x + 2y = –2 and 2x = 10 – 4y ound Up ound Up RRRRRound Up ound Up ound Up ound Up ound Up RRRRRound Up ound Up ound Up From this Topic you’ve seen that it doesn’t matter whether you use the graphing or substitution method — if the system is inconsistent, you won’t get any solutions. 238238238238238 Section 5.1 Section 5.1 Section 5.1 — Systems of Equations Section 5.1 Section 5.1 TTTTTopicopicopicopicopic 5.1.55.1.5 5.1.55.1.5 5.1.5 California Standards: Students solve ae ae ae ae a 9.0:9.0:9.0:9.0:9.0: Students solv Students solv Students solv Students solv system of tw tw tw tw two linear o linear o linear system of system of o linear system of o linear system of tions in two vo vo vo vo variaariaariaariaariabbbbbleslesleslesles tions in tw tions in tw equa equa equations in tw equa tions in tw equa algalgalgalgalgeeeeebrbrbrbrbraicall y and are ae ae ae ae abbbbblelelelele y and ar y and ar aicall aicall aically and ar y and ar aicall et the answwwwwererererer et the ans et the ans to interprprprprpret the ans to inter to inter et the ans to inter to inter gggggrrrrraaaaaphicall phicallyyyyy..... Students are phicall phicall phicall able to solve a system of two linear inequalities in two variables and to sketch the solution sets. What it means for you: You’ll learn how to spot systems of equations with infinitely many solutions. Key words: dependent coincide infinite substitution system of linear equations Dependent Systems Dependent Systems Dependent Systems Dependent Systems Dependent Systems Dependent Systems Dependent Systems Dependent Systems Dependent Systems Dependent Systems In Topic 5.1.4 you saw systems of equations with no solution. In this Topic you’ll see systems that have infinitely many solutions. A Dependent System Has Infinitely Many Solutions A Dependent System Has Infinitely Many Sol
utions A Dependent System Has Infinitely Many Solutions A Dependent System Has Infinitely Many Solutions A Dependent System Has Infinitely Many Solutions Some systems of two linear equations have an infinite number of solutions — in other words, there are an infinite number of points (x, y) that satisfy both of the equations in the system. Every solution of one equation is also a solution of the other equation. A system of equations with an infinite number of solutions is said to be dependent. pendent System — the Lines Coincide pendent System — the Lines Coincide phing a De GrGrGrGrGraaaaaphing a De phing a De pendent System — the Lines Coincide phing a Dependent System — the Lines Coincide pendent System — the Lines Coincide phing a De Example Example Example Example Example 11111 Solve this system of equations by graphing: y + x = 4 2y + 2x = 8 Solution Solution Solution Solution Solution Step 1: Graph both equations in the same coordinate plane. Line of first equation: y + x = 4 y = –x + 4 x 0 2 y 4 2 The line goes through (0, 4) and (2, 2). Line of second equation: 2y + 2x = 8 2y = –2x + 8 y = –x + 4 x 0 2 y 4 2 The line goes through (0, 4) and (2, 2). y-axis -axis 6 –5 –4 –3 –2 0 –1 –1 –2 –3 –4 –5 –6 Check it out: Every point on the line y + x = 4 is on the line 2y + 2x = 8. Step 2: Read off the coordinates of the point of intersection. The two equations represent the same line, so there are infinitely many points of intersection. Therefore the system of equations has infinitely many solutions — the system is dependent. The lines of equations in a dependent system coincide. Section 5.1 Section 5.1 Section 5.1 — Systems of Equations Section 5.1 Section 5.1 239239239239239 tement tement ue Sta es a TTTTTrrrrrue Sta ue Sta es a he Substitution Method Givvvvves a es a he Substitution Method Gi TTTTThe Substitution Method Gi he Substitution Method Gi tement ue Statement tement ue Sta es a he Substitution Method Gi Here’s the same problem you saw in Example 1, but this time using the substitution method. Example Example Example Example Example 22222 Solve this system of equations by substitution: y + x = 4 (Equation 1) 2y + 2x = 8 (Equation 2) Solution Solution Solution Solution Solution Step 1: Rearrange one equation so that one of the variables is expressed in terms of the other. Here, it’s easiest to solve for x or y in Equation 1. y + x = 4 y = –x + 4 (Equation 3) Step 2: Substitute –x + 4 for y in Equation 2. Then solve to find x. 2y + 2x = 8 2(–x + 4) + 2x = 8 (Equation 4) –2x + 8 + 2x = 8 8 = 8, which is a true statement. Therefore Equation 4 holds for any value of x. So the system of equations has infinitely many solutions — the system is dependent. Guided Practice 1. How many solutions does a dependent system have? Say whether the systems in Exercises 2–5 are dependent. 2. x – y = 5 and 2x = 2y + 10 3. 2x + 2y = 2 and y = –x + 1 4. 4y – 5x = 12 and y = 5 4 x + 3 5. –2x + 7y = –14 and y = 2 7 x + 1 Independent Practice By graphing, say whether the systems in Exercises 1–4 are dependent. 1. 3x – y = –1 and –3x + y = 1 2. y – 3x = –1 and –2y + 6x = 2 3. y = –2x + 6 and y = 1 2 x – 4 4. 4y + 8x = 8 and y + 2x = 2 Check it out: You can substitute any value of x into Equation 4 and make a true statement because x disappears. So there are infinitely many solutions. Use substitution to say whether the systems in Exercises 5–8 are dependent. 5. x + y = 2 and 2x + 2y = 4 7 and 2y + x = 7 6. x + y = 57 and y = 6x + 120 8. y = – 4 3 x – 4 and 3y + 4x = –12 ound Up ound Up RRRRRound Up ound Up ound Up ound Up ound Up RRRRRound Up ound Up ound Up It would be crazy to try to list the solutions to dependent systems, because there are infinitely many. You can actually rearrange the equations and show that they’re really saying the same thing. 240240240240240 Section 5.1 Section 5.1 Section 5.1 — Systems of Equations Section 5.1 Section 5.1 TTTTTopicopicopicopicopic 5.1.65.1.6 5.1.65.1.6 5.1.6 California Standards: Students solve ae ae ae ae a 9.0:9.0:9.0:9.0:9.0: Students solv Students solv Students solv Students solv system of tw tw tw tw two linear o linear o linear system of system of o linear system of o linear system of tions in two vo vo vo vo variaariaariaariaariabbbbbleslesleslesles tions in tw tions in tw equa equa equations in tw equa tions in tw equa algalgalgalgalgeeeeebrbrbrbrbraicall y and are ae ae ae ae abbbbblelelelele y and ar y and ar aicall aicall aically and ar y and ar aicall et the answwwwwererererer et the ans et the ans to interprprprprpret the ans to inter to inter et the ans to inter to inter gggggrrrrraaaaaphicall phicallyyyyy..... Students are phicall phicall phicall able to solve a system of two linear inequalities in two variables and to sketch the solution sets. What it means for you: You’ll get more practice solving systems of equations and interpreting the solutions. Key words: dependent independent inconsistent system of linear equations Check it out: “Interpreting the solution” means stating whether the system is independent (1 solution), inconsistent (0 solutions), or dependent (infinitely many solutions). Check it out: The lines on the graph have identical slopes and different intercepts, so they’re parallel. tions tions Equa Equa Systems of Systems of tions Equations Systems of Equa tions Equa Systems of Systems of tions tions Equa Equa Systems of Systems of Equations tions Systems of Equa Equa tions Systems of Systems of ther Examples ther Examples — Fur— Fur— Fur— Fur— Further Examples ther Examples ther Examples — Fur— Fur— Fur— Fur— Further Examples ther Examples ther Examples ther Examples ther Examples There’s nothing new to learn in this Topic — just practice at using the methods you’ve learned to solve systems of equations. tions tions Equa Equa t Solving Systems of t Solving Systems of actice a e Pre Practice a actice a e Pre Pr MorMorMorMorMore Pr tions Equations t Solving Systems of Equa actice at Solving Systems of tions Equa t Solving Systems of actice a Example Example Example Example Example 11111 Find and interpret the solution of this system of equations2 Solution Solution Solution Solution Solution Step 1: Graph both equations in the same coordinate plane. Line of first equation: y + x = 2 y = –x + 2 x 0 4 y 2 2– The line goes through (0, 2) and (4, –2). Line of second equation: y + x = –2 y = –x – 2 x 0 2– y 2– 0 The line goes through (0, –2) and (–2, 0). Draw the graph-axis 1 2 3 4 5 6 x-axis – x – 2 –6 –5 –4 –3 –2 6 5 4 3 2 1 0 –1 –1 –2 –3 –4 –5 –6 Step 2: Read off the coordinates of the point of intersection. The lines are parallel, so there’s no point of intersection. The system of equations has no solutions — the system is inconsistent. Section 5.1 Section 5.1 Section 5.1 — Systems of Equations Section 5.1 Section 5.1 241241241241241 Example Example Example Example Example 22222 Solve this system of equations and state whether the system is independent, inconsistent, or dependent: y – 3x = –5 (Equation 1) 3x – 4y = –7 (Equation 2) Solution Solution Solution Solution Solution Step 1: Solve for y in Equation 1. y – 3x = –5 y = 3x – 5 (Equation 3) Step 2: Substitute 3x – 5 for y in Equation 2. Then solve to find the value of x. 3x – 4y = –7 3x – 4(3x – 5) = –7 3x – 12x + 20 = –7 –9x = –27 x = 3 Step 3: Substitute 3 for x in Equation 3 to find y. y = 3x – 5 y = 3(3 So x = 3, y = 4 is the solution of the system of equations. Check by substituting the solution in the original equations. y – 3x = –5 4 – 3(3) = –5 4 – 9 = –5 –5 = – 5 — True statement 3x – 4y = –7 3(3) – 4(4) = –7 9 – 16 = –7 –7 = –7 — True statement The solution is correct. The system is independent since it has one solution. 242242242242242 Section 5.1 Section 5.1 Section 5.1 — Systems of Equations Section 5.1 Section 5.1 Guided Practice Solve each system of equations in Exercises 1–6 by either substitution or graphing. If a system isn’t independent then say whether it is inconsistent or dependent. 1. 2a – 7b = –12 and –a + 4b = 7 2. 3x – y = –2 and x – y = 0 3. y + x = 6 and 3y – x = 6 4. y + 2x = –1 and 3y + 4x = 3 5. 5y – 4x = 20 and y – 4 5 x = 4 6. 5y + 3x = –5 and y – x = –1 Independent Practice Solve each system of equations in Exercises 1–20 by either substitution or graphing. If a system isn’t independent then say whether it is inconsistent or dependent. y = 3x + 6 and 1 2 2. 2x – y = –2 and –2x + y = 2 3. 2x + y = 5 and 2x + y = 1 4. x = 2y + 5 and –2x + 4y = 2 5. 2x – 4y = 4 and 2y – 2x = –5 6. 7x – y = 2 and 2y – 14x = –4 7. –3x + 9y = –39 and 2x – y = 21 8. 4x + 5y = 2 and y = – 4 5 x – 10 9. 5x – y = –13 and –3x + 8y = 67 10. –5x + 9y = –4 and –10x + 18y = –1 11. 6x + 6y = –6 and 11y – x = 25 12. –4x + 7y = 15 and 8x – 14y = 37 13. –7x + 5y = 20 and 14x – 10y = –40 14. 5x + 6y = 5 and –5x + 3y = 1 15. –4x + 7y = 9 and 5x – 2y = –18 16. 5x – y = –15 and –3x + 9y = 93 17. –4x + 9y = 57 and 5x + 5y = 75 18. –2x + 5y = 33 and 10x – 25y = 69 19. –3x + 4y = 41 and –12x + 16y = –1 20. –2x – 5y = 75 and 4x + 10y = –150 ound Up ound Up RRRRRound Up ound Up ound Up RRRRRound Up ound Up ound Up ound Up ound Up That was a long Section, with lots of different methods. In Section 5.2 you’ll learn one more way of solving systems of equations algebraically, and then in Section 5.3 you’ll use all these methods to solve some real-life problems. Section 5.1 Section 5.1 Section 5.1 — Systems of Equations Section 5.1 Section 5.1 243243243243243 TTTTTopicopicopicopicopic 5.2.15.2.1 5.2.15.2.1 5.2.1 California Standards: Students solve ae ae ae ae a 9.0:9.0:9.0:9.0:9.0: Students solv Students solv Students solv Students solv system of tw tw tw tw two linear o linear o linear system of system of o linear system of o linear system of tions in two vo vo vo vo variaariaariaariaariabbbbbleslesleslesles tions in tw tions in tw equa equa equations in tw equa tions in tw equa algalgalgalgalgeeeeebrbrbrbrbraicall aically y y y y and are able to aicall aicall aicall interpret the answer graphically. Students are a
ble to solve a system of two linear inequalities in two variables and to sketch the solution sets. What it means for you: You’ll learn about the elimination method and then use it to solve systems of linear equations in two variables. Key words: system of linear equations elimination method Check it out: The size (absolute value) is the important thing here. It doesn’t matter if the coefficients have opposite signs — for example, –5 and 5. Section 5.2 tions tions Equa Equa Systems of Systems of tions Equations Systems of Equa tions Equa Systems of Systems of tions tions Equa Equa Systems of Systems of Equations Systems of Equa tions Systems of Equa tions Systems of — Eliminationtiontiontiontion — Elimina — Elimina — Elimina — Elimina — Eliminationtiontiontiontion — Elimina — Elimina — Elimina — Elimina Here’s one final method of solving systems of linear equations. tion Method tion Method Solving Using the Elimina Solving Using the Elimina tion Method Solving Using the Elimination Method tion Method Solving Using the Elimina Solving Using the Elimina The graphing method used in Section 5.1 isn’t generally very accurate when the numbers involved are large or are fractions or decimals. It’s pretty difficult to get exact fractions or decimals from a graph. In these cases, it’s simpler to use the elimination method. The Elimination Method Multiply or divide one or both equations so that one variable has coefficients of the same size in both equations. Get rid of (eliminate) this variable by either adding the equations or subtracting one equation from the other. Example Example Example Example Example 11111 Solve this system of equations: –3y + 4x = 11 5y – 2x = 5 Solution Solution Solution Solution Solution Step 1: Make the coefficients of one variable have the same absolute value in each equation. In this case it is easier to make the x-coefficients the same size, rather than the y-coefficients. Multiplying the second equation by 2 will give you a –4x to match the 4x in the first equation. 2(5y – 2x = 5) 10y – 4x = 10 Now you have two equations whose x-coefficients have the same absolute value: –3y + 4x = 11 10y – 4x = 10 Step 2: Add the two equations OR subtract one from the other to eliminate a variable. In this case, the coefficients of x are opposites of each other, so adding the equations will eliminate x. –3y + 4x = 11 + 10y – 4x = 10 7y = 21 ﬁ y = 3 244244244244244 Section 5.2 Section 5.2 Section 5.2 — The Elimination Method Section 5.2 Section 5.2 Example 1 continueduedueduedued Example 1 contin Example 1 contin Example 1 contin Example 1 contin Step 3: Now choose either of the original equations and substitute in the value you have found. Substitute 3 for y, and solve for x. –3y + 4x = 11 –3(3) + 4x = 11 –9 + 4x = 11 4x = 20 x = 5 So x = 5, y = 3 is the solution to the system of equations. It’s always a good idea to check your solution in the other equation too. 5y – 2x = 5 5(3) – 2(5) = 5 15 – 10 = 5 5 = 5 — True statement Guided Practice Solve the following systems of equations by elimination. 1. 5y – 3x = 19 4y + 3x = –1 3. 3y – 2x = 13 2y + x = 4 5. 3y + 2x = 23 4y – x = 16 7. 4y + 3x = 1 5y – 2x = –16 9. 7y – 3x = –11 2y + x = –5 2. 5y + 3x = –5 y – x = –1 4. 3y – 2x = –4 2y + 3x = –7 6. 7y – 2x = –29 2y + 5x = 14 8. 6y – 5x = 9 5y + 3x = 29 10. 8y + 5x = –14 5y – 3x = –21 tion Method tion Method actice Using the Elimina PrPrPrPrPractice Using the Elimina actice Using the Elimina tion Method actice Using the Elimination Method tion Method actice Using the Elimina In this example, you’ve got to subtract the equations to eliminate a variable, rather than add them. Example Example Example Example Example 22222 Solve this system of equations: 5a + 3b = 19 3a + 2b = 12 Solution Solution Solution Solution Solution Step 1: Make the coefficients of one variable the same in both equations. To make the coefficients of variable b the same, multiply the first equation by 2 and the second equation by 3. 2(5a + 3b = 19) ﬁ 10a + 6b = 38 3(3a + 2b = 12) ﬁ 9a + 6b = 36 Now you have two equations that have the same coefficient of b. Section 5.2 Section 5.2 Section 5.2 — The Elimination Method Section 5.2 Section 5.2 245245245245245 Example 2 continueduedueduedued Example 2 contin Example 2 contin Example 2 contin Example 2 contin Step 2: Add or subtract the two equations to eliminate a variable. In this case, the coefficients of b are both positive, so subtracting one equation from the other will eliminate b. 10a + 6b = 38 – 9a + 6b = 36 a = 2 Step 3: Now choose either of the original equations and substitute in the value you found. Substitute 2 for a, and solve for b. 3a + 2b = 12 3(2) + 2b = 12 6 + 2b = 12 2b = 6 b = 3 Check it out: It’s always a good idea to write both parts of the solution together clearly. Therefore a = 2, b = 3 is the solution to the system of equations. Check your solution in the other equation: 5a + 3b = 19 5(2) + 3(3) = 19 10 + 9 = 19 19 = 19 — True statement You have a few choices to make when you’re using the elimination method — such as which variable to eliminate, which equation to subtract from the other, and which of the original equations to substitute your result into. Always try to make things easy for yourself by keeping the algebra as simple as possible. Example Example Example Example Example 33333 Solve: 3y + 6x = –6 4y – 2x = 7 Solution Solution Solution Solution Solution Step 1: Make the x-coefficients the same size by multiplying the second equation by 3. 3(4y – 2x = 7) 12y – 6x = 21 Now you have two equations with x-coefficients of 6 and –6: 3y + 6x = –6 and 12y – 6x = 21. 246246246246246 Section 5.2 Section 5.2 Section 5.2 — The Elimination Method Section 5.2 Section 5.2 Example 3 continueduedueduedued Example 3 contin Example 3 contin Example 3 contin Example 3 contin Step 2: Add these equations together to eliminate x. 3y + 6x = –6 + 12y – 6x = 21 15y = 15 ﬁ y = 1 Step 2: Substitute 1 for y in one of the original equations and solve for x. 4y – 2x = 7 4(1) – 2x = 7 4 – 2x = 7 –2x = 3 x = –1.5 So x = –1.5, y = 1 is the solution to the system of equations. Check the solution in the other equation: 3y + 6x = –6 3(1) + 6(–1.5) = –6 3 – 9 = –6 — True Example Example Example Example Example 44444 Solve this system of equations: − a = b and 15 Solution Solution Solution Solution Solution This example has fractional coefficients. To make the equations easier to solve, first multiply each equation by the LCM of the denominators to convert the fractional coefficients to integers. Check it out: LCM of 8, 2, and 4 is 8. LCM of 3, 5, and 15 is 15. 8 ⎛ ⎜⎜⎜ ⎝ 5 8 a 1 b− 2 = 1 4 ⎞ ⎟⎟⎟ ﬁ 5a – 4b = 2 ⎠ 15 ⎛ ⎜⎜⎜ ⎝ 2 3 a 3 b− = 5 2 15 ⎞ ⎟⎟⎟ ﬁ 10a – 9b = 2 ⎠ Now you can solve these equations using the same method as the previous examples: Step 1: Make the a-coefficients the same by multiplying the first equation by 2. 2(5a – 4b = 2) 10a – 8b = 4 Now you’ve got two equations that have the same a-coefficient: 10a – 8b = 4 and 10a – 9b = 2. Step 2: Subtract one equation from the other to eliminate a. 10a – 8b = 4 – 10a – 9b = 2 b = 2 Section 5.2 Section 5.2 Section 5.2 — The Elimination Method Section 5.2 Section 5.2 247247247247247 Check it out: Subtracting the second equation from the first means that there are no negative terms in the resulting equation. Check it out: You could also substitute into one of the equations in which the fractional coefficients have been converted to integers. Example 4 continueduedueduedued Example 4 contin Example 4 contin Example 4 contin Example 4 contin Step 3: Substitute 2 for b in one of the original equations and solve for a 5a = 10 ﬁ a = 2 So a = 2, b = 2 is the solution to the system of equations. Check the solution in the other equation: 2 3 a 3 b− = 5 2 15 4 ⇒ − = 3 6 5 2 15 ⇒ 20 15 18 − = 15 2 15 — True Guided Practice Solve the following systems of equations by elimination. 11. 2a – 5b = 3 4a – 4b = –12 13. 2y = –x – 11 2y – x = –5 15. 7a + 2b = 10 3a + b = 1 12. 4y – 5x = 7 2y – 3x = –1 14. 7y – 2x = 3 6y – 5x = – 4 16. 5a – 2b = 3 3a – 5b = 17 17 15 18 Independent Practice Solve the following systems of equations by elimination. 1. 10y – 2x = 3 6y + 4x = 7 2. 3m – 4c = 1 6m – 6c = 5 3. 0.4m – 0.9c = –0.1 0.3m + 0.2c = 0.8 7. 5y – 4x = –6.1 8y + 5x = –6.625 9. 0.7y – 0.4x = –4.7 0.9y + 0.5x = –3 a a b + = 4. y – 0.4x = –1.8 y + 0.2x = –0.6 8. 0.7y – 0.3x = 2.7 0.4y + 0.3x = 0.6 10. 0.3c + 0.1d = 0.2 0.4c + 0.6d = –0.2 11. Find x and y, by first finding a and c using the following systems of equations: 0.3a – 0.7b = –0.5 and 1.7a + 3b = 11.1 1.2c + 0.3d = 3.9 and –0.2c – 0.4d = 0.4 ax + cy = 6 and 3ax – 2cy = –42 ound Up ound Up RRRRRound Up ound Up ound Up ound Up ound Up RRRRRound Up ound Up ound Up The elimination method is usually more reliable than trying to solve complicated systems of equations graphically. Just be really careful with the signs when you’re subtracting. 248248248248248 Section 5.2 Section 5.2 Section 5.2 — The Elimination Method Section 5.2 Section 5.2 TTTTTopicopicopicopicopic 5.3.15.3.1 5.3.15.3.1 5.3.1 California Standards: Students solve ae ae ae ae a 9.0:9.0:9.0:9.0:9.0: Students solv Students solv Students solv Students solv system of tw tw tw tw two linear o linear o linear system of system of o linear system of o linear system of tions in two vo vo vo vo variaariaariaariaariabbbbbleslesleslesles tions in tw tions in tw equa equa equations in tw equa tions in tw equa algalgalgalgalgeeeeebrbrbrbrbraicall aically y y y y and are able to aicall aicall aicall interpret the answer graphically. Students are able to solve a system of two linear inequalities in two variables and to sketch the solution sets. Students applpplpplpplpplyyyyy Students a Students a 15.0: 15.0: 15.0: Students a 15.0: Students a 15.0: algalgalgalgalgeeeeebrbrbrbrbraic tec hniques hniques aic tec aic tec hniques to aic techniques hniques aic tec solve rate problems, work problems, and percent mixture problems. What it means for you: You’ll solve real-lif
e problems involving systems of linear equations. Key words: system of linear equations substitution method elimination method Check it out: 8.50c is the amount spent on CDs. 12.50d is the amount spent on DVDs. Section 5.3 tions tions Equa Equa Systems of Systems of tions Equations Systems of Equa tions Equa Systems of Systems of tions tions Equa Equa Systems of Systems of Equations Systems of Equa tions Equa Systems of tions Systems of tions tions pplica pplica — — — — — AAAAApplica tions pplications tions pplica tions tions pplica pplica — — — — — AAAAApplica tions pplications pplica tions Now it’s time to use all the systems of equations methods from Sections 5.1 and 5.2 to solve some real-life problems. e Pre Proboboboboblems lems lems e Pre Pr eal-Lif tions Help Solve Re Re Re Re Real-Lif eal-Lif tions Help Solv tions Help Solv Equa Equa Systems of Systems of lems eal-Life Pr Equations Help Solv Systems of Equa lems eal-Lif tions Help Solv Equa Systems of Systems of You can use systems of linear equations to represent loads of real-life problems. Once you’ve written a system of equations, you can then use either substitution or elimination to solve the system of equations and therefore solve the problem. h Equationtiontiontiontion h Equa h Equa les in Eac ou Need the Same TTTTTwwwwwo o o o o VVVVVariaariaariaariaariabbbbbles in Eac les in Eac ou Need the Same YYYYYou Need the Same ou Need the Same les in Each Equa h Equa les in Eac ou Need the Same Example Example Example Example Example 11111 A store has a year-end sale on CDs and DVDs. Each CD is reduced to $8.50, and the DVDs are going for $12.50 each. Akemi bought a total of 15 items (some CDs and some DVDs) for $163.50. Determine how many CDs and DVDs she bought. Solution Solution Solution Solution Solution One way to find the number of CDs and DVDs is to set up a system of two linear equations in two variables. Represent the unknown values with variables — Let c = number of CDs Akemi bought d = number of DVDs Akemi bought Since Akemi bought a total of 15 CDs and DVDs: c + d = 15 — this is y st equationtiontiontiontion st equa st equa our fir our fir — this is y — this is y our first equa — this is your fir st equa our fir — this is y Now, consider the fact that each CD cost $8.50, each DVD cost $12.50, and Akemi spent $163.50. This leads to the equation: 8.50c + 12.50d = 163.50 — this is y our second equationtiontiontiontion our second equa our second equa — this is y — this is y — this is your second equa our second equa — this is y You now have a system of two linear equations in two variables, c and d. Section 5.3 Section 5.3 Section 5.3 — Applications of Systems of Equations Section 5.3 Section 5.3 249249249249249 Check it out This part is nothing new — it is the elimination method you used in Section 5.2. Example 1 continueduedueduedued Example 1 contin Example 1 contin Example 1 contin Example 1 contin Now solve your system of equations: c + d = 15 8.50c + 12.50d = 163.50 Step 1: Make the coefficient of c the same in both equations. Do this by multiplying the first equation (c + d = 15) by 8.50. 8.50(c + d = 15) 8.50c + 8.50d = 127.50 Now you have two equations that have the same coefficient of c: 8.50c + 12.50d = 163.50 and 8.50c + 8.50d = 127.50 Step 2: Subtract one equation from the other to eliminate c. 8.50c + 12.50d = 163.50 – 8.50c + 8.50d = 127.50 4d = 36 ﬁ d = 9 Step 3: Substitute d = 9 into one of the original equations. Don’t forget Always remember to answer the original question clearly, in words. c + d = 15 c + 9 = 15 c = 6 So Akemi bought 6 CDs and 9 DVDs. Check the solution in the other equation: 8.50c + 12.50d = 163.50 8.50(6) + 12.50(9) = 163.50 163.50 = 163.50 — True Guided Practice 1. Pedro bought a total of 18 paperback and hardcover books for a total of $150. If each paperback was on sale for $6.50 and the hardcovers were on sale for $9.50 each, calculate how many paperbacks and how many hardcovers Pedro bought. 2. Three cans of tuna fish and four cans of corned beef cost $12.50. However, six cans of tuna fish and three cans of corned beef cost $15. How much does each type of can cost individually? 3. A school raised funds for its sports teams by selling tickets for a play. A ticket cost $5.75 for adults and $2.25 for students. If there were five times as many adults at the play than students and ticket sales raised $2480, how many adult and student tickets were sold? 4. Teresa bought five cups and four plates for $14.50. However, five plates and four cups would have cost $14.75. Find the cost of each item individually. 5. Simon has a total of 21 dimes and quarters in his coin bank. If the value of the coins is $4.05, how many coins of each type does Simon have? 250250250250250 Section 5.3 Section 5.3 Section 5.3 — Applications of Systems of Equations Section 5.3 Section 5.3 tions to Solve e e e e AgAgAgAgAge Questions e Questions e Questions tions to Solv tions to Solv Equa Equa Using Systems of Using Systems of e Questions Equations to Solv Using Systems of Equa e Questions tions to Solv Equa Using Systems of Using Systems of Example Example Example Example Example 22222 The sum of Jose’s and Elizabeth’s ages is 40 years. Five years ago, Elizabeth was four times as old as Jose. How old are Jose and Elizabeth now? Solution Solution Solution Solution Solution First form a system of two equations. Present Ages: Let x = Jose’s age y = Elizabeth’s age The sum of their ages is 40 ﬁ x + y = 40 Ages 5 years ago: x – 5 = Jose’s age y – 5 = Elizabeth’s age Elizabeth’s age was four times Jose’s age ﬁ y – 5 = 4(x – 5) ﬁ y – 5 = 4x – 20 ﬁ y – 4x = –15 So the system of equations is: x + y = 40 y – 4x = –15 Now solve the system of equations. The y-coefficients are the same, so subtract one equation from the other to eliminate y: y + x = 40 – y – 4x = –15 5x = 55 ﬁ x = 11 Now substitute 11 for x in an original equation. x + y = 40 11 + y = 40 y = 29 Therefore Jose is 11 years old and Elizabeth is 29 years old. Check the solution in the other equation. y – 4x = –15 29 – 4(11) = –15 –15 = –15 — True Section 5.3 Section 5.3 Section 5.3 — Applications of Systems of Equations Section 5.3 Section 5.3 251251251251251 Guided Practice 6. The sum of Julio’s and Charles’s ages is 44 years. Seven years from now, Julio will be twice Charles’s age now. How old is each one now? 7. A mother is five times as old as her daughter. In five years, the mother will be three times as old as her daughter. How old is each now? 8. Anthony is twice as old as Teresa. In 20 years, the sum of their ages will be 43. How old is each now? 9. The sum of Ivy's and Audrey's ages is 27. Nine years ago, Ivy was twice as old as Audrey. How old is each now? 10. Three years from now, a father will be 10 times as old as his daughter. One year ago, the sum of their ages was 36. How old is each now? Independent Practice 1. Eight goldfish and 12 angelfish cost $53. Five goldfish and eight angelfish cost $34.75. If all goldfish are the same price and all angelfish are the same price, find the price of each goldfish and each angelfish. 2. Seven pairs of shorts and nine shirts cost $227.95. Three pairs of shorts and five shirts cost $116.55. If all pairs of shorts are the same price and all shirts are the same price, how much does each pair of shorts and each shirt cost? 3. Find the values of x and y if the figure shown on the right is a rectangle. 3x – y 2x + y 6 (5x + y) inches (7x – 5) inches (2y + 7) inches 14 4. By finding the values of x and y, calculate the area of the rectangle on the left. All dimensions are in inches. (3x + 5) inches 5. Given that the triangle on the right is an isosceles triangle with a 40 cm perimeter, find the values of x and y 2x + 3y ound Up ound Up RRRRRound Up ound Up ound Up ound Up ound Up RRRRRound Up ound Up ound Up Systems of linear equations are really useful for working out real-life problems. In the rest of this Section you’ll see more everyday situations modeled as systems of equations. 252252252252252 Section 5.3 Section 5.3 Section 5.3 — Applications of Systems of Equations Section 5.3 Section 5.3 TTTTTopicopicopicopicopic 5.3.25.3.2 5.3.25.3.2 5.3.2 California Standards: Students solve ae ae ae ae a 9.0:9.0:9.0:9.0:9.0: Students solv Students solv Students solv Students solv system of tw tw tw tw two linear o linear o linear system of system of o linear system of o linear system of tions in two vo vo vo vo variaariaariaariaariabbbbbleslesleslesles tions in tw tions in tw equa equa equations in tw equa tions in tw equa algalgalgalgalgeeeeebrbrbrbrbraicall aically y y y y and are able to aicall aicall aicall interpret the answer graphically. Students are able to solve a system of two linear inequalities in two variables and to sketch the solution sets. Students applpplpplpplpplyyyyy Students a Students a 15.0: 15.0: 15.0: Students a 15.0: Students a 15.0: algalgalgalgalgeeeeebrbrbrbrbraic tec hniques hniques aic tec aic tec hniques to aic techniques hniques aic tec solve rate problems, work problems, and percent mixture problems. What it means for you: You’ll solve integer problems involving systems of linear equations. Key words: system of linear equations substitution method tions tions Equa Equa Systems of Systems of tions Equations Systems of Equa tions Equa Systems of Systems of tions tions Equa Equa Systems of Systems of Equations tions Systems of Equa tions Equa Systems of Systems of lems lems er Proboboboboblems er Pr er Pr — Inte— Inte— Inte— Inte— Integgggger Pr lems lems er Pr er Proboboboboblems — Inte— Inte— Inte— Inte— Integgggger Pr lems lems er Pr er Pr lems er Pr lems In this Topic you’ll solve systems of linear equations to figure out solutions to problems involving integers. er Proboboboboblems lems lems er Pr e Integgggger Pr er Pr e Inte e Inte tions to Solv tions to Solv Equa Equa Using Systems of Using Systems of lems tions to Solve Inte Equations to Solv Using Systems of Equa lems er Pr e Inte tions to Solv Equa Usin
g Systems of Using Systems of Example Example Example Example Example 11111 The sum of two integers is 53. The larger number is 7 less than three times the smaller one. Find the numbers. Solution Solution Solution Solution Solution First form a system of two equations. Let x = smaller number y = larger number The sum of the two integers is 53, so x + y = 53. The larger is 7 less than 3 times the smaller one, so y = 3x – 7. So the system of equations is x + y = 53 and y = 3x – 7. Now solve the system of equations. The variable y in the second equation is already expressed in terms of x, so it makes sense to use the substitution method. Substitute 3x – 7 for y in the first equation. x + y = 53 x + (3x – 7) = 53 4x – 7 = 53 4x = 60 x = 15 Now, substitute 15 for x in the equation y = 3x – 7. y = 3x – 7 y = 3(15) – 7 y = 45 – 7 y = 38 That means that the integers are 15 and 38. The solution should work in both equations: x + y = 53 ﬁ 15 + 38 = 53 ﬁ 53 = 53 — True y = 3x – 7 ﬁ 38 = 3(15) – 7 ﬁ 38 = 38 — True A useful check is to make sure that the answer matches the information given in the question: The sum of the two integers is 15 + 38 = 53. This matches the question. Three times the smaller integer is 15 × 3 = 45. The larger integer is 45 – 38 = 7 less than this. Section 5.3 Section 5.3 Section 5.3 — Applications of Systems of Equations Section 5.3 Section 5.3 253253253253253 Guided Practice 1. The sum of two numbers is 35 and the difference between the numbers is 13. Find the numbers. 2. The sum of two integers is 6 and the difference between the numbers is 40. Find the numbers. 3. The difference between two numbers is 50. Twice the higher number is equal to three times the lower number. Find the numbers. 4. The difference between two numbers is 11. If the sum of twice the higher number and three times the lower number is 137, find the numbers. 5. There are two numbers whose sum is 64. The larger number subtracted from four times the smaller number gives 31. Find the two numbers. Independent Practice 1. The length of a rectangle is three times the width. The perimeter is 24 inches. Find the length and the width. 2. In a soccer match, Team A defeated Team B by 3 goals. There were a total of 13 goals scored in the game. How many goals did each team score? 3. An outdoor adventure club has 35 members, who are all either rock climbers or skiers. There are 3 more rock climbers than there are skiers. How many climbers are there? 4. The tens digit of a two-digit number is three times the ones digit. If the sum of the digits is 12, find the two-digit number. 5. The sum of the digits of a two-digit number is 13. The number is 27 more than the number formed by interchanging the tens digit with the ones digit. Find the number. 6. Subtracting the ones digit from the tens digit of a positive two-digit number gives –2. Find the number given that the number is 18 less than the number formed by reversing its digits. (More than one answer is possible.) 7. The ones digit of a positive two-digit number is seven more than the tens digit. If the number formed by reversing the digits is 63 more than the original number, find the original number. (More than one answer is possible.) 8. In a two-digit number, the sum of the digits is 10. Find the number if it is 36 more than the number formed by reversing its digits. ound Up ound Up RRRRRound Up ound Up ound Up RRRRRound Up ound Up ound Up ound Up ound Up For questions like this, it doesn’t matter which letters you choose to represent the unknown quantities. It’s a good idea to pick letters that remind you in some way of what they represent. For example, in Example 1 involving CDs and DVDs, the letters c and d were sensible choices. 254254254254254 Section 5.3 Section 5.3 Section 5.3 — Applications of Systems of Equations Section 5.3 Section 5.3 TTTTTopicopicopicopicopic 5.3.35.3.3 5.3.35.3.3 5.3.3 California Standards: Students solve ae ae ae ae a 9.0:9.0:9.0:9.0:9.0: Students solv Students solv Students solv Students solv system of tw tw tw tw two linear o linear o linear system of system of o linear system of o linear system of tions in two vo vo vo vo variaariaariaariaariabbbbbleslesleslesles tions in tw tions in tw equa equa equations in tw equa tions in tw equa algalgalgalgalgeeeeebrbrbrbrbraicall aically y y y y and are able to aicall aicall aicall interpret the answer graphically. Students are able to solve a system of two linear inequalities in two variables and to sketch the solution sets. Students applpplpplpplpplyyyyy Students a Students a 15.0: 15.0: 15.0: Students a 15.0: Students a 15.0: algalgalgalgalgeeeeebrbrbrbrbraic tec hniques tototototo hniques hniques aic tec aic tec aic techniques hniques aic tec solvsolvsolvsolvsolveeeee rate problems, work problems, and perperperperpercent cent cent cent cent lems..... e pre proboboboboblems lems lems e pre pr mixtur mixtur mixture pr lems mixtur mixtur What it means for you: You’ll solve percent mix problems involving systems of linear equations. Key words: percent mix system of linear equations Don’t forget: 5 5% is 100 , so the number of gallons of real fruit juice in a gallons of apple drink is 0.05a. tions tions Equa Equa Systems of Systems of tions Equations Systems of Equa tions Equa Systems of Systems of tions tions Equa Equa Systems of Systems of Equations tions Systems of Equa Equa tions Systems of Systems of lems lems cent Mix Proboboboboblems cent Mix Pr cent Mix Pr — P— P— P— P— Pererererercent Mix Pr lems lems cent Mix Pr cent Mix Proboboboboblems — P— P— P— P— Pererererercent Mix Pr lems lems cent Mix Pr cent Mix Pr lems cent Mix Pr lems Percent mix problems are questions that involve two different amounts being mixed together to make a single mixture. tions tions Equa Equa lems Can Be Systems of cent Mix Proboboboboblems Can Be Systems of lems Can Be Systems of cent Mix Pr PPPPPererererercent Mix Pr cent Mix Pr tions Equations lems Can Be Systems of Equa tions Equa lems Can Be Systems of cent Mix Pr Example Example Example Example Example 11111 A high school sports coach decided to give a year-end party for all students at the school. The cafeteria manager decided to make 50 gallons of fruit juice drink by mixing some apple drink that contains 5% real fruit juice with strawberry drink that contains 25% real fruit juice. If the 50-gallon fruit juice drink is 10% real fruit juice, how many gallons of apple drink and strawberry drink did the cafeteria manager use? Solution Solution Solution Solution Solution First form a system of two equations. Let a = gallons of apple drink used s = gallons of strawberry drink used The total number of gallons is 50, so: a + s = 50 Now consider the total real fruit juice in the drink: (5% of a) + (25% of s) = 10% of 50 0.05a + 0.25s = 0.10(50) 0.05a + 0.25s = 5 5a + 25s = 500 a + 5s = 100 So the system of equations is: a + s = 50 a + 5s = 100 Now solve the system of equations. The coefficient of a is the same in both equations, so subtract one equation from the other to eliminate a. a + 5s = 100 – a + s = 50 4s = 50 ﬁ s = 12.5 Substitute 12.5 for s in one of the original equations. a + s = 50 a + 12.5 = 50 a = 37.5 That means that 37.5 gallons of apple drink was used and 12.5 gallons of strawberry drink. Section 5.3 Section 5.3 Section 5.3 — Applications of Systems of Equations Section 5.3 Section 5.3 255255255255255 Example 1 continueduedueduedued Example 1 contin Example 1 contin Example 1 contin Example 1 contin Check the solution in the other equation. a + 5s = 100 37.5 + 5(12.5) = 100 100 = 100 — True The answer must also match the information given in the question: 5% of 37.5 gallons of apple drink is real fruit juice. This is 0.05 × 37.5 = 1.875 gallons. 25% of 12.5 gallons of strawberry drink is real fruit juice. This is 0.25 × 12.5 = 3.125 gallons. The total fruit juice is 1.875 + 3.125 = 5 gallons. 5 gallons is 10% of the 50-gallon mixture, which matches the question. Guided Practice 1. A football coach calls a total of 50 plays in a game, some running plays and some passing plays. If 60% of the plays she calls are running plays, how many passing plays did she call? 2. During a fund-raiser, a girl collected a total of 60 quarters and halfdollars. If she raised $21.25, how many coins of each type did she collect? 3. A mother wants to make a quart of 2% fat milk for her children. She has a gallon of 1% fat milk and a gallon of whole milk (4% fat). How much of each should she mix to get a quart of 2% fat milk? 4. A ranger had 150 gallons of 60%-pure disinfectant that he mixed with 80%-pure disinfectant until the mixture was 70%-pure disinfectant. How much 80%-pure disinfectant was used by the ranger? 5. Two brands of tea are worth 60 cents per pound and 90 cents per pound respectively. How many pounds of each tea must be mixed to produce 138 pounds of a mixture that would be worth 80 cents per pound? 6. A pharmacist has a bottle of 10% boric acid and a bottle of 6% boric acid. A prescription requires 50 milliliters of a 7% boric acid solution. What volume of each solution should the pharmacist mix to get the desired solution? 7. In an industrial process, milk containing 4% butterfat is mixed with cream that contains 44% butterfat to obtain 250 gallons of 10% butterfat milk. How many gallons of 4% butterfat milk and 44% butterfat cream must be mixed to obtain the 250 gallons of 10% butterfat milk? 256256256256256 Section 5.3 Section 5.3 Section 5.3 — Applications of Systems of Equations Section 5.3 Section 5.3 Independent Practice 1. A company invested a total of $5000, some at 6% and the rest at 8% per year. If the total return from the investments after one year was $330, how much money was invested at each rate? 2. Stacey used her 20% discount card to buy 7 cans of tomatoes and 15 cans of mushrooms for $10.64. The next day Brian spent $7.21 on 17 cans of tomatoes and 5 cans of mushrooms in a “30% off ” sale. Find the undiscounted pre-sale price of each type of can. 3. Te
resa’s piggy bank has a total of 100 dimes and nickels. If the piggy bank has a total of $7.50, how many of each coin does she have? 4. Anthony invested $1000 in two stocks. Stock A increased in value by 20%, while Stock B decreased in value by 10%. If Anthony ended the year with $1160 worth of stocks, how much money did he initially invest in each? 5. Dr. Baines makes a 5% metal hydride by mixing Material A, which contains 3% hydride, with Material K, which contains 7% hydride. If the final mixture weighs 10 kg, how much of each material does he mix? 6. Apples cost $0.75 per pound and pears cost $0.40 per pound. How much of each would need to be mixed to make 5 pounds of fruit salad worth $0.50 per pound? 7. Maddie invested $5000, some at a 5% rate of return and the rest at a 2% rate of return per year. If the total return from the investments after one year was $200, how much money was invested at each rate? 8. Stephen collected a total of 75 quarters and half-dollars. If he collected $30.50, how many coins of each type did he collect? 9. Robert used his 10% discount card to buy 4 pizzas and 2 calzones for $45. The next day Audrey spent $15 on 1 pizza and 2 calzones in a “25% off ” sale. Find the undiscounted price of pizzas and calzones. 10. Fred invested $6000, some at a 10% annual return and the rest at a 4% annual return. If the total return from the investments after one year was $400, how much money was invested at each rate? ound Up ound Up RRRRRound Up ound Up ound Up ound Up ound Up RRRRRound Up ound Up ound Up Percent mix problems don’t have to involve liquids like the example in this Topic. You could mix together solids or even amounts of money — the solving method is always the same. Section 5.3 Section 5.3 Section 5.3 — Applications of Systems of Equations Section 5.3 Section 5.3 257257257257257 TTTTTopicopicopicopicopic 5.3.45.3.4 5.3.45.3.4 5.3.4 California Standards: Students solve ae ae ae ae a 9.0:9.0:9.0:9.0:9.0: Students solv Students solv Students solv Students solv system of tw tw tw tw two linear o linear o linear system of system of o linear system of o linear system of tions in two vo vo vo vo variaariaariaariaariabbbbbleslesleslesles tions in tw tions in tw equa equa equations in tw equa tions in tw equa algalgalgalgalgeeeeebrbrbrbrbraicall aically y y y y and are able to aicall aicall aicall interpret the answer graphically. Students are able to solve a system of two linear inequalities in two variables and to sketch the solution sets. Students applpplpplpplpplyyyyy Students a Students a 15.0: 15.0: 15.0: Students a 15.0: Students a 15.0: algalgalgalgalgeeeeebrbrbrbrbraic tec hniques tototototo hniques hniques aic tec aic tec aic techniques hniques aic tec solvsolvsolvsolvsolveeeee rrrrraaaaate pr lems lems te proboboboboblems te pr te pr lems, work lems te pr problems, and percent mixture problems..... What it means for you: You’ll solve rate problems involving systems of linear equations. Key words: rate system of linear equations Check it out: When the boat travels downriver, it’s helped along by the water current. When the boat travels upstream, it’s slowed down by the water current. tions tions Equa Equa Systems of Systems of tions Equations Systems of Equa tions Equa Systems of Systems of tions tions Equa Equa Systems of Systems of Equations tions Systems of Equa Equa tions Systems of Systems of lems lems te Proboboboboblems te Pr te Pr — R— R— R— R— Raaaaate Pr lems lems te Pr te Proboboboboblems — R— R— R— R— Raaaaate Pr lems lems te Pr te Pr lems te Pr lems Here’s the final application of systems of linear equations — rate problems deal with speed, distance, and time. tions TTTTToooooooooo tions tions Equa Equa lems Can Be Systems of te Proboboboboblems Can Be Systems of lems Can Be Systems of te Pr RRRRRaaaaate Pr te Pr Equations lems Can Be Systems of Equa tions Equa lems Can Be Systems of te Pr Example Example Example Example Example 11111 During a storm, a flood rescue team in a boat takes 3 hours to travel downriver along a 120-mile section of Nastie river. If it takes 4 hours to travel the same river section upriver, find the boat’s speed in still water and the speed of the water current. [Assume the boat has the same speed relative to the water and that the speed of the current remains constant.] Solution Solution Solution Solution Solution First form a system of two equations. Let x = boat’s speed in still water y = water current speed Downriver: When traveling downriver, the water speed adds to the boat speed. So boat speed downriver = x + y Use the formula Distance = Speed × Time: 120 miles = (x + y) × 3 hours 120 = 3(x + y) 120 3 = x + y 40 = x + y Upriver: When traveling upriver, the water speed acts against the boat speed. So boat speed upriver = x – y Use Distance = Speed × Time: 120 miles = (x – y) × 4 hours 120 = 4(x – y) = x – y 120 4 30 = x – y So the system of equations is: 40 = x + y 30 = x – y 258258258258258 Section 5.3 Section 5.3 Section 5.3 — Applications of Systems of Equations Section 5.3 Section 5.3 Example 1 continueduedueduedued Example 1 contin Example 1 contin Example 1 contin Example 1 contin Now solve the system of equations. The y-coefficients have opposite values, so add the equations to eliminate y. 40 = x + y + 30 = x – y 70 = 2x ﬁ x = 35 Now substitute 35 for x in an original equation. 40 = x + y 40 = 35 + y y = 5 So, the boat’s speed in still water (x) is 35 mph and the water current speed (y) is 5 mph. Check it out: The distance was measured in miles and the time was measured in hours, so the units must be miles per hour (mph). Check the solution in the other equation. 30 = x – y 30 = 35 – 5 30 = 30 — True Guided Practice 1. The Lee High School crew team takes 2 hours to row downriver along a 60-mile section of the Potomac River. It takes 4 hours to travel the same river section upriver. Find the boat’s speed in still water and the speed of the water current. 2. A plane travels 300 miles in 40 minutes against the wind. Flying with the wind, the same plane travels 200 miles in 20 minutes. Find the plane’s speed without wind and the wind speed. 3. Sarah rides her bicycle to work, a journey of five miles (one way). One day the journey to work is against the wind, and takes 30 minutes. The ride back, with the wind, takes 20 minutes. Find the wind speed, and work out how long the ride (one way) would take with no wind. 4. A red car and a blue car start at the same time from towns that are 16 miles apart, and travel towards each other. The red car is 7 mph faster than the blue car. After 15 minutes the cars are 7 miles apart. Find the speed of each car. Section 5.3 Section 5.3 Section 5.3 — Applications of Systems of Equations Section 5.3 Section 5.3 259259259259259 Independent Practice 1. It takes a canoeist 1.5 hours to paddle down a 10-mile stretch of river. If it takes 3 hours to travel the same river section upriver, find the canoeist’s speed in still water and the speed of the water current. Round answers to the nearest tenth. 2. A medical helicopter pilot flies 200 miles with a tailwind in 2 hours. On the return trip, it takes 2.5 hours to fly against the wind. Find the speed of the helicopter in no wind and the speed of the wind. 3. A small powerboat travels down a 24-mile stretch of river in one hour. If it takes 3 hours to travel the same river section upriver, find the powerboat’s speed in still water and the speed of the water current. 4. A plane flies 700 miles with a tailwind in 5 hours. On the return trip, it takes 6 hours to fly the same 700 miles against the wind. Find the speed of the plane in no wind and the speed of the wind. Round answers to the nearest tenth. 5. Two planes start at the same time from cities that are 2500 miles apart, and travel toward each other. The rate of one plane exceeds the rate of the other by 15 mph. After 4 hours the airplanes were 1000 miles apart. Find the rate of each airplane. 6. Kyle lives in Washington, DC and Robert lives in Boston. The cities are 440 miles apart. They start driving toward each other at the same time. Kyle is driving an average of 5 mph faster than Robert. After 5 hours the cars are 180 miles apart. Find the rate of each car. 7. Sheri and Peter start running towards each other at the same time, from their houses that are 4000 feet apart. Sheri is running 2 feet per minute faster than Peter. After 5 minutes, they are still 300 feet apart. Find the running speeds of Sheri and Peter. 8. Casey leaves home at 10 a.m. and drives at an average speed of 25 mph. Marshall leaves the same house 15 minutes later, and drives the same route, but twice as fast as Casey. At what time will Marshall pass Casey, and how far will they be from home when he does? 9. Pittsburgh is 470 miles from Chicago and 350 miles from Philadelphia. Trains leave Chicago and Philadelphia at the same time, but the Chicago train travels 40 mph faster than the Philadelphia one. Both trains reach Pittsburgh at the same time. Find each train’s speed, rounding answers to the nearest tenth. 10. At 07:00 a train leaves a station, traveling at 55 mph. At 07:30 an express train leaves the same station, traveling the same route at 70 mph. How long will it take the express train to overtake the other train, and how far will they be from the station when it does? ound Up ound Up RRRRRound Up ound Up ound Up RRRRRound Up ound Up ound Up ound Up ound Up The word problems in this Section all look quite different — but you use the same methods each time to solve them. If you’re having trouble with solving them, look back at Sections 5.1 and 5.2 and try going through some examples that don’t involve real-life situations. 260260260260260 Section 5.3 Section 5.3 Section 5.3 — Applications of Systems of Equations Section 5.3 Section 5.3 Chapter 5 Investigation g Business g Business en in an Eg en in an Eg eaking Ev eaking Ev BrBrBrBrBreaking Ev g Business en in an Egg Business eaking Even in an Eg g Business en in an Eg eaki
ng Ev BrBrBrBrBreaking Ev g Business g Business en in an Eg en in an Eg eaking Ev eaking Ev en in an Egg Business eaking Even in an Eg g Business en in an Eg eaking Ev g Business Even though this Investigation looks tough because it’s about money, it’s just systems of equations. Part 1: You decide to set up an egg-selling business. You plan to keep 12 hens in your yard and sell the eggs from your home. Free-Range Eggs 20¢ each The initial costs of setting up the business are $47 for a hen house and $3 for each hen. The hens will cost a total of 50 cents a day to feed and you think each hen will lay one egg per day. You plan to sell the eggs for 20 cents each. 1) How many days will it be before you break even? Assume you manage to sell all the eggs laid without any going to waste. 2) On one set of axes, draw graphs to show how the costs and amount earned will change over the first 50 days. The “break-even” point is when the amount of money you have earned equals the amount of money you’ve spent on the business. Part 2: After further research, you discover that this type of hen only lays eggs on four days out of five. However, you can buy special hens for $5 each that will not only lay one egg per day, but will lay two eggs every fourth day. You have a total of $98 to invest in setting up your business. How many of each type of hen should you buy to maximize your profits? Remember — the hen house will only accommodate up to 12 hens. Extension 1) Your mother is not happy about you using the yard for your business and demands 10% of your profit. With this in mind, calculate how much money you will make in the first year. 2) You are given $50 for your birthday. Will you make more money over one year if you buy another hen house and one hen or invest it at 6%? Investigate how receiving a different amount of money would affect the best option. Open-ended Extension Make up a business and describe its costs and the amount it charges for its products or services. Calculate when you will break even. Exchange business plans with a partner and each devise some changes that will affect the other person’s business. Investigate how your partner’s changes will affect your profits and break-even point. Your break-even point doesn’t necessarily have to be after a certain number of days — it might be after you sell a milkshake ce ce ce ce company cccccererererertain numb ompany ompany milkshak tain numbererererer of your products. For example, a milkshak milkshak tain numb tain numb ompany might have an initial cost of buying a milkshak ompany tain numb edients edients ingr ingr blender and cups. There will then be the cost of the ingr edients for each milkshake made. After you sell a certain ingredients edients ingr equal equal equal to the amount that you’ve spent so far. number of milkshakes you will have earned an amount equal equal ound Up ound Up RRRRRound Up ound Up ound Up ound Up ound Up RRRRRound Up ound Up ound Up If you come across any real-life situation that involves two or more equations, you’ll probably have to solve them using systems of equations — using all the skills you learned in this Chapter. estigaaaaationtiontiontiontion — Breaking Even in an Egg Business estigestig estig pter 5 Invvvvvestig pter 5 In ChaChaChaChaChapter 5 In pter 5 In pter 5 In 261261261261261 Chapter 6 Manipulating Polynomials Section 6.1 Adding and Subtracting Polynomials ................. 263 Section 6.2 Multiplying Polynomials ..................................... 274 Section 6.3 Dividing Polynomials ......................................... 283 Section 6.4 Special Products of Binomials .......................... 297 Section 6.5 Factors .............................................................. 302 Section 6.6 Factoring Quadratics ......................................... 310 Section 6.7 More on Factoring Polynomials ......................... 318 Section 6.8 More on Quadratics........................................... 325 Investigation Pascal’s Triangle ............................................... 331 262262262262262 Topic 6.1.1 Section 6.1 Polynomials Polynomials California Standards: 10.0: Students add, subtract, multiply, and divide monomials and polynomials. Students solve multistep problems, including word problems, by using these techniques. What it means for you: You’ll learn what polynomials are, and you’ll simplify them. Key words: monomial polynomial like terms degree Check it out: The word monomial comes from the Greek words: mono — meaning one nomos — meaning part or portion. So monomial means one term. Check it out: The word polynomial comes from the Greek words: poly — meaning many nomos — meaning part or portion. So polynomial means many terms. Polynomial — another math word that sounds a lot harder than it actually is. Read on and you’ll see that actually polynomials are not as complicated as you might think. A Monomial is a Single Term A monomial is a single-term expression. It can be either a number or a product of a number and one or more variables. For example, 13, 2x², and –x3yn4 are all monomials. A Polynomial Can Have More Than One Term A polynomial is an algebraic expression that has one or more terms (each of which is a monomial). For example, x + 1 and –3x² + 2x + 1 are polynomials. There are a couple of special types of polynomial: A binomial is a two-term polynomial, such as x² + 1. A trinomial is a polynomial with three terms, such as –3x² + 2x + 1. Guided Practice For each of the polynomials below, state whether it is a monomial, a binomial, or a trinomial. 1. 5x 3. 2y 5. 2y + 2 7. 7x³ 9. x² + 2x + 3 11. 4.3x – 8.9x2 + 4.2x3 13. 0.3x2y + xy2 + 4xy 15. a2 + b2 2. 9x² + 4 4. 2x²y 6. 5x – 2 8. x² + y 10. 3x2 + 4x – 8 12. 0.3x2y 14. 8.7 16. 97.9a – 14.2c Section 6.1 — Adding and Subtracting Polynomials 263 Use Like Terms to Simplify Polynomials Like terms are terms that have exactly the same variables — for example, –2x² and 5x² are like terms. Like terms always have the same variables, but may have different coefficients. A polynomial can often be simplified by combining all like terms. Example 1 Simplify the expression 2x² + 4y + 3x². Solution Notice that there are two like terms, 2x² and 3x²: You can combine the like terms: 2x² + 3x² = 5x² So 2x² + 4y + 3x² = 5x² + 4y Guided Practice Simplify each of the following polynomials. 17. x + 1 + 2x 19. 9x² + 4x + 7x 18. 3y + 2y 20. x² + x + x² Simplify each of the following polynomials, then state whether your answer is a monomial, binomial, or trinomial. 21. 3x2 + 4 – 8 + x2 23. 3x2y – 2x2y + 8 25. 4x3 + 7 – x3 + 4 – 3x3 – 11 27. 3xy + 4xy + 5x2y – 4xy2 22. 8x3 + x4 – 6x3 + 4 24. 7 – 2y + 3 – 10 26. 5x2 + 9x2 + 4 + 2y 28. 9x5 + 2x2 + 4x4 + 5x5 – 3x4 – x2 Finding the Degree of a Polynomial The degree of a polynomial in x is the size of the highest power of x in the expression. If you see the phrase “a fourth-degree polynomial in x,” you know that it will contain at least one term with x4, but it won’t contain any higher powers of x than 4. For example: has degree 1 2x + 1 y² + y – 3 has degree 2 x4 – x² has degree 4 — it’s a first-degree polynomial in x — it’s a second-degree polynomial in y — it’s a fourth-degree polynomial in x 264 Section 6.1 — Adding and Subtracting Polynomials Guided Practice State the degree of each of the following polynomials. 29. 3x + 5 31. x² + 2x + 3 30. 3x4 + 2 32. x + 4 + 2x² Simplify and state the degree of each of the following polynomials. 33. 2x + x2 + x – 3 34. 3a3 + 4a – 2a3 + 4a2 35. 4x3 + 4x8 – 3x8 + 2x3 36. 3y + 2y – 5y2 + 6y 37. b13 + 2b13 – 8 + 4 – 3b13 38. z3 + z3 – z6 + z7 + 3z7 39. c4 + c3 + c3 – c4 + c – 2c3 40. x – 2x9 – 8x4 + 13x2 Independent Practice For the polynomials below state whether they are a monomial, a binomial, or a trinomial. 1. 19a2 + 16 3. 42xy 2. 2c – 4a + 6 4. 16a2b + 4ab2 Simplify each of the following polynomials. 5. 0.7x2 + 9.8 – x2 6. 17x2 – 14x9 + 7x9 – 7x2 + 7x9 7. 0.8x4 + 0.3x2 + 9.6 – x2 – 9x4 + 1.6x2 State the degree of the following polynomials. 8. x – 9x6 + 4 9. 14x8 + 16x10 + 4x8 10. 2x2 – 4x4 + 7x5 11. 2x2 – 4x + 8 Simplify each polynomial, state the degree of the polynomial, and determine whether it is a monomial, a binomial, or a trinomial. 12. 93a2 + 169 – 4a – 81a2 + 7 13. 7.9x2 – 13x4 – 1.5x4 + 1.4x2 1 15. 16. 1 9 14. 5x9 – 6x9 + 4 + x9 – 3 – 1 2 x9 – x3 – x3 + 1 x6 – 3 x10 – 3 5 4 x9 + 7 9 x10 – 3 4 17. When a third degree monomial is added to a second degree binomial, what is the result? 3 x6 + 1 9 18. When a 4th degree monomial is added to a 6th degree binomial, what are the possible results? Round Up Round Up This Topic gets you started on manipulating polynomials, by simplifying them. In the next couple of Topics you’ll see how to add and subtract polynomials. Section 6.1 — Adding and Subtracting Polynomials 265 Topic 6.1.2 California Standards: 2.0: Students understand and use such operations as taking the opposite, finding the reciprocal, taking a root, and raising to a fractional power. They understand and use the rules of exponents. 10.0: Students add, subtract, multiply, and divide monomials and polynomials. Students solve multistep problems, including word problems, by using these techniques. What it means for you: You’ll add polynomials and multiply a polynomial by a number. Key words: polynomial like terms inverse Adding Polynomials Adding Polynomials Adding polynomials isn’t difficult at all. The only problem is that you can only add certain parts of each polynomial together. The Opposite of a Polynomial The opposite of a number is its additive inverse. The opposite of a positive number is its corresponding negative number, and vice versa. For example, –1 is the opposite of 1, and 1 is the opposite of –1. To find the opposite of a polynomial, you make the positive terms negative and the negative terms positive. Example 1 Find the opposites of the following polynomials: a) 2x – 1 b) –5x² + 3x – 1 Solution a) –2x + 1 b) 5x² – 3x + 1 Don’t forget: See Topic 1.2.
2 for more on additive inverses. Guided Practice Find the opposites of the following polynomials. 1. 2x + 1 3. x² + 5x – 2 5. 3x2 + 4x – 8 7. 4x4 – 16 9. 5x4 – 6x2 + 7 11. –0.9x3 – 0.8x2 – 0.4x – 1.0 2. –5x – 1 4. 3x² – 2x + 3 6. –8x2 – 4x + 4 8. 8x3 – 6x2 + 6x – 8 10. –2x4 + 3x3 – 2x2 12. –1.4x3 – 0.8x2 – 1 2 x 266 Section 6.1 — Adding and Subtracting Polynomials Adding Polynomials Adding polynomials consists of combining all like terms. There are a few ways of adding polynomials — two of the methods are shown in Example 2. Example 2 Find the sum of –5x² + 3x – 1, 6x² – x + 3, and 5x – 7. Solution Method A — Collecting Like Terms and Simplifying (–5x² + 3x – 1) + (6x² – x + 3) + (5x – 7) = –5x² + 3x – 1 + 6x² – x + 3 + 5x – 7 = –5x² + 6x² + 3x – x + 5x – 1 + 3 – 7 = x² + 7x – 5 Method B — Vertical Lining Up of Terms –5x² + 3x – 1 + 6x² – x + 3 + 5x – 7 x² + 7x – 5 Both methods give the same solution. Don’t forget: See Topic 1.2.4 for the definition of multiplication. Multiplying a Polynomial by a Number Multiplying a polynomial by a number is the same as adding the polynomial together several times. Example 3 Multiply x + 3 by 3. Solution (x + 3) × 3 = (x + 3) + (x + 3) + (x + 3 = 3x + 9 Don’t forget: See Section 1.2 for more about the distributive property. The simple way to do this is to multiply each term of the polynomial by the number. In other words, you multiply out the parentheses, using the distributive property of multiplication over addition. Section 6.1 — Adding and Subtracting Polynomials 267 Example 4 Multiply x² + 2x – 4 by 3. Solution 3(x² + 2x – 4) = (3 × x²) + (3 × 2x) – (3 × 4) = 3x² + 6x – 12 Guided Practice Add these polynomials and simplify the resulting expressions. 13. (4x2 – 2x – 1) + (3x2 + x – 10) 14. (11x4 – 5x3 – 2x) + (–7x4 + 3x3 + 5x – 3) 15. –5c3 – 3c2 + 2c + 1 4c2 – c – 3 6c3 + c + 4 16. 5x2 + 3x – 3 –4x2 – 3x + 5 –2x2 + x – 7 Multiply these polynomials by 4. 17. 10y2 – 7y + 5 19. (–x² + x – 4) 18. (x² – 3x + 3) 20. (2x² + 5x + 2) Independent Practice In Exercises 1-7, simplify the expression and state the degree of the resulting polynomial. 1. (2x2 + 3x – 7) + (7x2 – 3x + 4) 2. (x3 + x – 4) + (x3 – 8) + (4x3 – 3x – 1) 3. (–x6 + x – 5) + (2x6 – 4x – 6) + (–2x6 + 2x – 4) 4. (3x2 – 2x + 7) + (4x2 + 6x – 8) + (–5x2 + 4x – 5) 5. (0.4x3 – 1.1) + (0.3x3 + x – 1.0) + (1.1x3 + 2.1x – 2.0) 6. 7. – 4a3 – 2a + 3 8a4 – 2a3 – 4a + 8 7a4 – 4a – 7 1.1c2 + 1.4c – 0.48 –4.9c2 – 3.6c + 0.98 7.3c2 + 0.13 Multiply each polynomial below by –4. 8. 4a2 + 3a – 2 10. –6x3 – 4x2 + x – 8 Multiply each polynomial below by 2a. 12. 3a2 + a – 8 9. –c2 + 3c + 1 11. 24x3 + 16x2 – 4x + 32 13. –7a4 + 2a2 – 5a + 4 Round Up Round Up Adding polynomials can look hard because there can be several terms in each polynomial. The important thing is to combine each set of like terms, step by step. 268 Section 6.1 — Adding and Subtracting Polynomials Topic 6.1.3 California Standards: 2.0: Students understand and use such operations as taking the opposite, finding the reciprocal, taking a root, and raising to a fractional power. They understand and use the rules of exponents. 10.0: Students add, subtract, multiply, and divide monomials and polynomials. Students solve multistep problems, including word problems, by using these techniques. What it means for you: You’ll learn how to subtract polynomials. Key words: polynomial like terms inverse Subtracting Polynomials Subtracting Polynomials Subtracting one polynomial from another follows the same rules as adding polynomials — you just need to combine like terms, then carry out all the subtractions to simplify the expression. Subtracting Polynomials Subtracting polynomials is the same as subtracting numbers. To subtract Polynomial A from Polynomial B, you need to subtract each term of Polynomial A from Polynomial B. Then you can combine any like terms to simplify the expression. Example 1 Subtract Polynomial A from Polynomial B, where Polynomial A = x² + x and Polynomial B = x² + 4x. Solution Subtract each term of Polynomial A from Polynomial B: Polynomial B – Polynomial A = x² + 4x – (x²) – (x) = x² – x² + 4x – x = 0 + 3x = 3x Guided Practice 1. Subtract x2 – 4 from x2 + 8. 2. Subtract 3x – 4 from 8x2 – 5x + 4. 3. Subtract x + 4 from x2 – x. 4. Subtract x2 – 16 from x2 + 8. 5. Subtract x2 + x – 1 from x + 4. 6. Subtract –3x2 + 4x – 5 from x2 – 7. 7. Subtract –3x2 – 5x + 2 from –2x3 – x2 – 7x. Simplify: 8. (9a – 10) – (5a + 2) 9. (5a2 – 2a + 3) – (3a + 5) 10. (x3 + 5x2 – x) – (x2 + x) Section 6.1 — Adding and Subtracting Polynomials 269 Don’t forget: See Topic 1.2.5 for the definition of subtraction. Subtracting is Simply Adding the Opposite Another way to look at subtraction of polynomials is to go back to the definition of subtraction. When you subtract Polynomial A from Polynomial B, what you’re actually doing is adding the opposite of Polynomial A to Polynomial B. Example 2 Subtract –5x² + 3x – 8 from –7x² + x + 5. Solution –7x² + x + 5 – (–5x² + 3x – 8) = –7x² + x + 5 + 5x² – 3x + 8 = –7x² + 5x² + x – 3x + 5 + 8 = –2x² – 2x + 13 Alternatively, you can do subtraction by lining up terms vertically: Example 3 Subtract –5x² + 3x – 8 from –7x² + x + 5. OR –7x² + x + 5 + (5x² – 3x + 8) –2x² – 2x + 13 This is the opposite of –5x² + 3x – 8 Solution –7x² + x + 5 – (–5x² + 3x – 8) –2x² – 2x + 13 Guided Practice Simplify the expressions in Exercises 11–16. 11. (3a4 + 4) – (2a2 – 5a4) 12. (6x2 + 8 – 9x4) – (3x – 4 + x3) 13. (9c2 + 11c2 + 5c – 5) – (–10 + 4c4 – 8c + 3c2) 14. (8a2 – 2a + 5a) – (9a2 + 2a + 4) 15. 6x2 – 6 –(5x2 + 9) 16. 8a2 + 4a – 9 –(3a2 – 3a + 7) 17. Subtract 7a3 + 3a – 12 from 5a2 – a + 4 by adding the opposite expression. Use the vertical lining up method. 18. Subtract (8p3 – 11p2 – 3p) from 4p3 + 6p2 – 10 by adding the opposite expression. Use the vertical lining up method. 270 Section 6.1 — Adding and Subtracting Polynomials If you add a polynomial to its opposite, the result will always be 0. Example 4 Find the sum of –5x² + 3x – 1 and 5x² – 3x + 1. Check it out: You can use this method to check answers to “opposites” questions. Solution –5x² + 3x – 1 + (5x² – 3x + 1) = –5x² + 3x – 1 + 5x² – 3x + 1 = –5x² + 5x² + 3x – 3x – Independent Practice Subtract the polynomials and simplify the resulting expression. 1. (5a + 8) – (3a + 2) 2. (8x – 2y) – (8x + 4y) 3. (–4x2 + 7x – 3) – (2x2 – 4x + 6) 4. (3a2 + 2a + 6) – (2a2 + a + 3) 5. –3x4 – 2x3 + 4x – 1 – (–2x4 – x3 + 3x2 – 5x + 3) 6. 5 – [(2k + 3) – (3k + 1)] 7. – 10a2 + 4a – 1 – (7a2 + 4a) 8. (x2 + 4x + 6) – (2x2 + 2x + 4) Solve these by first simplifying the left side of the equations. 9. (2x + 3) – (x – 7) = 40 11. (2 – 3x) – (7 – 2x) = 23 10. (4x + 14) – (–10x – 3) = 73 12. (17 – 5x) – (4 – 3x) – (6 – x) = 19 Find the opposite of the polynomials below. 13. x2 + 2x + 1 15. 4b2 – 6bc + 7c 14. –a2 + 6a + 4 16. a3 + 4a2 + 3a – 2 17. The opposite of a fifth degree polynomial has what degree? 18. If a monomial is subtracted from another monomial, what are the possible results? 19. What is the degree of the polynomial formed when a 2nd degree polynomial is subtracted from a 1st degree polynomial? 20. A 3rd degree polynomial has a 2nd degree polynomial subtracted from it. What is the degree of the resulting polynomial? Round Up Round Up Watch out for the signs when you’re subtracting polynomials. It’s usually a good idea to put parentheses around the polynomial you’re subtracting, to make it easier to keep track of the signs. Section 6.1 — Adding and Subtracting Polynomials 271 Topic 6.1.4 California Standards: 2.0: Students understand and use such operations as taking the opposite, finding the reciprocal, taking a root, and raising to a fractional power. They understand and use the rules of exponents. 10.0: Students add, subtract, multiply, and divide monomials and polynomials. Students solve multistep problems, including word problems, by using these techniques. What it means for you: You’ll use addition and subtraction of polynomials to solve more complicated problems. Key words: polynomial like terms Don’t forget: Remember that the perimeter of a rectangle is the sum of the lengths of all four sides. Don’t forget: Be really careful in part b) — the minus sign applies to both terms inside the parentheses. Adding and Subtracting Adding and Subtracting Polynomials Polynomials This Topic just contains one big example that uses both addition and subtraction of polynomials. An Example Using Addition and Subtraction Example 1 Find a) the sum of and b) the difference between the perimeters of the rectangles shown below 2x² – 3 x x² + 2 + 1 P 2 x² – 1 Solution The first thing you need to do is find and simplify expressions for each of the perimeters. Perimeter of P1 = (3x² – 3x – 2) + (3x² – 3x – 2) + (2x² – 3) + (2x² – 3) = 3x² – 3x – 2 + 3x² – 3x – 2 + 2x² – 3 + 2x² – 3 = 3x² + 3x² + 2x² + 2x² – 3x – 3x – 2 – 2 – 3 – 3 = 10x² – 6x – 10 Perimeter of P2 = (x² + 2x + 1) + (x² + 2x + 1) + (x² – 1) + (x² – 1) = x² + 2x + 1 + x² + 2x + 1 + x² – 1 + x² – 1 = x² + x² + x² + x² + 2x + 2x + 1 + 1 – 1 – 1 = 4x² + 4x a) Perimeter of P1 + Perimeter of P2 = (10x² – 6x – 10) + (4x² + 4x) = 10x² – 6x – 10 + 4x² + 4x = 10x² + 4x² – 6x + 4x – 10 = 14x² – 2x – 10 b) Perimeter of P1 – Perimeter of P2 = (10x² – 6x – 10) – (4x² + 4x) = 10x² – 6x – 10 – 4x² – 4x = 10x² – 4x² – 6x – 4x – 10 = 6x² – 10x – 10 272 Section 6.1 — Adding and Subtracting Polynomials Guided Practice Find the perimeter of each of these shapes. 2 4(5 – 3 + 1) x x 1. 2. 2 x 5( – x + 3) 3. 2 + 4 x x2 6( + ) x 3( – 1) x 2( + 7) x x2 7( + ) x Independent Practice Simplify each of these expressions. 1. 4y – [3(y – x) – 5(x + y)] 2. 3[2y – 3(y – 1) + 2(–y – 1)] 3. 3(x – y) – 2(x – 3y) 4. –2[(x – y) – 2(x + 2y)] – [(–2x – y) – (x + 2y)] 5. (–3x3 – x2y + 2xy2 – 2y3) – (–5x3 + x2y – 3xy2 – 3y3) + (–x3 – 2x2 y + y3) 3 – 3x 2) – x 3( 2 ( 7 – x ) 1) x – 3(2 2 ( 9 – x ) 6. Find the difference between the perimeters of the trapezoid and the triangle shown. 3(2 x – 1) 5 – 1x 7. Simplify 5(4x
+ 3) – 4(3x – 1) + 3(x – 1). 8. Simplify (8x3 – 5x2 – 2x – 7) – (6x3 – 3x2 + x – 5) – (–3x2 + 3x – 2). 9. Simplify –2(3x3 – 2x) – 3(–x3 + 7) – (2x2 – 5x3 – 5). 10. Find the sum of the opposites of: –2x3 + 3x2 – 5x + 1 and 3x3 – 2x2 + 3x – 3 11. Find the difference between the opposites of: –2x2 – 3x + 5 and 3x2 + 2x – 4 12. Simplify –2(3x2p – 2xp + 1) + 4(2x2p – xp – 3) – 5(x2p – 2xp – 1). 13. If the perimeter of the figure shown is 90 inches, what are the dimensions of the figure? 2 ( x – 3 ) x 3(2 + 1) Round Up Round Up You’ve covered addition and subtraction of polynomials in just four short Topics. As you might expect, the next Section covers multiplication of polynomials, and then you’ll learn about division of polynomials in Section 6.3. Section 6.1 — Adding and Subtracting Polynomials 273 Topic 6.2.1 Section 6.2 Rules of Exponents Rules of Exponents California Standards: 2.0: Students understand and use such operations as taking the opposite, finding the reciprocal, taking a root, and raising to a fractional power. They understand and use the rules of exponents. 10.0: Students add, subtract, multiply, and divide monomials and polynomials. Students solve multistep problems, including word problems, by using these techniques. What it means for you: You’ll multiply and divide algebraic expressions using the rules of exponents. Key words: exponent Don’t forget: See Topic 1.3.1 for more on the rules of exponents. Don’t forget: Rules 3 and 4 mean that here you multiply each power from inside the parentheses by 2, the power outside the parentheses. You learned about the rules of exponents in Topic 1.3.1 — in this Topic, you’ll apply those same rules to monomials and polynomials. Use the Rules of Exponents to Simplify Expressions These are the same rules you learned in Chapter 1, but this time you’ll use them to simplify algebraic expressions: Rules of Exponents 1) xa·xb = xa+b 3) (xa)b = xab 2) xa ÷ xb = xa–b (if x π 0) 4) (cx)b = cbxb a b 5) x = b a x or ( b a ) x 6) x–a = 1 x a (if x π 0) 7) x0 = 1 Example 1 Simplify the expression (–2x2m)(–3x3m3). Solution (–2x2m)(–3x3m3) = (–2)(–3)(x2)(x3)(m)(m3) = 6x2+3·m1+3 = 6x5m4 Put all like variables together Use Rule 1 and add the powers Example 2 Simplify the expression (3a2xb3)2. Solution (3a2xb3)2 = 32·a2·2·x2·b3·2 = 9a4x2b6 Use Rules 3 and 4 274 Section 6.2 — Multiplying Polynomials Example 3 Simplify the expression − 10 − mv . Solution − 5 3 2 10 x m v − 4 2 5 x mv − 5 10 Separate the expression into parts that have only one variable Use Rule 2 and subtract the powers = 2xm2 From Rule 5, anything to the power 0 is just 1 Guided Practice Simplify each expression. 1. –3at(4a2t3) 3. (–2x2y3)3 5. (–3x2t)3(–2x3t2)2 7. 10 12 10 j k m 4 9. 2. (–5x3yt2)(–2x2y3t) 4. –2mx(3m2x – 4m2x + m3x3) 6. –2mc(–3m2c3 + 5mc) 8. 2 4 8 a b c 14 7 4 a b 4 10. 9 4 4 16 b a j 5 3 ( b a c 32 ) 2 Independent Practice Simplify. ⎛ 1. 1 ⎜⎜⎜ ⎝ 2 3 4 a b ⎞ 0 ⎟⎟⎟ ⎠ ⎛ 2. 2 ⎜⎜⎜ ⎝ 8 4 3 xy z − ⎞ 2 ⎟⎟⎟ ⎠ 3. 4a2(a2 – b2) 4. 4m2x2(x2 + x + 1) 5. a(a + 4) + 4(a + 4) 6. 2a(a – 4) – 3(a – 4) 7. m2n3(mx2 + 3nx + 2) – 4m2n3 8. 4m2n2(m3n8 + 4) – 3m3n10(m2 + 2n3. b a 6 19 b a Find the value of ? that makes these statements true. 10. 11. 10 17 2 7 12. 4 4 20 3 a m 7 a m 6 13. m?(m4 + 2m3) = m6 + 2m5 14. m4a6(3m?a8 + 4m2a?) = 3m7a14 + 4m6a9 15 16. 4 18 ? 8 7 x y c ? 10 x y c 6 = 2 9 yc 3 x Round Up Round Up You can apply the rules of exponents to any algebraic values. In this Topic you just dealt with monomials, but the rules work with expressions with more than one term too. Section 6.2 — Multiplying Polynomials 275 Topic 6.2.2 California Standards: 2.0: Students understand and use such operations as taking the opposite, finding the reciprocal, taking a root, and raising to a fractional power. They understand and use the rules of exponents. 10.0: Students add, subtract, multiply, and divide monomials and polynomials. Students solve multistep problems, including word problems, by using these techniques. What it means for you: You’ll learn how to multiply monomials and polynomials. Key words: polynomial monomial distributive property degree Polynomial Multiplication Polynomial Multiplication To multiply two polynomials together, you have to multiply every single term together, one by one. The Distributive Property and Polynomial Products In Topic 6.1.2 you saw the method for multiplying a polynomial by a number — you multiply each term separately by that number. This method’s based on the distributive property from Topic 1.2.7. In the same sort of way, when you multiply a polynomial by a monomial, you multiply each term separately by that monomial — again, using the distributive property. Example 1 Simplify the expression –2a(a + 3a2). Solution –2a(a + 3a2) is a product of the monomial –2a and the binomial (a + 3a2), so multiply each term of the binomial by the monomial. = –2a(a) + (–2a)(3a2) = –2a2 – 6a3 To find the product of two polynomials, such as (a – 2b)(3a + b), you use the distributive property twice: Example 2 Simplify the expression (a – 2b)(3a + b). Solution (a – 2b)(3a + b) = (a)(3a + b) + (–2b)(3a + b) = [(a)(3a) + (a)(b)] + [(–2b)(3a) + (–2b)(b)] = (3a2 + ab) + (–6ab – 2b2) = 3a2 + ab – 6ab – 2b2 = 3a2 – 5ab – 2b2 Use the distributive property twice 276 Section 6.2 — Multiplying Polynomials Example 3 Simplify (3x – 2m)(4x – 3m). Solution (3x – 2m)(4x – 3m) = 3x(4x – 3m) – 2m(4x – 3m) = 12x2 – 9mx – 8mx + 6m2 = 12x2 – 17mx + 6m2 Example 4 Simplify (v + 3)(4 + v). Solution (v + 3)(4 + v) = v(4 + v) + 3(4 + v) = 4v + v2 + 12 + 3v = v2 + 7v + 12 Guided Practice Expand and simplify each product, using the distributive method. Show all your work. 1. (m + c)(m + 2c) 3. (2x – 3)(2x + 5) 5. (3x – 5)(2x – 3) 2. (x – 3y)(x + 2y) 4. (a – 4b)(a + 3b) 6. (5x + 3y)(2x + 3y) Determine whether the following are correct for the products given. 7. (a + b)(a – b) = a2 – b2 9. (a – b)2 = a2 – 2ab + b2 8. (a + b)2 = a2 + b2 10. (a + b)(a + b) = a2 + 2ab + b2 You Can Multiply Polynomials with Lots of Terms Check it out: Multiply each term in one set of parentheses by every term in the second set of parentheses. Example 5 Simplify (x + 2)(x2 + 2x + 3). Solution (x + 2)(x2 + 2x + 3) = x(x2 + 2x + 3) + 2(x2 + 2x + 3) = x3 + 2x2 + 3x + 2x2 + 4x + 6 = x3 + 4x2 + 7x + 6 Section 6.2 — Multiplying Polynomials 277 The Highest Power Gives the Degree of a Polynomial Example 6 Simplify (x – 3)(2x2 – 3x + 2) and state the degree of the product. Solution (x – 3)(2x2 – 3x + 2) = x(2x2 – 3x + 2) – 3(2x2 – 3x + 2) = 2x3– 3x2 + 2x – 6x2 + 9x – 6 = 2x3 – 9x2 + 11x – 6 The term 2x3 has the highest power, so the degree is 3. Guided Practice Expand and simplify each product, and state the degree of the resulting polynomial. 11. (x + 3)(2x2 – 3x + 1) 13. (x2 – 3x + 4)(2x + 1) 15. (3x + 4)(–2x2 + x – 2) 12. (2y – 3)(–3y2 – y + 1) 14. (3y3 + 4y – 2)(4y – 1) 16. (2x – 3)2 Determine whether the following are correct for the products given. 17. (a2 + b2)(a – ab + b) = a3 + b3 18. (a + b)(a2 – ab + b2) = a3 + b3 19. (a – b)(a2 + ab + b2) = a3 – b3 20. (a2 – b2)(a + ab + b) = a3 – b3 You Can Also Use the Stacking Method You can find the product of 63 and 27 by “stacking” the two numbers and doing long multiplication: Don’t forget: The units, tens, hundreds, and thousands are in separate columns × 63 2 × 63 You can use the same idea to find the products of polynomials — just make sure you keep like terms in the same columns. Example 7 Expand and simplify the product (2x + 3y)(x + 5y). Don’t forget: Keep like terms in the same column. Solution 2x + 3yy × x + 5yy 10xy + 15y2 + 2x2 + 3xy + 15y2 2x2 + 13xy + 15y2 5y(2x + 3y) x(2x + 3y) 278 Section 6.2 — Multiplying Polynomials Example 8 Simplify (x – 2)(2x2 – 3x + 4). Solution 2x2 – 3x + 4 × x – 2 –4x2 + 6x – 8 2x3 – 3x2 + 4x –n8 2x3 – 7x2 + 10x – 8 Guided Practice –2(2x2 – 3x + 4) x(2x2 – 3x + 4) Use the stacking method to multiply these polynomials: 21. (3x + y)(x + 2y) 23. (3x2 + 2x + 3)(3x – 4) 25. (a + b)2 27. (a – b)(a + b) 29. (a + b)(a2 – ab + b2) 22. (4x + 5y)(2x + 3y) 24. (4x2 – 5x + 6)(4x + 5) 26. (a – b)2 28. (a – b)(a2 + ab + b2) 30. (a2 – b2)(a2 + b2) Independent Practice Expand and simplify each product, using the distributive method. Show all your work. 1. (2x + 8)(x – 4) 3. (x – 3)(2 – x) 5. (3x – 8)(x2 – 4x + 2) 2. (x2 + 3)(x – 2) 4. (2x + 7)(3x + 5) 6. (2x – 4y)(3x – 3y + 4) Use the stack method to multiply. Show all your work. 7. (x2 – 4)(x + 3) 9. (4x2 – 5x)(1 + 2x – 3x2) 8. (x – y)(3x2 + xy + y2) 10. (x + 4)(3x2 – 2x + 5) Use these formulas to find each of the products in Exercises 11–16. (a + b)2 = a2 + 2ab + b2 (a – b)2 = a2 – 2ab + b2 (a + b)(a – b) = a2 – b2 11. (x + 2)2 13. (2x – 3)2 15. (5x + 3c)(5x – 3c) 12. (3x – 1)(3x + 1) 14. (4x + y)2 16. (8c + 3)2 Round Up Round Up You’ve seen the distributive property lots of times already, but it’s easy to make calculation errors when you’re multiplying long polynomials — so be careful and check your work. Section 6.2 — Multiplying Polynomials 279 Topic 6.2.3 California Standards: 2.0: Students understand and use such operations as taking the opposite, finding the reciprocal, taking a root, and raising to a fractional power. They understand and use the rules of exponents. 10.0: Students add, subtract, multiply, and divide monomials and polynomials. Students solve multistep problems, including word problems, by using these techniques. What it means for you: You’ll multiply polynomials to solve problems involving area and volume. Key words: polynomial monomial distributive property Polynomial Multiplication Polynomial Multiplication — Area and Volume — Area and Volume Polynomial multiplication isn’t just about abstract math problems. Like everything in math, you can use it to work out problems dealing with everyday life. Find Areas by Multiplying Polynomials x x x x Example 1 Find the area of the space between the two rectangles: x (3 + 2) inches x x (5 + 6) inches x x Solution The length of the middle rectangle is 5x + 6 – 2x = (3x +
6) inches. The width of the middle rectangle is 3x + 2 – 2x = (x + 2) inches. x Area of space = area of large rectangle – area of small rectangle = (5x + 6)(3x + 2) – (3x + 6)(x + 2) = 15x2 + 10x + 18x + 12 – (3x2 + 6x + 6x + 12) = 15x2 + 28x + 12 – 3x2 – 12x – 12 = (12x2 + 16x) in2 Guided Practice 1. Find the area of a rectangle whose dimensions are (3x + 4) inches by (2x + 1) inches. 2. Find the area of the triangle shown on the right. 1 (The formula for the area of a triangle is Area = 2 bh.) 3. The height of a triangle is (3x – 2) inches and its base is (4x + 10) inches. Find the area of the triangle. 4. Find the area of a rectangle whose dimensions are (3 + 2x) inches by (5 + 6x) inches. 2x – 3 5x + 3 5. Find the area of a square with side length (a2 + b2 – c2) feet. 6. The area of a trapezoid is given by A = ½h(b1 + b2) where h is the height and b1 and b2 are the lengths of the parallel sides. Find the area of this trapezoid. x in. (x + 1) in. (x + 4) in. 280 Section 6.2 — Multiplying Polynomials Multiply Polynomials to Find Volumes Example 2 Find the volume of the box on the right. (4 + 6) inches x (6 – 4) inches x Solution Volume = Length × Width × Height (5 + 8) inches x = (5x + 8)(6x – 4)(4x + 6) = [5x(6x – 4) + 8(6x – 4)](4x + 6) Multiply the first two polynomials = (30x2 – 20x + 48x – 32)(4x + 6) = (30x2 + 28x – 32)(4x + 6) Simplify the first product = 4x(30x2 + 28x – 32) + 6(30x2 + 28x – 32) Multiply out again = 120x3 + 112x2 – 128x + 180x2 + 168x – 192 = 120x3 + 112x2 + 180x2 – 128x + 168x – 192 Commutative law = (120x3 + 292x2 + 40x – 192) in3 Example 3 Find the volume of a box made from the sheet below by removing the four corners and folding. 2x 2x 2x 2x 2x 2x 2x 2x n i 6 2x in 8 in Solution Volume = Length × Width × Height (8 – 4x) in (6 – 4x) in = (8 – 4x)(6 – 4x)(2x) = [8(6 – 4x) – 4x(6 – 4x)](2x) = (48 – 32x – 24x + 16x2)(2x) = (48 – 56x + 16x2)2x = 96x – 112x2 + 32x3 = (32x3 – 112x2 + 96x) in3 Multiply the first two polynomials Simplify the first product Multiply by the third polynomial Section 6.2 — Multiplying Polynomials 281 Guided Practice 7. Find the volume of a cube with side length (3x + 6) inches. 8. A concrete walkway around a swimming pool is 6 feet wide. If the length of the pool is twice the width, x feet, what is the combined area of the walkway and pool? 6 ft 6 ft x ft 6 ft 6 ft 2 ftx Use the rectangular prism shown to answer these questions. 9. Find the volume of the prism. 10. Find the surface area of the prism. 11. If the height of the prism was reduced by 10%, what would be the new volume of the prism? x ( + 7) ft (3 – 1) ft x (2 + 3) ft x Independent Practice Expand and simplify the following. 1. (3y + 5)3 2. (2y – 1)3 3. The area of a parallelogram is given by the formula A = bh, where b is the length of the base and h is the height of the parallelogram. Find the area of a parallelogram that has a base length of (2x2 + 3x – 1) cm and a height of (3x – 1) cm. 4. A gardener wants to put a walkway around her garden, as shown on the right. What is the area of the walkway? x x x x x (2 + 5) feet (7 + 3) feet x 5. Obike made a box from a 10 inch by 9 inch piece of cardboard by cutting squares of x units from each of the four corners. Find the volume of his box. x x x x 10 inches 6. Find the volume of the solid shown. 7. Find the volume of another triangular prism that has the same base measurements as this one but a height 25% less than the height shown here. x ( + 4) ft ) ft x ( + 6) ft Round Up Round Up For problems involving area, you’ll have to multiply two terms or polynomials together. For problems involving volume, you’ll have to multiply three terms or polynomials. 282 Section 6.2 — Multiplying Polynomials TTTTTopicopicopicopicopic 6.3.16.3.1 6.3.16.3.1 6.3.1 Section 6.3 y Monomials y Monomials vision b vision b DiDiDiDiDivision b y Monomials vision by Monomials y Monomials vision b DiDiDiDiDivision b y Monomials y Monomials vision b vision b vision by Monomials y Monomials vision b y Monomials California Standards: Students Students 10.0: 10.0: 10.0: Students Students add, subtract, 10.0: Students 10.0: dividevidevidevidevide di di multiply, and di di ynomials..... ynomials ynomials monomials and pol monomials and pol monomials and polynomials monomials and pol ynomials monomials and pol Students solve multistep problems, including word problems, by using these techniques. What it means for you: You’ll learn how to use the rules of exponents to divide a polynomial by a monomial. Key words: polynomial monomial exponent distributive property Don’t forget: You covered the rules of exponents in Topics 1.3.1 and 6.2.1. Those rules of exponents that you saw in Topic 6.2.1 really are useful. In this Topic you’ll use them to divide polynomials by monomials. y a Monomial y a Monomial ynomial b viding a Polololololynomial b ynomial b viding a P DiDiDiDiDividing a P viding a P y a Monomial ynomial by a Monomial y a Monomial ynomial b viding a P To divide a polynomial by a monomial, you need to use the rules of exponents. The particular rule that’s useful here is: a b = xa–b provided x π 0 x x Example Example Example Example Example 11111 Divide 2x² by x. Solution Solution Solution Solution Solution 2 2x x = 2x2–1 = 2x1 = 2x Example Example Example Example Example 22222 Divide 2x³y + xy² by xy. Don’t forget: Like with multiplying, you need to treat the powers of x and y separately. Solution Solution Solution Solution Solution 2 3 x y + 2 xy xy = = ⎛ ⎜⎜⎜⎜ ⎝ ⎛ ⎜⎜⎜⎜ ⎝ 2 3 x y xy ⎞ ⎟⎟⎟⎟ ⎠ + 2 3 x x ⋅ y y ⎞ ⎟⎟⎟⎟ ⎠ 2 ⎞ ⎟⎟⎟⎟ ⎠ ⎛ ⎜⎜⎜⎜ ⎝ xy xy ⎛ x ⎜⎜⎜⎜ + ⋅ ⎝ x ⎞ ⎟⎟⎟⎟ ⎠ 2 y y DiDiDiDiDivide eac vide eac vide eac h ter h ter m in the m in the vide each ter h term in the m in the vide eac h ter m in the ession by y y y y xy,,,,, using ession b ession b eeeeexprxprxprxprxpression b using using using ession b using opertytytytyty oper oper e pre proper e pre pr the distributiutiutiutiutivvvvve pr the distrib the distrib oper the distrib the distrib = (2x3–1 ◊ y1–1) + (x1–1 ◊ y2–1) = (2x2 ◊ 1) + (1 ◊ y1) = 2x2 + y Section 6.3 Section 6.3 Section 6.3 — Dividing Polynomials Section 6.3 Section 6.3 283283283283283 Example Example Example Example Example 33333 Simplify mc v 2 10 4 3 mc v . Solution Solution Solution Solution Solution 2 3 2 m c v − 10 4 3 mc mc v 2 = ⎛ ⎜⎜⎜⎜ ⎝ 3 2 m c v 2 − 2 mc v 2 ⎞ ⎟⎟⎟⎟ ⎠ − ⎛ ⎜⎜⎜⎜ ⎝ − 2 3 m c v 4 − 2 mc v 2 ⎞ ⎟⎟⎟⎟ ⎠ + ⎛ ⎜⎜⎜⎜ ⎝ ⎞ 10 2− mc v33 4 2 ⎟⎟⎟⎟mc v ⎠ = (–1◊m3–1) – (–2◊m2–1◊c3–2) + (–5◊c4–2◊v3–1) = –m2 + 2mc – 5c2v2 Guided Practice Simplify each of these quotients. Don’t forget: You can cancel anything that is on both the top and the bottom of the fraction. See Topic 6.3.3. 1. 9m3c2v4 ÷ (–3m2cv3) − 10 3. 3 2 m x − 5 + 15 12 5 5. −14 . 4. 6. − 12 − 12 df k 3 − 12 dk + 28 8 8 ca d k 4 Independent Practice Simplify each of these quotients − 12 1. 3. 12 x 2 − 15 − 16 2 2 2 m c v 4 4. 2 3 xy xyz 5. Divide 15x5 – 10x3 + 25x2 by –5x2. 6. Divide 20a6b4 – 14a7b5 + 10a3b7 by 2a3b4. 7. Divide 4m5x7v6 – 12m4c2x8v4 + 16a3m6c2x9v7 by –4m4x7v4. Find the missing exponent in the quotients. − 2 x y ? 9. 8 xy 4x ound Up ound Up RRRRRound Up ound Up ound Up ound Up ound Up RRRRRound Up ound Up ound Up This leads on to the next few Topics, where you’ll divide one polynomial by another polynomial. First, you’ll learn how to find the multiplicative inverse of a polynomial in Topic 6.3.2. 284284284284284 Section 6.3 Section 6.3 Section 6.3 — Dividing Polynomials Section 6.3 Section 6.3 TTTTTopicopicopicopicopic 6.3.26.3.2 6.3.26.3.2 6.3.2 California Standards: 2.0:2.0:2.0:2.0:2.0: Students under stand stand Students under Students under Students understand stand stand Students under tions as tions as h operaaaaations as h oper h oper and use suc and use suc tions as and use such oper and use suc tions as h oper and use suc taking the opposite,,,,, finding finding finding finding finding ocal, taking a root, ocal, ocal, ecipr ecipr the r the r eciprocal, the recipr the r ocal, ecipr the r and raising to a fractional power..... TTTTThehehehehey under stand stand y under y under stand y understand stand y under and use the rules of and use the rules of and use the rules of and use the rules of and use the rules of eeeeexponents xponents..... xponents xponents xponents Students Students 10.0: 10.0: Students add, subtract, 10.0: Students Students 10.0: 10.0: dividevidevidevidevide di di multiply, and di di ynomials..... ynomials ynomials monomials and pol monomials and pol monomials and polynomials monomials and pol ynomials monomials and pol Students solve multistep problems, including word problems, by using these techniques. What it means for you: You’ll learn how to find the multiplicative inverse of a polynomial, and how to use negative exponents. Key words: polynomial monomial reciprocal exponent ynomials and ynomials and PPPPPolololololynomials and ynomials and ynomials and ynomials and ynomials and PPPPPolololololynomials and ynomials and ynomials and NeNeNeNeNegggggaaaaatititititivvvvve Pe Pe Pe Pe Pooooowwwwwererererersssss NeNeNeNeNegggggaaaaatititititivvvvve Pe Pe Pe Pe Pooooowwwwwererererersssss Before you divide one polynomial by another polynomial, you need to know how to write the multiplicative inverse of a polynomial. ocal of a P a P a P a P a Polololololynomial ynomial ynomial ocal of ocal of ecipr ecipr Finding the R Finding the R ynomial eciprocal of Finding the Recipr ynomial ocal of ecipr Finding the R Finding the R The reciprocal of a polynomial is its multiplicative inverse. To find the reciprocal of Polynomial A, you divide the number 1 by Polynomial A. Example Example Example Example Example 11111 Find the reciprocal of each of these polynomials: b) –5x² + 3x – 1 a) 2x – 1 Solution Solution Solution Solution Solution a) 1 1x − 2 b) 2− 5 x 1 + − 3 x 1 By the definition of division, to divide by a polynomial you need to multiply by the reciprocal (the inverse under multiplication) of that polynomial. Guided Practice Find the multiplicative inverse of each of
these expressions. 1. ab 2. a2 3. 2x + 4b 4. 3x + 1 5. 8x3 – 16x2 + 4 7. 2x2y4 8. 2 Section 6.3 Section 6.3 Section 6.3 — Dividing Polynomials Section 6.3 Section 6.3 285285285285285 ocals as Negggggaaaaatititititivvvvve Exponents e Exponents e Exponents ocals as Ne ocals as Ne ecipr ecipr rite R ou Can WWWWWrite R rite R ou Can YYYYYou Can ou Can e Exponents eciprocals as Ne rite Recipr e Exponents ocals as Ne ecipr rite R ou Can This rule of exponents gives you another way to express fractions: x–a = 1 x a provided x π 0 Example Example Example Example Example 22222 Simplify the following expressions: a) 2–3 b) 2x ◊ (3y)–1 c) 6b3 ◊ (3ab)–2 b) 2x ◊ (3y)– Solution Solution Solution Solution Solution a) 2–) 6b3 ◊ (3ab)–2 = 3 ⋅ 6 b 1 ab Guided Practice Simplify the following expressions. 10. (2a)–2 12. (4x2y)–2 14. (2x3y)–2(3x)3 16. (15a3b4x4)(3a2b2x3)–1 18. (2a2b3c2d3)2(– 1 3 ab2cd2)–2 20. (2abc)3(3a2bc)–2(a2b3c4) 11. (3ab)–3 13. (4xy)–1(2x2) 15. (16x5y7z12)(4x3y2z8)–1 17. (40ax2y4z5)2(4xy2z4)–2 19. (3xyz)2(2xyz)(6xyz)–1 21. (4xy)(8x2y3)(x2y5)–1 286286286286286 Section 6.3 Section 6.3 Section 6.3 — Dividing Polynomials Section 6.3 Section 6.3 ynomials as Frrrrractions actions actions ynomials as F rite Polololololynomials as F ynomials as F rite P ou Can WWWWWrite P rite P ou Can YYYYYou Can ou Can actions actions ynomials as F rite P ou Can A polynomial raised to a negative power can also be written as a fraction. Check it out If a pair of parentheses is raised to a negative power, its entire contents go to the bottom of the fraction. Example Example Example Example Example 33333 Simplify the following expressions: a) (a + b)◊(b – c)–1 b) (m – c)(m + c)◊(m – c)–1 c) (x – 1)–1◊(x + 1)–1 Solution Solution Solution Solution Solution a) (a + b)◊(b – c)–) (m – c)(m + c)◊(m – c)–1 = ( − + m c m c )( − m c ) ( ) = m + c Check it out There’s more on canceling fractions in 6.3.3. c) (x – 1)–1◊(x + 1)–( 1 ) Guided Practice Simplify the following expressions. 23. (w + v) × (w – v)–1 25. (a – x)–2(a – x) 22. (x + y) × (x + y)–1 24. (a – b)–1(a – b)(a + b) 26. (a – b)(a – b)(a + b)–1(a – b)–1 27. (a + x)(a – x)(a – b)–1 28. (t – s)2(t – s)–1(t + s )( 4 ) 29. (7y + 6z)(7y – 5z)–1(7y + 6z)–2 k j 4 )( − 41 ( + − 31. 30 Independent Practice Find the multiplicative inverse of the following. 1. 1.2 + 3.3a 4. 8m2 – 5mz + 4 2. 3x2 + 5y3 17 a − . 3 9 5. Simplify the following expressions. 3. –18.6 – 1.5x x . 4 3 y + 6. . 8 9 4 7. 15a2(–3a2)–1 9. (2y + 3)–2 × (2y + 3) 11. (2a2b4)–1(4ab)–1(16a3b4) 13. 2x(x – 1)(x – 1) –1(2) –1 15. (z – 5)2(3)–2(15)(z – 5)–1 8. (–2x)2(6ax)–2 10. 4ab × (3a)–1 × (4b)–1 12. (a2b–2)3(a–1b0)4(a–2b–3)–2 14. (3z)–1(z – 2)–2(z2)3(z – 2) 16. 20(z – 3)–2(z + 3)(z – 3)3(6)–1 ound Up ound Up RRRRRound Up ound Up ound Up ound Up ound Up RRRRRound Up ound Up ound Up Now you’ve got all the tools you need. You’ll start dividing one polynomial by another polynomial using two different methods in Topics 6.3.3 and 6.3.4. Section 6.3 Section 6.3 Section 6.3 — Dividing Polynomials Section 6.3 Section 6.3 287287287287287 TTTTTopicopicopicopicopic 6.3.36.3.3 6.3.36.3.3 6.3.3 California Standards: Students Students 10.0: 10.0: 10.0: Students Students add, subtract, 10.0: Students 10.0: dividevidevidevidevide di di multiply, and di di ynomials..... ynomials ynomials monomials and pol monomials and pol monomials and polynomials monomials and pol ynomials monomials and pol Students solve multistep problems, including word problems, by using these techniques. What it means for you: You’ll divide one polynomial by another polynomial by factoring. Key words: polynomial monomial factor exponent ynomials ynomials vision by Py Py Py Py Polololololynomials vision b vision b DiDiDiDiDivision b ynomials ynomials vision b vision by Py Py Py Py Polololololynomials ynomials ynomials vision b vision b DiDiDiDiDivision b ynomials vision b ynomials actoring actoring — F— F— F— F— Factoring actoring actoring — F— F— F— F— Factoring actoring actoring actoring actoring Now you’re ready to divide a polynomial by another polynomial. The simplest way to do this is by factoring the numerator and the denominator. essions essions actions Helps to Simplify Expr Canceling Frrrrractions Helps to Simplify Expr actions Helps to Simplify Expr Canceling F Canceling F essions actions Helps to Simplify Expressions essions actions Helps to Simplify Expr Canceling F Canceling F If a numerical fraction has a common factor in the numerator and denominator, you can cancel it — for example, = 5 10 ⋅ In the same way, if there are common factors in the numerator and denominator of an algebraic fraction, you can cancel them. Example Example Example Example Example 11111 Simplify ( x + 1 2 )( + x ( + x 1 ) 3 ) Solution Solution Solution Solution Solution + ( x 1 2 )( + x ( 3 ) + x 1 ) = 2x + 3 This technique’s really useful for dividing polynomials when the polynomials have already been factored: Check it out: See Chapter 8 for more about canceling fractions. Example Example Example Example Example 22222 Divide (x + 4)(1 – x)(3x + 2) by (1 – x). Solution Solution Solution Solution Solution (x + 4)(1 – x)(3x + 2) ÷ (1 – xx + 4)(3x + 2) 288288288288288 Section 6.3 Section 6.3 Section 6.3 — Dividing Polynomials Section 6.3 Section 6.3 Guided Practice Simplify each expression. + ( x ( 9 )( − x x 4 − 4 ) ) 1. + x 2 ( + )( 1 3 2. ( 2 x 1 2 )( + x − x 1 ) x 4 2 )( − 1 ) ( ( 5 x + x + )( 2 7 2 )( x x + + )( 7 2 3 5 )( x x + + 3 ) 2 ) 3. 4. ( − 3 2 + 3 x )( − . 6. ( 2 x + 7 x )( 2 − + 25 )( )( + 1 x ) 7. + 1 )( z 2 10 z − 1 ) ( ÷ 9. Divide (x + 3)(x + 4) by ( x + 3 ) 8 . 10. Divide 3 x x + by actorededededed actor actor ynomial Can Be F the Polololololynomial Can Be F ynomial Can Be F the P vides Evenlenlenlenlenlyyyyy,,,,, the P the P vides Ev vides Ev It Di IfIfIfIfIf It Di It Di ynomial Can Be Factor It Divides Ev actor ynomial Can Be F the P vides Ev It Di If you can divide a polynomial evenly, that means there is no remainder. This means that it must be possible to factor the polynomial (and it means that the divisor is a factor of the polynomial). Example Example Example Example Example 33333 Given that (x + 1) divides evenly into (x² – 4x – 5), find (x² – 4x – 5) ÷ (x + 1). Solution Solution Solution Solution Solution (x² – 4x – 5) ÷ (x + 1) Check it out: In Example 3, the numerator, x2 – 4x – 5, has been factored, giving (x + 1)(x – 5). See Section 6.6 for more on factoring quadratics )( x – 5) ou know thaw thaw thaw thaw that (t (t (t (t (x + 1) is a f actor actor ou kno ou kno YYYYYou kno + 1) is a f + 1) is a f actor + 1) is a factor actor ou kno + 1) is a f vides evvvvvenlenlenlenlenlyyyyy vides e vides e e told it di e told it di because you’ou’ou’ou’ou’rrrrre told it di because y because y e told it divides e vides e e told it di because y because y ) 5 Cancel (x + 1) fr Cancel ( Cancel ( om the top and bottom om the top and bottom + 1) fr + 1) fr om the top and bottom + 1) from the top and bottom Cancel ( Cancel ( om the top and bottom + 1) fr Section 6.3 Section 6.3 Section 6.3 — Dividing Polynomials Section 6.3 Section 6.3 289289289289289 Check it out: See Sections 6.6 and 6.8 for more on factoring quadratics. Guided Practice Simplify the quotients by canceling factors. 11. + x 4 + x 8 2 2 2x 14. x + 12 x 2 x 17. 2 + − x +( 2 x 20 ) 5 12. 2x 12 x + 4 16 15. x 9 x− 12 2 3 x 18 13. 20 2 z + + 4 5 z z 16. x 5 2 x − − 15 9 19 20. Find the ratio of the surface area to the volume of a cube with side length b. 21. Divide 4x – 12 by x2 – 2x – 3. Independent Practice Name the two factors that would divide into each expression below. 1. x2 + 7x 4. 6x2 + 9x 2. 2x – 8 5. x2 + 8x + 15 3. 6a – 15 6. a2 – 81 Simplify the quotients by canceling factors. 7. x 2 x − 2 − 4 10. a 2 a − − 8 64 8. − x 2 − x 6 5 15 4 11. − + 2a a 16 2 9. a 6 2 a 10 + + 42 70 a 12. 3 x x 2 + + x 1 13 14 15 12 6 16 12 2 x 17. 15 − − x 2 − 2 25 x 19. Find the ratio of the surface area to the volume of the rectangular prism shown. 2 + − x − 2 x 8 18. 2 x 2 b 2b b ound Up ound Up RRRRRound Up ound Up ound Up RRRRRound Up ound Up ound Up ound Up ound Up This method’s most useful for “divides evenly” questions — if a question mentions remainders, the long division method in Topic 6.3.4 is probably better. There are two methods for polynomial division, and you should use the one that makes the most sense for the question you’re doing. 290290290290290 Section 6.3 Section 6.3 Section 6.3 — Dividing Polynomials Section 6.3 Section 6.3 TTTTTopicopicopicopicopic 6.3.46.3.4 6.3.46.3.4 6.3.4 California Standards: Students Students 10.0: 10.0: 10.0: Students Students add, subtract, 10.0: Students 10.0: dividevidevidevidevide di di multiply, and di di ynomials..... ynomials ynomials monomials and pol monomials and pol monomials and polynomials monomials and pol ynomials monomials and pol Students solve multistep problems, including word problems, by using these techniques. What it means for you: You’ll divide one polynomial by another polynomial by long division. Key words: polynomial monomial divisor dividend Don’t forget: The dididididividend vidend vidend vidend is the vidend expression that is being divided. The dididididivisor visor visor visor is the expression visor you are dividing by. ynomials ynomials vision by Py Py Py Py Polololololynomials vision b vision b DiDiDiDiDivision b ynomials ynomials vision b vision by Py Py Py Py Polololololynomials ynomials ynomials vision b vision b DiDiDiDiDivision b ynomials ynomials vision b vision vision — Long Di — Long Di vision — Long Division vision — Long Di — Long Di vision vision — Long Di — Long Di — Long Division vision — Long Di vision — Long Di When a polynomial can’t be factored, there has to be some other way to divide it by another polynomial. The next method to try is the long division method. visions visions xact Di xact Di or None or None vision Method — f vision Method — f Long Di Long Di visions xact Divisions or Nonexact Di vision Method — for None Lo
ng Division Method — f visions xact Di or None vision Method — f Long Di Long Di The long division method for dividing polynomials is really similar to the long division method for integers. The aim is to find out how many groups of the divisor can be subtracted from the dividend. Example Example Example Example Example 11111 Calculate 6 2x − − 2 11 x − x 5 11 . Solution Solution Solution Solution Solution First, consider the leading term of the divisor and the leading term of the dividend. 6x² divided by 2x is 3x — in other words 2x will go into 6x² 3x times. So 3x goes above the line. Leading term of dividend Leading term of divisor 3x 6 ² – 11 – 11 x x 2 – 5x Then subtract the product of 3x and (2x – 5) from the dividend. 3x groups of (2x – 5) is (6x² – 15x). Subtract 3 x groups of (2 – 5) x from the dividend, leaving a remainder of 4 – 11. x 2 – 5x 3x 6 ² – 11 – 11 x x x – (6 ² – 15 ) x 4 – 11 x Next see how many times the leading term of the divisor will go into the leading term of what is left of the dividend. 4x ÷ 2x = 2, so you now need to subtract 2 groups of (2x – 5). 2 – 5x 3 + 2 x 6 ² – 11 – 11 x x x – (6 ² – 15 ) x 4 – 11 x – (4 – 10) x –1 2 groups of (2 – 5) = x(4 – 10). x Subtracting that from what is left of the dividend leaves a remainder of –1. 6 2x So − − 2 x 11 − x 5 11 = (3x + 2) remainder –1 Section 6.3 Section 6.3 Section 6.3 — Dividing Polynomials Section 6.3 Section 6.3 291291291291291 Guided Practice Divide using the long division method. 1. 4. 7. 2 2x 5 + − 9 x + 5 x + 16 2a 3 + 16 + a a 4 2 x x − + 25 5 2. 5. 7 2a + a + 6 a 23 + 2x 8. − 24 − x x 4 + 48 9. Divide (a3 – 6a – 4) by (a + 2). 10. Divide (4b2 + 22b + 12) by (2b + 1). 4 b 3. 6. 5 − − 2x 15 ynomials TTTTToooooooooo ynomials actor Higher-De-De-De-De-Degggggrrrrree Pee Pee Pee Pee Polololololynomials ynomials actor Higher actor Higher ou Can F YYYYYou Can F ou Can F ou Can Factor Higher ynomials actor Higher ou Can F If you’ve got polynomials of degree higher than 2 (such as cubic equations), it’s not always clear how to factor them. You can use long division to help factor expressions. For example, if you divide Polynomial A by Polynomial B and get an answer with remainder zero, then Polynomial B is a factor of Polynomial A. Example Example Example Example Example 22222 Divide (9m³ – 3m² – 26m – 8) by (3m + 4). Solution Solution Solution Solution Solution 3 + 4m – 26 – 8 m m m – (9m m – (–15m –15 ² – 26 – 8 ³ + 12 ²) m ² – 20 ) m –6 – 8m m –(–6 – 8) 0 Check it out: Factoring a polynomial fully can be useful for drawing graphs or solving equations. So (9m³ – 3m² – 26m – 8) ÷ (3m + 4) = 3m2 – 5m – 2 This also means that (9m³ – 3m² – 26m – 8) = (3m + 4)(3m2 – 5m – 2). If (3m2 – 5m – 2) can also be factored, then you can fully factor (9m³ – 3m² – 26m – 8). 292292292292292 Section 6.3 Section 6.3 Section 6.3 — Dividing Polynomials Section 6.3 Section 6.3 Guided Practice Simplify each quotient by dividing using the long division method. 15 11. + 2x 3 x − 4 17 x + 4 12 13. 3 − 6 y 2 y 19 y 2 + − 14 5 − 10 y 14 15. Divide 15a2 + 7ab – 2b2 by 5a – b. 16. Divide 2x4 – 2x3 + 3x – 1 by 2x3 + 1. Find the remaining factors of the polynomial, given that: 17. (x + 3) is a factor of x2 – x – 12. 18. (x + 4) is a factor of x3 + 4x2 – 25x – 100. Independent Practice 1. Divide (x2 + 16x + 49) by (x + 4). 2. Divide (x2 + 3x – 6) by (x + 4). Carry out the following divisions using the long division method. 3 2 x 3. − − 2 x − x 13 3 − 6 x 4 2a 4. 2x 2a 7. + 10 + a − 1 a 1 − 6 8. 2 − 2x 3 x x 11 − 2 + 9 9. 2 x 15 + − x 2 + 5 x 10. 2a 3 1 − − 4 a + 1 a 11 2a 12. + − a 35 − a 3 4 36 13. 10 2x + 2 x + 10 x 21 + 3 14. 3 a − − 7 22 a a − − 33 a 7 15. The width of a rectangle is (x – 3) cm and the area is (3x2 – 10x + 3) cm2. What is the length? 16. The base of a triangle is (x + 2) meters and the area is (x3 – 6x – 4) meters squared. What is the height of the triangle? (Area of a triangle is A = ½bh.) Find the remaining factors of the polynomial, given that: 17. (y + 5) is a factor of y3 + 9y2 + 23y + 15. 18. (y – 2) is a factor of y3 – 7y2 + 16y – 12. ound Up ound Up RRRRRound Up ound Up ound Up ound Up ound Up RRRRRound Up ound Up ound Up It can sometimes take a while to calculate using long division — but sometimes it’s the only way of working out a polynomial division. If the division looks simple you should try using factoring first. Section 6.3 Section 6.3 Section 6.3 — Dividing Polynomials Section 6.3 Section 6.3 293293293293293 TTTTTopicopicopicopicopic 6.3.56.3.5 6.3.56.3.5 6.3.5 California Standards: Students Students 10.0: 10.0: 10.0: Students Students add, subtract, 10.0: Students 10.0: dividevidevidevidevide di di multiply, and di di ynomials..... ynomials ynomials monomials and pol monomials and pol monomials and polynomials monomials and pol ynomials monomials and pol ultisteppppp ultiste Students solve me me me me multiste ultiste Students solv Students solv Students solv ultiste Students solv luding worororororddddd luding w luding w inc inc lems,,,,, inc lems lems prprprprproboboboboblems including w luding w inc lems prprprprproboboboboblems lems,,,,, b b b b by using these y using these y using these lems lems y using these lems y using these hniques..... hniques hniques tectectectectechniques hniques What it means for you: You’ll use factoring and long division to solve real-life problems involving polynomial division. Key words: polynomial monomial divisor dividend Check it out: You’ll cover factoring quadratics later, in Section 6.6. Check it out: Which method you choose really just depends on how easy you find long division compared with factoring. ynomials ynomials vision by Py Py Py Py Polololololynomials vision b vision b DiDiDiDiDivision b ynomials ynomials vision b vision by Py Py Py Py Polololololynomials ynomials ynomials vision b vision b DiDiDiDiDivision b ynomials ynomials vision b tions tions pplica pplica — — — — — AAAAApplica tions pplications tions pplica — — — — — AAAAApplica tions tions pplica pplica pplications tions pplica tions You can use most of the skills you’ve learned in this Section to solve geometric problems involving polynomials. lems Often Invvvvvolvolvolvolvolve Pe Pe Pe Pe Polololololynomials ynomials ynomials lems Often In olume Proboboboboblems Often In lems Often In olume Pr VVVVVolume Pr olume Pr ynomials ynomials lems Often In olume Pr Example Example Example Example Example 11111 The volume of the box shown is (40x³ + 34x² – 5x – 6) cm³. Write an expression for the height in cm, h, of the box, if it has width (2x + 1) cm and length (4x + 3) cm. (2x Solution Solution Solution Solution Solution Start by writing out the formula for volume: + 1) c m h cm (4x + 3) cm volume = length × width × height height = volume × length width h = 40 3 x ( 4 + 2 34 + x )( There are a couple of different ways to tackle a division problem like this: 1) You could use long division to divide the volume by one of the factors, leaving you with a quadratic expression that you can then factor by trial and error — you already know one of the factors. or 2) You could multiply together the two factors you know to get a quadratic expression. You can then use long division to divide the volume by the quadratic expression. In this case, the long division is a bit more complicated than those we tackled on the previous page, but this method does the division in one step rather than two. 294294294294294 Section 6.3 Section 6.3 Section 6.3 — Dividing Polynomials Section 6.3 Section 6.3 Example Example Example Example Example 22222 Method 1: Use long division to find the height, h, of the box shown in Example 1. Solution Solution Solution Solution Solution OK, so you already have the expression h = 3 40 x ( 4 2 + 34 x + )( . Use long division to divide the volume by one of the factors: 4 + 3 x – (40 ³ + 30 ²) x – 2 x x 10 ² + 40 ³ + 34 4 ² + 3 ) –8 – 6x x – (–8 – 6) 0 Then you just need to factor the quadratic (10x² + x – 2). You know that one of the factors is (2x + 1), from the original question. By trial and error, you can work out that (10x² + x – 2) will factor into (2x + 1)(5x – 2). So the height of the box must be (5x – 2) cm. Alternatively, you could do another long division to divide (10x² + x – 2) by (2x + 1). Example Example Example Example Example 33333 Method 2: Use the multiplying factors method to find the height, h, of the box shown in Example 1. Check it out: Again, you start with the leading term of each polynomial. Solution Solution Solution Solution Solution Start by multiplying together the factors you know: (4x + 3)(2x + 1) = 8x² + 4x + 6x + 3 = 8x² + 10x + 3 Then use long division to divide the volume by the product of the two factors: 8 ² + 10 + 3 x x 5 – 2 40 ³ + 34 40 ³ + 50 ² + 15 ) x x x –16 ² – 20 – 6 – (–16 ² – 20 – 6) x x x x So the answer is h = (5x – 2) cm. 0 Section 6.3 Section 6.3 Section 6.3 — Dividing Polynomials Section 6.3 Section 6.3 295295295295295 Guided Practice 1. If the area of a rectangle is (6y2 + 29y + 35) square units, and its length is (3y + 7) units, find the width of the rectangle. 2. The volume of a rectangular box is (60x3 + 203x2 + 191x + 36) cubic inches. Find the height of the box if the area B of the base is (15x2 + 47x + 36) square inches. [Hint: V = Bh] 3. The area of a circle is (px2 – 10px + 25p) m2. Show that the radius is (x – 5) m. 4. The volume of a prism is (x3 + 8x2 + 19x + 12) m3. Find the area of the base if the height is (x + 4) m. 5. A rectangular prism has volume (2b3 – 7b2 + 2b + 3) m3, height (b – 1) m, and length (2b + 1) m. Find its width. Independent Practice 1. The width of a rectangle is (3x + 4) m. If the area of the rectangle is (6x2 + 5x – 4) m2, what is the length? 2. The area, A, of a triangle is given by the formula 2A = bh, where b is the base length and h is the height. Find the length of the base if A = (3x2 – 16x + 16) ft2 and h = (2x – 8) ft. 3. Find the height of a rectangle with area (2x2 – 9x + 9) ft2 and width (2x – 3) ft. 4. The volume
of a prism is (2x3 + x2 – 3x) m3. If the area of the base is (x2 – x) m2, what is the height of the prism? 5. A rectangular prism has volume (b3 + 9b2 + 26b + 24) m3, width (b + 2) m, and length (b + 4) m. Find its height. 6. The volume of a prism is (144s3 + 108s2 – 4s – 3) m3. If the area of the base is (36s2 – 1) m2, what is the height? 7. The volume of a cylinder is (2pp3 + 7pp2 + 8pp + 3p) in3. Find the area of the base if the height is (2p + 3) in. 8. The volume of a cylinder is (18py3 – 3py2 – 28py – 12p) ft3. Find the height of the cylinder if the area of the base is (9py2 + 12py + 4p) ft2. ound Up ound Up RRRRRound Up ound Up ound Up ound Up ound Up RRRRRound Up ound Up ound Up In this Section you’ve seen two good ways of dividing one polynomial by another polynomial — factoring and long division. 296296296296296 Section 6.3 Section 6.3 Section 6.3 — Dividing Polynomials Section 6.3 Section 6.3 Topic 6.4.1 California Standards: 2.0: Students understand and use such operations as taking the opposite, finding the reciprocal, taking a root, and raising to a fractional power. They understand and use the rules of exponents. 10.0: Students add, subtract, multiply, and divide monomials and polynomials. Students solve multistep problems, including word problems, by using these techniques. 11.0 Students apply basic factoring techniques to second and simple thirddegree polynomials. These techniques include finding a common factor for all terms in a polynomial, recognizing the difference of two squares, and recognizing perfect squares of binomials. What it means for you: You’ll learn how to use special cases of binomial multiplication to save time in calculations. Key words: binomial difference of two squares Check it out: (a + b)² expands to give a perfect square trinomial — see Topic 6.8.2. Section 6.4 Special Products of Special Products of Two Binomials Two Binomials This Topic is all about special cases of binomial multiplication. Knowing how to expand these special products will save you time when you’re dealing with binomials later in Algebra I. Remember These Three Special Binomial Products Given any real numbers a and b, then: (a + b)2 = a2 + 2ab + b2 (a – b)2 = a2 – 2ab + b2 (a + b)(a – b) = a2 – b2 When You Expand (a + b)², You Always Get an ab-Term (a + b)2 = (a + b)(a + b) = a2 + ab + ba + b2 Using the distributive property = a2 + 2ab + b2 You can relate this equation to the area of a square: a a a2 ab ab b2 b b (a + b)2 is the same as the area of this large square — add the areas of the two smaller squares, a2 and b2, and the two rectangles, 2 × ab. Example 1 Expand and simplify (2x + 3)2. Solution Put the expression in the form (a + b)(a + b): (2x + 3)2 = (2x + 3)(2x + 3) = 4x2 + 6x + 6x + 9 = 4x2 + 12x + 9 2x 2x 4x2 6x (2x + 3)2 is the same as the area of the large square — add the areas of the two smaller squares, 4x2 and 9, and the two rectangles, 2 × 6x. 6x 3 9 3 Section 6.4 — Special Products of Binomials 297 Guided Practice Find and simplify each product. 1. (x + 5)2 4. (9y + z)2 2. (2y + 3)2 5. (n2 + 4n)2 3. (3y + 1)2 6. (3t + 6b)2 Find the areas of the squares below. 7. 8. 9. 10 Check it out: (a – b)² also expands to give a perfect square trinomial — see Topic 6.8.2. When You Expand (a – b)², the ab-Term is Negative (a – b)2 = (a – b)(a – b) = a2 – ab – ba + b2 Using the distributive property = a2 – 2ab + b2 You can also relate this equation to the area of a smaller square: a (a – b)2 a (a – b)2 is the same as the area of the darker square. To find the area of the darker square, you can subtract two rectangles of area ab — but then you have to add back on an area of b2 (this small square). b b2 b Example 2 Expand (3y – 2)2. Solution Put the expression in the form (a – b)(a – b): (3y – 2)2 = (3y – 2)(3y – 2) = (3y)2 – 2(3y × 2) + 22 = 9y2 – 2(6y) + 22 = 9y2 – 12y + 4 In the diagram on the right, (3y – 2)2 is the same as the area of the darker square. 3y (3y – 2)2 3y 2 4 2 298 Section 6.4 — Special Products of Binomials Guided Practice Find and simplify each product. 11. (m – 4)2 14. (2a – 5)2 17. (x2 – y)2 12. (y – 2)2 15. (7d – f)2 13. (4k – 5)2 16. (xy – 3)2 18. (9x – 3y)2 19. (3x2 – 4)2 20. Find the area of the square shown. (3x – 1) in. Check it out: This is called a difference of two squares, and it can also be used in factoring expressions — see Topic 6.8.1. There’s No ab-Term When You Expand (a + b)(a – b) (a + b)(a – b) = a2 – ab + ba – b2 Using the distributive property = a2 – ab + ab – b2 = a2 – b2 The fact that no ab-term is left at all makes it unusual, but also very useful if you remember it. Example 3 Multiply out (4m + 3)(4m – 3). Solution The expression is already in the form (a + b)(a – b), so you can convert it to the form a2 – b2: (4m + 3)(4m – 3) = (4m)2 – 32 = 16m2 – 9 Guided Practice Find and simplify each product. 21. (m – v)(m + v) 23. (3y + x)(3y – x) 22. (x + 5)(x – 5) 24. (k – 6t)(k + 6t) 25. (3x – 9y)(3x + 9y) 26. (6x + 6y)(6x – 6y) 27. (x + 3x2)(x – 3x2) 29. (x2 – x)(x2 + x) 28. (9p2 – 2)(9p2 + 2) 30. (7a2 + b)(7a2 – b) Section 6.4 — Special Products of Binomials 299 You Can Use These Standard Equations as Shortcuts The good thing about knowing these standard equations is that you don’t need to do all the work each time — you can save time by using the three special products. Example 4 Find the area of a square which has side lengths of (3x + 4) inches. Solution Using (a + b)2 = a2 + 2ab + b2 and putting in (3x + 4) instead of (a + b) you get: 3x 3x 9x2 12x Check it out: In this case, a2 = (3x)2, 2ab = 2 × (3x × 4), and b2 = 42. Area of square = (3x + 4)2 = 9x2 + (2 × 12x) + 16 = (9x2 + 24x + 16) in2 12x 16 4 4 Example 5 Multiply (4x – 7) by (4x – 7). Solution Using (a – b)2 = a2 – 2ab + b2 and putting in (4x – 7) instead of (a – b) you get: (4x – 7)2 = 16x2 – (2 × 28x) + 49 = 16x2 – 56x + 49 Example 6 Find the area of this rectangle: (5y + 4) cm Solution Using (a + b)(a – b) = a2 – b2 and putting in 5y for a and 4 for b you get: (5y – 4) cm Area = (5y + 4)(5y – 4) = (5y)2 – 42 = (25y2 – 16) cm2 Check it out: Using the equation here is a little shorter than expanding the parentheses. 300 Section 6.4 — Special Products of Binomials Independent Practice Find the areas of these shapes. 3. 4. 2a – b 3a – b 2 +a b 3a – b Find and simplify each product: 5. (2r – 3)(2r + 3) 6. (x3 + 2)(x3 – 2) 7. (6x2 – 1)2 8. (z2 – z)2 9. (x2 – 3x)(x2 + 3x) 10. (2x2 – 3x)(2x2 + 3x) 11. Find the area of this shaded region5x – 4) in. (5x + 4) in. The area of a circle with radius r is given by the formula A = pr2. Find the areas of these circles, giving your answers in terms of p: 12. A circle with radius (3x + 4). 13. A circle with radius (2x – 7). 14. A circle with radius (2a + b). 15. Find the coefficient of ab in the product (5a – 4b)2. 16. Find the coefficient of mc in the product (4m – 3c)(4m + 3c). Find and simplify each product: 17. (3x + 7)2 – (3x – 7)2 18. (2xa + 1)2 19. (km – xm)2 20. Nicole has a circle of card with a radius of (4x + 5) cm. She makes the circle into a ring by cutting a circular hole in the middle with a radius of (4x – 5) cm. Find the area of the ring, leaving your answer in terms of p. Round Up Round Up The main reason for learning these special products is to make your life easier when you multiply two binomials or factor quadratics. You’ll come across them throughout the rest of Algebra I. Section 6.4 — Special Products of Binomials 301 TTTTTopicopicopicopicopic 6.5.16.5.1 6.5.16.5.1 6.5.1 Section 6.5 Monomials Monomials s of s of actor actor FFFFFactor Monomials s of Monomials actors of Monomials s of actor FFFFFactor Monomials Monomials s of s of actor actor s of Monomials actors of Monomials s of Monomials actor California Standards: 11.0 Students applpplpplpplpply basic y basic y basic 11.0 Students a 11.0 Students a y basic 11.0 Students a y basic 11.0 Students a fffffactoring tec hniques to hniques to actoring tec actoring tec hniques to actoring techniques to hniques to actoring tec second- and simple third-d-d-d-dsecond- and simple thir second- and simple thir second- and simple thir second- and simple thir ynomials..... TTTTThese dededededegggggrrrrree pol hese hese ynomials ynomials ee pol ee pol hese ee polynomials hese ynomials ee pol lude finding lude finding hniques inc tectectectectechniques inc hniques inc lude finding hniques include finding lude finding hniques inc or all or all actor f actor f a common f a common f actor for all a common factor f or all a common f or all actor f a common f ynomial, ynomial, ms in a pol ms in a pol terterms in a pol terter ynomial, ms in a polynomial, ter ynomial, ms in a pol recognizing the difference of two squares, and recognizing perfect squares of binomials..... What it means for you: You’ll learn how to find factors of a monomial. Key words: factor monomial greatest common factor prime factor Don’t forget: A monomial is a single term expression made up of a number or a product of a number and variables — such as 13, 2x2 or –x3yn4. In previous Topics you’ve already done lots of manipulation of polynomials — but you can often make manipulations easier by breaking down polynomials into smaller chunks. In this Topic you’ll break down monomials by factoring. s — and So Do Monomials s — and So Do Monomials actor s Has Havvvvve Fe Fe Fe Fe Factor actor s Has Ha Number Number s — and So Do Monomials actors — and So Do Monomials Numbers Ha s — and So Do Monomials actor Number Number Sometimes a number can be written as the product of two or more smaller numbers. Those smaller numbers are called factors of that number. For example, 6 = 2 × 3 — so 2 and 3 are factors of 6. The same is true for monomials. Unless they’re prime numbers, monomials can be written as the product of two or more numbers or letters. Those smaller numbers or letters are called factors of that monomial. For example, 3xy = 3 × x × y — so 3, x, and y are factors of 3xy. Guided Practice Write down all of the factors of each of these numbe
rs: 1. 8 4. 16 3. 15 6. 24 2. 10 5. 11 Write down each monomial as a product of the smallest possible factors: 7. 3x 10. 5xy 9. 6p 12. 20mn 8. 7z 11. 12uv h Monomial h Monomial Eac Eac visor of visor of he GCF is a Di TTTTThe GCF is a Di he GCF is a Di h Monomial Each Monomial visor of Eac he GCF is a Divisor of h Monomial Eac visor of he GCF is a Di The greatest common factor (GCF) of a set of monomials is the largest possible divisor of all monomials in the set. Check it out: Factors that can’t be factored themselves, like 2, 3, 5, x and y, are called prime factors. It’s these prime factors you need to use here. Example Example Example Example Example 11111 Find the greatest common factor of 12x²y², 18x³y², and 30x4y4. Solution Solution Solution Solution Solution Start by writing down each monomial as a product of the smallest factors possible: 12 ² ² = 2 × 2 × 3 × × × × 18 ³ ² = 2 × 3 × 3 × × × × × 30 Then list all the numbers that are factors of all three terms: 2, 3, x, x, y, y. These are called the common factors. 302302302302302 Section 6.5 Section 6.5 Section 6.5 — Factors Section 6.5 Section 6.5 Example 1 continued The greatest common factor is the product of all the common factors: GCF = = 6x²y² In other words, 6x²y² is the largest possible divisor of 12x²y², 18x³y², and 30x4y4. Guided Practice Use the method from Example 1 to write down the greatest common factor of each set of products: 13. 12, 24, 42 14. 9ab2, 15a2b2, 12ab 15. 6m2cv2, 10m2c2v, 4m2c2v3 16. 5mx2t, 15m2xt2, 20mxt 17. 9x3y2, 27x2y3 o or MMMMMororororore e e e e MMMMMonomials onomials onomials o or ind the GCF of TTTTTwwwwwo or o or ind the GCF of y to FFFFFind the GCF of ind the GCF of y to Another WWWWWaaaaay to y to Another Another onomials onomials o or ind the GCF of y to Another Another You can also find the GCF of two or more monomials by simply multiplying together the GCFs of each of the different parts. Example Example Example Example Example 22222 Find the greatest common factor of 12x²y², 18x³y², and 30x4y4. Solution Solution Solution Solution Solution The GCF of 12, 18, and 30 is 6. The GCF of x², x³, and x4 is x². The GCF of y² and y4 is y². So, the GCF of 12x²y², 18x³y², and 30x4y4 is 6 × x² × y² = 6x²y². Check it out: You could also calculate all of these GCFs using the method in Example 1. Guided Practice Use the method from Example 2 to write down the greatest common factor of each set of products: 18. b3m2cv, bm2v 19. 2(m + 1), –3(m + 1), (m + 1)2 20. 8(v – 1)2, 4(v – 1)3, 12(v – 1)2 21. 6x2yz, 15xz 22. 21x4y4z4, 42x3y4z5, 14x6y3z2 Section 6.5 Section 6.5 Section 6.5 — Factors Section 6.5 Section 6.5 303303303303303 Independent Practice Write down all of the factors of each of these numbers: 1. 25 4. 67 3. 36 6. 70 2. 12 5. 80 Write each of these as a product of prime factors: 7. 48 10. 66 8. 72 11. 450 9. 120 12. 800 Check it out: You can use either method of finding the greatest common factor here — it’s up to you. Write each monomial as a product of the smallest possible factors: 13. 66z2 16. –98a2b 19. 3x2yz 15. –102x3y 18. –80rs5 21. –100f 2gh4 14. 4b3d2 17. 64y3z3 20. 16pq3r3 Write down the greatest common factor of each set of products: 22. 18, 36 24. 95, 304 26. 21p2q, 35pq2 28. –60r2s2t2, 45r3t3 30. 14a2b3, 20a3b2c2, 35ab3c2 32. 14a2b2, 18ab, 2a3b3 23. 84, 75 25. 17a, 34a2 27. 12an2, 40a4 29. 18, 30, 54 31. 18x2, 30x3y2, 54y3 33. 32m2n3, 8m2n, 56m3n2 34. The area of a rectangle is 116 square inches. What are its possible whole number dimensions? 35. The area of a rectangle is 1363 square centimeters. If the measures of the length and width are both prime numbers, what are the dimensions of the rectangle? 36. Marisela is planning to have 100 tomato plants in her garden. In what ways can she arrange them in rows so that she has the same number of plants in each row, at least 5 rows of plants, and at least 5 plants in each row? 37. A walkway is being paved using 2-ft-by-2-ft paving stones. If the length of the walkway is 70 ft longer than the width and its area is 6000 ft2, how many paving stones make up the length and the width of the walkway? ound Up ound Up RRRRRound Up ound Up ound Up RRRRRound Up ound Up ound Up ound Up ound Up Factoring is the best way of working out which smaller parts make up a number or monomial. In the next few Topics you’ll use factoring to break down full expressions, which makes it much easier to do tricky jobs like solving some kinds of equations. 304304304304304 Section 6.5 Section 6.5 Section 6.5 — Factors Section 6.5 Section 6.5 TTTTTopicopicopicopicopic 6.5.26.5.2 6.5.26.5.2 6.5.2 California Standards: 11.0 Students applpplpplpplpply basic y basic y basic 11.0 Students a 11.0 Students a y basic 11.0 Students a y basic 11.0 Students a fffffactoring tec hniques to hniques to actoring tec actoring tec hniques to actoring techniques to hniques to actoring tec second- and simple third-d-d-d-dsecond- and simple thir second- and simple thir second- and simple thir second- and simple thir ynomials..... TTTTThese dededededegggggrrrrree pol hese hese ynomials ynomials ee pol ee pol hese ee polynomials hese ynomials ee pol lude finding lude finding hniques inc tectectectectechniques inc hniques inc lude finding hniques include finding lude finding hniques inc or all or all actor f actor f a common f a common f actor for all a common factor f or all a common f or all actor f a common f ynomial ynomial ms in a pol ms in a pol terterms in a pol terter ynomial, ms in a polynomial ter ynomial ms in a pol recognizing the difference of two squares, and recognizing perfect squares of binomials. What it means for you: You’ll learn how to find factors of a polynomial. Key words: factor monomial polynomial greatest common factor actorsssss actor actor Simple F Simple F Simple Factor actor Simple F Simple F actorsssss actor actor Simple F Simple F Simple Factor actor Simple F Simple F ynomials ynomials ofofofofof P P P P Polololololynomials ynomials ynomials ofofofofof P P P P Polololololynomials ynomials ynomials ynomials ynomials The Topics in this Section are all about making your work a lot easier. Factoring polynomials can make them simpler, which means they’re easier to manipulate. actored ed ed ed ed TTTTToooooooooo actor actor ynomials Can (Sometimes) Be F PPPPPolololololynomials Can (Sometimes) Be F ynomials Can (Sometimes) Be F ynomials Can (Sometimes) Be Factor actor ynomials Can (Sometimes) Be F When a polynomial can be expressed as the product of two or more numbers, monomials, or polynomials, those smaller numbers, monomials, or polynomials are called factors of that polynomial. For example, if Polynomial A can be written as either: 2x² + x or x(2x + 1), this means that x is a factor of Polynomial A. It also means that (2x + 1) is a factor of Polynomial A. To find the factors of a polynomial, start by finding the greatest common factor of all the terms. Example Example Example Example Example 11111 Factor 6m³ – 4m². Solution Solution Solution Solution Solution × The factors that are present in both terms are 2, m, and m — so 2, m, and m are factors of both terms. So the GCF = 2 × m × m = 2m² Next you need to rewrite the expression with the factor taken out: 6m³ = 2 × m × m) × 3 × m = 2m² × 3m 4m² = 2 × 2 × m × m = (2 × m × m) × 2 = 2m² × 2 So 6m³ – 4m² = (2m² × 3m) – (2m² × 2) = 2m²(3m – 2) It’s always worth checking your factoring by multiplying out again: 2m²(3m – 2) = (2 × m² × 3 × m) – (2 × m² × 2) = (6 × m2+1) – (4 × m²) = 6m3 – 4m² Section 6.5 Section 6.5 Section 6.5 — Factors Section 6.5 Section 6.5 305305305305305 Guided Practice In each polynomial, find the greatest common factor of the terms. 1. 12y2 – 3y 2. a3 + 3a 3. 14a3 – 28a2 + 56a 4. 16y2 – 24y3 5. 60x3 + 24x2 + 16x 6. 18y5 + 6y4 + 3y2 Use your answers from Exercises 1 – 6 to factor the following: 7. 12y2 – 3y 8. a3 + 3a 9. 14a3 – 28a2 + 56a 10. 16y2 – 24y3 11. 60x3 + 24x2 + 16x 12. 18y5 + 6y4 + 3y2 actorsssss actor actor aking Out F e Method for or or or or TTTTTaking Out F aking Out F e Method f An An An An An AlterAlterAlterAlterAlternananananatititititivvvvve Method f e Method f aking Out Factor actor aking Out F e Method f Again, find the GCF by multiplying together the GCFs of each of the different parts — the coefficients and the variables. Then just rewrite the expression with the GCF taken out, as before. Example Example Example Example Example 22222 Factor 6m³ – 4m². Solution Solution Solution Solution Solution The greatest common factor of 6 and 4 is 2. Since both terms contain m, the lowest power will be a factor. So GCF = 2 × m × m = 2m² To write 6m³ – 4m² as a factored expression, multiply and divide by 2m², as shown: Check it out: Example 2 covers the same problem as Example 1, but this time using the GCF of each of the different parts. ⎛ 6m³ – 4m² = 2m2 6 ⎜⎜⎜⎜ ⎝ = 2m2(3m – 2) ⎞ ⎟⎟⎟⎟ ⎠ Guided Practice Factor each polynomial below. 13. x2 – 4x 15. 24x3 – 15x2 + 6x 17. 4a3 – 6a2 + 6a 19. 6b3 – 3b2 + 12b 14. x2 – x 16. 8x3 + 2x2 + 4x 18. 14b2 + 7b – 21 20. a4 + a5 + 5a3 306306306306306 Section 6.5 Section 6.5 Section 6.5 — Factors Section 6.5 Section 6.5 actors s s s s TTTTToooooooooo actor actor ynomial F e Out Polololololynomial F ynomial F e Out P ou Can TTTTTakakakakake Out P e Out P ou Can YYYYYou Can ou Can ynomial Factor actor ynomial F e Out P ou Can At the start of the Topic you saw that 2x² + x can be written as x(2x + 1). This means that both x (a monomial) and 2x + 1 (a polynomial) are factors of 2x² + x. The next few examples are about finding polynomial factors of the form (ax + b). Example Example Example Example Example 33333 Factor (d – 1)x² + (d – 1)x + (d – 1). Solution Solution Solution Solution Solution Each term is a product of (d – 1) and something else, so (d – 1) is a common factor. To write (d – 1)x² + (d – 1)x + (d – 1) as a factored expression, put the (d – 1) outside parentheses and divide everything inside the parentheses by (d – 1), as shown: (d – 1)x² + (d – 1)x + (d – 1) = (
d – 1) ⎛ ⎜⎜⎜⎜ ⎝ d – 1)(x2 + x + 1) ⎞ ⎟⎟⎟⎟ ⎠ Example Example Example Example Example 44444 Show that (a + b) is a factor of ac + bc + ad + bd. Solution Solution Solution Solution Solution You’ve been given the factor, so try writing the polynomial as a factored expression. If you can do that, you’ll have shown that (a + b) is a factor. Take the (a + b) outside parentheses, as above: + ) ad + a ( b ) + bd + a ( b ) ⎞ ⎟⎟⎟⎟ ⎠ = + a ( ) ( ) b b a + ac + ac + bc + ad + bd ⎛ ⎜⎜⎜ ⎝ ⎛ ⎜⎜⎜ ⎝ ⎛⎛ ⎜⎜⎜ ⎝ + + ac ( a bcc ) bc + ( a b + + bd ) b ad ( ⎞ ⎟⎟⎟⎟ ⎠ ⎞ ⎟⎟⎟⎟ ⎠ = (a + b)(c + d) Therefore (a + b) is a factor. Section 6.5 Section 6.5 Section 6.5 — Factors Section 6.5 Section 6.5 307307307307307 Check it out: You’re aiming to cancel (a + b) from the top. Check it out: Note that grouping the fractions differently helps to show (a + b) as a factor. Example Example Example Example Example 55555 Factor and simplify the following expression: (x – 2)(x + 2)x + (x – 2)x + (x – 2)(x + 2). Solution Solution Solution Solution Solution Each term is a product of (x – 2) and something else, so (x – 2) is a common factor. To write (x – 2)(x + 2)x + (x – 2)x + (x – 2)(x + 2) as a factored expression, write the (x – 2) outside parentheses, then divide all terms by (x – 2): (x – 2)(x + 2)x + (x – 2)x + (x – 2)(x + 2) = (x – 2) ( x ⎛ ⎜⎜⎜ ⎝ − + x )( )( − 2 ) + 2 ) ⎞ ⎟⎟⎟⎟ ⎠ = (x – 2)[(x + 2)x + x + (x + 2)] = (x – 2)(x2 + 2x + x + x + 2) = (x – 2)(x2 + 4x + 2) Guided Practice Factor and simplify. 21. x(2x + 1) + 3(2x + 1) 22. 3y2(2 – 3x) + y(2 – 3x) + 5(2 – 3x) 23. 2x4(5x – 3) – x2(5x – 3) + (5x – 3) 24. 2a(3a – 1) + 6(3a – 1) 25. (4 – x)x2 + (4 – x)2x + (4 – x)1 26. Show that (x + 3) is a factor of x2 + 2x + 3x + 6. 27. Show that (y + 2) is a factor of y2 + y + 2y + 2. 28. Show that (3x – 4) is a factor of 6x2 + 9x – 8x – 12. Factor and simplify the following expressions. 29. (x + 3)x + (x + 3)(x – 1) + (x + 3)x – 2(x + 3) 30. (2x – 1)2x + (2x – 1)(2x – 1) + (2x – 1)x – (2x – 1) 31. (x3 + x2)(x + 1) – (x3 + x2)(x2 + x) + (x3 + x2)(x2 + 2x + 3) 308308308308308 Section 6.5 Section 6.5 Section 6.5 — Factors Section 6.5 Section 6.5 Independent Practice In each polynomial, find the greatest common factor of the terms. 1. 8x2 – 12yx 2. 81x3 + 54x2 3. 21x3yz – 35x2y + 70xyz Factor each polynomial below. 4. 2x – 6 5. 6x2 – 12x 6. 5(c + 1) – 2y(c + 1) 7. 6y2 – 12y3 + 18y 8. x2(k + 3) + (k + 3) 9. k(y – 3) – m(y – 3) 10. (x + 1)2 – 2(x + 1)3 11. m(y – 5)2 – (y – 5) 12. (x2 – 2x) + (4x – 8) 13. (y2 + 3y) + (3y + 9) 14. (2my – 3mx) + (–4y + 6x) 15. –2m(x + 1) + k(x + 1) 16. x5 + 3x3 + 2x4 18. 3x3 – 6x2 + 9x 20. 2m3n – 6m2n2 + 10mn 17. 8y2x + 4yx2 + 4y2x2 19. 4x5 – 4x3 + 16x2 21. Show that (x + 4) is a factor of x2 – 5x + 4x – 20. 22. Show that (2x + 5) is a factor of 2x2 – 2x + 5x – 5. 23. Show that (3x – 1) is a factor of 6x2 + 3x – 2x – 1. 24. Show that (a – b) is a factor of 2ac – ad – 2bc + bd. Factor and simplify the following expressions. 25. (4x2 + 3)(2x + 1) + (4x2 + 3)(2x + 2) + 8(4x2 + 3) 26. 20(4a + 3b) – (x + 1)(4a + 3b) + (x2 + x + 8)(4a + 3b) + (4a + 3b) ound Up ound Up RRRRRound Up ound Up ound Up RRRRRound Up ound Up ound Up ound Up ound Up The greatest common factor is really useful when you’re trying to factor polynomials, because it’s always the best factor to use. In the next Section you’ll see that you can also factor more complicated polynomials like quadratics. Section 6.5 Section 6.5 Section 6.5 — Factors Section 6.5 Section 6.5 309309309309309 TTTTTopicopicopicopicopic 6.6.16.6.1 6.6.16.6.1 6.6.1 Section 6.6 actoring Quadraaaaaticsticsticsticstics actoring Quadr actoring Quadr FFFFFactoring Quadr actoring Quadr FFFFFactoring Quadr actoring Quadraaaaaticsticsticsticstics actoring Quadr actoring Quadr actoring Quadr California Standards: Students applpplpplpplpply basic y basic y basic Students a Students a 11.0: 11.0: y basic 11.0: Students a 11.0: y basic Students a 11.0: fffffactoring tec hniques to hniques to actoring tec actoring tec hniques to actoring techniques to hniques to actoring tec second- second- second- and simple thirdsecond- second- ynomials..... TTTTThese dededededegggggrrrrree pol hese hese ynomials ynomials ee pol ee pol hese ee polynomials hese ynomials ee pol lude finding lude finding hniques inc hniques inc tectectectectechniques inc lude finding hniques include finding lude finding hniques inc or all or all actor f actor f a common f a common f actor for all a common factor f or all a common f or all actor f a common f ynomial, ynomial, ms in a pol ms in a pol terterms in a pol terter ynomial, ms in a polynomial, ter ynomial, ms in a pol recognizing the difference of two squares, and recognizing perfect squares of binomials. What it means for you: You’ll learn how to factor simple quadratic expressions. Key words: quadratic polynomial factor binomial Check it out: The order of the parentheses doesn’t matter — you could have (x + 3) first. In Section 6.5 you worked out common factors of polynomials. Factoring quadratics follows the same rules, but you have to watch out for the squared terms. actorsssss o or More Fe Fe Fe Fe Factor actor actor o or Mor oducts of TTTTTwwwwwo or Mor o or Mor oducts of oducts of ynomials as Pr PPPPPolololololynomials as Pr ynomials as Pr ynomials as Products of actor o or Mor oducts of ynomials as Pr A quadratic polynomial has degree two, such as 2x2 – x + 7 or x2 + 12. Some quadratics can be factored — in other words they can be expressed as a product of two linear factors. Suppose x2 + bx + c can be written in the form (x + m)(x + n). Then: x2 + bx + c = (x + m)(x + n) b, c, m, and n are numbers = x(x + n) + m(x + n) = x2 + nx + mx + mn Expand out the parentheses using the distributive property So, x2 + bx + c = x2 + (m + n)x + mn Therefore b = m + n and c = mn So, to factor x2 + bx + c, you need to find two numbers, m and n, that multiply together to give c, and that also add together to give b. Example Example Example Example Example 11111 Factor x2 + 5x + 6. Solution Solution Solution Solution Solution The expression is x2 + 5x + 6, so find two numbers that add up to 5 and that also multiply to give 6. The numbers 2 and 3 multiply together to give 6 and add together to give 5. You can now factor the quadratic, using these two numbers: x2 + 5x + 6 = (x + 2)(x + 3) To check whether the binomial factors are correct, multiply out the parentheses and then simplify the product: (x + 2)(x + 3) = x(x + 3) + 2(x + 3) Using the distrib opertytytytyty oper oper Using the distributiutiutiutiutivvvvve pre pre pre pre proper Using the distrib Using the distrib oper Using the distrib = x2 + 3x + 2x + 6 = x2 + 5x + 6 This is the same as the original expression, so the factors are correct. 310310310310310 Section 6.6 Section 6.6 Section 6.6 — Factoring Quadratics Section 6.6 Section 6.6 Check it out: The numbers –3 and +2 multiply together to give –6 and add together to give –1. Check it out: The numbers –2 and –3 multiply together to give +6 and add together to give –5. Check it out: The numbers +4 and –2 multiply together to give –8 and add together to give +2. Example Example Example Example Example 22222 Factor x2 – x – 6. Solution Solution Solution Solution Solution Find two numbers that multiply to give –6 and add to give –1, the coefficient of x. Because c is negative (–6), one number must be positive and the other negative. x2 – x – 6 = (x – 3)(x + 2) Check whether the binomial factors are correct: (x – 3)(x + 2) = x(x + 2) – 3(x + 2) Using the distrib opertytytytyty oper oper Using the distributiutiutiutiutivvvvve pre pre pre pre proper Using the distrib Using the distrib oper Using the distrib = x2 + 2x – 3x – 6 = x2 – x – 6 This is the same as the original expression, so the factors are correct. Example Example Example Example Example 33333 Factor x2 – 5x + 6. Solution Solution Solution Solution Solution Find two numbers that multiply to give +6 and add to give –5, the coefficient of x. Because c is positive (6) but b is negative, the numbers must both be negative. x2 – 5x + 6 = (x – 2)(x – 3) Check whether the binomial factors are correct: (x – 2)(x – 3) = x(x – 3) – 2(x – 3) Using the distrib Using the distrib Using the distrib Using the distributiutiutiutiutivvvvve pre pre pre pre proper oper oper opertytytytyty Using the distrib oper = x2 – 3x – 2x + 6 = x2 – 5x + 6 This is the same as the original expression, so the factors are correct. Example Example Example Example Example 44444 Factor x2 + 2x – 8. Solution Solution Solution Solution Solution Find two numbers that multiply to give –8 and add to give +2, the coefficient of x. x2 + 2x – 8 = (x + 4)(x – 2) Check whether the binomial factors are correct: (x + 4)(x – 2) = x(x – 2) + 4(x – 2) Using the distrib opertytytytyty oper oper Using the distributiutiutiutiutivvvvve pre pre pre pre proper Using the distrib Using the distrib oper Using the distrib = x2 – 2x + 4x – 8 = x2 + 2x – 8 This is the same as the original expression, so the factors are correct. Section 6.6 Section 6.6 Section 6.6 — Factoring Quadratics Section 6.6 Section 6.6 311311311311311 Guided Practice Factor each expression below. 1. a2 + 7a + 10 3. x2 – 17x + 72 5. b2 + 2b – 24 7. x2 – 15x + 54 9. t2 + 16t + 55 11. x2 – 3x – 18 13. x2 – 2x – 15 15. x2 – 4 17. 4x2 – 64 19. x2 – 49a2 2. x2 + 7x + 12 4. x2 + x – 42 6. a2 – a – 42 8. m2 + 2m – 63 10. p2 + 9p – 10 12. p2 + p – 56 14. n2 – 5n + 4 16. x2 – 25 18. 9a2 – 36 20. 4a2 – 100b2 Independent Practice Find the value of ? in the problems below. 1. x2 + 3x – 4 = (x + 4)(x + ?) 3. x2 + 16x – 17 = (x + ?)(x – 1) 5. a2 + 6a – 40 = (a – 4)(a + ?) Factor each expression below. 2. a2 – 2a – 8 = (a + ?)(a + 2) 4. x2 – 14x – 32 = (x + 2)(x + ?) 6. x2 – 121 9. x2 + 2x + 1 12. d2 + 21d + 38 15. a2 – 16a + 48 18. a2 + 5a – 24 7. 4c2 – 64 10. x2 + 8x + 16 13. x2 – 13x + 42 16. x2 + 18x + 17 19. b2 – 19b – 120 8. 16a2 – 225 11. b2 – 10b + 25 14. a2 – 18a + 45 17. x2 – 24x + 80 20. x2 + 14x – 72 21. Determine whether (x +
3) is a factor of x2 – 2x – 15. 22. If 2n3 – 5 is a factor of 12n5 + 2n4 – 30n2 – 5n, find the other factors. 23. If (8n – 3) is a factor of 8n3 – 3n2 – 8n + 3, find the other factors. 24. If (2x + 5) is a factor of 2x3 + 15x2 + 13x – 30, find the other factors. 25. If (a – 1) is a factor of a3 – 6a2 + 9a – 4, find the other factors. 26. If (x – 2) is factor of x3 + 5x2 – 32x + 36, find the other factors. What can you say about the signs of a and b when: 27. x2 + 9x + 16 = (x + a)(x + b), 28. x2 – 4x + 6 = (x + a)(x + b), 29. x2 + 10x – 75 = (x + a)(x + b), 30. x2 – 4x – 32 = (x + a)(x + b)? ound Up ound Up RRRRRound Up ound Up ound Up RRRRRound Up ound Up ound Up ound Up ound Up The method in this Topic only works for quadratic expressions that have an x2 term with a coefficient of 1 (so it’s usually written just as x2 rather than 1x2). In the next Topic you’ll see how to deal with other types of quadratics. 312312312312312 Section 6.6 Section 6.6 Section 6.6 — Factoring Quadratics Section 6.6 Section 6.6 TTTTTopicopicopicopicopic 6.6.26.6.2 6.6.26.6.2 6.6.2 California Standards: Students applpplpplpplpply basic y basic y basic Students a Students a 11.0: 11.0: y basic 11.0: Students a 11.0: y basic Students a 11.0: fffffactoring tec hniques to hniques to actoring tec actoring tec hniques to actoring techniques to hniques to actoring tec second- second- second- and simple thirdsecond- second- ynomials..... TTTTThese dededededegggggrrrrree pol hese hese ynomials ynomials ee pol ee pol hese ee polynomials hese ynomials ee pol lude finding lude finding hniques inc hniques inc tectectectectechniques inc lude finding hniques include finding lude finding hniques inc or all or all actor f actor f a common f a common f actor for all a common factor f or all a common f or all actor f a common f ynomial, ynomial, ms in a pol ms in a pol terterms in a pol terter ynomial, ms in a polynomial, ter ynomial, ms in a pol recognizing the difference of two squares, and recognizing perfect squares of binomials. What it means for you: You’ll learn how to factor more complicated quadratic expressions. Key words: quadratic polynomial factor trial and error actoring Quadraaaaaticsticsticsticstics actoring Quadr actoring Quadr FFFFFactoring Quadr actoring Quadr actoring Quadraaaaaticsticsticsticstics actoring Quadr actoring Quadr FFFFFactoring Quadr actoring Quadr + + ² + ² + — — — — — ax² + + c ² + bx + + ² + — — — — — ax² + + + ² + ² + ² + bx + + c ² + + The method in Topic 6.6.1 for factoring a quadratic expression only works if the x²-term has a coefficient of 1. It’s a little more complicated when the x²-coefficient (a) isn’t 1 — but only a little. om Each h h h h TTTTTererererermmmmm om Eac om Eac actor fr actor fr e Out a Common F ou Can TTTTTakakakakake Out a Common F e Out a Common F ou Can YYYYYou Can ou Can actor from Eac e Out a Common Factor fr om Eac actor fr e Out a Common F ou Can If you see a common factor in each term, such as a number or a variable, take it out first. Example Example Example Example Example 11111 Factor 3x² + 15x + 18. Solution Solution Solution Solution Solution 3x² + 15x + 18 = 3(x² + 5x + 6) The expression in parentheses can be factored using the method in Topic 6.6.1: = 3(x + 2)(x + 3) But — if the expression in parentheses still has a π 1, then the expression will need to be factored using the second method, shown in Example 2. Guided Practice Factor each expression completely. 1. 3x2 + 15x + 12 3. 2t2 – 22t + 60 5. 4x2 + 32x + 64 7. 5m2 + 20m + 15 9. –2k2 – 20k – 32 11. 2 + x – x2 13. 3x2y – 33xy – 126y 15. 100 + 75x – 25x2 17. 32a2 – 8x2a2 19. 2x2m2 + 28x2m – 102x2 2. 4y2 – 12y – 112 4. 3r2 – 75 6. 7p2 + 70p + 63 8. 6x2 + 42x + 60 10. –3m2 – 30m – 72 12. –3x2 – 84x – 225 14. 10x2 + 290x + 1000 16. 100n2y + 100ny – 5600y 18. 21a – 80 – a2 20. 3a2b2x2 + 30a2bx + 63a2 Section 6.6 Section 6.6 Section 6.6 — Factoring Quadratics Section 6.6 Section 6.6 313313313313313 + c b b b b by y y y y TTTTTrial and Er ² + bx + actor ax² + rial and Errrrrrororororor rial and Er rial and Er + + ² + ² + actor actor ou Can F YYYYYou Can F ou Can F ou Can Factor rial and Er + ² + actor ou Can F If you can’t see a common factor, then you need to get ax² + bx + c into the form (a1x + c1)(a2x + c2), where a1 and a2 are factors of a, and c1 and c2 are factors of c. a = a1a2 ax² + bx + c = (a1x + c1)(a2x + c2) a1c2 + a2c1 = b c = c1c2 Note that if a number is positive then its two factors will be either both positive or both negative. If a number is negative, then its two factors will have different signs — one positive and one negative. These facts give important clues about the signs of c1 and c2. Example Example Example Example Example 22222 Factor 3x² + 11x + 6. Solution Solution Solution Solution Solution Write down pairs of factors of a = 3: = 1 and a22222 = a11111 = 1 = 1 3x² + 11x + 6 Write down pairs of factors of c = 6: = 1 and c22222 = c11111 = 1 = 1 = 2 and c22222 = c11111 = 2 = 2 Now find the combination of factors that gives a1c2 + a2c1 = b = 11. Put the x-terms into parentheses first, with the pair of coefficients 3 and 1: (3x )(x ). Now try all the pairs of c1 and c2 in the parentheses and find the possible values of a1c2 + a2c1 (and a1c2 – a2c1): (3x 1)(x 6) multiplies to give 18x and x, which add/subtract to give 19x or 17x. (3x 6)(x 1) multiplies to give 3x and 6x, which add/subtract to give 9x or 3x. (3x 2)(x 3) multiplies to give 9x and 2x, which add/subtract to give 1111111111x or 7x. (3x 3)(x 2) multiplies to give 6x and 3x, which add/subtract to give 9x or 3x. So (3x 2)(x 3) is the combination that gives 11x (so b = 11). Now fill in the + / – signs. Both c1 and c2 are positive (since c = c1c2 and b = a1c2 + a2c1 are positive), so the final factors are (3x + 2)(x + 3). Check by expanding the parentheses to make sure they give the original equation: (3x + 2)(x + 3) = 3x2 + 9x + 2x + 6 = 3x2 + 11x + 6 That’s what you started with, so (3x + 2)(x + 3) is the correct factorization. Check it out: Each pair of coefficients c1 and c2 has TWO possible positions. 314314314314314 Section 6.6 Section 6.6 Section 6.6 — Factoring Quadratics Section 6.6 Section 6.6 Check it out: You can consider separately whether the values of c1 and c2 should be positive or negative. Guided Practice Factor each polynomial. 21. 2x2 + 5x + 2 23. 2y2 + 7y + 3 25. 4x2 + 12x + 9 27. 3x2 + 13x + 12 29. 4a2 + 16a + 7 31. 8a2 + 46a + 11 33. 9x2 + 12x + 4 35. 4b2 + 32b + 55 37. 10a2 + 23a + 12 39. 4x2 + 34x + 16 22. 2a2 + 13a + 11 24. 4x2 + 28x + 49 26. 6x2 + 23x + 7 28. 2x2 + 11x + 5 30. 2p2 + 14p + 12 32. 3g2 + 51g + 216 34. 4a2 + 36a + 81 36. 3x2 + 22x + 24 38. 6t2 + 23t + 20 40. 15b2 + 96b + 36 us Sign us Sign s a Min eful if TTTTTherherherherhere’e’e’e’e’s a Min s a Min eful if eful if Be Car Be Car us Sign s a Minus Sign Be Careful if us Sign s a Min eful if Be Car Be Car Example Example Example Example Example 33333 Factor 6x² + 5x – 6. Solution Solution Solution Solution Solution Write down pairs of factors of a = 6 and a22222 = 1 a11111 = 6 = 1 = 6 = 2 and a22222 = a11111 = 2 = 2 6x² + 5x – 6 Write down pairs of factors of c = –6 (ignoring the minus sign for now): = 1 and c22222 = c11111 = 1 = 1 = 2 and c22222 = c11111 = 2 = 2 Put the x-terms into parentheses first, with the first pair of possible values for a1 and a2, 6 and 1: (6x )(1x ). Now try all the pairs of c1 and c2 in the parentheses and find a1c2 + a2c1 and a1c2 – a2c1 like before: (6x 1)(x 6) multiplies to give 36x and x, which add/subtract to give 37x or 35x. (6x 6)(x 1) multiplies to give 6x and 6x, which add/subtract to give 12x or 0x. (6x 2)(x 3) multiplies to give 18x and 2x, which add/subtract to give 20x or 16x. (6x 3)(x 2) multiplies to give 12x and 3x, which add/subtract to give 15x or 9x. None of these combinations works, so try again with (2x )(3x ): (2x 1)(3x 6) multiplies to give 12x and 3x, which add/subtract to give 15x or 9x. (2x 6)(3x 1) multiplies to give 2x and 18x, which add/subtract to give 20x or 16x. (2x 3)(3x 2) multiplies to give 4x and 9x, which add/subtract to give 13x or 55555x. 5x is what you want, so you can stop there — so (2x 3)(3x 2) is the right combination. Now fill in the + / – signs to get b = +5. One of c1 and c2 must be negative, to give c = –6, so the final factors are either (2x + 3)(3x – 2) or (2x – 3)(3x + 2). The x-term of (2x + 3)(3x – 2) will be 9x – 4x = 5x, whereas the x-term for (2x – 3)(3x + 2) will be 4x – 9x = –5x. So the correct factorization is (2x + 3)(3x – 2). Section 6.6 Section 6.6 Section 6.6 — Factoring Quadratics Section 6.6 Section 6.6 315315315315315 Guided Practice Factor each polynomial. 41. 2x2 + 3x – 2 43. 5k2 + 13k – 6 45. 6b2 – 23b + 7 47. 3k2 – 2k – 1 49. 18 + 5x – 2x2 51. 9x2 + 12x + 4 53. 3x2 – 7x – 6 55. 6x2 + 2x – 20 57. 6y2 + y – 12 42. 3y2 – y – 2 44. 3x2 – x – 10 46. 2x2 – 5x + 2 48. 3v2 – 16v + 5 50. 28 + x – 2x2 52. 7a2 – 26a – 8 54. 12x2 + 5x – 2 56. 18x2 + x – 4 58. 9m2 – 3m – 20 tic to WWWWWororororork Fk Fk Fk Fk Faster aster aster tic to Use the + and – Signs in the Quadraaaaatic to tic to Use the + and – Signs in the Quadr Use the + and – Signs in the Quadr aster aster tic to Use the + and – Signs in the Quadr Use the + and – Signs in the Quadr Looking carefully at the signs in the quadratic that you are factoring can help to narrow down the choices for a1, a2, c1, and c2. Example Example Example Example Example 44444 Factor 3x² + 11x + 6. Solution Solution Solution Solution Solution Here c = 6, which is positive — so its factors c1 and c2 are either both positive or both negative. But since b = 11 is positive, you can tell that c1 and c2 must be positive (so that a1c2 + a2c1 is positive). Example Example Example Example Example 55555 Factor 3x² – 11x + 6. Solution Solution Solution Solution Solution Here c is also positive, so c1 and c2 are either both positive or both negative. But since b = –11 is negative, you can tell that c1 and c2 must be negative (so that a1c2 + a2c1 is
negative). 316316316316316 Section 6.6 Section 6.6 Section 6.6 — Factoring Quadratics Section 6.6 Section 6.6 Example Example Example Example Example 66666 Factor 6x² + 5x – 6. Solution Solution Solution Solution Solution In this expression, c is negative, so one of c1 and c2 must be positive and the other must be negative. So instead of looking at both the sums and differences of all the different combinations a1c2 and a2c1, you only need to look at the differences. Guided Practice Factor each expression. 59. 2n2 + n – 3 61. 4a2 + 4a + 1 63. 9y2 + 6y + 1 65. 5x2 – x – 18 67. 6t2 + t – 1 60. 2x2 – 5x – 3 62. 3x2 – 4x + 1 64. 4t2 + t – 3 66. 9x2 – 6x + 1 68. b2 + 10b + 21 Independent Practice Factor each polynomial. 1. 5k2 – 7k + 2 3. 12t2 – 11t + 2 5. 10 + h – 3h2 7. 6 + 5x – 6x2 2. 4k2 – 15k + 9 4. 9 – 7k – 2k2 6. 15x2 – 14x – 8 8. 3x4 + 8x2 – 3 9. If the area of a rectangle is (6x2 + 25x + 14) square units and the length is (3x + 2) units, find the width w in terms of x. 10. The area of a parallelogram is (12x2 + 7x – 10) cm2, where x is positive. If the area is given by the formula Area = base × height, find the base length and the height of the parallelogram, given that they are both linear factors of the area. Factor each of these expressions completely. 11. x2ab – 3abx – 18ab 12. (d + 2)x2 – 7x(d + 2) – 18(d + 2) 13. (x + 1)m2 – 2m(x + 1) + (x + 1) 14. –2dk2 – 14dk + 36d 15. Five identical rectangular floor tiles have a total area of (15x² + 10x – 40) m². Find the dimensions of each floor tile, if the length of each side can be written in the form ax + b, where a and b are integers. ound Up ound Up RRRRRound Up ound Up ound Up ound Up ound Up RRRRRound Up ound Up ound Up Now you can factor lots of different types of polynomials. In the next Section you’ll learn about another type — quadratic expressions containing two different variables. Section 6.6 Section 6.6 Section 6.6 — Factoring Quadratics Section 6.6 Section 6.6 317317317317317 TTTTTopicopicopicopicopic 6.7.16.7.1 6.7.16.7.1 6.7.1 California Standards: Students applpplpplpplpply basic y basic y basic Students a Students a 11.0: 11.0: y basic 11.0: Students a 11.0: y basic Students a 11.0: fffffactoring tec hniques to hniques to actoring tec actoring tec hniques to actoring techniques to hniques to actoring tec second- second- second- and simple thirdsecond- second- ynomials..... TTTTThese dededededegggggrrrrree pol hese hese ynomials ynomials ee pol ee pol hese ee polynomials hese ynomials ee pol lude finding lude finding hniques inc hniques inc tectectectectechniques inc lude finding hniques include finding lude finding hniques inc or all or all actor f actor f a common f a common f actor for all a common factor f or all a common f or all actor f a common f ynomial, ynomial, ms in a pol ms in a pol terterms in a pol terter ynomial, ms in a polynomial, ter ynomial, ms in a pol recognizing the difference of two squares, and recognizing perfect squares of binomials. What it means for you: You’ll learn how to factor quadratic expressions containing two variables. Key words: quadratic factor trial and error Check it out: Following the rules from Topic 6.6.2, both of the terms must be positive — so the + signs can be put in the parentheses from the start. Section 6.7 actoring Quadraaaaaticsticsticsticstics actoring Quadr actoring Quadr FFFFFactoring Quadr actoring Quadr FFFFFactoring Quadr actoring Quadraaaaaticsticsticsticstics actoring Quadr actoring Quadr actoring Quadr in TTTTTwwwwwo o o o o VVVVVariaariaariaariaariabbbbbleslesleslesles in in in in in TTTTTwwwwwo o o o o VVVVVariaariaariaariaariabbbbbleslesleslesles in in in in So far, most of the quadratics you’ve factored have had only one variable — but the same rules apply if there are two variables. actorededededed actor actor Also Be F Also Be F les Can tics with TTTTTwwwwwo o o o o VVVVVariaariaariaariaariabbbbbles Can les Can tics with Quadraaaaatics with tics with Quadr Quadr Also Be Factor les Can Also Be F actor Also Be F les Can tics with Quadr Quadr In Section 6.6 you saw that a quadratic expression such as x² + 2x + 1 can be written as two factors — in this case (x + 1)(x + 1). The same is true of an expression such as x² + 2xy + y² — it can be written as (x + y)(x + y). The method for factoring an expression like this is the same as before: Example Example Example Example Example 11111 Factor the following expression: x² + 4xy + 3y² Solution Solution Solution Solution Solution x² + 4xy + 3y² = (x + )(x + ) To fill the gaps you need two numbers or expressions that will multiply together to make 3y² and add together to make 4y. Try out some sets of numbers or expressions that multiply to make 3y²: 3y² and 1 add together to make 3y² + 1 3y and y add together to make 4y So x² + 4xy + 3y² = (x + 3y)(x + y). Example Example Example Example Example 22222 Factor the following expression: 3p² + 5pq + 2q² Solution Solution Solution Solution Solution 3p² + 5pq + 2q² = (3p + )(p + ) Filling these gaps is a little more complicated. You need two numbers or expressions that will multiply together to make 2q² and, when multiplied by the 3p and p respectively, add together to make 5pq. Try out some sets of numbers or expressions: (3p + q)(p + 2q) — this would give pq-terms of 6pq and pq, which add to make 7pq (3p + 2q)(p + q) — this would give pq-terms of 3pq and 2pq, which add to make 5pq 5pq is what you need, so 3p² + 5pq + 2q² = (3p + 2q)(p + q). 318318318318318 Section 6.7 Section 6.7 Section 6.7 — More on Factoring Polynomials Section 6.7 Section 6.7 Guided Practice Factor each polynomial below. 1. x2 + 3xy + 2y2 3. a2 + 9ab + 18b2 5. d2 + 21md + 20m2 7. x2 – xy – 2y2 9. a2 – ab – 12b2 2. x2 + 14xy + 40y2 4. p2 + 7pq + 12q2 6. k2 + 8pk + 12p2 8. x2 – 2xz – 8z2 10. x2 – 5xy + 6y2 The areas of the rectangles below are the products of two binomials with integer coefficients. Find the possible length and width of each rectangle. 11. Area = (x2 + 3xa + 2a2) ft2 12. Area = (y2 – 9yb + 14b2) in2 13. Area = (3x2 – 4xb + b2) ft2 14. Area = (5a2 – ab – 18b2) m2 15. Area = (4c2 + 4cd + d2) ft2 16. Area = (12x2 – 5xy – 2y2) in2 e More Pe Pe Pe Pe Possibilities ossibilities ossibilities e Mor or is TTTTTricricricricrickkkkky ify ify ify ify if TTTTTherherherherhere are are are are are Mor e Mor or is rial and Errrrrror is or is rial and Er TTTTTrial and Er rial and Er ossibilities ossibilities e Mor or is rial and Er Example Example Example Example Example 33333 Factor the following expression: 2m² – 11mp + 5p² Solution Solution Solution Solution Solution 2m² can be factored into 2m and m, so: 2m² – 11mp + 5p² = (2m – )(m – ) To fill the gaps you need two terms in p that will multiply together to make 5p², and when multiplied by 2m and m respectively, will add together to make –11mp. Try out some sets of parentheses that multiply to make 5p²: (2m – 5p)(m – p) — this would give mp-terms of –2mp and –5mp, which add to make –7mp (2m – p)(m – 5p) — this would give mp-terms of –10mp and –mp, which add to make –11mp The second one gives the –11mp needed, so 2m² – 11mp + 5p² = (2m – p)(m – 5p). Section 6.7 Section 6.7 Section 6.7 — More on Factoring Polynomials Section 6.7 Section 6.7 319319319319319 Check it out: Both p-terms must be negative, so that the mp-terms add up to a negative term and the p-terms multiply to give a positive p²-term. Example Example Example Example Example 44444 Factor the following expression: 9x² + 6xz – 8z² Solution Solution Solution Solution Solution The z²-term is negative, so one of the z-terms will be negative and the other positive. So that means there are a lot more combinations to try out. 9x² can be made by either 3x × 3x or 9x × x, and –8z² can be made by any of –2z × 4z, 2z × –4z, 8z × –z, and –8z × z. Try out some sets of parentheses: (9x – 4z)(x + 2z) — this would give 18xz and –4xz, which add to make +14xz (9x + 2z)(x – 4z) — this would give –36xz and 2xz, which add to make –34xz (9x – 2z)(x + 4z) — this would give 36xz and –2xz, which add to make +34xz (9x + 4z)(x – 2z) — this would give –18xz and 4xz, which add to make –14xz (3x – 4z)(3x + 2z) — this would give 6xz and –12xz, which add to make –6xz (3x – 2z)(3x + 4z) — this would give 12xz and –6xz, which add to make +6xz You can stop here because +6xz is the expression you are trying to get. So 9x² + 6xz – 8z² = (3x – 2z)(3x + 4z). Guided Practice Factor each of the polynomials below. 17. 2x2 – 5xy – 3y2 19. 3x2 + 17xy + 10y2 21. 3g2 + 7gh + 4h2 23. 4f 2 – 16gf + 15g2 25. 8m2 – 2mh – 15h2 18. 3m2 – 7mp + 2p2 20. 4x2 + 9xy + 2y2 22. 2a2 + 9ab + 9b2 24. 49w2 + 7wz – 6z2 26. 6x2 + 17xy + 12y2 The areas of the rectangles below are the products of two binomials with integer coefficients. Find the possible dimensions of the rectangles. 27. Area = (2m2 + 3mn – 2n2) ft2 28. Area = (a2 + 8ab + 15b2) in.2 29. Area = (3x2 + 10xy + 8y2) m2 30. Area = (15x2 – 29xy + 14y2) ft2 31. Area = (6a2 + 11ab – 10b2) ft2 32. Area = (6c2 + 10cd + 4d2) m2 320320320320320 Section 6.7 Section 6.7 Section 6.7 — More on Factoring Polynomials Section 6.7 Section 6.7 Independent Practice Factor each of these polynomials. 1. p2 – 7pq + 10q2 3. c2 + 3cd – 40d2 5. y2 – 8xy + 15x2 7. 2r2 + 11rk + 14k2 2. g2 + 4hg – 21h2 4. m2 + 8mn – 20n2 6. 5p2 + 26pq + 5q2 8. 3x2 – 8xy – 3y2 Simplify and factor the following polynomials. 9. (x2 + 13x + 8) – (3x2 + 10x – 3) – (x2 + 2x + 6) + (4x2 – 5x – 2) 10. (5x2 + 2x + 4) – (6x2 – 3x + 7) + (4x2 – x – 4) 11. (4x2 – 6xy – 10y2) – (2x2 – 8xy + 2y2) 12. (6x2 + 3xy + 8y2) – (3x2 – 12xy – 10y2) 13. (2t2 – 8tz – 5z2) – (4t2 + 2tz – 15z2) + (5t2 + tz – 40z2) 14. (6x2 – 4xy – 25y2) – (2x2 + 4xy + 25y2) – (2x2 + 4xy – 18y2) The areas of the circles below are products of p and a binomial squared. Find the radius of each circle. 15. Area of circle is (9px2 + 24pxy + 16py2) in.2 16. Area of the circle is (4px2 – 20pax + 25pa2) m2 17. Area of the circle is (9py2 + 48pzy + 64pz2) ft2 18. Area of the circle is (4pm2 + 20pmn + 25pn2
) ft2 The areas of the parallelograms below are the products of two binomials with integer coefficients. Find the dimensions of each parallelogram if a = 10 and b = 5. 19. Parallelogram with area (8a2 – 10ab + 3b2) ft2 20. Parallelogram with area (4a2 – 9b2) ft2 21. Parallelogram with area (12a2 + 11ab – 5b2) ft2 22. Parallelogram with area (40a2 – 51ab – 7b2) ft2 Factor the following polynomials. 23. 6a2z2k + 20z2abk + 16b2z2k 24. 36a2b2c – 15ab3c – 6b4c 25. 25x3y2 – 5x2y3 – 90xy4 26. 16a2z2c + 16abz2c + 4b2z2c Factor and simplify completely. 27. 12x2(x + 2) + 25xy(x + 2) + 12y2(x + 2) 28. 18a2b2(a – 1) – 33ab3(a – 1) – 30b4(a – 1) ound Up ound Up RRRRRound Up ound Up ound Up RRRRRound Up ound Up ound Up ound Up ound Up This is a long process, so it’s easy to make mistakes. You should always check your answer by multiplying out the parentheses again. If you don’t get the expression you started with, you must have gone wrong somewhere. That means you’ll need to go back a stage in your work and try a different combination of factors. Section 6.7 Section 6.7 Section 6.7 — More on Factoring Polynomials Section 6.7 Section 6.7 321321321321321 TTTTTopicopicopicopicopic 6.7.26.7.2 6.7.26.7.2 6.7.2 California Standards: Students applpplpplpplpply basic y basic y basic Students a Students a 11.0: 11.0: y basic 11.0: Students a 11.0: y basic Students a 11.0: fffffactoring tec hniques to hniques to actoring tec actoring tec hniques to actoring techniques to hniques to actoring tec second- and simple third-d-d-d-dsecond- and simple thir second- and simple thir second- and simple thir second- and simple thir ynomials..... TTTTThese dededededegggggrrrrree pol hese hese ynomials ynomials ee pol ee pol hese ee polynomials hese ynomials ee pol hniques inc hniques inc tectectectectechniques inc lude finding lude finding hniques include finding lude finding hniques inc lude finding or all or all actor f actor f a common f a common f actor for all a common factor f or all a common f or all actor f a common f ynomial, ynomial, ms in a pol ms in a pol terterms in a pol terter ynomial, ms in a polynomial, ter ynomial, ms in a pol recognizing the difference of two squares, and recognizing perfect squares of binomials. What it means for you: You’ll learn how to factor third-degree expressions. Key words: polynomial degree factor Don’t forget: See Topic 6.6.2 for more on common factors. d-Degggggrrrrreeeeeeeeee d-De d-De actoring TTTTThirhirhirhirhird-De actoring actoring FFFFFactoring d-De actoring actoring TTTTThirhirhirhirhird-De d-Degggggrrrrreeeeeeeeee d-De d-De actoring actoring FFFFFactoring d-De actoring ynomials ynomials PPPPPolololololynomials ynomials ynomials PPPPPolololololynomials ynomials ynomials ynomials ynomials Quadratics are second-degree polynomials because they have an x2-term. Now you’ll factor polynomials with an x3-term too. ynomials in Stagggggeseseseses ynomials in Sta d-Degggggrrrrree Pee Pee Pee Pee Polololololynomials in Sta ynomials in Sta d-De actor TTTTThirhirhirhirhird-De d-De actor FFFFFactor actor ynomials in Sta d-De actor To factor a third-degree polynomial, the first thing you should do is look for a common factor. Separate any obvious common factors, then try to factor the remaining expression (the part inside the parentheses). Example Example Example Example Example 11111 Factor x³ + 7x² + 12x completely. Solution Solution Solution Solution Solution All of the terms in x³ + 7x² + 12x contain x, so x is a factor: x³ + 7x² + 12x = x ⎛ ⎜⎜⎜⎜ ⎝ 3 x x + 27 x x + x 12 x = x(x2 + 7x + 12) ⎞ ⎟⎟⎟⎟ ⎠ Now look at the factor in the parentheses — this is a quadratic expression that it may be possible to factor. In this case it’s possible to factor it, using the method from Section 6.6: x³ + 7x² + 12x = x(x2 + 7x + 12) = x(x + 3)(x + 4) Guided Practice Factor completely these polynomials. 1. 4y3 + 26y2 + 40y 3. 6x3 – 7x2 – 20x 5. 6a3b2 + 33a2b2 + 15ab2 7. 10a3b2c2 + 45a2b2c2 + 20ab2c2 9. 12b4y + 38b3y + 30b2y 11. 36h3k4 + 12h2k4 – 63hk4 13. 189b6c4 – 60b5c4 – 96b4c4 15. 12x4y2 + 78x3y2 + 108x2y2 17. 18b4c – 87b3c + 105b2c 2. 24x3 – 33x2 + 9x 4. 12x3 – 18x2 – 12x 6. 6k3j3 – 10k2j3 – 4kj3 8. 16x4z2 + 12x3z2 – 4x2z2 10. 24c5t2 + 132c4t2 + 144c3t2 12. 40x6y5 + 11x5y5 – 2x4y5 14. 24x2b2c2 + 44xb2c2 – 140b2c2 16. 12a4b2 + 46a3b2 + 40a2b2 18. 12a4b2c – 46a3b2c + 40a2b2c 322322322322322 Section 6.7 Section 6.7 Section 6.7 — More on Factoring Polynomials Section 6.7 Section 6.7 Example Example Example Example Example 22222 Factor –2x³ – 2x² + 4x completely. Solution Solution Solution Solution Solution Again, x is a factor. Each term has an even coefficient — so you can take out a factor of 2. And given that two of the terms are negative, you can take out a factor of –2 instead of 2. This is helpful because it means that the coefficient of x² becomes 1 — which makes the quadratic expression much easier to factor, as you saw in Section 6.6. –2x³ – 2x² + 4x = –2x ⎛ ⎜⎜⎜⎜ ⎝ 2x(x2 + x – 2) Now factor the quadratic: –2x³ – 2x² + 4x = –2x(x2 + x – 2) = –2x(x – 1)(x + 2) ⎞ ⎟⎟⎟⎟ ⎠ Note that in Example 2, if 2x had been factored out instead of –2x, the result would have been: –2x³ – 2x² + 4x = 2x 3 ⎛ ⎜⎜⎜⎜ ⎝ − ⎞ ⎟⎟⎟⎟ ⎠ = 2x(–x2 – x + 2) = 2x(1 – x)(x + 2) or 2x(x – 1)(–x – 2) ...which is also correct, but is a little trickier to factor. Guided Practice Factor completely the polynomials below. 19. –4x2y2 – 6xy2 + 4y2 21. –12a3b3 – 30a3b2 + 18a3b 23. –4a3b – 26a2b – 36ab 25. –16a3b2 + 16ab2 27. –147x3y + 84x2y – 12xy 29. 18b4c – 66b3c + 60b2c 31. –42c4d – 28c3d + 14c2d 33. –162a2b3 + 2a2b 35. –60b2d6 + 9b2d5 + 6b2d4 20. –8xy4 + 4xy3 + 60xy2 22. –8w3k – 42w2k – 10wk 24. –90b4c2 – 174b3c2 – 48b2c2 26. –45x4y2 – 50x3y2 – 5x2y2 28. –162c5d + 288c4d – 128c3d 30. –50y4z – 130y3z + 60y2z 32. –54y3z2 + 21y2z2 + 3yz2 34. 16a4b2 + 176a3b2 + 484a2b2 36. –12b4f2 + 68b3f 2 – 96b2f 2 Section 6.7 Section 6.7 Section 6.7 — More on Factoring Polynomials Section 6.7 Section 6.7 323323323323323 Independent Practice Factor these polynomials completely. 1. –8a3 + 78a2 + 20a 3. 14y2d2 + 7yd2 – 42d2 5. 250x2y4 + 100x2y3 + 10x2y2 2. 18x2y2 + 51xy2 + 15y2 4. 30x3y3 – 35x2y3 – 100xy3 6. 120a3b3c3 + 28a2b3c3 – 8ab3c3 7. Which of the following is equivalent to 18a2x2 + 3a2x – 3a2? (i) 3a2(2x + 1)(2x – 1) (ii) 3a2(3x – 1)(2x + 1) 8. Which of the following is equivalent to 18a2x2 – 32a2? (i) 2a2(3x + 4)(3x – 4) (ii) 2a2(3x – 4)(3x – 4) 9. Which of the following is a factor of 35x2 + 64ax + 21a2? (i) 5x + 3a (iii) 7x + 5a (ii) 5x + 7a (iv) 7x + 7a 10. Which of the following is a factor of 8y2a2 – 50ya3 – 42a4? (i) 2a3 (iii) 4y – 7a (ii) 8a2 (iv) 4y + 3a Find the value of ? in the problems below. (The symbol “∫” means that the equation is true for all values of the variables.) 11. 81j4 – 36j? + 4j2 ∫ j2(9j – 2)2 12. 18w3 – 48w? + 32w ∫ 2w(3w – 4)2 13. 12a3b + 70a?b + 72ab ∫ 2ab(2a + 9)(3a + 4) 14. –12a4b? – 58a3b3 – 70a2b3 ∫ –2a2b3(2a + 5)(3a + 7) 15. –56a4b4 + 12a3b4 + 8a2b4 ∫ –?a2b4(2a – 1)(7a + 2) 16. A cylinder has a base with dimensions that are binomial factors. If the volume of the cylinder is (75px3 + 30px2 + 12px) in.3 and the height is 3x in., find the radius of the base, in terms of x. (V = pr2h) 17. A rectangular prism has a base with dimensions that are binomial factors. If the volume of the prism is (30x3 – 28x2 – 16x) in3 and the height is 2x in., find the dimensions of the base, in terms of x. 18. The product of three consecutive odd integers is x3 + 6x2 + 8x. Find each of the three integers in terms of x. 19. The volume of a rectangular box is (6x3 + 17x2 + 7x) cubic inches, and its height is x inches. Find the dimensions of the base of the box, w and l, in terms of x, given that w and l can be expressed in the form (ax + b), where a and b are integers. 20. A cylinder has a base with dimensions that are binomial factors. If the volume of the cylinder is (36px3 – 96px2 + 64px) ft3 and the height is 4x ft, find the radius of the base, in terms of x. (V = pr2h) ound Up ound Up RRRRRound Up ound Up ound Up RRRRRound Up ound Up ound Up ound Up ound Up Take care when you’re factoring cubic expressions — you need to take it step by step. Often you’ll be able to factor out a term containing x — then you’ll be left with a normal quadratic inside the parentheses. Look back at Section 6.6 if you’re having trouble with the quadratic part. 324324324324324 Section 6.7 Section 6.7 Section 6.7 — More on Factoring Polynomials Section 6.7 Section 6.7 TTTTTopicopicopicopicopic 6.8.16.8.1 6.8.16.8.1 6.8.1 California Standards: Students applpplpplpplpply basic y basic y basic Students a Students a 11.0: 11.0: y basic 11.0: Students a 11.0: y basic Students a 11.0: fffffactoring tec hniques to hniques to actoring tec actoring tec hniques to actoring techniques to hniques to actoring tec second- second- second- and simple thirdsecond- second- dededededegggggrrrrree pol ynomials..... TTTTThese hese hese ynomials ynomials ee pol ee pol ee polynomials hese hese ynomials ee pol lude lude hniques inc tectectectectechniques inc hniques inc lude finding a hniques include lude hniques inc common factor for all terms in a polynomial, rrrrrecoecoecoecoecognizing gnizing gnizing gnizing gnizing ence of tw tw tw tw twooooo ence of ence of the difffffferererererence of the dif the dif the dif ence of the dif squareseseseses, and recognizing squar squar squar squar perfect squares of binomials..... What it means for you: You’ll use the difference of two squares to factor quadratics. Key words: difference of two squares quadratic Section 6.8 ence of ence of he Difffffferererererence of he Dif he Dif TTTTThe Dif ence of ence of he Dif TTTTThe Dif he Difffffferererererence of ence of ence of he Dif he Dif ence of he Dif ence of o Squareseseseses o Squar o Squar TTTTTwwwwwo Squar o Squar TTTTTwwwwwo Squar o Squareseseseses o Squar o Squar o Squar Being able to recognize the difference of two squares is really useful — it helps you factor quadratic expressions. actor Quadraaaaaticsticst
icsticstics actor Quadr actor Quadr es to F es to F o Squar ence of TTTTTwwwwwo Squar o Squar ence of Use Difffffferererererence of ence of Use Dif Use Dif es to Factor Quadr o Squares to F actor Quadr es to F o Squar ence of Use Dif Use Dif A difference of two squares is one term squared minus another term squared: m2 – c2. You can use this equation to factor the difference of two squares: m2 – c2 = (m + c)(m – c) The difference of two squares m2 – c2 = The sum of the two terms (m + c) × The difference (m – c) between the two terms Example Example Example Example Example 11111 Factor x2 – 9. Don’t forget: The difference of two squares equation was derived in Topic 6.4.1. Solution Solution Solution Solution Solution Substitute x2 for m2 and 9 for c2 in the difference of two squares equation and you get: x2 – 9 = (x + 3)(x – 3) The square root of 9 is 3. Don’t forget the opposite signs. Guided Practice Factor each expression completely. 1. x2 – 16 3. c2 – 49 5. x2 – y2 7. 64 – c2 9. 11x2 – 176 11. 7x2 – 63 13. 3m2n – 12n 15. 162a – 2a3 2. a2 – 25 4. a2 – 100 6. 81 – x2 8. 144 – y2 10. 3y2 – 300 12. 5a2 – 125 14. 6a3 – 216a 16. 7x3y – 7xy3 Section 6.8 Section 6.8 Section 6.8 — More on Quadratics Section 6.8 Section 6.8 325325325325325 Each h h h h TTTTTererererermmmmm Eac Eac oot of k Out the Square Re Re Re Re Root of oot of k Out the Squar WWWWWororororork Out the Squar k Out the Squar oot of Eac Eac oot of k Out the Squar Example Example Example Example Example 22222 Factor 4x2 – 25b2. Solution Solution Solution Solution Solution 4x2 = (2x)2 , so the square root of 4x2 is 2x. 25b2 = (5b)2 , so the square root of 25b2 is 5b. Put the values into the difference of two squares equation: m2 – c2 = (m + c)(m – c) 4x2 – 25b2 = (2x + 5b)(2x – 5b) Square root of 4x2 Square root of 25b2 Guided Practice Factor each expression completely. 17. a2 – 4b2 19. 9a2 – 64x4 21. 49x2 – 4y2 23. 2c3 – 98c 25. 36x2b2 – 49a2c2 27. 75m3n2 – 108a2b2m 18. 4y2 – 81x2 20. 4a2 – 16x2 22. 25y4 – 81c8 24. 3m2n3 – 12a2n 26. 16xbc2 – xba2 28. 18a2b – 242c2b Independent Practice Factor each polynomial completely. 1. 5m3n – 80mn 3. 343m3 – 252mn2 5. 3a3b – 3ab3 7. 4(3x + 2)a2b2 – 9(3x + 2) 9. 3(9a2 – 64b2)a + 8b(9a2 – 64b2) 10. (2x + 5b)4x2 – (2x + 5b)25b2 2. 54a3b – 24ab 4. 50b3 – 18bc2 6. (2x + 1)x2 – (2x + 1) 8. 4a2b2 – 100b2 The areas of the rectangles below are the products of two binomials. Find the two binomials. 11. Area = (81x2 – 100y2) ft2 13. Area = (100a2 – b2) ft2 12. Area = (16a2 – 9y2) ft2 14. Area = (4a2b2 – 25x2) cm2 1 2 (x2 – 8b2) in.2. Find expressions for its base 15. A triangle has area and height dimensions, given that they are binomial factors of the area. Find the base and height when x = 10 and b = 2, if the height is greater than the length. ound Up ound Up RRRRRound Up ound Up ound Up ound Up ound Up RRRRRound Up ound Up ound Up If you see any quadratic expression in the form m2 – c2, you can use the difference of two squares to factor it as (m + c)(m – c), without needing to do all the math. 326326326326326 Section 6.8 Section 6.8 Section 6.8 — More on Quadratics Section 6.8 Section 6.8 TTTTTopicopicopicopicopic 6.8.26.8.2 6.8.26.8.2 6.8.2 California Standards: Students applpplpplpplpply basic y basic y basic Students a Students a 11.0: 11.0: y basic 11.0: Students a 11.0: y basic Students a 11.0: fffffactoring tec hniques to hniques to actoring tec actoring tec hniques to actoring techniques to hniques to actoring tec second- second- second- and simple thirdsecond- second- dededededegggggrrrrree pol ynomials..... TTTTThese hese hese ynomials ynomials ee pol ee pol ee polynomials hese hese ynomials ee pol lude lude hniques inc tectectectectechniques inc hniques inc lude finding a hniques include lude hniques inc common factor for all terms in a polynomial, recognizing the difference of two squares, gnizing perfectectectectect gnizing perf and rrrrrecoecoecoecoecognizing perf gnizing perf gnizing perf binomials..... binomials binomials es of es of squar squar es of binomials squares of binomials es of squar squar What it means for you: You’ll learn about how to factor special quadratics called perfect square trinomials. Key words: perfect square trinomial quadratic binomial factor Don’t forget: If this seems a bit unfamiliar, take a look at Topic 6.4.1 on special binomial products. rinomials rinomials ect Square e e e e TTTTTrinomials ect Squar ect Squar PPPPPerferferferferfect Squar rinomials rinomials ect Squar ect Square e e e e TTTTTrinomials PPPPPerferferferferfect Squar rinomials rinomials ect Squar ect Squar rinomials ect Squar rinomials Perfect square trinomials are quadratic expressions of the form (m + c)2 or (m – c)2. ect Square e e e e TTTTTrinomial rinomial rinomial ect Squar a Binomial is a Perferferferferfect Squar ect Squar a Binomial is a P he Square ofe ofe ofe ofe of a Binomial is a P a Binomial is a P he Squar TTTTThe Squar he Squar rinomial rinomial ect Squar a Binomial is a P he Squar You can use one of two equations to work out the square of a binomial: (m + c)2 = m2 + 2mc + c2 Square of binomial = First term squared + (m + c)2 m2 Twice the product of both terms 2mc + Second term squared c2 Or, if the second term in the binomial is subtracted: (m – c)2 = m2 – 2mc + c2 Square of binomial = First term squared – (m – c)2 m2 Twice the product of both terms 2mc + Second term squared c2 ect Square e e e e TTTTTrinomials rinomials rinomials ect Squar actor Perferferferferfect Squar ect Squar actor P actor P tions to F tions to F Use the Equa Use the Equa rinomials tions to Factor P Use the Equations to F rinomials ect Squar actor P tions to F Use the Equa Use the Equa Example Example Example Example Example 11111 Factor x2 + 2xy + y2. Solution Solution Solution Solution Solution Substitute x2 for m2, 2xy for 2mc, and y2 for c2 in the first equation above: (m + c)2 = m2 + 2mc + c2 ﬁ x2 + 2xy + y2 = (x + y)2 Sometimes you need to factor each term in the expression to get it into the correct form. Section 6.8 Section 6.8 Section 6.8 — More on Quadratics Section 6.8 Section 6.8 327327327327327 Example Example Example Example Example 22222 Factor 4x2 – 12xy + 9y2. Solution Solution Solution Solution Solution Factor each term: 4x2 – 12xy + 9y2 4x2= (2x)2 12xy = 2(2x·3y) 9y2= (3y)2 So: 4x2 – 12xy + 9y2 = (2x)2 – 2(2x·3y) + (3y)2 Substitute (2x)2 for m2, 2(2x·3y) for 2mc, and (3y)2 for c2 in the second perfect square trinomial equation: (m – c)2 = m2 – 2mc + c2 ﬁ (2x)2 – 2(2x·3y) + (3y)2 = (2x – 3y)2 Guided Practice Factor each expression completely. 1. m2 + 2m + 1 3. 25y2 + 10y + 1 5. 9x2 – 6x + 1 7. 16x2 – 24x + 9 9. 25k2 – 20kt + 4t2 2. 4r2 – 4ry + y2 4. 9k2 + 6ky + y2 6. k2 + 4k + 4 8. 4r2x2 + 4rkx + k2 10. m2r2 – 6mr + 9 Independent Practice Factor each expression completely. 1. 16a2b2 + 24ab + 9 3. 9c2d2 + 36cd + 36 5. 162x2y2m + 144xym + 32m 6. 108k3z2 – 180k2z + 75k 2. 25x2y2 – 40xya + 16a2 4. 49m2n2 – 70mn + 25 Find the radius of each of the circles below, given that the area, A, is the product of a binomial squared and p. 7. A = (pa2 + 2pa + p) ft2 9. A = (49pm2 – 56pmn + 16pn2) ft2 10. A = (81px2y2 – 90pxyz + 25pz2) ft2 8. A = (25py2 + 30py + 9p) ft2 The volume, V, of each cylinder below is the product of the height, p, and the radius squared. Find the radius in each case: 11. V = (98x2p + 84xp + 18p) cm3, height = 2 cm 12. V = (147pa2 – 84pab + 12pb2) m3, height = 3 m 13. V = (36px3 + 48px2m + 16xm2) m3, height = 4x m 14. V = (243x4p – 270px3b + 75x2b2) ft3, height = 3x2 ft ound Up ound Up RRRRRound Up ound Up ound Up ound Up ound Up RRRRRound Up ound Up ound Up The phrase “perfect square trinomials” makes this Topic sound much harder than it actually is. They’re really just a special case of the normal quadratic equations that you know and love. 328328328328328 Section 6.8 Section 6.8 Section 6.8 — More on Quadratics Section 6.8 Section 6.8 TTTTTopicopicopicopicopic 6.8.36.8.3 6.8.36.8.3 6.8.3 California Standards: Students applpplpplpplpply basic y basic y basic Students a Students a 11.0: 11.0: y basic 11.0: Students a 11.0: y basic Students a 11.0: fffffactoring tec hniques to hniques to actoring tec actoring tec hniques to actoring techniques to hniques to actoring tec second- second- second- and simple thirdsecond- second- ynomials..... TTTTThese dededededegggggrrrrree pol hese hese ynomials ynomials ee pol ee pol hese ee polynomials hese ynomials ee pol lude finding lude finding hniques inc hniques inc tectectectectechniques inc lude finding hniques include finding lude finding hniques inc or all or all actor f actor f a common f a common f actor for all a common factor f or all a common f or all actor f a common f ynomial, ynomial, ms in a pol ms in a pol terterms in a pol terter ynomial, ms in a polynomial, ter ynomial, ms in a pol recognizing the difference of two squares, and recognizing perfect squares of binomials. What it means for you: You’ll group like terms to factor polynomials. Key words: common factor like terms Check it out: It’s difficult to see any common factors in the expression given in Example 1, so it’s a good idea to group terms together as a first step. ouping ouping y Gr y Gr actoring b actoring b FFFFFactoring b ouping y Grouping actoring by Gr ouping y Gr actoring b FFFFFactoring b ouping ouping y Gr y Gr actoring b actoring b y Grouping actoring by Gr ouping y Gr ouping actoring b Grouping like terms means that you can more easily see whether there are common factors in a polynomial — then you can factor them out. actorsssss actor actor ms to See Common F oup Like e e e e TTTTTerererererms to See Common F ms to See Common F oup Lik GrGrGrGrGroup Lik oup Lik ms to See Common Factor actor ms to See Common F oup Lik Sometimes you need to group terms together before you can see any common factors in an expression — then you can use the distributive property, ab + ac = a(b + c), to factor them out. Example Example Example Example Example 11111 Factor by grouping 3y + 5t
y – 6k –10tk. Solution Solution Solution Solution Solution Group 3y + 5ty and –6k – 10tk together in parentheses: (3y + 5ty) + (–6k – 10tk) 3y and 5ty have a common factor of y. –6k and –10tk have a common factor of –2k. Factor out the common factors: (3y + 5ty) + (–6k –10tk) = y(3 + 5t) – 2k(3 + 5t) Now you can see there’s another common factor to factor out: (3 + 5t) Using the distributive property: y(3 + 5t) – 2k(3 + 5t) = (y – 2k)(3 + 5t) Example Example Example Example Example 22222 Factor completely 8rt – 6ckt + 3ckm – 4rm. Solution Solution Solution Solution Solution Rearrange the expression and group in parentheses: 8rt – 6ckt + 3ckm – 4rm = (8rt – 4rm) + (–6ckt + 3ckm) 8rt and –4rm have a common factor of 4r. 6ckt and 3ckm have a common factor of 3ck. Section 6.8 Section 6.8 Section 6.8 — More on Quadratics Section 6.8 Section 6.8 329329329329329 Example 2 continueduedueduedued Example 2 contin Example 2 contin Example 2 contin Example 2 contin Take out the common factors: (8rt – 4rm) + (–6ckt + 3ckm) = 4r(2t – m) + 3ck(–2t + m) = 4r(2t – m) – 3ck(2t – m) Using the distributive property: 4r(2t – m) – 3ck(2t – m) = (4r – 3ck)(2t – m) Another common factor to factor out is (2t – m). Guided Practice Factor each expression by grouping. 1. km + 2k – 2m – 4 3. 3x + 9 – 4kx – 12k 5. 1 – k + t – tk 7. 6ky + 15t – 10y – 9kt 9. 6rx2 + 15xy – 2rxy – 5y2 2. tx – ty – mx + my 4. 1 + 3y – 5k – 15ky 6. kt – 2k + 3t – 6 8. 8hx + 10h – 12tx – 15t 10. 10tx – 3k – 15t + 2kx Independent Practice Factor each expression completely. 1. 2x2 + x + 8x + 4 3. 4x2 + 14x + 14x + 49 5. 12n2 + 21n + 8n + 14 7. n2 – 16n + 20n – 320 9. 2x2 + 5xy + 4xy + 10y2 11. 12a2 + 9ab – 28ab – 21b2 13. 4a2 – 6ab + 6ab – 9b2 2. 6x2 + 9x + 4x + 6 4. 2x2 + 5x + 6x + 15 6. 6x2 + 8x + 15x + 20 8. 3c2 – c + 6c – 2 10. 3m2 + 3mn – mn – n2 12. 2x3 + 2x2y + 3xy2 + 3y3 14. 4b2 – 20bx – 2xb + 10x2 Find a value of ? so that the expression will factor into two binomials. 15. 20n2 – 25n + ?n – 20 17. ?c2 – 12c + 2c – 4 16. 8xy – 4xz + 4wy – ?wz 18. 3a2 – ?a + 6a – 2 19. The area of a rectangle is the product of two binomials (with integer coefficients). If the area of the rectangle is (3a2 + a + 3a + 1) m2, find the dimensions of the rectangle. 20. The area of a square is the square of a binomial. If the area of the square is (4x2 + 2x + 2x + 1) in.2, find the side length of the square. 21. The area of a circle is the product of p and the radius squared. If the radius is a binomial and the area of the circle is (9px2 + 15pxb + 15pxb + 25pb2) in2, find the radius. ound Up ound Up RRRRRound Up ound Up ound Up RRRRRound Up ound Up ound Up ound Up ound Up That’s the end of a Section full of neat little ways of making math a lot less painful. You’ll often need to use the methods for difference of two squares, perfect square trinomials, and factoring by grouping — so look back over the Topics in this Section to make sure you understand them. 330330330330330 Section 6.8 Section 6.8 Section 6.8 — More on Quadratics Section 6.8 Section 6.8 Chapter 6 Investigation s trianglelelelele s triang s triang ascal’ ascal’ PPPPPascal’ ascal’s triang s triang ascal’ PPPPPascal’ s trianglelelelele s triang s triang ascal’ ascal’ ascal’s triang s triang ascal’ Pascal’s triangle was originally developed by the ancient Chinese. However, the French mathematician Blaise Pascal was the first person to discover the importance of all the patterns it contains. 1 Part 1: Look at the numbers in the triangle. Write down a rule that could be used to predict each number in the triangle. Use your rule to predict the next row of the triangle. 1 1 1 4 Part 2: Find and simplify the following: • (x + 1)2 (x + 1)3 (x + 1)4 1 1 6 Hint: this is just (x + 1)(x + 1)2 1 1 1 2 33 6 10 20 15 4 1 5 10 5 15 1 1 6 1 How are your answers linked to Pascal’s Triangle? Use the triangle to predict the expansion of (x + 1)5. Test your prediction. Extension Investigate other patterns in Pascal’s Triangle. Here are some ideas: Look at the numbers in the diagonal lines of the triangle. Investigate “hockey stick” shapes, like the one shown on the right. Find the sum of each row of numbers. What do you notice? Find rows in which the second number is prime. What is special about these rows? 2 33 6 10 20 15 5 15 10 Open-ended Extension If there are two children in a family, there can either be two girls, two boys, or a girl and a boy. The probability of each combination and the ratios of the probabilities are shown in the table: noitanibmoC First child Girl Boy Second child Result Girl Boy Girl Boy Two girls One of each One of each Two boys ytilibaborP seitilibaborpfooitaR slrig2 yob1,lrig1 syob2 1 4 1 2 1 4 1 2 1 Investigate the link between Pascal’s Triangle and the probabilities of having different combinations of boys and girls in families with different numbers of children. Remember, it doesn’t matter which order the boys and girls are born in. ound Up ound Up RRRRRound Up RRRRRound Up ound Up ound Up ound Up ound Up ound Up ound Up Although it just looks like a funny pile of shapes and numbers, there are a lot of real-life problems that can be solved using the patterns in Pascal’s Triangle. estigaaaaationtiontiontiontion — Pascal’s Triangle 331331331331331 estigestig estig pter 6 Invvvvvestig pter 6 In ChaChaChaChaChapter 6 In pter 6 In pter 6 In Chapter 7 Quadratic Equations and Their Applications Section 7.1 Solving Quadratic Equations ....................... 333 Section 7.2 Completing the Square ............................... 342 Section 7.3 The Quadratic Formula and Applications .... 355 Section 7.4 Quadratic Graphs ........................................ 362 Section 7.5 The Discriminant ......................................... 372 Section 7.6 Motion Tasks and Other Applications .......... 379 Investigation The Handshake Problem............................. 386 332332332332332 Topic 7.1.1 California Standards: 11.0: Students apply basic factoring techniques to second- and simple thirddegree polynomials. These techniques include finding a common factor for all terms in a polynomial, recognizing the difference of two squares, and recognizing perfect squares of binomials. 14.0: Students solve a quadratic equation by factoring or completing the square. What it means for you: You’ll solve quadratic equations by factoring. Key words: quadratic factor zero property Section 7.1 Solving Quadratic Solving Quadratic Equations by Factoring Equations by Factoring In this Topic you’ll use all the factoring methods that you learned in Chapter 6 to solve quadratic equations. Quadratic Equations Have Degree 2 Quadratic equations contain a squared variable, but no higher powers — they have degree 2. These are all quadratic equations, as the highest power of the variable is 2: (i) x2 – 3x + 2 = 0 (ii) 4x2 + 12x – 320 = 0 (iii) y2 + 4y – 7 = 2y2 – 2y The general form of a quadratic equation is ax2 + bx + c = 0, where a, b, and c are numbers, and a is not 0. For example, in (i) above, a = 1, b = –3, and c = 2, while in (ii), a = 4, b = 12, and c = –320. Example (iii) above is a quadratic in y, while the others are quadratics in x. Guided Practice The general form of a quadratic equation is ax2 + bx + c = 0, where a, b, and c are numbers. Identify a, b, and c in these equations. 1. –x2 + 5x – 6 = 0 3. 4x2 – 12x + 9 = 0 5. –x2 – 4x – 4 = 0 7. 6y2 + 28y + 20 = 5 – 6y2 9. 4(x2 – 5x) = –25 11. 7x(7x + 4) + 4x3 + 3 = 2(2x3 + 1) 2. 6x2 + 31x + 35 = 0 4. 16x2 – 8x + 1 = 0 6. 64x2 + 48x + 9 = 0 8. 4x2 + 6x + 1 = 3x2 + 8x 10. 3x(3x + 4) + 8 = 4 Solving is Finding Values That Make the Equality True An equation is a statement saying that two mathematical expressions are equal. For example, 7 + 2 = 9, 4x + 2 = 14, and x2 – 3x + 2 = 0 are all equations. If an equation contains a variable (an unknown quantity), then solving the equation means finding possible values of the variable that make the equation a true statement. Section 7.1 — Solving Quadratic Equations 333 Example 1 Find a solution of the equation x2 – 3x + 2 = 0. Solution If you evaluate the above equation using, say, x = 3, then you get: 32 – (3 × 3) + 2 = 0 This is not a true statement (since the left-hand side equals 2). So x = 3 is not a solution of the equation. But if instead you substitute x = 1, then you get: 12 – (3 × 1) + 2 = 0 This is a true statement. So x = 1 is a solution of the equation x2 – 3x + 2 = 0. Guided Practice Determine which of the two values given is a solution of the equation. 12. –x2 + 5x – 6 = 0 for x = –3 and x = 3 13. 6x2 + 31x + 35 = 0 for x = – 5 3 and x = 3 14. 4x2 – 12x + 9 = 0 for x = 1 and x = 2 2 15. 16x2 – 8x + 1 = 0 for x = – 1 4 and x = 1 16. –x2 – 4x – 4 = 0 for x = 2 and x = –2 17. 64x2 + 48x + 9 = 0 for x = – 3 5 3 4 8 and x = 3 8 Zero Property — if xy = 0, then x = 0 or y = 0 (or Both) One way to solve a quadratic equation is to factor it and then make use of the following property of zero: Check it out: This says that if the product of two expressions is zero (so you get zero when you multiply them together), then at least one of those expressions must itself be zero. Zero Property If the product mc = 0, then either: (i) m = 0, c = 0, (ii) (iii) both m = 0 and c = 0. Example 2 Solve x2 + 2x – 15 = 0 by factoring. Solution x2 + 2x – 15 = (x – 3)(x + 5) So if (x – 3)(x + 5) = 0, then by the zero property, either (x – 3) = 0 or (x + 5) = 0. So either x = 3 or x = –5. 334 Section 7.1 — Solving Quadratic Equations Don’t forget: See Chapter 6 for more information on factoring. Example 3 Solve 2x2 + 3x – 20 = 0 by factoring. Solution 2x2 + 3x – 20 = (2x – 5)(x + 4) = 0 So either x = 5 2 or x = –4. Guided Practice Solve each of these quadratic equations by using the zero property. 18. (2x + 7)(3x + 5) = 0 20. 49x2 – 1 = 0 22. 4x2 + 8x + 3 = 0 24. 4x2 – 11x – 3 = 0 26. 2x2 + 11x + 12 = 0 28. 3x2 – 17x – 28 = 0 19. (x – 5)(x – 1) = 0 21. 64a2 – 25 = 0 23. 2x2 – 17x – 9 = 0 25. 10x2 – x – 2 = 0 27. 10x2 – 27x + 5 = 0 29. 2x2 – x – 28 = 0 Using Fac
toring to Solve Quadratic Equations 1) First arrange the terms in the quadratic equation so that you have zero on one side. 2) Then factor the nonzero expression (if possible). 3) Once done, you can use the zero property to find the solutions. Example 4 Solve x2 – 6x – 7 = 0. Solution The right-hand side of the equation is already zero, so you can just factor the left-hand side: x2 – 6x – 7 = (x + 1)(x – 7) So (x + 1)(x – 7) = 0. Using the zero property, either x + 1 = 0 or x – 7 = 0. So either x = –1 or x = 7. Section 7.1 — Solving Quadratic Equations 335 Example 5 Solve x2 + 2x – 11 = –3. Solution This time, you have to arrange the equation so you have zero on one side. By adding 3 to both sides, the right-hand side becomes 0. x2 + 2x – 11 = –3 ﬁ x2 + 2x – 8 = 0 Now you can factor the left-hand side: x2 + 2x – 8 = (x + 4)(x – 2) = 0 Since you have two expressions multiplied to give zero, you can use the zero property. That is, either x + 4 = 0 or x – 2 = 0. So either x = –4 or x = 2. Example 6 Solve 3x2 + 168 = 45x. Solution Once again, the first thing to do is get zero on one side: 3x2 + 168 = 45x ⇒ 3x2 – 45x + 168 = 0 The left-hand side can be factored, which means you can rewrite this as: 3(x2 – 15x + 56) = 0, or 3(x – 7)(x – 8) = 0 So using the zero property, either x – 7 = 0 or x – 8 = 0. So either x = 7 or x = 8. Guided Practice Solve each of these equations. 30. x2 – 2x – 15 = 0 32. k2 + 10k + 24 = 0 34. 8k2 – 14k = 49 36. 6y2 + 28y + 20 = 5 – 6y2 38. 4(x2 – 5x) = –25 40. x(x + 4) + 9 = 5 42. 6x(3x – 4) – 7 = –15 44. 7x(7x + 2) + 4x3 + 3 = 2(2x3 + 1) 45. (2x + 9)2(x + 3)(x + 1)–1(x + 3)–1(x + 1) = 0 46. 2x(3x + 3) + 4(x + 1) = 1 + 2x + 2x2 31. x2 – 7x – 18 = 0 33. 4m2 + 4m – 15 = 0 35. 15k2 + 28k = –5 37. 4x2 + 6x + 1 = 3x2 + 8x 39. 3x(3x + 4) + 8 = 4 41. x(x – 5) + 3 = –3 43. 3 = 2 – 12x(3x – 1) 336 Section 7.1 — Solving Quadratic Equations Independent Practice The general form of a quadratic equation is ax2 + bx + c = 0, where a, b, and c are numbers. Identify a, b, and c in the quadratic equations below. 1. 4x2 + 20x + 9 = 0 3. 2x2 + 5x = 35 + 14x 5. y(2y + 7) = 9(2y + 7) 2. x2 – 9x + 8 = 0 4. x(2x + 3) = 5(2x + 3) 6. (x + 2)(x – 2) = 3x Use the zero product property to solve these equations. 7. (2y + 9)(2y – 3) = 0 8. (2a + 5)(a – 11) = 0 9. (y – 3)2(y – 5)(y – 3)–1 = 0 10. (y – 4)3(2y – 9)2(y – 4)–3(y – 7)(2y – 9)–1 = 0 Solve the following equations. 11. 4x2 + 20x + 9 = 0 13. 2x2 + 5x = 35 + 14x 15. y(2y + 7) = 9(2y + 7) 17. 2x(2x – 5) = 3(2x – 5) 12. x2 – 9x + 8 = 0 14. x(2x + 3) = 5(2x + 3) 16. (x + 2)(x – 2) = 3x 18. 2x(3x – 1) + 7 = 7(2 – 3x) 19. The product of two consecutive positive numbers is 30. Find the numbers. 20. The product of two consecutive positive odd numbers is 35. Find the numbers. 21. The area of a rectangle is 35 ft2. If the width of the rectangle is x ft and the length is (3x + 16) ft, find the value of x. 22. The area of a rectangle is 75 cm2. If the length of the rectangle is (4x + 25) cm and the width is 2x cm, find the dimensions of the rectangle. 23. Eylora has x pet goldfish and Leo has (4x – 25). If the product of the numbers of Eylora’s and Leo’s goldfish is 21, how many goldfish does Leo have? 24. Scott fixed x computers and Meimei fixed (5x – 7) computers. If the product of the number each fixed is 6, who fixed more computers? Round Up Round Up Don’t forget that you need to rearrange the equation until you’ve got a zero on one side before you can factor a quadratic. Not all quadratics can be factored like this, as you’ll see in the next Topic. Section 7.1 — Solving Quadratic Equations 337 Topic 7.1.2 California Standards: 11.0: Students apply basic factoring techniques to second- and simple thirddegree polynomials. These techniques include finding a common factor for all terms in a polynomial, recognizing the difference of two squares, and recognizing perfect squares of binomials. 14.0: Students solve a quadratic equation by factoring or completing the square. What it means for you: You’ll solve quadratic equations by taking square roots. Key words: quadratic factor square root zero property Quadratic Equations Quadratic Equations — Taking Square Roots — Taking Square Roots Some quadratics can be solved by taking square roots. But to use this method properly, you have to remember something about squares and square roots. The Square Root Method If you square the numbers m and –m, you get the same answer, since m2 = (–m)2 (= p, say). If you take the square root of p, there are two possible answers, m or –m. In other words, if m2 = p, then m = ± p . Example 1 uses the above property to find the two possible solutions of a quadratic equation. Example 1 Solve the equation x2 = 25. Solution Take the square root of both sides to get x = ±5. In Example 1, you need only put the “±” on one side of the equation. Here’s why: Take the square root of both sides of the original equation to get x 2 25= . But when you take square roots, you have to allow for both sides to be either positive or negative. So actually, there are four possibilities here: x = 5, –x = 5, x = –5, or –x = –5 However, x = 5 and –x = –5 are the same, as are –x = 5 and x = –5. So in fact it’s enough to put the “±” sign on just one side of the equation. 338 Section 7.1 — Solving Quadratic Equations Check it out: This is the same problem as in Example 1. You Can Also Solve this Equation by Factoring You could also solve the above equation using the method of factoring from the previous pages. But to use the factoring method, you have to have an equation of the form: “something = 0” — then you can use the zero property. Example 2 Solve the equation x2 = 25 by factoring. Solution x2 = 25 x2 – 25 = 0 (x + 5)(x – 5) = 0 So using the zero property, either x + 5 = 0 or x – 5 = 0. So either x = 5 or x = –5. That is, x = ±5. Guided Practice Find the square roots of the expressions below. Show your work. 1. 49 4. 128 7. x2 + 4x + 4 2. 64 5. 4t2 8. a2 + 16a + 64 3. 256 6. 16t4 9. 4y2 + 12y + 9 Solve the equations below by finding the square root. 10. x2 = 4 13. 3a2 = 75 11. x2 = 9 14. a2 = 81 12. 2x2 = 32 15. 5x2 = 180 16. Use the zero product property and factoring to verify your answers to Exercises 10–15 above. More Square Root Examples Example 3 Find the solution set of 3x2 – 7 = 101. Solution Here, you can get x2 on its own on one side of the equation, with no x’s on the other side. This allows you to take the square root, as above. 3x2 – 7 = 101 3x2 = 108 x2 = 36 x = ± 36 That is, x = ±6 — or x Œ {6, –6} Section 7.1 — Solving Quadratic Equations 339 Check it out: x = ±6 and x Œ {6, –6} mean the same thing. Check it out: You can’t get x2 on its own because if you multiply out these parentheses, you get x2 – 14x + 49. So if you put x2 on one side, you’ll have 14x on the other side, which is no good. Example 4 Solve (x – 7)2 = 64. Solution This time, you can’t get x2 on its own (with no x’s on the other side), but you already have (x – 7)2 alone, which is just as good. (x – 7)2 = 64 x – 7 = ± 64 So x = 15 or x = –1. Check Your Answers by Using the Original Equation You can always check your answers by substituting your solutions into the original equation to see if you get a true statement. This example shows how you’d check the solution reached in Example 4. Example 5 Show that x = 15 and x = –1 are solutions of the equation (x – 7)2 = 64. Solution Do this by substituting x = 15 and x = –1 into the equation (x – 7)2 = 64, and seeing if you get true statements. Put x = 15: (15 – 7)2 = 64 82 = 64 64 = 64 Put x = –1: (–1 – 7)2 = 64 (–8)2 = 64 64 = 64 These are both true statements, so x = 15 and x = –1 are both solutions of the equation. Guided Practice Find the square roots of the expressions below. 17. c2 + 6c + 9 20. 9x2 – 24x + 16 23. 49 + 28y + 4y2 18. x2 + 14x + 49 21. 9x2 + 30x + 25 24. 4x2 + 4bx + b2 19. x2 – 6x + 9 22. 25 – 30k + 9k2 25. k2 – 8kx + 16x2 Solve the following equations by using the square root method. 26. k2 = 1 29. m2 = 432 32. (k – 6)2 = 72 27. x2 = 49 30. (x + 3)2 = 81 33. (v + 7)2 = 147 28. p2 = 125 31. (x – 5)2 = 121 34. x2 + 4x + 4 = 36 340 Section 7.1 — Solving Quadratic Equations A Couple More Examples Example 6 Solve 4x2 – 20x + 25 = 9 by taking square roots of both sides. Solution 4x2 – 20x + 25 = 9 4 2x x− 20 + = 25 9 2 = x − ( 2 9 5 ) 2x – 5 = ±3 Since (2x – 5)2 = 4x2 – 20x + 25 So 2x = 8 or 2x = 2 — which means that x = 4 or x = 1. Example 7 Solve y2 = 75, giving your answer in its simplest form. Don’t forget: See Topics 1.3.2 and 1.3.3 for more information on expressions involving radicals. Solution y2 = 75 = ± y = ± y 75 ⋅ 25 3 ⋅ 3 = ± 25 y y = ±5 3 Independent Practice Find the square roots of the expressions below. 1. 25k2 – 60kr + 36r2 2 12 Solve the following equations by using the square root method. k 4 m 3. 25 25 20 20 + + − + p p 4 4 m 2 2 16 + 9 4. 4x2 – 4x + 1 = 16 5. 9x2 + 12x + 4 = 169 6. k2 – 14k + 49 = 9 16 7. 4x2 – 12x + 9 = 16 8. 9x2 – 6x + 1 = 4 9. The sides of a square are each (2x – 16) cm long. Find the value of x that would give a square with an area of 108 cm². 10. The product of the number of CDs that Donna and Keisha have is 16a2 + 56a + 49. If both have the same number of CDs, find how many CDs Donna has, in terms of a. Round Up Round Up Don’t forget that square roots result in two possible solutions. Also, no matter how you’ve solved a quadratic, it’s a good idea to check your solutions by substituting them back into the original equation. Section 7.1 — Solving Quadratic Equations 341 Topic 7.2.1 Section 7.2 Completing the Square Completing the Square California Standard: 14.0: Students solve a quadratic equation by factoring or completing the square. What it means for you: You’ll form perfect square trinomials by adding numbers to binomial expressions. Key words: completing the square perfect square trinomial binomial Check it out: The trinomial (with 3 terms) can be written as the square of a binomial (a binomial has 2 terms). “Completing the square” is
another method for solving quadratic equations — but before you solve any equations, you need to know how completing the square actually works. Writing Perfect Square Trinomials as Perfect Squares An expression such as (x + 1)2 is called a perfect square — because it’s (something)2. In a similar way, an expression such as x2 + 2x + 1 is called a perfect square trinomial (“trinomial” because it has 3 terms). This is because it can be written as a perfect square: x2 + 2x + 1 = (x + 1)2 Any trinomial of the form x2 + 2dx + d 2 is a perfect square trinomial, since it can be written as the square of a binomial: x2 + 2dx + d 2 = (x + d)2 Converting Binomials to Perfect Squares The binomial expression x2 + 4x is not a perfect square — it can’t be written as the square of a binomial. However, it can be turned into a perfect square trinomial if you add a constant (a number) to the expression. Example 1 Convert x2 + 4x to a perfect square trinomial. Solution To do this you have to add a number to the original expression. First look at the form of perfect square trinomials, and compare the coefficient of x with the constant term (the number not followed by x or x2): x2 + 2dx + d 2 = (x + d)2 The coefficient of x is 2d, while the constant term is d 2. So the constant term is the square of half of the coefficient of x. To convert x2 + 4x to a perfect square trinomial, add the square of half of 4 — that is, add 22 = 4, to give x2 + 4x + 4 = (x + 2)2 342 Section 7.2 — Completing the Square Don’t forget: b is the coefficient of x. Check it out: The solution is equivalent to (x + 4)2. Check it out: In Example 3, y2 – 12y + 36 could also be written as (y – 6)2. Guided Practice By adding a constant, convert each of these binomials into a perfect square trinomial. 1. x2 + 14x 3. x2 + 2x 5. y2 + 20y 2. x2 – 12x 4. x2 – 8x 6. p2 – 16p Completing the Square for x2 + bx Completing the square for x2 + bx To convert x2 + bx into a perfect square trinomial, add ⎛ ⎜⎜⎜ The resulting trinomial is x ⎝ ⎞ 2 ⎠ . b+ ⎟⎟⎟2 ⎛ ⎜⎜⎜ ⎝ b 2 ⎞ 2 ⎟⎟⎟ . ⎠ Example 2 Form a perfect square trinomial from x2 + 8x. Solution Here, b = 8, so to complete the square you add ⎛ ⎜⎜⎜ ⎝ 8 2 ⎞ 2 ⎟⎟⎟ . ⎠ This gives you x2 + 8x + 16. Example 3 What must be added to y2 – 12y to make it a perfect square trinomial? Solution This time, b = –12. To complete the square you add −⎛ ⎜⎜⎜ ⎝ 12 2 ⎞ 2 ⎟⎟⎟ ⎠ = (–6)2 = 36. So 36 must be added (giving y2 – 12y + 36). Section 7.2 — Completing the Square 343 Don’t forget: Remember the negative sign in front of the 10 coefficient. Example 4 Suppose x2 – 10x + c is a perfect square trinomial, and is equal to (x + k)2. What are the values of c and k? Solution Here the coefficient of x is –10. So to form a perfect square trinomial, the constant term has to be the square of half of –10, so c = (–5)2 = 25. Therefore x2 – 10x + c = x2 – 10x + 25 = (x + k)2. Now multiply out the parentheses of (x + k)2. (x + k)2 = x2 + 2kx + k2 This has to equal x2 – 10x + 25, which gives x2 – 10x + 25 = x2 + 2kx + k2. Equate the coefficients of x: the coefficient of x on the left-hand side is –10, while on the right-hand side it is 2k. So –10 = 2k, or k = –5. Comparing the constant terms in a similar way, you find that 25 = k2 , which is also satisfied by k = –5. Guided Practice Find the value of k that will make each expression below a perfect square trinomial. 7. x2 – 7x + k 9. x2 + 6mx + k 8. q2 + 5q + k 10. d 2 – 2md + k Form a perfect square trinomial from the following expressions by adding a suitable term. 11. x2 + 10x 13. y2 + 2y 15. y2 – 18y 17. y2 + 12y 19. 6x + 9 21. 25 – 20y 23. 4a2 + 12ab 12. x2 – 16x 14. x2 + bx 16. a2 – 2a 18. y2 + 36y 20. 1 – 8x 22. 4y2 + 4yb 24. 9a2 + 16b2 The quadratics below are perfect square trinomials. Find the value of c and k to make each statement true. 25. x2 – 6x + c = (x + k)2 27. 4x2 + 12x + c = (2x + k)2 29. 4a2 – 4ab + cb2 = (2a + kb)2 26. x2 + 16x + c = (x + k)2 28. 9x2 + 30x + c = (3x + k)2 30. 9a2 – 12ab + cb2 = (3a + kb)2 344 Section 7.2 — Completing the Square If the Coefficient of x2 isn’t 1, Add a Number With an expression of the form ax2 + bx, you can add a number to make an expression of the form a(x + k)2. Example 5 If 3x2 – 12x + m is equal to 3(x + d)2, what is m? Solution Multiply out the parentheses of 3(x + d)2 to get: 3(x + d)2 = 3x2 + 6dx + 3d2 So 3x2 – 12x + m = 3x2 + 6dx + 3d2 Equate the coefficients of x, and the constants, to get: –12 = 6d and m = 3d2 The first equation tells you that d = –2. And the second tells you that m = 3(–2)2, or m = 12. So 3x2 – 12x + 12 = 3(x – 2)2. Completing the square for ax2 + bx The expression ax2 + bx can be changed to a trinomial of the form a(x + k)2. ⎛ ⎜⎜⎜ ⎝ b 4 a ⎛ ⎜⎜⎜ The resulting trinomial is a x ⎝ To do this, add ⎞ 2 ⎟⎟⎟ ⎠ ⎞ ⎟⎟⎟2 Example 6 Convert 2x2 + 10x to a perfect square trinomial. Solution Here, a = 2 and b = 10, so you add This gives you 2 2x x+ 10 25 + . 2 ⎛ ⎜⎜⎜ ⎝ 10 2 1 2 ⎞ 2 ⎟⎟⎟ = ⎠ ⎛ ⎜⎜⎜ ⎝ 1 2 100 4 ⎞ ⎟⎟⎟ = ⎠ 25 2 Section 7.2 — Completing the Square 345 Check it out: + 2x x+ 10 2 can also be 25 2 ⎛ ⎜⎜⎜ ⎝ ⎞ 2 ⎟⎟⎟ . ⎠ 5 2 written as 2 x + Guided Practice The quadratics below are of the form a(x + d)2. Find the value of m and d in each equation. 31. 5x2 + 10x + m = 5(x + d)2 33. 2x2 – 28x + m = 2(x + d)2 35. 4x2 + 32x + m = 4(x + d)2 37. 20x2 – 20x + m = 5(2x + d)2 39. 27x2 + 36x + m = 3(3x + d)2 32. 4x2 – 24x + m = 4(x + d)2 34. 3x2 – 30x + m = 3(x + d)2 36. 20x2 + 60x + m = 5(2x + d)2 38. 27x2 + 18x + m = 3(3x + d)2 40. 16x2 – 80x + m = 4(2x + d)2 Add a term to convert each of the following into an expression of the form a(x + k)2. 41. 2x2 – 12x 43. 6y2 – 60y 45. 5x2 + 245 47. 36x + 12 42. 3a2 + 12a 44. 4x2 – 48x 46. 8x2 + 2 48. 120x + 100 Independent Practice Find the value of c that will make each expression below a perfect square trinomial. 1. x2 + 9x + c 3. x2 + 12xy + c 2. x2 – 11x + c 4. x2 – 10xy + c Complete the square for each quadratic expression below. 5. x2 – 6x 7. b2 – 10b 9. c2 – 12bc 6. a2 – 14a 8. x2 + 8xy 10. x2 + 4xy Find the value of m and d in each of the following. 11. 5x2 – 40x + m = 5(x + d)2 13. 3x2 – 6x + m = 3(x + d)2 15. 4x2 + 24x + m = 4(x + d)2 12. 2x2 + 20x + m = 2(x + d)2 14. 3x2 – 30x + m = 3(x + d)2 16. 7x2 – 28x + m = 7(x + d)2 Add a term to convert each of the following into an expression of the form a(x + k)2. 17. 3x2 – 30x 19. 18x2 – 48x 21. 27x2 + 12 18. 2x2 + 8x 20. 5x2 + 180 22. 36x2 + 196 23. The length of a rectangle is twice its width. If the area can be found by completing the square for (18x2 + 60x) ft2, find the width of the rectangle. Round Up Round Up OK, so now you know how to add a number to a binomial to make a perfect square trinomial. In the next Topic you’ll learn how to convert any quadratic expression into perfect square trinomial form — and then in Topic 7.2.3 you’ll use this to solve quadratic equations. 346 Section 7.2 — Completing the Square Topic 7.2.2 California Standards: 14.0: Students solve a quadratic equation by factoring or completing the square. What it means for you: You’ll convert any quadratic expression into perfect square trinomial form. Key words: completing the square perfect square trinomial binomial More on Completing More on Completing the Square the Square In Topic 7.2.1, you converted binomial expressions like ax2 + bx to perfect square trinomials by adding a number. In this Topic you’ll take a more general quadratic expression like ax2 + bx + c and write this in the form a(x + k) 2 + m. Writing x2 + bx + c in the form (x + k)2 + m In earlier Topics, you converted an expression of the form x2 + bx to a ⎛ ⎜⎜⎜ ⎝ perfect square trinomial by adding ⎞ 2 ⎟⎟⎟ to it. ⎠ b 2 Example 1 Convert x2 + 4x to a perfect square trinomial. Solution Here b = 4, so to convert this to a perfect square trinomial, you add ⎛ ⎜⎜⎜ ⎝ 4 2 ⎞ 2 ⎟⎟⎟ = = . 2 4 ⎠ 2 So x2 + 4x + 4 = (x + 2)2. Here x2 + 4x + 4 is a perfect square trinomial. Another way to think about this is to say that your original expression was equal to an expression of the form (x + k)2 + m. Check it out: Example 2 is almost the same problem as in Example 1 — it’s just asked in a different way. Example 2 Express x2 + 4x in the form (x + k)2 + m. Solution x2 + 4x + 4 = (x + 2)2. Therefore x2 + 4x = (x + 2)2 – 4. Guided Practice Express each of the following in the form (x + k)2 + m. 1. x2 + 6x 4. x2 + 24x 7. x2 + 18x 2. x2 + 20x 5. x2 – 22x 8. x2 – 14x 3. x2 + 12x 6. x2 – 10x 9. x2 – 16x Section 7.2 — Completing the Square 347 More general quadratic expressions x2 + bx + c can also be written in the form (x + k)2 + m. Writing x2 + bx + c in the form (x + k)2 + m 1) First take just the first two terms (x2 + bx) and convert this to a perfect square trinomial by adding the square of half of b: 2 x + + bx ⎛ ⎜⎜⎜ ⎝ b 2 ⎞ ⎛ 2 ⎟⎟⎟ = + ⎜⎜⎜ x ⎠ ⎝ ⎞ 2 ⎟⎟⎟ ⎠ b 2 2) Rewrite this in the form x 2 ⎛ ⎜⎜⎜ + = + ⎝ bx x 3) Add c to both sides: x 2 + + = + c bx x ⎛ ⎜⎜⎜ ⎝ ⎞ 2 ⎟⎟⎟ − ⎠ ⎛ ⎜⎜⎜ ⎝ ⎞ 2 ⎟⎟⎟ ⎠ b 2 b 2 ⎞ 2 ⎟⎟⎟ − ⎠ ⎛ ⎜⎜⎜ ⎝ ⎞ 2 ⎟⎟⎟ + ⎠ b 2 b 2 c Check it out: Compare this to (x + k)2 + m, and you see that ⎛ b= − ⎜⎜⎜ ⎝ 2 b= 2 and m + c. ⎞ 2 ⎟⎟⎟ ⎠ k Example 3 Express x2 + 4x + 1 in the form (x + k)2 + m. Solution ⎛ 1) Again, b = 4, so add 4 ⎜⎜⎜ ⎝ 2 expression to find a perfect square trinomial — that is, x2 + 4x + 4 = (x + 2)2. ⎞ 2 ⎟⎟⎟ = 22 = 4 to the first two terms of your ⎠ So the first two terms of your original expression can be expressed: x2 + 4x = (x + 2)2 – 4 2) Now add c (= 1) to both sides of this equation to get: x2 + 4x + 1 = (x + 2)2 – 3 Example 4 Write x2 – 6x + 3 in the form (x + k)2 + m. Solution 1) Here, b = –6, so add −⎛ 6 ⎜⎜⎜ ⎝ 2 ⎞ 2 ⎟⎟⎟ ⎠ = (–3)2 = 9 to x2 – 6x for a perfect square trinomial — that is, x2 – 6x + 9 = (x – 3)2. So the first two terms of the original quadratic can be expressed: x2 – 6x = (x – 3)2 – 9 2) Add c (= 3) to both sides of this equation to get: x2 – 6x + 3 = (x – 3)2 – 6 348 Section 7.2 — Completing the Square Guided Practice Express the following in the form (x + k)2 + m [or a(x + k)2 + m]. 10. x2 + 4x +
8 12. x2 + 8x + 5 14. x2 + 3x + 5 16. x2 + 4x + 5 18. x2 – 20x + 20 20. 4x2 + 16x + 4 11. x2 + 6x + 14 13. x2 – 12x + 8 15. x2 – 5x – 7 17. x2 – 6x + 30 19. x2 + 22x + 54 21. 2x2 + 20x + 25 Writing ax2 + bx + c in the Form a(x + k)2 + m This is the most general case you can meet — writing ax2 + bx + c as an expression of the form a(x + k)2 + m. The method’s very similar to the one used in the previous examples. Example 5 Write 2x2 + 6x + 7 in the form a(x + k)2 + m. Solution 1) Factor your expression by taking the coefficient of x2 outside ( parentheses: 2x2 + 6x + ) The first two terms in parentheses (x2 + 3x) can be made into a perfect square trinomial in the usual way — by adding ⎛ ⎜⎜⎜ ⎝ ⎞ 2 9 ⎟⎟⎟ = . ⎠ 4 3 2 This means that x2 + 3x + )3 — that is, x2 + 3x = x +( )3 = x +( 2 2 So x2 + 3x + +( 2 )3 2 9 4 – 9 4 . or, combining the fractions, x2 + 3x + = x +( 7 2 2 )3 2 + 5 4 3) Therefore the original expression (2x2 + 6x + 7) can be written: 2 ⎡ x +( ⎢ ⎢ ⎣ +⎡ ⎢ ⎣ 2 ⎤ ⎥ + ⎦ 3 2 5 2 Section 7.2 — Completing the Square 349 Check it out: Now you can deal with the expression in parentheses exactly as before. Check it out: This is exactly the same process as in Examples 3 and 4 on the previous page. Check it out: Completing the square can be used to find the highest point of an object’s path as it flies through the air. You’ll see this in more detail in Sections 7.4 and 7.6. Example 6 Write 4x2 – 4x + 3 in the form a(x + k)2 + m. Solution 1) Factor the expression: 4x2 – 4x + 3 = 4(x2 – x + 3 4 ) 2) Convert the first two terms from inside the parentheses into a perfect square trinomial: x2 – x + 1 4 = (x – 1 2 )2 So rewriting the above, you get: x2 – x = (x – Add 3 4 to both sides: x2 – x + 3 4 = (x – 1 2 )2 + )2 – 1 4 1 2 1 2 3) Finally, use your expression from 1) above: 4x2 – 4x + 3 = 4 x −( ⎡ ⎢ ⎢ ⎣ (x – 1 2 )2 + 2 Guided Practice Express the following in the form (x + k)2 + m [or a(x + k)2 + m]. 22. 2x2 – 9x + 4 24. 3x2 – 12x + 14 26. –x2 – 3x + 4 28. –3x2 – 9x + 21 23. 2x2 + 9x + 5 25. –x2 + 2x + 3 27. –2x2 + 6x + 10 29. –4x2 + 120x – 80 Independent Practice Express the following in the form (x + k)2 + m. 1. x2 + 4x 3. x2 + 10x 5. x2 – 1 x 4 2. x2 – 6x 4. x2 – 20x 6. x2 + 2 x 3 Express the following in the form of a(x + k)2 + m 7. –5x2 – 90x – 150 9. 4x2 – 3x + 12 11. 6x2 + 48x + 16 8. 2x2 + 5x + 10 10. 5x2 + 20x + 4 12. 3x2 + 42x + 49 13. A square has an area of x2 + 14x + k. Find the value of k. 14. A circle has an area of px2 + 18px + kp. Find the value of k. Round Up Round Up There’s been a lot of build-up to actually solving quadratic equations using the completing the square method — but it’s coming up next, in Topic 7.2.3. 350 Section 7.2 — Completing the Square Topic 7.2.3 California Standards: 14.0: Students solve a quadratic equation by factoring or completing the square. What it means for you: You’ll use completing the square to solve quadratic equations. Key words: completing the square perfect square trinomial binomial Check it out: The aim is to get a perfect square involving x on one side of the equation. Solving Equations by Solving Equations by Completing the Square Completing the Square This is what the whole Section has been leading up to. The process of completing the square is a really useful method that can help solve quadratic equations. You can Solve Quadratics by Completing the Square The best way to show this is with an example: Example 1 Solve x2 – 10x + 21 = 0 by completing the square. Solution To do this, you apply some of the techniques from the earlier parts of this chapter. 1) Take the constant (= 21) onto the right-hand side of the equation. Then convert what remains on the left-hand side (x2 – 10x) to a perfect square trinomial by adding the square of half the coefficient of x (to both sides of the equation) — so add (–5)2 = 25. x2 – 10x = –21 x2 – 10x + 25 = –21 + 25 x2 – 10x + 25 = 4 2) Now you can use the square root method to solve the equation. x2 – 10x + 25 = 4 (x – 5)2 = 4 x – 5 = ±2 x – 5 = 2 or x – 5 = –2 x = 7 or x = 3 Section 7.2 — Completing the Square 351 Example 2 Solve the following quadratic equation: 4x2 – 9x + 2 = 0 Solution The best thing to do here is to divide the equation by 4 first. Then you need to solve: x2 – 9 4 x + 1 2 = 0 1) x2 – x2 – 9 4 9 4 x = – 1 2 Move the constant to the other side x + 2⎛ ⎞ 9 ⎟⎟⎟⎟ = – ⎜⎜⎜ ⎠ ⎝ 8 1 2 + 2⎛ ⎞ 9 ⎟⎟⎟⎟ = ⎜⎜⎜ ⎠ ⎝ 8 49 64 Add a number to both sides to get a perfect square trinomial x−⎛ ⎜⎜⎜ ⎝ 2 ⎞ ⎟⎟⎟⎟ ⎠ 9 8 = 49 64 Rewrite the perfect square trinomial as a square So now you have to solve: x−⎛ ⎜⎜⎜ ⎝ 2 ⎞ ⎟⎟⎟⎟ ⎠ 9 8 = 49 64 2) Use the square root method. x−⎛ ⎜⎜⎜ ⎝ 2 ⎞ ⎟⎟⎟⎟ ⎠ 9 8 = 49 64 or x – 9 8 = – x = 16 8 = 2 or x = 2 8 = 7 8 1 4 Guided Practice Solve the following equations by completing the square. 1. x2 – 6x + 5 = 0 3. x2 – 10x – 200 = 0 5. x2 + 14x + 33 = 0 7. x2 – 2x – 8 = 0 9. x2 + 18x – 19 = 0 2. x2 + 4x – 5 = 0 4. x2 – 12x – 64 = 0 6. x2 – 8x + 12 = 0 8. x2 – 16x + 39 = 0 10. x2 – 20x – 44 = 0 11. x2 + 22x + 21 = 0 12. 2x2 – 12x + 10 = 0 13. 3a2 + 12a – 15 = 0 14. 2y2 + 20y + 18 = 0 352 Section 7.2 — Completing the Square Check it out: You could complete the square at this stage, but it’s easier if the coefficient of x2 is 1. Example 3 Solve 5x2 – 2x – 3 = 3x2 – 6x + 13. Solution Rearrange the equation first so that all like terms are combined: 5x2 – 2x – 3 = 3x2 – 6x + 13 2x2 + 4x – 16 = 0 Now divide through by 2 so that the coefficient of x2 is 1. x2 + 2x – 8 = 0 Take the constant to the other side — then convert what’s left to a perfect square. x2 + 2x = 8 x2 + 2x + 1 = 8 + 1 (x + 1)2 = 9 Use the completed square to solve the original equation. (x + 1)2 = 9 x + 1 = ±3 So x + 1 = 3 or x + 1 = –3 x = 2 or x = –4 Example 4 Find t if 24t – 3t 2 – 48 = –9. Solution 24t – 3t2 – 48 = –9 3t2 – 24t + 39 = 0 t2 – 8t + 13 = 0 t2 – 8t = –13 ( ) = –13 + 8 2 (t – 4)2 = –13 + 16 (t – 4)2 = 3 t2 – 8t + 8 2 2 2 ( ) Add a number to both sides to get a perfect square trinomial t – 4 = ± 3 So t = 4 + 3 or t = 4 – 3 Section 7.2 — Completing the Square 353 Guided Practice Solve each of these quadratic equations using the method of completing the square: 15. x2 – 2x – 15 = 0 16. x2 + 2x – 24 = 0 17. x2 + 14x + 45 = 0 18. x2 – 15x + 56 = 0 19. 5x2 + 12x + 38 = 2x2 + 36x + 2 20. –150 + 2b2 = 20b 21. 5a(a + 2) – 120 = 0 22. 5y(y + 2) – 13 = 2(y2 – y + 1) 23. 5x2 + 44x – 49 = 2x(1 – x) 24. 4x(x – 11) – 4 = 4(x – 12) 25. 6a(a + 3) = 6(a – 1) + 216 26. 5y(y – 6) + 51 = 131 27. 7b2 + 14b – 32 = 5b2 + 26b 28. 3(x2 – 5) – 9 = 111 – 12x 29. 8(x + 6)2 – 128 = 6(x + 6)2 30. 4(x + 5)2 – 200 = 2(x + 5)2 Independent Practice Solve by completing the square. 1. x2 + 2x – 3 = 0 3. a2 + 8a – 9 = 0 2. y2 – 4y – 12 = 0 4. b2 + 10b – 24 = 0 5. x2 – 12x + 20 = 0 6. 3x2 – 42x + 39 = 0 7. 2a2 + 12a + 10 = 0 8. 5b2 + 20b – 105 = 0 9. 7d2 – 14d – 21 = 0 10. 2x2 + 48x – 50 = 0 11. 3c2 + 66c + 63 = 0 12. 3y(y – 8) + 2 = 254 13. 6c(c – 4) = 2(18c + 168) 14. 5(a2 + 14) = 5 – 70a 15. y2 – 2py – 15p2 = 0 16. x2 – 3kx + 2k2 = 0 17. An expression for the area of a rectangle is (2x2 + 4x) cm2 and is equal to 6 cm2. Find the value of x. 18. An expression for the area of a circle is (4py2 – 16py) ft2 and is equal to 20p ft2. Find the value of y. Round Up Round Up Make sure you understand all the examples — then once you understand the method, get plenty of practice. Completing the square means first forming a perfect square trinomial, and then using the new form to solve the original equation. 354 Section 7.2 — Completing the Square Topic 7.3.1 Section 7.3 The Quadratic Formula The Quadratic Formula California Standards: 19.0: Students know the quadratic formula and are familiar with its proof by completing the square. 20.0: Students use the quadratic formula to find the roots of a seconddegree polynomial and to solve quadratic equations. What it means for you: You’ll use the quadratic formula to solve quadratic equations — and you’ll derive the quadratic formula itself. Key words: quadratic formula completing the square You can also use the quadratic formula to solve quadratic equations. It works every time. Quadratic Equations can be in Any Variable The standard form for a quadratic equation is ax2 + bx + c = 0 (a π 0) Any quadratic equation can be written in this form by, if necessary, rearranging it so that zero is on one side. A lot of the quadratic equations you will see may contain a variable other than x — but they are still quadratic equations like the one above, and can be solved in the same way. Example 1 Find the solutions of the following quadratic equations: a) x2 – 4x + 3 = 0 b) y2 – 4y + 3 = 0 Solution a) Here, a = 1, b = –4, and c = 3. This equation factors to give (x – 3)(x – 1) = 0 So using the zero property: (x – 3) = 0 or (x – 1) = 0, or x = 3 or x = 1 b) Here, the variable is y rather than x — but that does not affect the solutions. Again a = 1, b = –4, and c = 3, so it is the same equation as in a), and will have the same solutions: y = 3 or y = 1 You can see that the two quadratic equations are really the same — only the variables have changed. Section 7.3 — The Quadratic Formula and Applications 355 The Quadratic Formula Solves Any Quadratic Equation The solutions to the quadratic equation ax2 + bx + c = 0 are given by the quadratic formula: − ± b x = − 4 ac 2 b 2 a You can derive the quadratic formula by completing the square: 2 ax ax + + = bx 2 0 c + = − c bx +⎡ ⎢ x ⎣ +⎡ ⎢⎢ x ⎣ +⎡ ⎢ ac b 2 4 a − 4 ac 2 a 4 b 2 2 Dividing the equation by a Completing the square 2 Rearranging ± ac a Include positive and negative roots = − + b So x ac −2 4 b 2 a = − − b or x ac − ac −2 b 2 a Examples Using the Quadratic Formula Example 2 Solve x2 – 5x – 14 = 0 using the quadratic formula. Solution Start by writing down the values of a, b, and c: a = 1, b = –5, and c = –14 Now very carefully substitute these values into the quadratic formula ac 14 ) ± 5 = x + 56 = 25 2 ± 5 81 2 = 9 ±5 2 So x = + 5 9 2 = 7 or x = − 5 9 2 = − 2 Check it out: The quadratic formul
a gives you the solutions of any quadratic equation in terms of a, b, and c. It’s sometimes called the general solution of the quadratic equation ax2 + bx + c = 0. Don’t forget: See Section 7.2 for a reminder about completing the square. Don’t forget: Be very careful if there are minus signs. Check it out: You can check these answers by substituting them into the original equation: 72 – (5 × 7) – 14 = 0 and (–2)2 – (5 × (–2)) – 14 = 0. So both solutions are correct. 356 Section 7.3 — The Quadratic Formula and Applications It’s really important to practice with the quadratic formula — you have to be able to use it correctly. Example 3 Solve 2x2 – 3x – 2 = 0 using the quadratic formula. Solution a = 2, b = –3, and c = –2 Putting these values into the quadratic formula... = x = x − ± b − 4 ac 16 4 = ± 3 25 4 = ± 33 5 4 So x = + 3 5 4 = 2 or Example 4 Solve 2x2 – 11x + 13 = 0. Solution a = 2, b = –11, and c = 13 Putting these values into the quadratic formula... = x = x = x ac 2 4 − − ± b b 2 a − − ± − ) 11 ( ( ) 11 ⋅ 2 2 − 121 104 4 2 − ⋅ ⋅ ) 4 2 13 ( 11 = ±± 17 4 ± 11 So x = +11 4 17 or x = 17 −11 4 Section 7.3 — The Quadratic Formula and Applications 357 Check it out: Leave these answers as radical expressions unless you’re told otherwise. Guided Practice Use the quadratic formula to solve each of the following equations. 1. x2 – 2x – 143 = 0 3. x2 + 2x – 1 = 0 5. 2x2 – 5x + 2 = 0 7. 2x2 – 7x – 3 = 0 9. 18x2 + 3x – 1 = 0 2. 2x2 + 3x – 1 = 0 4. x2 + 3x + 1 = 0 6. 3x2 – 2x – 3 = 0 8. 6x2 – x – 1 = 0 10. 4x2 – 5x + 1 = 0 11. The equation 2x2 – 7x – 4 = 0 factors to (2x + 1)(x – 4) = 0. Using the zero product property we can find that x = – 1 2 or x = 4. Verify this using the quadratic formula. 12. The height of a triangle is 4 ft more than 4 times its base length. If the triangle’s area is 5 2 ft2, find the length of its base. Independent Practice Use the quadratic formula to solve each of the following equations. 1. 5x2 – 11x + 2 = 0 3. 7x2 + 6x – 1 = 0 5. 10x2 + 7x + 1 = 0 7. 5x2 – 2x – 3 = 0 9. 4t2 + 7t – 2 = 0 11. 2x2 – x = 1 13. 2x2 + 7x = 4 15. 4x2 – 13x + 3 = 0 17. 25x2 – 9 = 0 19. 10x2 + 1 = 7x 2. 2x2 + 7x + 3 = 0 4. x2 – 7x + 5 = 0 6. 3y2 – 8y – 3 = 0 8. 4x2 + 3x – 5 = 0 10. 6m2 + m – 1 = 0 12. 3x2 – 5x = 2 14. 4x2 + 17x = 15 16. 4x2 – 1 = 0 18. 4x2 + 15x = 4 20. 16x2 + 3 = 26x Solve these equations by factoring and using the zero product property, then verify the solutions by solving them with the quadratic formula. 21. x2 + 4x + 4 = 0 23. x2 – x – 12 = 0 25. 6x2 + 29x = 5 22. 4y2 – 9 = 0 24. 2x2 – 3x – 9 = 0 26. 7x2 + 41x = 6 27. The length of a rectangle is 20 cm more than 4 times its width. If the rectangle has an area of 75 cm2, find its dimensions. 28. The equation h = –14t2 + 12t + 2 gives the height of a tennis ball t seconds after being hit. How long will the ball take before it hits the ground? Round Up Round Up The quadratic formula looks quite complicated, but don’t let that put you off. If you work through the derivation of the formula on p356 then you should see exactly why it contains all the elements it does. 358 Section 7.3 — The Quadratic Formula and Applications Topic 7.3.2 California Standards: 20.0: Students use the quadratic formula to find the roots of a seconddegree polynomial and to solve quadratic equations. What it means for you: You’ll model real-life problems using quadratic equations and then solve them using the quadratic formula. Key words: quadratic formula completing the square Check it out: The mathematical model in Example 1 is the statement that the numbers should be represented by x and (x + 9). It’s very important to be clear here to avoid confusion later. Check it out: In fact, this equation factors. But if you didn’t spot that, use the quadratic formula as in the example. Check it out: Careful — you haven’t finished yet. Now you need to go back to your initial model and interpret the solutions of your quadratic. Applications of Quadratics Applications of Quadratics Sometimes you will have to make a “mathematical model” first, and then solve the equations you get from it. Then when you get your solutions, you have to interpret them. Modeling Means Writing Your Own Equations Here’s an example of modeling a real-life situation as a quadratic equation: Example 1 The difference between a pair of numbers is 9. Find all such pairs of numbers that have a product of 220. Solution There are two different numbers in each pair — call the lower number x. Then the higher number is x + 9. Write the information from the question in the form of an equation: x(x + 9) = 220 This is a quadratic equation, so rearrange it to the form ax2 + bx + c = 0. x(x + 9) = 220 x2 + 9x = 220 x2 + 9x – 220 = 0 Write down a, b, and c: a = 1, b = 9, and c = –220 Now you can use the quadratic formula ac 2 b 2 a 2 ) ( 9 220 − ± + 81 880 2 20 or x = =22 = −40 2 = 2 − ± 9 2 11 . − So x = 961 == 31 − ±9 2 So there are two possible values for x (where x is the lower of the two numbers): x = –20 or x = 11 The higher of the two numbers is found by adding 9 to each of these values. So there are two possible pairs of numbers, and they are: –20 and –11 and 11 and 20 Section 7.3 — The Quadratic Formula and Applications 359 Example 2 Find the dimensions of the rectangle whose length is 7 inches more than twice its width, and whose area is 120 in2. Solution Let x = width in inches. Then the length is 2x + 7 inches. From the question: x(2x + 7) = 120 Rearrange this quadratic into standard form: 2x2 + 7x – 120 = 0 Now use the formula with a = 2, b = 7, and c = –120 ac − ⋅ 4 2 ⋅ 2 2 + 960 2 b a 2 2 7 49 4 ⋅ − ( 120 ) = − ± 7 1009 4 = − ±7 . 31 76 4 So x = 6.19 or x = –9.69 But the width cannot be negative, so you can ignore x = –9.69. So the width must be x = 6.2 in. (to 1 decimal place), and the length is then 2x + 7 = (2 × 6.19) + 7 = 19.4 in. (to 1 decimal place). Guided Practice 1. The difference between two numbers is 7. Find all possible pairs of such numbers if the product of the two numbers is 198. 2. Find the dimensions of a rectangular garden whose length is 10 meters more than three times its width, if the area is 77 m2. 3. Twice the square of a number is equal to eight times the number. Find the number. 4. The sum of the squares of two consecutive odd integers is 74. Find the numbers. 5. The sum of the squares of two consecutive even integers is 340. Find the possible numbers. 6. The length of a rectangular field is 10 meters less than four times its width. Find the dimensions if its area is 750 square meters. 7. When 15 and 19 are each increased by t, the product of the resulting numbers is 837. Find the value(s) of t. 8. A mother is three times as old as her daughter. Four years ago the product of their ages was 256. Find their current ages. 360 Section 7.3 — The Quadratic Formula and Applications Independent Practice 1. A man is five times older than his son. In three years’ time, the product of their ages will be 380. Find their ages now. 2. Lorraine is 10 years older than Ahanu. In three years’ time the product of their ages will be 600. Find Ahanu and Lorraine’s ages now. 3. A picture of 10 inches by 7 inches is in a frame whose area (including the space for the picture) is 154 square inches. Find the dimensions of the frame if the gap between the edge of the picture and the frame is the same all the way around. x x 10 in. 7 in. x x 4. Jennifer has a picture of her boyfriend Zach measuring 10 inches by 8 inches. She frames the picture in a frame that has an area of 224 square inches (including the space for the picture). Find the dimensions of the picture frame if the gap between the edge of the picture and the frame is the same all the way around. 5. A wire of length 50 feet is bent to form a rectangular figure that has no overlap. If the area of the figure formed is 144 square feet, find the dimensions of the figure. 6. A piece of wire 22 yards long is bent to form a rectangular figure whose area is 28 square yards. Find the dimensions of the figure, given that there is no overlap in the wire. 7. Show that the sum of the solutions of 4x2 – 4x – 3 = 0 is equal to 1 (= − b a ). 8. Show that the product of the solutions of x2 – 7x + 10 = 0 is equal to 10 (= ). c a 9. Using the quadratic formula, show that the sum of the solutions of the general quadratic equation ax2 + bx + c = 0 is equal to − b a product of the roots is , and that the . c a Round Up Round Up Quadratic equations pop up a lot in Algebra I. If you know the quadratic formula then you’ll always be able to solve them by just substituting the values into the formula. Section 7.3 — The Quadratic Formula and Applications 361 Topic 7.4.1 California Standards: 21.0: Students graph quadratic functions and know that their roots are the x-intercepts. What it means for you: You’ll learn about the shape of various quadratic graphs. Key words: quadratic parabola concave vertex line of symmetry root Section 7.4 Graphs of Graphs of Quadratic Functions Quadratic Functions So far in this Chapter you’ve solved quadratic equations in several different ways. In this Section you’ll see how the graphs of quadratic functions can be plotted using the algebraic methods you’ve already seen. The Graphs of Quadratic Functions are Parabolas If you plot the graph of any quadratic function, you get a curve called a parabola. The graphs of y = ax2 (for various values of a) on the right show the basic shape of any quadratic graph. y = x2 a ( = 1) y = –x2 a ( = –1) -3 -4 • The parabola’s either a u-shaped or n-shaped curve depending on the sign of a. The graph of y = ax2 is concave up (u-shaped — it opens upwards) when a > 0, but concave down (n-shaped — it opens downwards) when a < 0. y y = ax2 y = x3 2 a ( = 3) 5 4 3 2 1 -2 -1 0 1 2 3 x 4 -1 -2 -3 -4 -5 1 2 y = – x2 1 a ( = – ) 2 All quadratic graphs have one vertex (maximum or minimum point). For the curves shown above, the vertex is at the origin (0, 0). All quadratic graphs have a vertical line of symmetry. For the
graphs above, the line of symmetry is the y-axis. Notice that a bigger value of ΩaΩ results in a steeper (narrower) parabola. For example, the graph of y = 3x2 is steeper than the graph of y = x2. The basic shape of all quadratic graphs (that is, for any quadratic function y = ax2 + bx + c) is very similar to the ones above. They’re all concave up or concave down depending on the sign of a (concave up if a > 0 and concave down if a < 0). However, the graph can be stretched or squashed, and in a different place relative to the x- and y-axes, depending on the exact values of a, b, and c. 362 Section 7.4 — Quadratic Graphs Guided Practice Match the equations with their graphs on the right. B 1. y = –3x2 x2 – 2 2. y = 1 4 3. y = 2x2 + 3 x2 – 1 4. y = – 1 2 5. y = 2x2 A –6 –5 –4 –3 –1 0 1 –2 –3 –4 –5 –6 –7 y = ax2 + c is Like y = ax2 but Moved Up or Down by c This diagram shows the graphs of y = x2 + c, for three values of c: y = x 2 + c y = x2 y = x2 + 1 y 5 4 3 2 1 -4 -3 -2 -1 0 1 2 3 -1 -2 -3 -4 -5 y = x2 – 4 The top and bottom parabolas in the diagram are both the same shape as the graph of y = x2. The only differences are: (i) the graph of y = x2 + 1 is 1 unit x 4 higher up the y-axis. (ii) the graph of y = x2 – 4 is 4 units lower down the y-axis. The graph of y = x2 – 4 crosses the x-axis when y = 0 — that is, when x2 – 4 = 0 (or x = ±2). In fact, the x-intercepts of any quadratic graph y = ax2 + bx + c are called the roots of the function, and they correspond to the solutions of the equation ax2 + bx + c = 0. The graph of y = x2 + 1 does not cross the x-axis at all. This is because x2 + 1 = 0 does not have any real solutions. So the graph of a quadratic function may cross the x-axis twice (y = x2 – 4), may touch the x-axis in one place (y = x2), or may never cross it (y = x2 + 1). It all depends on how many roots the quadratic function has. However, the graph will always have a y-intercept — the graph will always cross the y-axis at some point. Section 7.4 — Quadratic Graphs 363 Guided Practice Describe the graphs of the quadratics below in relation to the graph of y = x2. 6. y = x2 + 1 8. y = 2x2 + 2 7. y = x2 – 3 9. y = 1 4 x2 – 5 10. y = –x2 + 1 11. y = –2x2 – 4 The graphs in Exercises 12 and 13 are transformations of the graph of y = x2. Find the equation of each graph. 12. 132 –1 0 1 –2 x 1 2 –2 –1 0 1 –2 –3 –4 Independent Practice Match the equations with their graphs on the right. B A 1. y = x2 – 1 2. y = –x2 – 1 3. y = 3x2 4. y = – 1 4 x2 5. y = –x2 + 3 –6 –5 –4 –3 –2 y 6 5 4 3 2 1 –1 0 1 –2 –3 –4 –5 – Describe the graphs of the quadratics below in relation to the graph of y = x2. 6. y = 1 2 x2 + 1 8. y = –2x2 + 3 7. y = –4x2 9. y = 1 3 x2 Round Up Round Up Now you know how the a and c parts of the equation y = ax2 + c affect the graph. In the next Topic you’ll learn how to draw some quadratic graphs yourself. 364 Section 7.4 — Quadratic Graphs Topic 7.4.2 California Standards: 21.0: Students graph quadratic functions and know that their roots are the x-intercepts. 22.0: Students use the quadratic formula or factoring techniques or both to determine whether the graph of a quadratic function will intersect the x-axis in zero, one, or two points. What it means for you: You’ll graph quadratic functions by finding their roots. Key words: quadratic parabola intercept vertex line of symmetry root Drawing Graphs of Drawing Graphs of Quadratic Functions Quadratic Functions In this Topic you’ll use methods for finding the intercepts and the vertex of a graph to draw graphs of quadratic functions. Find the Roots of the Corresponding Equations In general, a good way to graph the function y = ax2 + bx + c is to find: (i) the x-intercepts (if there are any) — this involves solving a quadratic equation, (ii) the y-intercept — this involves setting x = 0, (iii) the vertex. Example 1 Sketch the graphs of y = x2 – 3x + 2 and y = –2x2 + 6x – 4. Solution (i) To find the x-intercepts of the graph of y = x2 – 3x + 2, you need to solve: x2 – 3x + 2 = 0 This quadratic factors to give: (x – 1)(x – 2) = 0 Using the zero property, x = 1 or x = 2. So the x-intercepts are (1, 0) and (2, 0). (ii) To find the y-intercept, put x = 0 into y = x2 – 3x + 2. This gives y = 2, so the y-intercept is at (0, 2). (iii) The x-coordinate of the vertex is always halfway between the x-intercepts. So the x-coordinate of the vertex is given by And the y-coordinate of the vertex is: ⎛ ⎜⎜⎜ ⎝ 3 2 So the vertex is at ,−( 3 2 1 4 ). ⎟⎟⎟ − ×( 3 2 ⎞ ⎠ )+ = − 2 3 2 1 4 Also, the parabola’s line of symmetry passes through the vertex. So here, the line of symmetry is the line x = 3 2 . Section 7.4 — Quadratic Graphs 365 Example 1 continued The next function is the same as in the previous example, only multiplied by –2. The coefficient of x2 is negative this time, so the graph is concave down. (i) To find the x-intercepts of the graph of y = –2x2 + 6x – 4, you need to solve: –2x2 + 6x – 4 = 0 This quadratic factors to give –2(x – 1)(x – 2) = 0. So using the zero property, x = 1 or x = 2. This means the x-intercepts are at (1, 0) and (2, 0). (ii) Put x = 0 into y = –2x2 + 6x – 4 to find the y-intercept. The y-intercept is (0, –4). (iii) The vertex is at x = . 3 2 ⎛ ⎜⎜⎜ So the y-coordinate of the vertex is at: –2 × ⎝ ,( Therefore the coordinates of the vertex are 3 2 ×( 2 ⎞ ⎟⎟⎟⎟ + 6 3 ⎠ 2 ), and the line of symmetry is again x = Here are both graphs drawn on the same axes: Check it out: In the equation y = x2 – 3x + 2 the coefficient of x2 is positive (= 1), so the parabola will be u-shaped2 -1 0 1 2 3 x 4 -1 -2 -3 - line of symmetry 366 Section 7.4 — Quadratic Graphs Guided Practice Exercises 1–4 are about the quadratic y = x2 – 1. 1. Find the x–intercepts (if there are any). 2. Find the y–intercepts (if there are any). 3. Find the vertex. 4. Using the vertex, x-intercepts, and y-intercepts, graph the quadratic. Exercises 5–8 are about the quadratic y = (x – 1)2 – 4. 5. Find the x–intercepts (if there are any). 6. Find the y–intercepts (if there are any). 7. Find the vertex. 8. Using the vertex, x-intercepts, and y-intercepts, graph the quadratic. Independent Practice For each of the quadratics in Exercises 1–7, follow these steps: i) Find the x–intercepts (if any), ii) Find the y–intercepts (if any), iii) Find the vertex, iv) Using the vertex, x-intercepts, and y-intercepts, graph the quadratic. 1. y = x2 – 2x 3. y = –4x2 – 4x + 3 5. y = x2 + 4x + 4 7. y = –9x2 – 6x + 3 2. y = x2 + 2x – 3 4. y = x2 – 4 6. y = –x2 + 4x + 5 Describe the characteristics of quadratic graphs of the form y = ax2 + bx + c that have the following features, or say if they are not possible. 8. No x-intercepts 10. Two x-intercepts 12. No y-intercepts 14. Two y-intercepts 9. One x-intercept 11. Three x-intercepts 13. One y-intercept 15. Three y-intercepts 16. Which quadratic equation has the following features? Vertex (3, –4), x-intercepts (1, 0), (5, 0), and y-intercept (0, 5) 17. Which quadratic equation has the following features? Vertex (0, 16), x-intercepts (4, 0), (–4, 0), and y-intercept (0, 16) Round Up Round Up A quadratic function has the general form y = ax2 + bx + c (where a π 0). When you draw the graph of a quadratic, the value of a determines whether the parabola is concave up (u-shaped) or concave down (n-shaped), and how steep it is. Changing the value of c moves the graph in the direction of the y-axis. Note that if a = 0, the function becomes y = bx + c, which is a linear function whose graph is a straight line. Section 7.4 — Quadratic Graphs 367 Topic 7.4.3 California Standards: 21.0: Students graph quadratic functions and know that their roots are the x-intercepts. 22.0: Students use the quadratic formula or factoring techniques or both to determine whether the graph of a quadratic function will intersect the x-axis in zero, one, or two points. What it means for you: You’ll graph quadratic functions by first completing the square of the equation. Key words: quadratic completing the square parabola intercept vertex line of symmetry root Quadratic Graphs and Quadratic Graphs and Completing the Square Completing the Square If there are no x-intercepts, then it’s impossible to find the vertex by saying that the vertex is halfway between the x-intercepts (like you saw in Topic 7.4.2). But you can use the method of completing the square. Write the Equation in the Form y = (x + k)2 + p Example 1 Sketch the graph of y = x2 – 6x + 10. Solution You could try to find the x-intercepts by factoring the equation: x2 – 6x + 10 = 0 This time, the left-hand side doesn’t factor. So to find the solutions you could try the quadratic formula with a = 1, b = –6, and c = 10. − ± b = x 4 ac − 10 6 ± −4 2 However, since you cannot take the square root of a negative number, this tells you that the quadratic function has no real roots — the equation can’t be solved using real numbers. This means that the graph of y = x2 – 6x + 10 never crosses the x-axis. But this doesn’t mean that you can’t find the vertex — you just have to use a different method. The trick is to write the equation of the quadratic in the form y = (x + k)2 + p — you need to complete the square. So take the first two terms of the quadratic, and add a number to make a perfect square. 368 Section 7.4 — Quadratic Graphs Example 1 continued x2 – 6x + 2 )6 −( 2 x2 – 6x + 10 = (x – 3)2 – + 10 = (x – 3)2 )6 −( 2 2 x2 – 6x + 10 = (x – 3)2 + 1 Therefore the function you need to sketch is y = (x – 3)2 + 1. Now, the minimum value that (x – 3)2 takes is 0 (since a squared number cannot be negative). Therefore the minimum value of y = (x – 3)2 + 1 is 0 + 1 = 1. This minimum value occurs at x = 3 (the value for x where (x – 3)2 = 0). So the coordinates of the vertex of the parabola are (3, 1). As before, the line of symmetry passes through the vertex — so the line of symmetry is x = 3 . The graph of the quadratic function y = (x + k)2 + p has its vertex at (–k, p). The line of symmetry of the graph is x = –k. Check it out: The coeffic
ient of x2 is positive, so this parabola is concave up. Example 1 continued The graph of y = x2 – 6x + 10 = (x – 3)2 + 1: line of symmetry = 3 x y 6 5 4 3 2 1 vertex (3, 1) -1 1 2 3 4 5 6 7 -1 x 8 Section 7.4 — Quadratic Graphs 369 Don’t forget: See Section 7.2 for more about completing the square. Check it out: The coefficient of x2 is negative, so this is a concavedown parabola. Find the Vertex by Completing the Square Example 2 Complete the square for 4x2 – 12x + 11. Then find the vertex and line of symmetry of y = 4x2 – 12x + 11. Solution This is a concave-up parabola, since the coefficient of x2 is positive. 4x2 – 12x + 11 = 4 = 4 = 4 2 2 2 11 4 ⎡ ⎢ x ⎣ ⎡ −( ⎢ x ⎢ ⎣ ⎡ −( x⎢⎢ ⎢ ⎣ ⎤ − + ⎥ 3 x ⎦ ) −( ) + ⎤ ⎥ ⎥ ⎦ 11 4 = 4 x −( 2 ) + 2 3 2 2 )3 2 is 0, the minimum value of Since the minimum value of x −( 4 x −( )3 2 2 + 2 must be 0 + 2 = 2. This minimum value occurs at x = 3 2 . So the vertex of the graph of y = 4x2 – 12x + 11 is at and the line of symmetry is x = 3 2 . 2,( 3 2 ) , As before, put x = 0 to find the y-intercept — this is at y = 11. Example 3 Write 4x – x2 – 7 in the form a(x + k)2 + m, and sketch the graph. Solution 4x – x2 – 7 = –x2 + 4x – 7 Factor out –1 to make completing the square easier. = –[x2 – 4x + 7] = –[(x – 2)2 + 3] = –(x – 2)2 – 3 But (x – 2)2 is never negative — the minimum value it takes is 0. So –(x – 2)2 can never be positive, and the maximum value it can take is 0. This means that the maximum value of –(x – 2)2 – 3 must be –3, which it takes when x – 2 = 0 — that is, at x = 2. So the vertex of the graph is at (2, –3). And the line of symmetry is x = 2. As always, find the y-intercept by putting x = 0. This is at y = –7. y 1 -1 0 -1 1 2 3 4 5 x 6 -2 -3 -4 -5 -6 -7 -8 -9 370 Section 7.4 — Quadratic Graphs Guided Practice Sketch the graph of each function below, stating the y-intercept and x-intercepts (where appropriate). Use the method of completing the square to find the coordinates of the vertex, and the line of symmetry. 1. y = x2 – 12x + 20 3. y = x2 – 2x – 3 5. y = –x2 – 2x + 3 7. y = –2x2 – 8x + 10 9. y = x2 – 4x + 12 2. y = x2 + 8x + 12 4. y = x2 – 4x – 5 6. y = –x2 – x + 6 8. y = 2x2 + x – 6 10. y = 3x2 + 6x + 6 Independent Practice In Exercises 1–2, use the information that a ball was thrown vertically into the air from a platform 3 2 m above sea level. The relationship between the height in meters above sea level, h, and the number of seconds since the ball was thrown, t, was found to be h = –5t2 + 6t + 3 2 . 1. After how many seconds did the ball reach its maximum height? 2. What was the ball’s maximum height above sea level? The first 8 seconds in the flight of a paper airplane can be modeled by the quadratic h = 1 8 t2 – t + 4, where h is the height in feet and t is the time in seconds. Use this information to answer Exercises 3–4. 3. In the first 8 seconds of its flight, when did the airplane reach its minimum height? 4. What was the minimum height of the plane in the first 8 seconds of its flight? A ball is thrown vertically into the air from a platform. The relationship between the ball’s height in meters, h, and the number of seconds, t, since the ball was thrown was found to be h = –5t2 + 10t + 15. Use this information to answer Exercises 5–8. 5. After how many seconds did the ball reach its maximum height? 6. What was the maximum height of the ball? 7. At what height was the ball initially thrown? 8. When did the ball hit the ground? Round Up Round Up Take a look at Section 7.2 if all this stuff about completing the square seems unfamiliar. Completing the square is a really useful way of graphing quadratics because it gives you the vertex of the graph straightaway. Section 7.4 — Quadratic Graphs 371 Topic 7.5.1 California Standards: 22.0: Students use the quadratic formula or factoring techniques or both to determine whether the graph of a quadratic function will intersect the x-axis in zero, one, or two points. What it means for you: You’ll learn what the discriminant is, and how you can use it to tell how many roots a quadratic equation has. Key words: discriminant quadratic intercept root Section 7.5 Quadratic Equations Quadratic Equations and the Discriminant and the Discriminant In this Section you’ll meet a particular part of the quadratic formula that tells you how many solutions a quadratic equation has. This Section carries straight on from the previous Section on graphing quadratic functions. Use the Quadratic Formula to Find x-Intercepts A general quadratic equation has the form: ax2 + bx + c = 0 The solutions to this equation are given by the quadratic formula: − ± b = x 4 ac −2 b 2 a The quadratic formula can be used to help draw the graph of a quadratic function y = ax2 + bx + c. By finding where y = 0 (that is, by solving ax2 + bx + c = 0), you can find the x-intercepts of the parabola. But it’s sometimes impossible to get an answer from the quadratic formula. When b2 – 4ac is negative, the square root in the formula cannot be taken. This means the graph never crosses the x-axis. Don’t forget: See Section 7.3 for more on the quadratic formula. = x b − ± b 2 4– 2 a ac When b2 – 4ac < 0, there are no x-intercepts — the parabola never crosses the x-axis. In fact, the value of b2 – 4ac tells you a lot about the solutions of a quadratic equation, and about the x-intercepts of the corresponding quadratic graph. b2 – 4ac Tells You How Many Roots a Quadratic Has For the quadratic equation ax2 + bx + c = 0, the expression b2 – 4ac is called the discriminant. The discriminant’s used to determine the number of roots of the function y = ax2 + bx + c (and so the number of x-intercepts of the graph of y = ax2 + bx + c). 372 Section 7.5 — The Discriminant b2 – 4ac > 0 Means the Quadratic Has 2 Distinct Roots If b2 – 4ac > 0 (if b2 – 4ac is positive), then the equation ax2 + bx + c = 0 has two distinct real solutions, and the function y = ax2 + bx + c has two distinct real roots. In other words, the roots are real and unequal. Example 1 Describe the nature of the solutions of the equation 2x2 + 3x – 2 = 0. Then find the values of the roots, and sketch the graph of y = 2x2 + 3x – 2. Solution First write down the values of a, b, and c: 2x2 + 3x – 2 = 0 so a = 2, b = 3, and c = –2 So the discriminant is: b2 – 4ac = 32 – [4 × 2 × (–2)] = 9 – (–16) = 9 + 16 = 25 Since b2 – 4ac is positive, the equation 2x2 + 3x – 2 = 0 has two distinct (unequal) real solutions. This in turn means that the function y = 2x2 + 3x – 2 has two real roots — its graph crosses the x-axis in two places. To work out the actual values of the roots, use the quadratic formula: 4 ac −2 b 2 a 25 − ± 2 or 1 2 So the graph of y = 2x2 + 3x – 2 meets the x-axis in two places: (–2, 0) and ( 1 2 , 0). To sketch the graph, you also need the y-intercept: y = 2x2 + 3x – 2 = 2(0)2 + 3(0) – 2 = –2 y 5 4 3 2 1 x -4 -3 -2 -1 -1 1 2 3 -2 -3 -4 Section 7.5 — The Discriminant 373 Check it out: There are two distinct (unequal), real roots — so there are two distinct x-intercepts on the graph. Example 2 Find the nature of the solutions of the quadratic equation x2 + 5x + 2 = 0. Solution Again, the best thing to do first is write down the values of a, b, and c: x2 + 5x + 2 = 0 so a = 1, b = 5, and c = 2 Then the discriminant is: b2 – 4ac = 52 – (4 × 1 × 2) = 25 – 8 = 17 The discriminant is positive, so there are two unequal, real solutions of x2 + 5x + 2 = 0. You could use the quadratic formula to find the actual solutions: x = − ± 5 17 2 = –5 + 17 2 or –5 – 17 2 So the x-intercepts of the graph of y = x2 + 5x + 2 are at: ⎛ ⎜⎜⎜ ⎝ –5+ 17 2 ⎞ ⎟⎟⎟ and ⎠ ⎛ ⎜⎜⎜ ⎝ 0, –5– 17 2 ⎞ ⎟⎟⎟ ⎠ 0, Guided Practice Describe the nature of the solutions of each quadratic equation, and find the values of the solutions. 1. x2 + x – 12 = 0 3. x2 + 5x + 4 = 0 5. 2x2 + 5x + 2 = 0 7. 3x2 + 7x – 6 = 0 9. 2(5x2 + 1) = 9x 11. x(10x + 7) = –1 2. x2 + 2x – 3 = 0 4. 3x2 – 7x + 4 = 0 6. x2 + 3x – 1 = 0 8. 2x2 + 9x = 5 10. 6(2x2 + 1) + 17x = 0 12. 2x(4x + 7) = –3 Use the x- and y–intercepts of the quadratics below to decide in which quadrant the vertex of each equation would be. 13. y = x2 + x – 12 15. y = –x2 + 2x + 3 14. y = x2 + 2x – 3 16. y = 3x2 – 12x – 15 374 Section 7.5 — The Discriminant b2 – 4ac = 0 Means the Quadratic Has 1 Double Root If b2 – 4ac = 0, then the solutions of ax2 + bx + c = 0 (and the roots of y = ax2 + bx + c) are real and equal. Graphically, equal roots (also called a double root) mean the graph of the quadratic function just touches the x-axis, but does not cross it. Example 3 Determine the nature, and find the value(s), of the solution(s) of the equation x2 – 6x + 9 = 0. Sketch the graph of y = x2 – 6x + 9. Solution First write down the values of a, b, and c: x2 – 6x + 9 = 0 so a = 1, b = –6, and c = 9 So the discriminant is: b2 – 4ac = (–6)2 – 4 × 1 × 9 = 36 – 36 = 0 This time, the discriminant is zero, so y = x2 – 6x + 9 has a double root. In other words, its graph just touches the x-axis without actually crossing it. As always, to work out where this double root is, use the quadratic formula: − 4 ac 1 1 2 3 4 5 -1 x So the graph of y = x2 – 6x + 9 touches the x-axis at (3, 0). Guided Practice Describe the nature of the solutions of the quadratic equations, and find the value(s) of the solution(s). 17. x2 + 4x + 4 = 0 19. x2 + 6x = –9 21. 5x(5x – 6) = –9 23. 4x2 = –5(4x + 5) 25. 8x(x + 3) + 18 = 0 18. 9x2 – 6x + 1 = 0 20. 3x(x – 10) + 75 = 0 22. 8x(2x + 3) = –9 24. 4x(x + 1) + 1 = 0 26. 24x(2x + 9) + 243 = 0 Section 7.5 — The Discriminant 375 Check it out: b2 – 4ac = 0, so the quadratic formula simplifies to −b a2 . b2 – 4ac < 0 Means the Quadratic Has No Real Roots If b2 – 4ac < 0 (if b2 – 4ac is negative), then ax2 + bx + c = 0 has no real solutions, and y = ax2 + bx + c has no real roots. If there are no real roots, then there are no x-intercepts — the graph of the quadratic does not intersect (cross or touch) the x-axis. Example 4 Describe the graphs of: (a) y = x2 + 2x + 3, and (b) y = –2x2 + 4x – 5 Solution (a) H
ere a = 1, b = 2, and c = 3, so: b2 – 4ac = 22 – 4 × 1 × 3 = 4 – 12 = –8 So the graph of y = x2 + 2x + 3 never intersects the x-axis. Since a > 0, the graph is concave up (u-shaped) and stays above the x-axis. -5 -4 -3 -2 (b) Here a = –2, b = 4, and c = –5 , so: b2 – 4ac = 42 – 4 × (–2) × (–5) = 16 – 40 = –24 y (ab) = –2 ² + 4 – 5 y x -1 6 5 4 3 2 1 -1 -2 -3 -4 -5 -6 -7 So the graph of y = –2x2 + 4x – 5 never intersects the x-axis either. But this time, since a < 0, the graph is concave down (n-shaped) and stays below the x-axis. Guided Practice Use the discriminant to verify that there are no real number solutions for the quadratic equations below. 27. x2 + 3x + 4 = 0 29. 3x2 + 8 = 9x 31. 2x(2x + 3) + 3 = 0 33. 3x2 + 5x + 3 = 0 28. 5x2 – 4x + 3 = 0 30. 2x2 = 7(x – 1) 32. 3x2 + 1 = 3x 34. 2x(x – 3) = –5 376 Section 7.5 — The Discriminant Check it out: If k = 0, the x2 term would disappear, leaving a linear equation. The graph crosses the x-axis in two places, so the equation can’t be linear. Using the Discriminant Example 5 Suppose the graph of y = kx2 + x – 6 intersects the x-axis at two distinct points (where k is some constant). What are the possible values of k? Solution Since the graph of y = kx2 + x – 6 has two distinct x-intercepts, the discriminant (b2 – 4ac) must be positive. So write down your a, b, and c: a = k, b = 1, c = –6 Now use the fact that the discriminant is positive: b2 – 4ac > 0 12 – 4k(–6) > 0 1 + 24k > 0 24k > –1 k > – 1 24 So k can be any real number greater than – 1 24 , but not zero: – 1 24 < k < 0 or k > 0. Example 6 At how many points does the graph of y = x2 – 2x + 3 intersect the x-axis? Solution Here, a = 1, b = –2, c = 3 So b2 – 4ac = (–2)2 – 4 × 1 × 3 = 4 – 12 = –8 The discriminant is negative, so y = x2 – 2x + 3 = 0 has no real roots, and so the graph of y = x2 – 2x + 3 does not intersect the x-axis. Example 7 Find the values of k for which y = 5x2 – 3x + k has a double root. Solution As always, it’s a good idea to begin by writing down your a, b, and c: a = 5, b = –3, c = k So b2 – 4ac = (–3)2 – 4 × 5 × k = 9 – 20k The quadratic has a double root where the discriminant equals zero. This is the case when 9 – 20k = 0, or when k = 9 20 . So y = 5x2 – 3x + k has a double root only when k = 9 20 . Section 7.5 — The Discriminant 377 Example 8 Describe the nature of the roots of y = x2 + 2 x + 1 2 . Solution This time, a = 1, b = 2 , and c = 1 2 , so b2 – 4ac = ( 2 ) The discriminant is zero, so the function has a double root. This root is at x − ± b = − 4 ac Independent Practice Determine the number of roots of the following functions, and find the values of any real roots. 1. y = x2 – 2x – 3 3. y = x2 – 3x – 1 5. y = 4x2 – 4x + 1 7. y = 2x2 – 4x + 3 9. y = 2x2 + x – 6 11. y = x2 – x + 5 2. y = x2 + 4x + 3 4. y = x2 + x – 1 6. y = x2 – 8x + 16 8. y = x2 + x + 3 10. y = 4x2 – 12x + 9 12. y = x2 – 2x – 2 13. Find the possible values of k if y = 3x2 – kx + 3 has a double root. 14. Find the possible values of p if 2x2 – 5x + p = 0 has two real solutions. 15. If x = –1 is a root of y = 3x2 + x – k, find k and the other root. 16. If x = 1 2 is a solution to kx2 + 9x – 5 = 0, find both solutions. 17. Find the possible values of p if y = px2 – 7x – 7 has no real roots. State the number of times that the graphs of the following quadratic functions intercept the x-axis: 18. y = x2 – 3x – 28 20. y = 4x2 + 2x + 1 22. y = 5x2 + 3x + 1 19. y = 4x2 + 4x + 1 21. y = 2x2 – x – 1 23. Find all possible values of k when y = 3x² – 2kx + k has a double root. 24. When y = 3x² – 2kx + k has a root of x = 2, find k and the other root. Round Up Round Up Remember — the discriminant is just one part of the quadratic formula. If all you need to know is how many roots a function has, you don’t need to use the full formula — just the discriminant. 378 Section 7.5 — The Discriminant Topic 7.6.1 Section 7.6 Motion Tasks Motion Tasks California Standards: 23.0: Students apply quadratic equations to physical problems, such as the motion of an object under the force of gravity. What it means for you: You’ll model objects under the force of gravity using quadratic equations, and then solve the equations. Key words: quadratic gravity vertex parabola intercept completing the square Quadratic equations have applications in real life. In particular, you can use them to model objects that are dropped or thrown up into the air. Quadratic Functions Describe Motion Under Gravity Example 1 The height of a stone thrown up in the air is modeled by the equation h = 80t – 16t 2, where t represents the time in seconds since the stone was thrown and h is the height of the stone in feet. After how many seconds is the stone at a height of 96 feet? Explain your answer. Solution The stone reaching a height of 96 feet is represented by h = 96, so you need to solve 80t – 16t 2 = 96. Rewriting this in the form ax2 + bx + c = 0 (using t instead of x) gives: 16t2 – 80t + 96 = 0 t2 – 5t + 6 = 0 Divide through by 16 (t – 2)(t – 3) = 0 Factor the quadratic equation t – 2 = 0 or t – 3 = 0 Solve using the zero property t = 2 or t = 3 So the stone is at a height of 96 feet after 2 seconds (on the way up), and again after 3 seconds (on the way down). Example 2 Use the same information from Example 1. After how many seconds does the stone hit the ground? Explain your answer. Solution When the stone hits the ground, h = 0. So solve 80t – 16t 2 = 0. 5t – t 2 = 0 t(5 – t) = 0 t = 0 or t = 5 Divide through by 16 Solve using the zero property But t = 0 represents when the stone was thrown, so the stone must land at t = 5 — after 5 seconds. Section 7.6 — Motion Tasks and Other Applications 379 Example 3 Use the same information from Example 1. Calculate h at t = 7. Explain your answer. Solution At t = 7, h = (80 × 7) – (16 × 72) = 560 – 784 = –224 Negative values of h suggest that the stone is beneath ground level. This can’t be true — the height can’t be less than zero feet. But the stone landed after 5 seconds — so after t = 5, the function h = 80t – 16t 2 doesn’t describe the motion of the stone. Example 4 Use the same information from Example 1. What is the maximum height of the stone? Justify your answer. Solution The graph of h = 80t – 16t 2 is a parabola, and the maximum height reached is represented by the vertex of the parabola, which you can find by completing the square. Check it out: The formula h = 80t – 16t2 only describes the stone’s motion for a certain range of t. It can’t describe what happens before the stone is thrown or after it lands. Don’t forget: See Section 7.4 for more about plotting quadratic graphs. h = 80t – 16t2 = –16(t2 – 5t) ⎡ = –16 t −( ⎢ ⎢ ⎣ 5 2 2 ) − ⎤ ⎥ ⎥ ⎦ 25 4 = –16 t −( 2 )5 2 + 100 y 100 90 80 70 60 50 40 30 20 10 The stone reaches its maximum height after half its flight time — at t = 2.5 s. x 1 2 3 4 5 So the maximum height is 100 feet (which is reached at t = 5 2 = 2.5 s). Guided Practice 1. In a Physics experiment, a ball is thrown into the air from an initial height of 24 meters. Its height h (in meters) at any time t (in seconds) is given by h = –5t2 + 10t + 24. Find the maximum height of the ball and the time t at which it will hit the ground. 2. A firework is propelled into the air from the ground. Its height after t seconds is modeled by h = 96t – 16t2. The firework needs to explode at a height of 128 feet from the ground. After how long will it first reach this height? If the firework fails to explode, when will it hit the ground? 380 Section 7.6 — Motion Tasks and Other Applications Plotting a Graph Makes Problems Easier to Understand Example 5 The height above the ground in feet (h) of a ball after t seconds is given by the quadratic function h = –16t2 + 32t + 48. Explain what the h-intercept and t-intercepts mean in this situation, and find the maximum height reached by the ball. Check it out: From the graph, you can see that the ball reaches its maximum height before half its flight time has passed. Solution To get a clearer picture of what everything means, it helps to draw a graph. The intercept on the vertical axis (the h-axis) is found by putting t = 0: h = 48. h 80 70 60 50 40 30 20 10 The intercepts on the horizontal axis (the t-axis) are found by solving h = 0 — that is, –16t2 + 32t + 48 = 0. –1 0 –2 1 2 3 t t2 – 2t – 3 = 0 (t – 3)(t + 1) = 0 t = 3 or t = –1 In this situation, the intercept on the vertical axis (the h-intercept) represents the initial height of the ball when it was thrown (at t = 0). So here, the ball was thrown from 48 feet above the ground. The intercepts on the horizontal axis (the t-intercepts) represent the times at which the ball was at ground level. However, the function only describes the motion of the ball between t = 0 (when it was thrown) and t = 3 (when it lands). So the t-intercept at t = 3 represents the point when the ball lands. The t-intercept at t = –1 doesn’t have any real-life significance here. To find the maximum height, you need to find the vertex of the parabola — so complete the square: –16t2 + 32t + 48 = –16[t2 – 2t – 3] = –16[(t – 1)2 – 4] = –16(t – 1)2 + 64 The vertex of this parabola occurs where t = 1, and so the vertex is at (1, 64). This means the maximum height of the ball is 64 feet. Section 7.6 — Motion Tasks and Other Applications 381 Check it out: Put t = 1 into the quadratic to find the maximum value. Independent Practice A baseball is hit from homebase. Its height in meters is modeled by the equation h = 25t – 5t2, where t is the time in seconds. 1. After how many seconds will the ball be at a height of 20 meters? 2. What height will the baseball reach before it starts descending? A rocket is fired into the air. Its height in feet at any time is given by the equation h = 1600t – 16t2, where t is the time in seconds. 3. Find the height of the rocket after 2 seconds. 4. After how many seconds will the rocket be 30,000 feet above the ground? 5. After how many seconds will the rocket hit the ground? 6. As a skydiver steps out of a plane, she
drops her watch. The distance in feet, h, that the watch has fallen after t seconds is given by the equation h = 16t2 + 4t. After how many seconds will the watch have fallen 600 feet? The height in feet of an object projected upwards is modeled by the equation h = 100t – 16t2. 7. How long after being projected is the object 100 feet above the ground? 8. What is the greatest height reached by the object? The area of a rectangle is given by the formula A = x(20 – x) cm, where x is the width. 9. Find x when A = 84 cm2. 10. What value of x maximizes the area A? James and Mei are each standing on diving boards, and each throw a ball directly upwards. The height of each ball above the pool in feet, h, is plotted against the time in seconds, t, since it was thrown. 11. The height of James’s ball can be calculated using the equation h = –16t² + 30t + 10. From what height above the pool does James throw his ball? 12. The height of Mei’s ball can be calculated using the equation h = –16t² + 32t + 20. After how many seconds does her ball reach its maximum height? 13. Calculate the difference in maximum heights of the balls, to 1 decimal place. Round Up Round Up Plotting a graph is often a good idea when you’re working on motion problems — because then you can see at a glance when an object reaches the ground, or when it reaches its maximum height. 382 Section 7.6 — Motion Tasks and Other Applications Topic 7.6.2 California Standards: 23.0: Students apply quadratic equations to physical problems, such as the motion of an object under the force of gravity. What it means for you: You’ll model money problems using quadratic equations, and then solve the equations. Key words: economic profit quadratic vertex parabola completing the square Economic Tasks Economic Tasks As well as the motion tasks you saw in Topic 7.6.1, you can use quadratic equations to model real-life problems involving money. Applications of Quadratics to Economics The best way to introduce quadratic equations modeling money problems is to show you an example: Example 1 The owner of a restaurant wishes to graph the annual profit of his restaurant against the number of people he employs. He calculates that the annual profit in thousands of dollars (P) can be modeled by the formula P = –0.3x2 + 4.5x, where x is the number of people employed. According to the owner’s formula, how many full-time members of staff does the restaurant have to employ to make a profit of $15,000? Solution You have a formula for the profit P, and you have to find when this equals 15 (since the formula gives you the profit in thousands of dollars). So you need to solve the quadratic equation –0.3x2 + 4.5x = 15. Rewriting this in the form ax2 + bx + c = 0 gives: 0.3x2 – 4.5x + 15 = 0 x2 – 15x + 50 = 0 (x – 10)(x – 5) = 0 x = 10 or x = 5 Divide through by 0.3 Solve using the zero property This means that the restaurant can employ either 5 people or 10 people and make a profit of $15,000. Section 7.6 — Motion Tasks and Other Applications 383 Example 2 Use the same information from Example 1. According to the owner’s formula, how many full-time members of staff should the restaurant employ to make maximum profit? Solution To find the maximum profit, you need to find the maximum value of the quadratic P = –0.3x2 + 4.5x. To do this, you can complete the square: P = –0.3x2 + 4.5x = –0.3(x2 – 15x ⎡ ⎛ ⎢ ⎜⎜⎜ ⎢ ⎝ ⎣ ⎛ ⎜⎜ x⎜⎜ ⎝ x − 15 2 2 ⎞ ⎟⎟⎟ ⎠ − 15 2 2 ⎞ ⎟⎟⎟ ⎠ 2 ⎞ ⎟⎟⎟ ⎠ ⎤ ⎥ ⎥ ⎦ − 15 2 2 ⎞ ⎛ ⎟⎟⎟ ⎜⎜⎜ ⎠ ⎝ + ( . 0 3 + ( 3 10 225 4 ) ) 225 4 = − . 0 3 ⎛ ⎜⎜⎜ ⎝ x − 15 2 = − . 0 3 ⎛ ⎜⎜⎜ ⎝ − x 2 ⎞ ⎟⎟⎟ ⎠ 15 2 + 135 8 So the vertex of the parabola is at ( 15 2 , 135 8 ) , which (in theory) means that the restaurant should employ 7.5 people to make the maximum possible profit of $16,875. Clearly, the restaurant can’t employ 7.5 people — a good idea now is to draw the graph so that you can answer this question more realistically. Find the x-intercepts by solving P = 0: P = –0.3x2 + 4.5x = –0.3(x2 – 15x) = –0.3x(x – 15) = 0 at x = 0 and x = 15 So the graph looks like this 20 P (7.5, 16.875) 15 10 10 11 12 13 14 15 16 Number of employees x 384 Section 7.6 — Motion Tasks and Other Applications Check it out: Work out the profit like this: (–0.3 × 72) + (4.5 × 7) = (–0.3 × 82) + (4.5 × 8) = 16.8 Example 2 continued You can see from the symmetry of the graph (the line of symmetry is x = 7.5) that the maximum possible profit while employing a whole number of people is at x = 7 and x = 8, at which points the profit is $16,800. So, if the restaurant employs more than 8 people, profits decrease, possibly because there is not enough work for more than 8 people to do efficiently. Guided Practice 1. The profit p in cents per 10-minute period earned from driving a taxicab is given by p = 80x – 3x2, where x is the speed in mph. What speed would yield a profit of 512 cents per 10-minute period? 2. An investor kept track of her portfolio profit, P, at time, t, measured in years after she began investing. If P = 4000t2 – 28000t + 3000 represents her profit, after how many years will she have made $150,000 profit? 3. The amount of money a customer is willing to spend at a store is related to t, the number of minutes they have to wait before being served. If M = –t2 + 8t + 17 represents the money a customer spends, how long will it take before a customer decides to leave the store without spending any money? Independent Practice Leo produces x pounds of salsa. The ingredients cost 0.1x2 – 30 dollars and he makes 2x dollars revenue from the sale of his salsa. 1. What is Leo’s maximum possible profit? 2. How many pounds of salsa would Leo need to sell to break even? The value in dollars, V, of a certain stock can be modeled by the equation V = –16t2 + 88t + 101, where t represents the time in months. 3. What was the original value of the stock? 4. What was the maximum value of the stock? 5. When did the stock reach the maximum value? 6. When did the stock become worthless? The value, V, of Juan’s investment portfolio can be modeled by the equation V = 16t2 – 256t + 16,000, where t is the time in months. 7. What was the original value of Juan’s portfolio? 8. What was the minimum value of Juan’s portfolio? 9. When will Juan’s investment portfolio be worth $16,576.00? Round Up Round Up Usually when you graph quadratic problems involving money, the vertex of the graph shows you the point where there’s maximum profit. Section 7.6 — Motion Tasks and Other Applications 385 Chapter 7 Investigation he Handshake Pre Pre Pre Pre Proboboboboblemlemlemlemlem he Handshak he Handshak TTTTThe Handshak he Handshak TTTTThe Handshak he Handshake Pre Pre Pre Pre Proboboboboblemlemlemlemlem he Handshak he Handshak he Handshak This Investigation shows that Math techniques can be useful everywhere — even at parties. There are 45 people at a party. If each person shakes hands with every other person, how many handshakes will there be altogether? Suppose there are n people at a party. How many handshakes will there be in terms of n? Things to think about: • How many handshakes does each person make? How many handshakes would there be if there were smaller groups of people? Are there any patterns? Extension 1) The triangular numbers form the following sequence: 1, 3, 6, 10, 15, 21, ... If the first triangular number is 1, what will the nth triangular number be? 2) There are n people at a round table. Each person shakes hands with everyone else at the table, except for the person on their left and the person on their right. How many handshakes will there be in total? 3) Three diagonals of a hexagon are shown on the right. (The sides of the shape are not diagonals.) How many diagonals will a regular polygon with n sides have? Explain your answer. Open-ended Extension 1) At a high school reunion everyone shakes hands with everyone else once. Towards the end of the evening, a math teacher arrives and shakes hands with just the students that he actually taught. There were 3107 handshakes in total. How many people were at the reunion before the math teacher arrived and how many of the people present had the math teacher taught? Is your answer the only one possible? Explain your reasoning. 2) Look back at the original handshake problem. Change the problem in some way and investigate the effect on the number of handshakes. For example, you could investigate the number of handshakes if each person shakes both the left and right hands of each other person with their own left and right hand. How many handshakes would there be for groups of extraterrestrials with h hands? ound Up ound Up RRRRRound Up ound Up ound Up ound Up ound Up RRRRRound Up ound Up ound Up In this Investigation you’ve recorded results, identified patterns and applied what you’ve learned to new situations. There’s a lot you can say about even quite simple-looking math problems. estigaaaaationtiontiontiontion — The Handshake Problem estigestig estig pter 7 Invvvvvestig pter 7 In 386386386386386 ChaChaChaChaChapter 7 In pter 7 In pter 7 In Chapter 8 Rational Expressions and Functions Section 8.1 Rational Expressions .......................................... 388 Section 8.2 Multiplying and Dividing Rational Expressions .... 394 Section 8.3 Adding and Subtracting Rational Expressions .... 403 Section 8.4 Solving Equations with Fractions ........................ 412 Section 8.5 Relations and Functions ...................................... 420 Investigation Transforming Functions ....................................... 435 387387387387387 Topic 8.1.1 California Standards: 12.0: Students simplify fractions with polynomials in the numerator and denominator by factoring both and reducing them to the lowest terms. What it means for you: You’ll find out about the conditions for rational numbers to be defined. Key words: rational numerator denominator undefined Don’t forget: Take another look at Chapter 1 to remind yourself about rational numbers. Check it out: p is called the numerator and q is called th
e denominator. Section 8.1 Fractions and Fractions and Rational Expressions Rational Expressions In this Topic you’ll find out about the necessary conditions for rational numbers to be defined. Rational Expressions Can Be Written as Fractions A rational expression is any expression that can be written in the form of a fraction — that means it has a numerator and a denominator. Examples of rational expressions are . Rational expressions are written in the form p q , where q π 0. An Expression is Undefined if the Denominator is Zero If the denominator is equal to zero, then the expression is said to be undefined (see Topic 1.3.4). So, for example, x 2 x − 1 + 1 Example 1 is defined whenever x is not equal to –0.5. Determine the value of x for which the expression 7 x + is undefined. 2 Solution It is undefined when the denominator x + 2 equals zero. This means that the expression is undefined when x = –2. Example 2 Determine the value(s) of x for which the expression 2 2x − is undefined. 4 Solution It’s undefined when the denominator x2 – 4 equals zero. So, solve x2 – 4 = 0 to find the values of x: x2 – 4 = (x – 2)(x + 2) = 0 x – 2 = 0 or x + 2 = 0 ﬁ x = 2 or x = –2 Therefore, 388 Section 8.1 — Rational Expressions 2 2x − is undefined when x = ±2. 4 Example 3 Determine the value(s) of x for which the expression x 7 8 x− + is undefined. 15 2 x Solution Factor the denominator to give: x 7 )( 3 x − ) 5 − ( x If the denominator equals zero, the expression is undefined. This happens when either (x – 3) or (x – 5) equals zero. So, the expression is undefined when x = 3 or x = 5. Guided Practice Determine the value(s) of the variables that make the following rational expressions undefined. − 3 2 + 4 8 3 1y − + − 2. x x 3. 1. 4 x x 5 4 12 5. − 2 11 − 3 y 4 2y + 6. k 4 + 26 11 k 5 − 2 k 1 30 + 7 28 8 45 9. 3 a 3 Independent Practice Determine the value(s) of the variables that make the following rational expressions undefined. x 1. 23 . 2 m 4 − 18 k 5. 3 k 3 6 13 + 2 x 5 30 x 32 7. Jane states that the rational expression 3 x + 3 x − − is defined 2 1 x 1 2 when x is any real number. Show that Jane is incorrect. Round Up Round Up This Topic about the limitations on rational numbers will help you when you’re dealing with fractions in later Topics. In Topic 8.1.2 you’ll simplify rational expressions to their lowest terms. Section 8.1 — Rational Expressions 389 Topic 8.1.2 California Standards: 12.0: Students simplify fractions with polynomials in the numerator and denominator by factoring both and reducing them to the lowest terms. What it means for you: You’ll learn about equivalent fractions and how to simplify fractions to their lowest terms. Key words: equivalent rational simplify common factor Check it out: The two shapes on the far right represent x being divided up into the fractions 5 and 10 6 12 . Equivalent Fractions Equivalent Fractions Saying that two rational expressions are equivalent is just a way of saying that two fractions represent the same thing. Equivalent Fractions Have the Same Value A ratio is a comparison of two numbers, often expressed by a fraction — for example, a b . A proportion is an equality between two ratios. Four quantities a, b, c, and d are in proportion if a b = c d . Fractions like these that represent the same rational number or expression are often called equivalent fractions 10 12 You can determine whether two fractions are equivalent by using this rule: The rational expressions a b and c d are equivalent if ad = bc. Example 1 Prove that x 5 6 and x 10 12 are equivalent. Solution 5x·12 = 60x and 6·10x = 60x This is ad in the rule above This is bc in the rule above So, the two rational expressions are equivalent. Guided Practice Prove that the following pairs of rational expressions are equivalent. 1. m 54 6 and m 18 2 2. 1 3 and 2 6 x x 3. 12 − x 3 9 and 4 − x 3 390 Section 8.1 — Rational Expressions Check it out: All of the common factors have been canceled, leaving the simplest equivalent fraction. Don’t forget: The GCF of two numbers is the largest number that is a factor of both of them. is the simplest form of 56 64x Check it out: 7 8x because 1 is the only common factor of the numerator and denominator. Simplify Fractions by Canceling Common Factors A rational expression can be written in its lowest terms by reducing it to the simplest equivalent fraction. This is done by factoring both the numerator and denominator and then canceling the common factors — that means making sure its numerator and denominator have no common factors other than 1. For example: 66 78 1 ⋅ 6 11 = ⋅ 1 6 13 = 11 13 Example 2 Reduce the expression 56 64x to its lowest terms. Solution The greatest common factor (GCF) of 56 and 64 is 8. This means that: 56 x 64 = So, 56 64x and 7 8x are equivalent fractions. Numbers are not the only things that can be canceled — variables can be canceled too. For example: mc cv = ⋅ Guided Practice Reduce each of the following rational expressions to their lowest terms. 4. 21 28 7. d 10 30 5. 12 18 8. − 4 10 2 m c 3 2 m c 6. bx b x2 9 )( 5 5 )( − + ) m m ) 10. 11. + + 5 15 x x 3 12. 2 ( m m + )( Section 8.1 — Rational Expressions 391 Check it out: The greatest common factor of (x + 3) and 6 is 1. So, x + 3 is the simplest form. 6 Check it out: Note that m – 3 = –1(3 – m) so you can substitute this into the expression. Some Harder Examples to Think About Factoring the numerator and denominator is the key to doing this type of question. Breaking down a complicated expression into its factors means you can spot the terms that will cancel. Example 3 Simplify the expression 2 x 6 x − 9 − . 18 Solution Factor the numerator and denominator, then cancel common factors: 2 x 6 x 1 − 9 − = 18 ( x − + 3 )( Example 4 Simplify the expression Solution Factor the numerator and denominator completely )( m m 3 )( 1 − m 3 3 )( ( − ⋅ − m m 1 3 )( ( )) = − ) Cancel the common factor (3 – m) Example 5 Reduce this expression to its lowest terms: Solution Factor both the numerator and denominator 10 x 2 − + x 15 21 10 x 2 − + x 15 21 x = = ) 15 ) 21 10 ⋅ − + 1 )( ( x x x 5 + ⋅ + 1 1 )( ( = Cancel the common factors 392 Section 8.1 — Rational Expressions Guided Practice 13. Show how you can simplify the rational expression Simplify the following rational expressions. 14. 4 2 k − − k 16 15. 2 − 2 m c − 2 mc c 16 20 − 2 3 + k k 26 − − 3 13 k 10 )( m m m )( + 5 ) + 5 ) 17. 18. 2 m 2 k + − 6 k mk + − 2 mk m 2 2 19. Independent Practice 1. Simplify 2 2 k 2 c 2 − + ck 3 ck . Simplify ( ) . Reduce each of the following rational expressions to their lowest terms. 2 16 x − xv 2 2 − v − 24 4. 3. 2 v 2 y 3 y 18 − − 2 y 21 y 12 + − 30 48 6. 2 x 4 2 x 10 − − 4 x 24 + + 50 60 x 7. 3 k + 2 2 + 10 k + − 2 k 21 k 35 . 2 2 x x + + 5 4 xy xy − − 2 2 14 21 y y 9. 10. a 2 − a 3 12 − − 2 3 a 10 11. 12. Matthew simplified 2 2x 2 − − x + 6 x 21 in this way 13 ab ab )( + . Explain the error that Matthew has made and then simplify the expression correctly. Round Up Round Up If you were to condense everything from this Section into a couple of points, they would be: • Rational expressions are the same as fractions and are undefined if the denominator equals zero. • A rational expression can be reduced to its lowest equivalent fraction by dividing out common factors of the numerator and denominator. Section 8.1 — Rational Expressions 393 TTTTTopicopicopicopicopic 8.2.18.2.1 8.2.18.2.1 8.2.1 California Standards: Students Students 13.0: 13.0: Students add, subtract, 13.0: Students 13.0: Students 13.0: ultiplyyyyy, and divide rrrrraaaaational mmmmmultipl tional tional ultipl ultipl tional tional ultipl eeeeexprxprxprxprxpressions and functions essions and functions..... essions and functions essions and functions essions and functions e both e both Students solv Students solv Students solve both e both Students solv e both Students solv y and y and tionall tionall computa computa y and tionally and computationall y and tionall computa computa hallenging hallenging ptually cy cy cy cy challenging ptuall ptuall conce conce hallenging conceptuall hallenging conce ptuall conce prprprprproboboboboblems b y using these y using these lems b lems b y using these lems by using these y using these lems b hniques..... hniques hniques tectectectectechniques hniques What it means for you: You’ll multiply rational expressions by factoring and canceling. Key words: rational nonzero common factor Section 8.2 Multiplying Multiplying Multiplying Multiplying Multiplying Multiplying Multiplying Multiplying Multiplying Multiplying essions essions tional Expr tional Expr RRRRRaaaaational Expr essions tional Expressions essions tional Expr RRRRRaaaaational Expr essions essions tional Expr tional Expr tional Expressions essions tional Expr essions In Section 8.1 you learned about simplifying rational expressions. In this Topic you’ll learn to multiply rational expressions, but then you’ll use the same simplification methods to express your solutions in their simplest forms. essions essions tional Expr ying by Ry Ry Ry Ry Raaaaational Expr tional Expr ying b ying b Multipl Multipl essions tional Expressions Multiplying b essions tional Expr ying b Multipl Multipl Given any nonzero expressions m, c, b, and v mb cv m c = That is, the product of two rational expressions is the product of the numerators divided by the product of the denominators. These expressions can often be quite complicated, so simplify them as much as you can before multiplying. First, factor the numerators and denominators (if possible). Then find any factors that are common to both the top line (the numerator) and the bottom line (the denominator) and cancel them before multiplying. Example Example Example Example Example 11111 Simplify ab + 4 b 2 a 5 a ⋅ 10 + 2 a b b . Solution Solution Solution Solution Solution Step 1: Factor the numerators and denominators ) ) ⋅ Step 2: Cancel all the common factors and multiply 2a 394394394394394 Section 8.2 Section 8.2 Section 8.2 — Multiplyi
ng and Dividing Rational Expressions Section 8.2 Section 8.2 Example Example Example Example Example 22222 Multiply and simplify Solution Solution Solution Solution Solution Step 1: Factor the numerators and denominators if possible ( ) 33 2 2 a a )( 3 2 a )( − 2 a a )( ( − ⋅ − 2 1 )( ( − + + − ( ( = = a a ) ⋅ ⋅ Step 2: Cancel all the common factors and multiply − )( ( 2 a a − ⋅ − 1 2 1 )( ( = − 1 2 Guided Practice Multiply and simplify the rational expressions. 1. 14 15 mc ck ⋅ 30 28 kp pt + . 5 a ⋅ ab + 4 b b 10 + 2 a b 5. 2 a − v t p ⋅ tv −5 t 5 v 2. − ( p 2 ( + 2 )( p + 3 ) p 1 ) ⋅ 4. ( p + p + 1 )( 3 + p 4 ) 6 10 + + 11 3 4 You can extend this concept to the multiplication of any number of rational expressions, with any number of variables. Example Example Example Example Example 33333 Simplify 14 16 x − − x + 2 12 Solution Solution Solution Solution Solution Factor the numerators and denominators, and cancel all the common factors and multiply )( − + 1 x ( 2 ( )( −7 2x 2 or )( − x2 7 ) 1 3 )) 1 Section 8.2 Section 8.2 Section 8.2 — Multiplying and Dividing Rational Expressions Section 8.2 Section 8.2 395395395395395 Check it out: Either of these answers is acceptable. The second is the simpler of the two forms though. Example Example Example Example Example 44444 Simplify 12 15 ⋅ ax 2 + + a 8 15 a − + − a 3 4 x ⋅ 12 2 x x 2 + + x 9 18 + + . 6 x 9 Solution Solution Solution Solution Solution Factor the numerators and denominators, then cancel all the common factors and multiply. Check it out: You can leave your answer in either “factored” or “multiplied out” form — they’re both equally valid. = = 1 ( x 2 a ( + )( 3 x − 3 a )( + )( 3 )( + + )( 6 3 (( ) + ) )( − 3 )( a + 6 ) − 3 ) ax or 18 Guided Practice Multiply and simplify the rational expressions. 7 20 8 12 3 2 x x + + xy 3 + − 2 xy . 2 − x y − − xy ) Independent Practice Multiply and simplify the rational expressions ab + − x + − 2 x 2 + ab 5 + 2 ab 7 + a 6 + a 3 2 3. 5. 2 a 2 b 2 + − ab 2 b + − 3 ab + − ab 6 a − + 3 ab + ak + ak 27 + + 6 x 60 ) 8 ⋅ 6 ) 12 ( ( x x + + )( 1 )(( 15 36 18 20 42 ⋅ 2 x 7. 3 − 2 x + 10 x 9 + 24 ound Up ound Up RRRRRound Up ound Up ound Up RRRRRound Up ound Up ound Up ound Up ound Up Usually the most difficult thing when solving problems like these is factoring the numerators and denominators. Look for “difference of two squares” expressions, perfect squares, and minus signs that you can factor outside the parentheses. 396396396396396 Section 8.2 Section 8.2 Section 8.2 — Multiplying and Dividing Rational Expressions Section 8.2 Section 8.2 TTTTTopicopicopicopicopic 8.2.28.2.2 8.2.28.2.2 8.2.2 California Standards: Students Students 13.0: 13.0: Students add, subtract, 13.0: Students 13.0: Students 13.0: multiply, and dididididivide r vide raaaaational tional tional vide r vide r tional tional vide r eeeeexprxprxprxprxpressions and functions essions and functions..... essions and functions essions and functions essions and functions e both e both Students solv Students solv Students solve both e both Students solv e both Students solv y and y and tionall tionall computa computa y and tionally and computationall y and tionall computa computa ptually cy cy cy cy challenging hallenging hallenging ptuall ptuall conce conce conceptuall hallenging conce hallenging ptuall conce prprprprproboboboboblems b y using these y using these lems b lems b y using these lems by using these y using these lems b hniques..... hniques hniques tectectectectechniques hniques What it means for you: You’ll divide rational expressions by factoring and canceling. Key words: rational reciprocal common factor viding viding DiDiDiDiDividing viding viding viding viding DiDiDiDiDividing viding viding essions essions tional Expr tional Expr RRRRRaaaaational Expr essions tional Expressions essions tional Expr RRRRRaaaaational Expr essions essions tional Expr tional Expr tional Expressions essions tional Expr essions Dividing by rational expressions is a lot like multiplying — you just have to do an extra step first. ocal ocal ecipr ecipr y the R y the R ying b ying b viding is the Same as Multipl DiDiDiDiDividing is the Same as Multipl viding is the Same as Multipl ocal eciprocal y the Recipr ying by the R viding is the Same as Multiplying b ocal ecipr y the R ying b viding is the Same as Multipl Given any nonzero expressions m, c, b, and v: m ÷ = ⋅ = c mv cb m c b v v b That is, to divide m c by b v , multiply by the reciprocal of b v . You can extend this concept to the division of any rational expression. Don’t forget: See Topic 1.2.2 for more on reciprocals. Suppose you pick a number such as 10 and divide by, say, 1 2 The question you’re trying to answer is “How many times does 1 2 . go into 10?” or “How many halves are in 10?” Division Equivalent to 10 ÷ 1 2 = 20 Check it out: The middle numbers in these two columns are reciprocals of each other. 10 ÷ 10 ÷ 10 ÷ 1 3 1 4 1 n = 30 = 40 10 × 2 = 20 10 × 3 = 30 10 × 4 = 40 = 10n 10 × n = 10n So, 10 divided by a fraction is equivalent to 10 multiplied by the reciprocal of that fraction. Dividing anything by a rational expression is the same as multiplying by the reciprocal of that expression. So you can always rewrite an expression a ÷ b in the form a (where b is any nonzero expression). ⋅ =1 b a b Section 8.2 Section 8.2 Section 8.2 — Multiplying and Dividing Rational Expressions Section 8.2 Section 8.2 397397397397397 actorsssss actor actor ou Should Still Cancel Common F YYYYYou Should Still Cancel Common F ou Should Still Cancel Common F ou Should Still Cancel Common Factor actor ou Should Still Cancel Common F Example Example Example Example Example 11111 2 − k 25 2 k ÷ (k + 5). Simplify Solution Solution Solution Solution Solution 2 − k 25 2 k ÷ (k + 5) can be written as 2 − k 25 2 k ÷ k + 5 1 Rewrite the division as a multiplication by the reciprocal of the divisor. = 2 − k 25 2 k 1 k + · 5 Factor as much as you can. − ( k = )( Cancel any common factors between the numerators and denominators. − ( k = 5 )( Check your answer. Multiply your answer by (k + 5): = )( 25 2 k Example Example Example Example Example 22222 Simplify Solution Solution Solution Solution Solution Rewrite the division as a multiplication by the reciprocal of the divisor Factor all numerators and denominators. − − + − = − ⋅ )( 2 2 )( ) Cancel any common factors between the numerators and denominators )( 2 )( 398398398398398 Section 8.2 Section 8.2 Section 8.2 — Multiplying and Dividing Rational Expressions Section 8.2 Section 8.2 ✓ Guided Practice Divide and simplify each expression. 1. 2 2 2 b c d 3 bcd ÷ 2 bcd abc 3 10 ÷ 2 x − − x 4 − 2 25 )( b − 1 ) + ( b 2. 4 Check it out: It makes a difference which order you do the divisions because division is not associative, which means: (a ÷ b) ÷ c π a ÷ (b ÷ c) Check it out: In practice, this means that you can rewrite each division as a multiplication by the reciprocal of the divisor. At Once At Once essions essions Expr Expr vide Long Strings of vide Long Strings of ou Can Di YYYYYou Can Di ou Can Di At Once essions At Once Expressions vide Long Strings of Expr ou Can Divide Long Strings of At Once essions Expr vide Long Strings of ou Can Di Just like multiplication, you can divide any number of rational expressions at once, but it makes a big difference which order you do things in. If there are no parentheses, you always work through the calculation from left to right, so that Example Example Example Example Example 33333 Simplify . Solution Solution Solution Solution Solution Rewrite each division as a multiplication by the reciprocal of the divisor + − 22 Factor all numerators and denominators )( + )( ( )( 1 x x + 1 ) Cancel any common factors between the numerators and denominators )( + )( + )( Section 8.2 Section 8.2 Section 8.2 — Multiplying and Dividing Rational Expressions Section 8.2 Section 8.2 399399399399399 Parentheses override this order of operations, so you need to simplify any expressions in parentheses first: ⎛ c ⎜⎜⎜ ÷ ÷ ⎝ d a b ⎞ ⎟⎟⎟⎟ ⎠ Guided Practice − + − 3 k − 2 10 k 2 k Divide and simplify each expression 14 ⎜⎜⎜ ⎝ ) ( 12 14 + − )( + ) 3 ÷ 8 18 x ( x ( 6. 2 7. 8. ÷ 6 − 2 ⎞ ⎟⎟⎟⎟ ⎠ ) 4 ) 1 x + + 2 2 − 36 the Same TimeTimeTimeTimeTime t the Same t the Same vide a vide a y and Di y and Di ou Can Multipl YYYYYou Can Multipl ou Can Multipl vide at the Same y and Divide a ou Can Multiply and Di t the Same vide a y and Di ou Can Multipl Say you have an expression like this to simplify: a b c ÷ × d e f Again, you work from left to right, and anywhere you get a division, multiply by the reciprocal, so Example Example Example Example Example 44444 Simplify 2 p 2 p 2 + − 2 pq q − − 2 pq 3 q 2 × 2 p pq − + − pq 3 pq + 2 q . Solution Solution Solution Solution Solution Rewrite any divisions as multiplications by reciprocals. = 2 p 2 p 2 + − pq 2 q − − 2 pq 3 q 2 × 2 p pq − + pq pq + 2 q Factor all numerators and denominators. − + 2 q p )( 3 q p )( q p )( + )( q ) Cancel any common factors between the numerators and denominators. − p q p ( q p )( + )( q p 3 q p )( − + 400400400400400 Section 8.2 Section 8.2 Section 8.2 — Multiplying and Dividing Rational Expressions Section 8.2 Section 8. )( 1 q ) 1 our Stepspspspsps our Ste ustify All All All All All YYYYYour Ste our Ste ustify ustify ou Can J e Sure e e e e YYYYYou Can J ou Can J e Sur MakMakMakMakMake Sur e Sur ou Can Justify our Ste ustify ou Can J e Sur Example Example Example Example Example 55555 21 ÷ 2 a t + 2 2 a t + − at − at 12 t 14 t 2 = − a 2 + . 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r Distrib Distrib opertytytytyty oper oper e pre proper e pre pr Distributiutiutiutiutivvvvve pr Distrib Distrib oper Distrib Distrib CommCommCommCommCommutautautautautatititititivvvvve and e and e and e and e and associa associa associatititititivvvvve pr e pre proper e pre pr oper oper opertiestiestiestiesties associa associa oper ofofofofof m m m m multiplica ultiplica ultiplica ultiplicationtiontiontiontion ultiplica InInInInInvvvvverererererse and se and se and se and se and identity pr identity pr oper oper opertiestiestiestiesties identity proper identity pr identity pr oper Distrib Distrib Distributiutiutiutiutivvvvve pr e pre proper e pre pr oper oper opertytytytyty Distrib Distrib oper 21 − + + − 221 a 4 − − )( ( a a 1 2 − + )( ( )( 1 − a )( − at 12 14 + − t at + − a 6 + − 2 a ) 77 1 ) − a 1 )( 7 ) +11 a ) )( 1 − − ) 3 ) 3 )( )( ( + − − − )( 1 )( = Guided Practice Simplify these rational expressions. 2 t 10 12 11 14 21 ÷ 12 10 × 15 Section 8.2 Section 8.2 Section 8.2 — Multiplying and Dividing Rational Expressions Section 8.2 Section 8.2 401401401401401 Independent Practice Divide and simplify each expression. 2 2 − k + − m km − + km 3 2 2 m 2. 2 2 k 3 16 − 2 x 4. 6. 2 m ÷ − mv 10 10 11 12 3 ÷ 2 − x 16 + + 12. 2 y 13 Simplify these rational expressions. 14 15. − + 2 v 2 − vw 4 + vw vw + − vw 2 vw 55 4 vw vw − + w 3 w 8 2 2 4 16. 17. m 2 + 2 m n 2 + mn 2 − 3 mn mn 5 mn + − + mn 2 − 3 mn n 2 ÷ ⎛ ⎜⎜⎜⎜ ⎝ mn 5 mn + − mn 3 mn 2 + n − 22 n ÷ 2 − + 2 mn mn mn 3 mn 2 + n − 22 n ⎞ ⎟⎟⎟⎟ ⎠ 2 ÷ − + 2 mn mn 2 n 2 ound Up ound Up RRRRRound Up ound Up ound Up ound Up ound Up RRRRRound Up ound Up ound Up It’s really important that you can justify your work step by step, because division of rational expressions can involve lots of calculations that look quite similar. 402402402402402 Section 8.2 Section 8.2 Section 8.2 — Multiplying and Dividing Rational Expressions Section 8.2 Section 8.2 Topic 8.3.1 California Standards: 13.0: Students add, subtract, multiply, and divide rational expressions and functions. Students solve both computationally and conceptually challenging problems by using these techniques. What it means for you: You’ll add and subtract rational numbers with the same denominators. Key words: common denominator rational common factor Section 8.3 Fractions with Fractions with Identical Denominators Identical Denominators In Section 8.2 you dealt with multiplying and dividing rational expressions, so you can probably guess that adding and subtracting is next. First up are fractions with the same denominators, which are the easiest kind to add and subtract. Common Denominators Make Things Easier To add or subtract two rational expressions with the same denominator, just add or subtract the numerators, then put the result over the common denominator. Here’s a simple numerical example Reduce your answer to its simplest form. Add the numerators and divide the answer by the common denominator. Subtract the numerators and divide the answer by the common denominator. In general, given any expressions m, c, and v, where v is nonzero and Example 1 Simplify . Solution Add the numerators and divide by the common denominator Section 8.3 — Adding and Subtracting Rational Expressions 403 Example 2 Simplify , justifying each step. Solution Subtract the numeratorsk t = Guided Practice Simplify each expression. Divide the difference by the common denominator, t Use distributive property to simplify –(2k – 5) Commutative and associative properties of addition Combine like terms + 3 x + 4 x 2. 5 2 − 3 −. 3. 5 Cancel Any Common Factors Example 3 Add and simplify m 5 4 − m+ 7 4 4 . Solution Add the numerators and divide by the common denominator = 12 m 4 1 m( 4 3 1 4 = 3m – 1 = − ) 1 Factor the numerator and cancel any common factors 404 Section 8.3 — Adding and Subtracting Rational Expressions Example 4 Subtract and simplify Solution Subtract the numerators and divide by the common denominator )( Guided Practice Simplify each expression. 7. 3 + + 12 10. 2 4 b 9 Independent Practice Simplify each expression. 1 + + 11 x + + 2 4 x 28 3 2 35 9 x 2 3 + + 18 13 + + 4 17 12 − 12 6. ⎛ ⎜⎜⎜ ⎝ − 1 2 x − 2 x 4 1 ⎛ ⎞ ⎜⎜⎜⎜ ⎟⎟⎟ ⎠ ⎝ ⎞ ⎟⎟⎟⎟⎟ ⎠ ⎛ ⎜⎜⎜⎜ ⎝ 7 ⎞ ⎛ ⎟⎟⎟⎟ ⎜⎜⎜⎜ ⎠ ⎝ ( 23 10 20 + + xx + 23 ⎟⎟⎟⎟20 ⎠ Round Up Round Up When subtracting algebraic fractions, be sure to distribute the minus sign over the whole parenthesis that follows it, changing any plus signs to minus signs and vice versa. And make sure you reduce your answer to its simplest form — look for “difference of two squares” patterns. Section 8.3 — Adding and Subtracting Rational Expressions 405 Topic 8.3.2 California Standards: 13.0: Students add, subtract, multiply, and divide rational expressions and functions. Students solve both computationally and conceptually challenging problems by using these techniques. What it means for you: You’ll add and subtract rational numbers with different denominators. Key words: common denominator rational common factor least common multiple Don’t forget: See Topic 1.3.4 for more on equivalent fractions. Fractions with Fractions with Different Denominators Different Denominators For you to be able to add or subtract two rational expressions, they must have a common denominator. If the two fractions you start with have different denominators, you need to find equivalent fractions with a common denominator before you can add or subtract them. You Need to Find a Common Denominator In general, you can find a common denominator by simply multiplying the denominators together. You then convert each fraction to an equivalent fraction with this common denominator. Example 1 Add 2 3 1 + . 5 Solution Multiply the denominators to get a common denominator (3◊5). Convert each fraction into an equivalent fraction with the common denominator and respective numerators 2◊5 and 1◊3 10 = + 15 3 15 Once you have two fractions with the same denominator, add the numerators and divide by the common denominator. = 13 15 So — given any expressions m, c, b, and v, where c and v are nonzero and b c − = v c mv + bc cv mv − bc cv 406 Section 8.3 — Adding and Subtracting Rational Expressions Example 2 Subtract . Solution Find a common denominator: 4 × 5 = 20 Convert each fraction to an equivalent fraction with 20 as the denominator ) = 20 5 − m 20 and ) 12 = 20 + m 20 Don’t forget: See Topic 1.3.6 for more on the LCM. In Examples 1 and 2, the LCM of the denominators is simply their product. Subtract the equivalent fractions. 20 − m 20 5 − 12 + m 20 20 = ( 20 m + m 20 ) − − ) 5 ( 12 20 20 m − − 5 12 20 − m 20 Distribute the minus sign and simplify 25 8 m− 20 = = This method works well with fairly simple examples like these, but when you have more complicated problems to deal with, it can become a lengthy process. A better alternative is to find the least (or lowest) common multiple — the LCM. Guided Practice Use the method of multiplying denominators to simplify the following. 1. 4 − 16 12 Simplify each expression by first finding the LCM of the denominators. 5. 2 c − − c 5 8 24 + 2 8 c 3 + − c 2 12 2 3 y + 6. 17 10 + y 20 10 2 2 x 9. + + x 15 + 14 10 + 11 5 Section 8.3 — Adding and Subtracting Rational Expressions 407 Finding the Least Common Multiple (LCM) The least common multiple of two or more denominators is the smallest possible number (or simplest expression) that’s divisible by all denominators. To find the LCM: Factor the denominators completely. Multiply together the highest power of each factor that appears in any of the denominators. Any factor that appears in more than one denominator should only be included once in the LCM. Using the LCM as your common denominator makes the problem as simple as possible. Adding or Subtracting Fractions with Different Denominators When adding or subtracting fractions: 1) Find the least common multiple of the denominators. 2) Convert each fraction into an equivalent fraction with the LCM as its denominator. 3) Add or subtract the numerators and divide by the common denominator (LCM). 4) Factor both the numerator and denominator if possible, and cancel any common factors. Check it out: The highest power of x is x2. The highest power of y is y. The highest power of (x + 1) is (x + 1)2. So the LCM is x2y(x + 1)2. Example 3 Simplify − Solution Step 1: Find the LCM of x2y(x + 1) and x(x + 1)2: LCM = x2y(x + 1)2 Step 2: Convert each fraction to an equivalent fraction with x2y(x + 1)2 as the denominator )( )( ( xy ⋅ ( xy ( xy 408 Section 8.3 — Adding and Subtracting Rational Expressions Example 3 continued Step 3: Add the equivalent fractions. − y x 2 )( + + + xy y x )( 1 ) )+ 21 2 ( xx y x xy = xy − − + + y 2 ( x y x 2 + xy 2 ) 1 + ( 5 x ) Guided Practice Simplify each expression. 10. 4 − 6 2 k + 12 3 − 4 k 12 ) ( 11. 13 Add Lots of Fractions Using a Common Denominator If you want to add or subtract more than two fractions at once, you have to put all the numerators over a common denominator. Example 4 Simplify Solution Step 1: Find the LCM of 2x + 6, 4x – 4, and x2 + 2x – 3. Factor each denominator: 2x + 6 = 2 · (x + 3) 4x – 4 = 4 · (x – 1) = 22 · (x – 1) x2 + 2x – 3 = (x – 1)(x + 3) LCM = 4(x – 1)(x + 3) Section 8.3 — Adding and Subtracting Rational Expressions 409 Check it out: The highest power of 2 is 22 = 4. The highest power of (x – 1) is (x – 1). Highest power of (x + 3) is (x + 3). So the LCM is 4(x – 1)(x + 3). Example 4 continued Step 2: Convert each fraction to an equivalent fraction with 4(x – 1)(x + 3) as the denominator − ( 10 x − )( − )( )( )( + x 1 ) + = + 20 x ( x− x )( 1 1 ) + ( 44 ) 3 ) 3 Step 3: Add or subtract the equivalent fractions. − ) 3 ( 4 + x ( 20 − x 1 )( ) 1 + ) 3 x == = = = − 20 + + x ( 4 − 4 x 1 )( ( + − ( 10 − ) 1 x ( 10 + − 1 x x )( − + ) 1 ( 4 x ) 3 10 ( 4 ( 4 x x 12 x 20 ) 3 4 x ( − x )( 1 20 ( ) 3 − + + − x 4 10 + x −− ( 4 )( 1 − − x 18 6 + − x x 1 )( − + x ( 6 − )( − x )( 1 ( 22 ) Guided Practice Simplify each expression. 14 16 15 17 18 Check it out: Simplify your answer as much as possible —
factor the numerator and denominator and cancel any common factors. 19 410 Section 8.3 — Adding and Subtracting Rational Expressions Independent Practice Simplify each expression. 1 16 + − x 4 5 2 a 3 10 − 11 − 10 ab 5 2y 6. − + xy x − y 1 − + 1 xy 2 y 7 −( 1 ) t 2 9. 2 t 4 x + x 5 15 − − + 1 3 x x 8. 10. 2 y 3 + + 7 y 12 − 2 + − 3 y 4 2 y 11 12 13 14 15. y 2 + 2 x xy − 2 3 y xy + 16. 2 k 17 18. 2 x 19 20 21 22 Round Up Round Up Remember — you can only add or subtract fractions when they’ve got the same denominators. That means that the first step when you’re dealing with fractions with different denominators is to find a common denominator. Then you can use the same methods you saw in Topic 8.3.1. Section 8.3 — Adding and Subtracting Rational Expressions 411 Topic 8.4.1 Section 8.4 Solving Fractional Equations Solving Fractional Equations California Standards: 13.0: Students add, subtract, multiply, and divide rational expressions and functions. Students solve both computationally and conceptually challenging problems by using these techniques. What it means for you: You’ll solve equations involving rational numbers. Key words: common denominator rational common factor limitation undefined least common multiple This Topic uses all the skills that you’ve learned in the earlier Topics in this Section — but now you’ll be dealing with equations rather than just expressions. There are Limitations on Fraction Equation Solutions A fractional equation is undefined when the denominator of any of its expressions equals zero. This means that there are limitations on the possible solutions of a fractional equation. For example, for the fractional equation 1 − x 3 1 = 2 + the limitations 1 x on x are that 3x – 1 π 0 and x + 1 π 0, that is x πππππ 1 3 and x π π π π π –1. Guided Practice State the limitations on the possible solutions of these equations. 1. 5 − − x 6 2x = 1 x 2 13 . 2 x Solve Using the LCM of the Denominators When solving fractional equations where the rational expressions have different denominators, multiply the entire equation by the least common multiple (LCM) of the denominators. This allows all the denominators to be canceled out, which makes the equation easier to solve. Example 1 Solve the fractional equation + Solution To solve for x, find the least common multiple of the denominators. In this example, the LCM of the denominators is simply the denominators multiplied together, i.e. 8(3x – 1)(x + 1). 412 Section 8.4 — Solving Equations with Fractions Example 1 continued Multiply the entire equation by the LCM of the denominators, then divide out the common factors, which gives − )( )( − )( 1 1 + 1 ) 1 x Multiplying by the LCM of the denominators means that all the denominators in the equation cancel, leaving: 3(3x – 1)(x + 1) + 8(x + 1) = 2·8(3x – 1) (9x2 + 6x – 3) + (8x + 8) = 48x – 16 9x2 + 14x + 5 = 48x – 16 9x2 – 34x + 21 = 0 (9x – 7)(x – 3) = 0 x = 7 9 or x = 3 Guided Practice Check it out: This is a quadratic equation, so you could use the quadratic formula (see Topic 7.3.1) to work out the solutions instead of factoring. Check it out: Make sure that these solutions are allowed — that x π 1 3 and x π –1. Solve each of the following fractional equations. − 3 2 7. 3 6. = − − + 1 1 x x 10 − = x x 5 6 2 3 8. a − = 1 5 + a 3 10. 5 x + = 1 11 − x 1 11 Always Check that the Solutions Work It’s a good idea to check the solutions you have worked out to make sure that they are correct. The best way to check a solution is to put it back into the original equation and verify that both sides of the equation are equal. In the problem on the last page, the solutions for the fractional equation + were found to be x = 7 9 or x = 3. Check these solutions by putting them back into the equation, one at a time. Section 8.4 — Solving Equations with Fractions 413 Check it out: Remember that dividing a number by a fraction is the same as multiplying that number by the reciprocal of the fraction, for example: ÷ = × = a a c b ac b b c Example 1 continued First, check 16 16 7 9 is valid. Both sides are equal, which means that the solution x = Now, check x = 3 = Both sides are equal, which means that the solution x = 3 is correct. If you put the solution into the equation and find that the two sides are not equal, then your solution is incorrect. That means you’ll have to go back and check each stage of your work. Example 2 Solve = − , first stating any values of x for which the 1 equation is undefined. Solution First take a look at the equation and figure out the limitations. The equation is undefined when either 2x + 1 = 0 or x – 1 = 0. This happens when x = – 1 2 or x = 1. So the limitations on the solution are that x cannot equal – 1 2 or 1. Step 1: Find the least common multiple of the denominators: 2x + 1 and x – 1. The LCM for these denominators is (2x + 1)(x – 1). Step 2: Now, multiply both sides of the equation by (2x + 1)(x – 1) to eliminate the denominators from the rational expressions − )( )( 1 1 = − ⋅ ( 1 2 x + )( 1 x − ) 1 414 Section 8.4 — Solving Equations with Fractions Check it out: The whole equation can be divided by the common factor 2. Check it out: Make sure both solutions are allowed, that is x π – 1 2 , x π 1. Check it out: Both of these solutions are correct because the equation balances when the solutions are substituted for x. Example 2 continued Step 3: Reduce the equation to its lowest terms: 3·(x – 1) – 4·(2x +1) = –1·(2x + 1)(x – 1) 3x – 3 – 8x – 4 = –1·(2x2 – x – 1) –5x – 7 = –2x2 + x + 1 2x2 – 6x – 8 = 0 x2 – 3x – 4 = 0 Step 4: Then factor to find the solutions: (x – 4)(x + 1) = 0 So, x – 4 = 0 or x + 1 = 0 So the solutions are: x = 4 or x = –1 Now, check your solutions to make sure that they work. Put x = 4 into the equation: Put x = –1 into the equation Guided Practice Solve each of the following fractional equations. 12 36 + − 2 x 8 2 x 13 +( 2 c 1 ) 2 c 14. + 4 15 15 12 16 Section 8.4 — Solving Equations with Fractions 415 Independent Practice Solve each of the following fractional equations. 3. y 3 y 5. 6 1 + = x 12 2 x 4. Solve for y the following equation: − 1 1 7. Use your solution to Exercise 6 to find the value of y2 – 1. = − 2 3 y y 8 1 + 10 y − y 2 + + 8. Find the value of 5k – 7 if . One number is 8 less than another. The quotient formed by dividing the higher number by the lower number is Find the values of the two numbers. 15 11 . 10. One integer is 5 less than another. Find the numbers if the sum of their reciprocals is 17 66 . 11. The denominator of a fraction is 3 more than its numerator. If the sum of the fraction and its reciprocal is 29 10 , find the fraction. 12. The denominator of a fraction is 2 more than the numerator. A second fraction is created by subtracting 2 from the numerator and subtracting 3 from the denominator of the first fraction. Find the two fractions if the first fraction subtracted from the second fraction is 5 88 . 13. Solve the following system of equations by first rearranging them so that they are not in fractional form 15 x + 18 = 11 x + + 4 y 3 2 y Round Up Round Up The main thing to learn here is that fractional equations can be solved by first multiplying the entire equation by the least common multiple of the denominators, then canceling all common factors. The equation you’re left with can then be solved using methods from earlier Chapters. Remember to check that all your solutions actually work by putting them into the original equation. 416 Section 8.4 — Solving Equations with Fractions Topic 8.4.2 California Standards: 13.0: Students add, subtract, multiply, and divide rational expressions and functions. Students solve both computationally and conceptually challenging problems by using these techniques. What it means for you: You’ll model real-life situations and solve equations involving rational numbers. Key words: common denominator rational common factor limitation undefined least common multiple Don’t forget: Remember: Time = Distance Speed Check it out: You need to convert the time from minutes to hours because in the problem, speed is measured in miles per hour. Applications of Applications of Fractional Equations Fractional Equations Now it’s time to solve some more fractional equations. This time, the problems are about real-life situations, so you’ll have to model them as equations before you can start with the math working. You Might Be Given a Word Problem to Solve Here’s one long example showing a model of a real-life problem. Example 1 In a bike race, Natasha cycles along flat ground for 4 miles against a 2 mile per hour wind. After four miles, she turns around and follows the same route back to the start, this time with the wind behind her. If the journey takes 50 minutes, find how fast Natasha would travel if there were no wind. (Assume that she would travel at a constant speed without a wind.) Solution Step 1 — Write the equation. The time Natasha takes to complete the race, can be written as: Timethere + Timeback = 50 minutes Using this, an equation for Natasha’s race time can be written in terms of distance and speed. Distance Speed there there + Distance Speed back back = Time pp total tri 4 − s 2 + 4 + s 2 50 = = 60 5 6 Where: s = Natasha’s speed in no wind s – 2 = Natasha’s speed against the wind s + 2 = Natasha’s speed with the wind behind her s π ±2 (for the rational expressions to be defined) Section 8.4 — Solving Equations with Fractions 417 Example 1 continued Step 2 — Multiply by the LCM. The LCM of the denominators is 6(s – 2)(s + 2). Multiply the equation by the LCM of the denominators to give: 24(s + 2) + 24(s – 2) = 5(s – 2)(s + 2) 24s + 48 + 24s – 48 = 5s2 – 20 5s2 – 48s – 20 = 0 Step 3 — Factor and Solve for x. Now, factor the equation and solve for x: Don’t forget: Remember to check that the solutions work when put back into the original equation. (5s + 2)(s – 10) = 0 So, s = – 2 5 or s = 10 So Natasha’s speed if there were no wind would be 10 miles per hour. Although – 2 5 is a valid solution of
the algebraic problem, it isn’t a correct answer for this example because speed can only have a positive value. Guided Practice 1. On Monday, a distribution company shipped a load of oranges in crates, with a total weight of 124 lb. On Tuesday it shipped another load of oranges, also with a total weight of 124 lb. However, on Tuesday there was one crate fewer than on Monday, so each crate was 1 8 2. Rose spent $2.40 on pens. If each pen had cost 4 cents more, she would have been able to buy 10 fewer pens for the same money. How many pens did Rose buy? lb heavier. How many crates were shipped on Monday? 3. Dajanique bought x boxes of candy for a total of $1.26. She kept four boxes and sold the rest for a total of $1.40. If she sold each box for 3 cents more than it cost her, how many boxes did she buy? 4. A teacher spent $8.40 on sets of chapter tests. If each set of tests had been 2 cents less, the teacher would have gotten two extra sets for the same price. How many sets did the teacher get? 418 Section 8.4 — Solving Equations with Fractions Independent Practice 1. Vanessa bought a set of military medals through an antique dealer for $120. She gave two of the medals to her father and resold the rest, charging $4 more per medal than she paid. If Vanessa sold the medals for a total of $156, how much did she charge for each medal? 2. Hearst Castle is 180 miles from Ventura by road. Two coaches leave Ventura at the same time, both heading for Hearst Castle, but one averages 5 mph faster than the other. If the faster coach reaches Hearst Castle half an hour earlier than the slower coach, what is the average speed of the faster coach? 3. A cyclist completes a 150-mile race in a certain amount of time. She completes another 150-mile race a month later, but this time it takes an hour longer to cover the same distance and her average speed is 5 mph less than in the first race. Find the average speed for the cyclist during her first race. 4. Juan’s band produced a number of CDs to sell at a gig. The batch of CDs cost $170.00 to make and each CD was sold for $2.50 more than it cost to produce. Juan gave 2 CDs to his friends, but sold the rest for a total of $198.00. How many CDs did Juan produce? 5. Chen bought a bag of groceries weighing 15 pounds. His friend, Jo, bought a bag of groceries that also weighed 15 pounds, but contained one less item. The average weight per item for Jo’s groceries was ½ pound more than for Chen’s. How many items were in Jo’s grocery bag? 6. José ordered a box of fruits for his market stand costing $6. When his order arrived, José discovered that 20 fruits were rotten and threw them in the trash. He sold all of the remaining fruits for a total of $8, charging 3 cents more for each fruit than he paid for it. How many fruits did José order to begin with? 7. A wholesaler bought a batch of T-shirts for $77.00. She gave two of the T-shirts to her daughters and then sold the rest for a total of $90. If the wholesaler sold each T-shirt for $2 more than it cost her, how much did she pay for each T-shirt? Round Up Round Up The only difference between solving word problems and answering straightforward algebra questions is that you have to write the equations yourself for a word problem. After that, they are solved in exactly the same way. Section 8.4 — Solving Equations with Fractions 419 Topic 8.5.1 Section 8.5 Relations Relations California Standards: 16.0: Students understand the concepts of a relation and a function, determine whether a given relation defines a function, and give pertinent information about given relations and functions. 17.0: Students determine the domain of independent variables and the range of dependent variables defined by a graph, a set of ordered pairs, or a symbolic expression. What it means for you: You’ll find out what relations are, and some different ways of showing relations. Key words: relation ordered pair domain range Check it out: In an ordered pair (x, y), the first variable, x, is usually called the independent variable, and the second, y, is the dependent variable — it depends on the first. For example, in the relation y = 4x2 – 3x, the values of y in the range depend on the values of x in the domain. Check it out: The order you write the elements of the domain or range isn’t important, so you could write the range from Example 1 as {2, 4, 5, 7, 8}. Also, if an element is connected to more than one element in the other set, you only need to write it down once. Relations in Math are nothing to do with family members. They’re useful for describing how the x and y values of coordinate pairs are linked. A Relation is a Set of Ordered Pairs Before you can define a relation, you need to understand what “ordered pairs” are: An ordered pair is just two numbers or letters, (or anything else) written in the form (x, y). If x and y are both real numbers, ordered pairs can be plotted as points on a coordinate plane, where the first number in the ordered pair represents the x-coordinate and the second number represents the y-coordinate. y This point represents the ordered pair (–2, 1). x 1 2 3 3 2 1 –1 –1 –2 –3 –3 –2 A relation is any set of ordered pairs. Relations are represented using set notation, and can be named using a letter: for example: m = {(1, 4), (2, 8), (3, 12), (4, 16)}. Every relation has a domain and a range. Domain: the set of all the first elements (x-values) of each ordered pair, for example: domain of m = {1, 2, 3, 4} Range: the set of all the second elements (y-values) of each ordered pair, for example: range of m = {4, 8, 12, 16} An important point to note is that there may or may not be a reason for the pairing of the x and y values. Looking at the relation m, above, you can see that the x and y values are related by the equation y = 4x — but not all relations can be described by an equation. Example 1 State the domain and range of the relation r = {(1, 4), (3, 7), (3, 5), (5, 8), (9, 2)}. Solution Domain = {1, 3, 5, 9} Range = {4, 7, 5, 8, 2} 420 Section 8.5 — Relations and Functions Example 2 State the domain and range of the relation f = {(a, 2), (b, 3), (c, 4), (d, 5)}. Solution Domain = {a, b, c, d} Range = {2, 3, 4, 5} Ordered pairs are not the only way to represent relations. Guided Practice State the domain and range of each relation. 1. f(x) = {(1, 1), (–2, 1), (3, 5), (–3, 10), (–7, 12)} 2. f(x) = {(–1, –1), (2, 2), (3, –3), (–4, 4)} 3. f(x) = {(1, 2), (3, 4), (5, 6), (7, 8)} 4. f(x) = {(a, b), (c, d), (e, f), (g, h)} 5. f(x) = {(–1, 0), (–b, d), (e, 3), (7, –f)} 6. f(x) = {(a, –a), (b, –b), (–c, c), ( 1 2 , –j)} Mapping Diagrams Can Be Used to Represent Relations One way to visualize a relation is to use a mapping diagram. In the diagram, the area on the left represents the domain, while the area on the right represents the range. The arrows show which member of the domain is paired with which member of the range. This mapping diagram represents the relation t = {(2, v), (3, c), (6, m)}. 2 3 6 t c v m Domain Range Guided Practice State the domain and range of each relation. 7. 8. 2 4 6 8 10 12 4 36 100 16 64 144 1 2 –3 0 –1 9. a e i o u –8 –6 5 12 10 Section 8.5 — Relations and Functions 421 You Can Use Input-Output Tables In input-output tables, the input is the domain and the output is the range. This table represents the relation {(1, 1), (2, 3), (3, 6), (4, 10), (5, 15)}. Input Output 1 2 3 4 5 1 3 6 10 15 Guided Practice State the domain and range of each relation. 11. Input Output Input Output 12. 1 5 12 32 6 0.3 –2 –1 2 3 8 5 8 13 13. Input Output –3 –1 0 1 –26 0 1 2 Relations Can Be Plotted as Graphs Relations can be plotted on a coordinate plane, where the domain is represented on the x-axis and the range on the y-axis. Graphs are most useful when you have continuous sets of values for the domain and range, so that you can connect points with a smooth curve or straight line. This graph represents the relation {(x, y = x)} with domain {–2 £ x £ 8}. y 8 6 4 2 –2 2 4 6 8 x –2 Guided Practice State the domain and range of each relation. 14. y (–2, 3) 15. y (3, 9) x (2, –3) (–9, –3) x Check it out: The filled circles mean that the domain includes the values –2 and 8. A hollow circle would mean that the value was not included, and an arrowhead would mean that the domain continued to infinity in that direction. So, for example: y domain = {–2 < x < 8} 8 6 4 2 –2 2 4 6 8 –2 y ∞ domain = {– 8 6 ∞x < < } 4 2 –2 2 4 6 8 –2 x x 422 Section 8.5 — Relations and Functions Independent Practice Define each of the following terms. 1. Relation 2. Range 3. Domain In Exercises 4–15, state the domain and range of each relation. 4. {(x, x2) : x Œ {–1, 2, 4}} 5. y (0, 2) 6. y (0, 1) (–2, 0) x (2, 0) (–1, 0) x (1, 0) (0, –2) (0, –1) ( ) 7. f x = {( , { , . , }} 8. f(x) = {(x, x2 – 1) : x Œ {0, 1, 2}} 9. f(x) = {(x, –x2 + 3) : x Œ {–2, –1, 0, 1, 2}} 10. f(x) = {(x, 11. f(x) = {(x, x x − {–1, 0, 2}} ) : x Œ {–1, 0, 2, 3}} 12. y (–2, 4) 13. (–4, 6) y (3, 2) x (–3, –2) (4, 2) x (2, –2) 14 15. 1 –3 a e –1 4 d a b Round Up Round Up The important thing to remember is that a relation is just a set of ordered pairs showing how a domain set and a range set are linked. Section 8.5 — Relations and Functions 423 Topic 8.5.2 California Standards: 16.0: Students understand the concepts of a relation and a function, determine whether a given relation defines a function, and give pertinent information about given relations and functions. 18.0: Students determine whether a relation defined by a graph, a set of ordered pairs, or a symbolic expression is a function and justify the conclusion. What it means for you: You’ll find out what functions are, and you’ll say whether particular relations are functions. Key words: function relation ordered pair domain range Functions Functions A function is a special type of relation. Functions Map from the Domain to the Range A relation is any set of ordered pairs — without restriction. A function is a type of relation that has the f
ollowing restriction on it: A function is a set of ordered pairs (x, y) such that no two ordered pairs in the set have the same x-value but different y-values. That is, each member of the domain maps to a unique member of the range. Example 1 Determine whether each of the following relations is a function or not. Justify your answers. a) k = {(0, 0), (1, 1), (2, 4), (3, 9)} b) m = {(1, 2), (2, 5), (1, 4), (3, 6)} c) p = {(–1, –3), (0, –1), (1, 1), (2, 3)} d) v = {(–2, 5), (–1, 5), (0, 5), (1, 5), (5, 5)} Solution a) k is a function. No two different ordered pairs have the same x-value. b) m is not a function. The ordered pairs (1, 2) and (1, 4) have the same first entry, but different second entries. c) p is a function. No two different ordered pairs have the same x-value. d) v is a function. No two different ordered pairs have the same x-value. Guided Practice State whether each relation in Exercises 1–4 is a function or not. Explain your reasoning. 1. m = {(1, 1), (2, 8), (3, 27)} 2. b = {(a, 1), (b, 2), (c, 3), (a, 4)} 3. v = {(7, 1), (7, 2), (7, 7)} 4. t = {(1, 7), (2, 7), (7, 7)} 424 Section 8.5 — Relations and Functions Functions Can Be Represented in Different Ways Relations do not have to be written as lists of ordered pairs for you to be able to identify functions. Example 2 Determine whether each of the mappings below represents a function. a) b) c) 2 5 7 11 a b c k n 2 5 6 7 11 a b c k n 2 5 6 7 11 a b c n Domain Range Domain Range Domain Range Solution a) and b) are NOT functions, since 7 is mapped to two different values — (7, a) and (7, c) have the same x-value. c) IS a function. Each member of the domain only maps onto one member of the range. Guided Practice The mapping on the right shows the relation g(x). 5. State the domain and range of the relation. 6. Is the relation a function? Explain your answer. 7. Find g(0). The mapping on the right shows the relation h(x). 8. State the domain and range of the relation. 9. Is this relation a function? Explain your answer. 10. Find h(–3). –4 –3 2 5 p q r s t Don’t forget: Remember that this isn’t always the case — some functions cannot be expressed as an equation. Functions are Often Written as Equations Some functions can be expressed as an equation. For this to be possible, there must be a reason for the pairing between each member of the domain and each member of the range. The equation represents the way the members of the domain and range are paired. Section 8.5 — Relations and Functions 425 Example 3 Express the following function as an equation: f = {(1, 1), (2, 4), (3, 9), (4, 16)} Check it out: The symbol Œ is shorthand for “is a member of the set.” It just means that x takes only those values. Solution The relationship between the x-values and y-values is y = x2. The domain of the function is x Œ {1, 2, 3, 4}. The function can be written: f = {(x, y) such that y = x2 and x ŒŒŒŒŒ {1, 2, 3, 4}} You will often see functions written in the form y = x2, without any domain specified. By convention, you then take the domain to be all values of x for which the function is defined. Guided Practice Express the following functions in terms of an equation. 11. f = {(–4, 0), (–3, 1), (0, 4), (1, 5), (2, 6)} 12. g = {(–2, 5), (0, 1), (1, 2), (2, 5)} 13. h = {(–5, –4.5), (–3, –2.5), (1, 1.5), (3, 3.5), (5, 5.5)} 14. f = {(–3, –27), (–2, –8), (–1, –1), (0, 0), (1, 1), (2, 8)} 15. g = {(–2, 8), (0, 0), (1, 2), (2, 8), (3, 18)} 16. h = {(–3, 17), (–1, 1), (0, –1), (1, 1), (2, 7)} The Vertical Line Test Shows if a Graph is a Function By definition, a function cannot have any two ordered pairs that have the same first coordinate but different second coordinates, i.e. for each value of x there is only one possible value of y. Graphically, this means that no vertical line can intersect the graph of a function at more than one point. Vertical Line Test to determine whether a graph represents a function: Simply hold a straightedge parallel to the y-axis at the far left-hand side of the graph, then move it horizontally along the graph from left to right. If, at any position along the x-axis, it is possible for you to draw a vertical line that intersects with the graph more than once, then the graph does not represent a function. 426 Section 8.5 — Relations and Functions Example 4 Use the vertical line test to determine whether the following graphs represent functions. y y y x x x Solution The first two graphs pass the vertical line test — you cannot draw a vertical line that intersects with either graph at more than one point, so they represent functions. The third graph does not represent a function — the vertical line test fails. Guided Practice In Exercises 17–20, use the vertical line test to determine whether each graph represents a function or not. 17. 19. y y 18. 20. x x y y x x Section 8.5 — Relations and Functions 427 Independent Practice 1. Define a function and give an example. In Exercises 2–7, use the given relation with the domain x = {–2, –1, 0, 1, 2} to generate sets of ordered pairs. Use them to determine whether the relation is a function or not. 2. m = {(x, x2 – 4)} 3. t = {(x, x + 2)} 4. k = {(x, y = (x – 2)(x + 2)} 5. p = {(x, y = ± 4 2− x } 6. j = {(x, y = 2x – 1)} 7. b = {(x, y = x ± (3x – 4)} 8. In the equation x2 + y2 = 9, is y a function of x? Explain your reasoning. Using the graph on the right, answer Exercises 9–12 about relation f. 9. State the domain and range of the relation. 10. Is the relation a function? 11. Find the value(s) of f(0). 12. Find the value(s) of f(3). 3 2 1 –1 –( ) 1 2 3 x Use the vertical line test to determine if the graphs below are functions. 13. y 14. y x x For Exercises 15–16, find the range (y) for each relation when the domain is {–4, –2, 0, 2, 4}, and determine whether the relation is a function. 15. y = 1 2 16. y = x – 6 x + 1 For Exercises 17–19, find the domain (x) for each relation when the range is {–6, –3, 0, 3, 6}, and determine whether the relation is a function. 17. y = 1 3 x – 2 18. y = x + 5 19. y = ± x 2 20. Are all quadratics of the form y = ax2 + bx + c and y = –ax2 + bx + c functions? Explain your answer. 21. Are circles functions? Explain your answer. Round Up Round Up Functions are special types of relations. That means that all functions are relations — but not all relations are functions. A relation is only a function if it maps each member of the domain to only one member of the range. 428 Section 8.5 — Relations and Functions Topic 8.5.3 California Standards: 16.0: Students understand the concepts of a relation and a function, determine whether a given relation defines a function, and give pertinent information about given relations and functions. 17.0: Students determine the domain of independent variables and the range of dependent variables defined by a graph, a set of ordered pairs, or a symbolic expression. What it means for you: You’ll see some different ways of representing functions. Key words: function domain range Check it out: The notation m(x) is read as “m of x” or “the value of function m at x.” Function Notation Function Notation Functions are often written using function notation. Think of a Function as a “Rule” or “Machine” The function m is a rule that assigns to each value x of its domain a distinct value m(x) of its range. Think of m as a machine that processes x according to some rule and outputs the value m(x). Suppose m(x) = x2. You could draw the following diagram to represent how the function processes any given value of x. Function machine: m(x) = x2 Input x = 3 Domain m x( ) = x2 2= 3 Output m x( ) = 9 Range m(x) = x2 takes a value of x from the domain, in this case x = 3, squares it, 32 = 9, and outputs a value of the range, 9. When you are referring to a particular value of a function, you replace the x with the relevant number — so in the example above, the value of m when x = 3 would be denoted m(3). Similarly, you can substitute expressions into a function, so for example you could find m(x + 3) = (x + 3)2 by replacing x with (x + 3) in the function. Section 8.5 — Relations and Functions 429 Example 1 Given the function f (x) = 2x2 – 1, find a) f (–1), b) f (3), and c) f (x + a). Solution a) f (–1) = 2 · (–1) (–1) = 1 b) f (3) = 2 · 32 – 1 = 2 · 9 – 1 = 18 – 1 = 17 \ f (3) = 17 c) f (x + a) = 2 · (x + a)2 – 1 = 2(x2 + 2ax + a2) – 1 = 2x2 + 4ax + 2a2 – 1 Example 2 If P(x) = x3 + 1, find the range of P(x) when the domain of P(x) is the set {–2, –1, 0, 1, 2}. Solution The range is the set of all values of P(x) for which x Œ {–2, –1, 0, 1, 2}. So, range = {P(–2), P(–1), P(0), P(1), P(2)} = {(–2)3 + 1, (–1)3 + 1, (0)3 + 1, (1)3 + 1, (2)3 + 1} = {–7, 0, 1, 2, 9} Guided Practice For Exercises 1–6, let f (x) = 3x – 1 and g(x) = x2 – 2. Find each of the values indicated. 1. f (–2) 4. f (a – 1) 2. f ⎛ ⎜⎜⎜ ⎝ ⎞ ⎟⎟⎟ ⎠ 1 3 3. g(a) 5. f (3) · g (–2) 6. g(k) – f (k) Independent Practice 1. The function f(t) = 5 9 (t – 32) is used to convert temperatures from degrees Fahrenheit to degrees Celsius. Find f(212). 2. Supposing f (x) = –2x2 + x – 5, evaluate f (x) for x Œ{–1, 2, b}. In Exercises 3–5, find f (x + h) for each of the given functions. 3. f (x) = 2x – 1 5. f (x) = –3x + h 4. f (x) = x2 – x – 2 Use the functions f(x) = 2x + 1 and g(x) = x2 + 2x + 1 to find the value of each expression in Exercises 6–9. 6. f(–1) + g(2) 8. f(x + h) + g(x + h) 7. g(–1) – f(3) 9. f(x + h) – g(x + h) Round Up Round Up There’s nothing too complicated here — the main thing is to recognize that if you see the notation f(something), then it’s likely to be a function. 430 Section 8.5 — Relations and Functions Topic 8.5.4 California Standards: 16.0: Students understand the concepts of a relation and a function, determine whether a given relation defines a function, and give pertinent information about given relations and functions. 17.0: Students determine the domain of independent variables and the range of dependent variables defined by a graph, a set of ordered pairs, or a symbolic expression
. What it means for you: You’ll show whether two functions are equal. Key words: function ordered pair domain range More on Functions More on Functions This Topic’s all about the special rules for telling whether two functions are equal. Equality of Functions A function f is equal to another function g if, and only if, the set of ordered pairs of f is identical to the set of ordered pairs of g. It isn’t enough for the two functions to be represented by equivalent equations [for example, y = 2x + 4 and y = 2(x + 2)]. For the two sets of ordered pairs to be identical, the two functions must also have the same domain (and therefore the same range). Example 1 a) Determine whether the following two functions are equal: m = {(x, y), such that y = x2 – 4 and x Œ {–1, 0, 2, 3}} b = {(x, y), such that y = (x – 2)(x + 2) and x Œ {–1, 0, 2, 3}} b) State the range of m. Solution a) Find each set of ordered pairs by substituting each value of x into the equation for y. m = {(–1, [–1]2 – 4), (0, 02 – 4), (2, 22 – 4), (3, 32 – 4)} = {(–1, –3), (0, –4), (2, 0), (3, 5)} b = {(–1, [–1 – 2][–1 + 2]), (0, [0 – 2][0 + 2]), (2, [2 – 2][2 + 2]), (3, [3 – 2][3 + 2])} = {(–1, –3), (0, –4), (2, 0), (3, 5)} Since each ordered pair for m is in b and vice versa, then the functions m and b are equal. b) The range of m = range of b = {–3, –4, 0, 5} Section 8.5 — Relations and Functions 431 Example 2 The domain of both of the following functions is the set of all real numbers. p = {(x, y = x3 + 3x2 + 3x + 1)} q = {(x, y = (x + 1)3} Determine whether the two functions are equal. Solution Both functions have the same domain, so if the equations that generate the y-values are equal, then the functions must be equal. Expand (x + 1)3: (x + 1)3 = (x + 1)(x + 1)(x + 1) = (x2 + 2x + 1)(x + 1) = x3 + 3x2 + 3x + 1 So, p and q are equal. Guided Practice Determine whether the pairs of relations/functions below are equal. 1. A = {(x, y), such that y = 2x + 8 and x Œ {–1, 0, 1, 2}} B = {(x, y), such that 2y = 4x + 16 and x Œ {–1, 0, 1, 2}} 2. A = {(x, y), such that y = x2 and x Œ {1, 4, 9, 16}} B = {(x, y), such that y2 = x and x Œ {1, 4, 9, 16}} Determine whether the pairs of relations/functions below are equal. The domain of each relation/function is the set of real numbers. 3. M = {(x, y = (x + 4)2)} N = {(x, y = x2 + 4x + 4)} 4. M = {(x, y = x2 + 1)} N = {(x, y = (x + 1)2)} 5. M = {(x, y = x4(x + 1)(2x – 3))} N = {(x, y = 2x6 – x5 – 3x4)} 6. M = {(x, y = x(3x – 2)(3x + 1))} N = {(x, y = 9x3 + 3x2 – 2x)} More Function Examples Example 3 Suppose v(x) State any restrictions on x. b) Find v(–3). Solution a) The function is undefined when its denominator (x – 1) is 0, so x πππππ 1. b) Substitute x = –3: 432 Section 8.5 — Relations and Functions 1 = + 9 1 − 4 10 = − = − 4 5 2 2 v(–3) = )− + 3 ( − − 3 1 So, v(–3) = − 5 2 Example 4 Determine the range of the function represented by the graph on the right. Explain your answer. f x( ) x 1 (– , –3) 2 –3 Solution The lowest value of f(x) is –3. The values of f(x) get infinitely large as x gets larger (in both the positive and negative x directions) — as indicated by the arrowheads on the graph. Range = {f(x) : f(x) ≥≥≥≥≥ –3} Example 5 Check it out: A colon means “such that,” so you would read this solution as “the set of all f(x) such that f(x) is greater than or equal to –3.” Determine the domain of f (x) = 2 for which f (x) is defined. Explain your thinking. 6 x − , given that it contains all real x Solution Any expression n is defined for all real values of n ≥ 0. So, for the function to be defined, 2x – 6 ≥ 0. ﬁ 2x ≥ 6 ﬁ x ≥ 3 So, the domain of the function = {x : x ŒŒŒŒŒ R and x ≥≥≥≥≥ 3} Don’t forget: R is the set of all real numbers. Example 6 Given that g(x) = x2 – 1, find ( in terms of x and h. Solution = = ⎡ ⎣⎢ ( x + − 2 h ) ⎦⎥ − −⎡ ⎤ x 1 ⎣⎢ 2 ⎤ 1 ⎦⎥ 2 x + 2 xh Substitute in the functions g(x) and g(x + h) 1 Expand and simplify + 2 = xh h 2xx h = 2x + Section 8.5 — Relations and Functions 433 Guided Practice In Exercises 7–10, find the values in terms of h (and x, where appropriate). 7. Supposing f (x) = x2 – 2, find f (h + 2) – f (2). 8. If f (x) = 2x2 + 4x – 6, find f (h – 3) + f (2h). ( ) f x ( f x 9. Supposing f (x) = x2, find + − . 10. If m(x) = 2x – 3, find the value of Determine the domain for the functions in Exercises 11 and 12, given that they contain all real numbers for which the functions are defined. 11 16 4 12 Independent Practice For the functions in Exercises 1–4, state any restrictions and find f(–1). 1. f(x) = 1 x + 3x + 2 3. f(x) = 2 x + 6 2. f(x) = 2 2 x x 3 4. f(x) = x 2 − − 9 x 6 − + 11 x 4 1− Determine the range of the functions in Exercises 5–6. 6. 1 2 3 x 4 –4 –3 –2 5. –4 –3 –2 y 4 3 2 1 –1 –1 –2 –3 –4 y 4 3 2 1 –1 –1 –2 –3 –4 1 2 3 x 4 7. If f(x) = x2 + 2x – 1, find f(x + h) – f(x). 8. If g(x) = 2x2 + x – 4, find g(x + h) – g(x). 9. If f(x) = (x + 1)2 + x, find f(2h) – f(2). ( ) f x ( f x 10. If f(x) = x2 + 2x, find . + − ) . 11. If f(x) = x2 – 2x + 1, find 12. Give an example of two equal functions. Round Up Round Up There are two methods to determine whether two functions are equal. If you’re given a small number of values for the domain, you can substitute each value of the domain into both functions and see if you get identical ordered pairs. The alternative is to show that the two functions can be represented by equivalent equations — and if the domains are also equal, the functions will be equal. 434 Section 8.5 — Relations and Functions Chapter 8 Investigation ming Functions ming Functions ansforororororming Functions ansf ansf TTTTTrrrrransf ming Functions ming Functions ansf ansforororororming Functions TTTTTrrrrransf ming Functions ming Functions ansf ansf ming Functions ansf ming Functions In this Investigation you’ll practice interpreting information about functions using graphs. On a set of axes, draw the graph of the following function: Investigate the functions: = = f f {( , x y ) such that y = x and x ∈ R } {( , x y ) such that y = + x , a x ∈ R , and a ∈ R } , and f = {( , x y ) such that y = + x , a x ∈ R , and a ∈ R }. Write down a general rule for what happens. Extension 1) The graph of the function f = {( , x y ) such that y = x } original function y-axis can be stretched or compressed. The diagram on the right shows a horizontal compression and stretch. x-axis Investigate how the function must be changed to stretch or compress the graph. Write down a general rule. 2) Investigate how the function f = {( , x y ) such that y = x } must be changed so that the graph is reflected in the x-axis. Open-ended extension 1) The graph on the right has been produced by performing a series of different transformations on the graph of f = {( , x y ) such that y = x } . Identify the new function that is represented by the graph. 2) Perform a series of transformations on the graph of = , recording each one. such that y , x y {( = } x ) f Identify the function that your graph represents. -8 -7 -6 -5 -4 -3 -1 0 -1 -2 -3 -4 -5 -6 -7 - Check that you are correct by creating a table of x- and y-values for your function and plotting the points. ound Up ound Up RRRRRound Up ound Up ound Up ound Up ound Up RRRRRound Up ound Up ound Up The rules about transforming the absolute value function also apply to other graphs of functions — so this Investigation is really useful for all sorts of math problems. estigaaaaationtiontiontiontion — Transforming Functions estigestig estig pter 8 Invvvvvestig pter 8 In ChaChaChaChaChapter 8 In pter 8 In pter 8 In 435435435435435 436436436436436436436436436436 Appendixes Glossary ................................................... 438 Formula Sheet ................................................... 440 Index ................................................... 442 437437437437437 Glossary Symbols • infinity < > ∴ therefore ⊥ is less than is greater than is perpendicular to is parallel to is less than or equal to is greater than or equal to is not equal to the irrational numbers ΩΩΩΩΩΩΩΩΩΩ £ ≥ π I N the natural numbers Q the rational numbers Δ the empty set « the intersection of sets » the union of sets Ã is a subset of À is not a subset of Œ is an element of œ is not an element of ﬁ implies R the real numbers W the whole numbers Z the integers A aaaaabsolute v alue alue bsolute v bsolute v alue the distance between zero and a number on the bsolute value alue bsolute v number line (the absolute value of a is written \|a\|) ession aic exprxprxprxprxpression ession aic e algalgalgalgalgeeeeebrbrbrbrbraic e aic e ession a mathematical expression containing at ession aic e least one variable tion) tion) ultiplica ultiplica dition and m ties (of ad ad ad ad addition and m dition and m ties (of ties (of oper oper associatititititivvvvve pre pre pre pre proper associa associa tion) ultiplication) dition and multiplica operties (of tion) ultiplica dition and m ties (of oper associa associa for any a, b, c: a + (b + c) = (a + b) + c a(bc) = (ab)c B base base base in the expression bx, the base is b base base binomial binomial binomial a polynomial with two terms binomial binomial C losed intervvvvval al al al al an interval that includes its endpoints ccccclosed inter losed inter losed inter losed inter losure pre pre pre pre proper dition and dition and umber ad umber ad eal-n eal-n ties (of r r r r real-n ties (of ties (of oper oper losur losur ccccclosur operties (of dition and umber addition and eal-number ad dition and umber ad eal-n ties (of oper losur mmmmmultiplica tion) tion) ultiplica ultiplica tion) when two real numbers are added or ultiplication) tion) ultiplica multiplied, the result is also a real number actor actor common f common f actor a number or expression that is a factor of two common factor actor common f common f or more other numbers or expressions commcommcommcommcommutautautautautatititititivvvvve pre pre pre pre proper tion) tion) ultiplica ultiplica dition and m ties (of ad ad ad ad addition and m dition and m ties (of ties (of op
er oper tion) for ultiplication) dition and multiplica operties (of tion) ultiplica dition and m ties (of oper any a, b: a + b = b + a and ab = ba completing the squareeeee the process of changing a quadratic completing the squar completing the squar completing the squar completing the squar expression into a perfect square trinomial compound inequality compound inequality compound inequality two inequalities combined using either compound inequality compound inequality “and” (a conjunction) or “or” (a disjunction) D a monomial) a monomial) ee (of ee (of dededededegggggrrrrree (of a monomial) the sum of the powers of the variables ee (of a monomial) a monomial) ee (of ynomial ynomial a pol a pol dededededegggggrrrrree of ee of ee of ynomial the largest degree of a polynomial’s a polynomial ee of a pol ynomial a pol ee of terms tor tor denomina denomina tor the bottom expression of a fraction denominator tor denomina denomina tions tions equa equa pendent system of dededededependent system of pendent system of tions a system of equations with equations pendent system of equa tions pendent system of equa infinitely many possible solutions discriminant discriminant discriminant for a quadratic equation ax2 + bx + c = 0, the discriminant discriminant discriminant is b2 – 4ac distrib distrib distributiutiutiutiutivvvvve pr e pre proper e pre pr oper oper ty (of ty (of ty (of m m m m multiplica ultiplica ultiplica tion o tion o tion ovvvvver ad er ad er ad dition) dition) operty (of ultiplication o er addition) dition) distrib distrib oper ty (of ultiplica tion o er ad dition) for any a, b, c: a(b + c) = ab + ac 438438438438438 Glossar Glossaryyyyy Glossar Glossar Glossar tion or function) a relaelaelaelaelation or function) tion or function) a r a r domain (of domain (of tion or function) the set of all possible domain (of a r tion or function) a r domain (of domain (of “inputs” of a relation or function E tions tions alent equa equivvvvvalent equa alent equa equi equi tions equations that have the same solution alent equations tions alent equa equi equi set equivvvvvalent fr actions actions alent fr alent fr equi equi actions fractions are equivalent if they have the alent fractions actions alent fr equi equi same value xponent eeeeexponent xponent xponent in the expression bx, the exponent is x xponent F fffffactoring actoring actoring actoring writing a polynomial as a product of two or more actoring factors fffffactor actor actor actor a number or expression that can be multiplied to get actor another number or expression — for example, 2 is a factor of 6, because 2 × 3 = 6 function function function a rule for transforming an “input” into a unique “output” function function G actor (GCF) actor (GCF) test common f gggggrrrrreaeaeaeaeatest common f test common f actor (GCF) largest expression that is a test common factor (GCF) actor (GCF) test common f common factor of two or more other expressions; all other common factors will also be factors of the GCF ouping symbols gggggrrrrrouping symbols ouping symbols ouping symbols symbols that show the order in which ouping symbols mathematical operations should be carried out — such as parentheses and brackets I tion) tion) ultiplica ultiplica dition and m identities (of ad ad ad ad addition and m dition and m identities (of identities (of tion) ultiplication) dition and multiplica tion) ultiplica dition and m identities (of identities (of the additive identity is 0 (zero) — 0 can be added to any other number without changing it; the multiplicative identity is 1 — any number can be multiplied by 1 without changing tions tions equa equa inconsistent system of inconsistent system of tions a system of equations with equations inconsistent system of equa tions equa inconsistent system of inconsistent system of no solutions inteinteinteinteintegggggererererersssss the numbers 0, ±1, ±2, ±3,...; the set of all integers is denoted Z sets) sets) section (of section (of inter inter sets) the intersection of two or more sets is section (of sets) intersection (of sets) section (of inter inter the set of elements that are in all of them; intersection is denoted by « inininininvvvvvererererersessessessesses a number’s additive inverse is the number that can be added to it to give 0 (the additive identity); a number’s multiplicative inverse is the number that it can be multiplied by to give 1 (the multiplicative identity) umbersssss the set of all numbers that cannot be umber umber tional n iririririrrrrrraaaaational n tional n tional number umber tional n written as a fraction p q , where pŒZ and qŒN; the set of all irrational numbers is denoted I L least common m least common m ultiple (L ultiple (L ultiple (LCM)CM)CM)CM)CM) the smallest expression that least common multiple (L least common m least common m ultiple (L has two or more other expressions as factors likliklikliklike ter e ter e ter e termsmsmsmsms two or more terms that contain the same variables, e ter and where each variable is raised to the same power in every term and and tion in tion in linear equa linear equa and y an equation that can be written in the tion in x and linear equation in and tion in linear equa linear equa form Ax + By = C (where A and B are not both zero) M monomial monomial monomial an expression with a single term monomial monomial N turtur umbersssss the set of numbers 1, 2, 3,...; umber turtural nal nal nal nal number umber nananananatur umber the set of all natural numbers is denoted N nnnnnumer tor umeraaaaator tor umer umer tor the top expression of a fraction umer tor ession umeric exprxprxprxprxpression ession umeric e umeric e nnnnnumeric e ession a number or an expression containing only ession umeric e numbers (and therefore no variables) O open intervvvvvalalalalal an interval that does not contain its endpoints open inter open inter open inter open inter ed pair orororororderderderderdered pair ed pair ed pair a pair of numbers or expressions written in the ed pair form (x, y); they can be used to identify a point in the coordinate plane P allel lines parparparparparallel lines allel lines allel lines lines with equal slopes; lines in the same plane allel lines that never meet e trinomial e trinomial ect squar perfperfperfperfperfect squar ect squar e trinomial a trinomial that can be written as the ect square trinomial e trinomial ect squar square of a binomial pendicular lines pendicular lines perperperperperpendicular lines pendicular lines lines whose slopes multiply together to give pendicular lines the product –1; lines that intersect at 90° point-slope fororororormmmmmulaulaulaulaula an equation of a line of the form point-slope f point-slope f point-slope f point-slope f y – y1 = m(x – x1), where m is the slope and (x1, y1) are the coordinates of a particular point lying on that line polynomial polynomial polynomial a monomial or sum of monomials polynomial polynomial actorizationtiontiontiontion a factorization of a number where each actoriza actoriza prime f prime f prime factoriza actoriza prime f prime f factor is a prime number umber umber prime n prime n umber a number that can only be divided by itself and 1 prime number prime n umber prime n oduct oduct prprprprproduct oduct the result of multiplying numbers or expressions oduct together equality equality ties of ties of oper prprprprproper oper equality ties of equality operties of oper equality ties of addition property of equality: if a = b, then a + c = b + c multiplication property of equality: if a = b, then ac = bc subtraction property of equality: if a = b, then a – c = b – c division property of equality: if a = b, then a ÷ c = b ÷ c Q tic equationtiontiontiontion a polynomial equation of degree 2 tic equa tic equa quadraaaaatic equa quadr quadr tic equa quadr quadr R rrrrradical adical adical adical an expression written with a radical symbol adical (for example, 2 ) adicand adicand rrrrradicand adicand the number inside a radical symbol adicand tion or function) tion or function) a relaelaelaelaelation or function) a r a r e (of e (of rrrrrangangangangange (of tion or function) the set of all possible e (of a r tion or function) a r e (of “outputs” of a relation or function rrrrraaaaational e tional e tional e tional exprxprxprxprxpression ession ession ession an expression written as a fraction tional e ession umbersssss the set of all numbers that can be written as umber umber tional n rrrrraaaaational n tional n tional number umber tional n p q numbers is denoted Q , where p∈Z and q∈N; the set of all rational a fraction umbersssss denoted R, all numbers of the number line rrrrreal n umber umber eal n eal n eal number umber eal n rrrrrecipr ocal ocal ecipr ecipr ocal the multiplicative inverse of an expression eciprocal ocal ecipr rrrrrelaelaelaelaelationtiontiontiontion any set of ordered pairs (the first number or expression in each pair can be thought of as the relation’s “input,” the second as the relation’s “output”) an equationtiontiontiontion an equation’s roots are its solutions an equa an equa rrrrroots of oots of oots of oots of an equa an equa oots of S umber sign of a n a n a n a n a number umber sign of sign of umber whether a number is positive or negative sign of umber sign of slope slope slope the steepness of a line; the ratio of the vertical “rise” to the slope slope horizontal “run” between any two points on a line slope-intercececececept fpt fpt fpt fpt fororororormmmmmulaulaulaulaula an equation of a line of the form slope-inter slope-inter slope-inter slope-inter y = mx + b, where m is the slope and b is the y-intercept square re re re re rootootootootoot if p2 = q, then p is a square root of q; if p is positive squar squar squar squar it is the principal square root of q, but if p is negative it is the minor square root of q subset subset subset a subset of a set is a set whose elements are all subset subset contained in the set sumsumsumsumsum the result of adding numbers or exp
ressions together tions tions equa equa system of system of tions two or more equations equations system of equa tions equa system of system of T tertermsmsmsmsms the parts that are added to form an expression terter ter trinomial trinomial trinomial a polynomial with three terms trinomial trinomial U sets) sets) union (of union (of sets) the union of two or more sets is the set of union (of sets) sets) union (of union (of elements that are in at least one of them; union is denoted by » V vvvvvariaariaariaariaariabbbbblelelelele a letter that is used in place of a number W umbersssss the set of numbers 0, 1, 2, 3,...; umber umber hole n hole n wwwwwhole n hole number umber hole n the set of all whole numbers is denoted W X -intercececececeptptptptpt the x-coordinate of a point where a graph meets -inter -inter x-inter -inter the x-axis Y -intercececececeptptptptpt the y-coordinate of a point where a graph meets -inter -inter y-inter -inter the y-axis Z zzzzzererererero pro pro pro pro product pr oduct pr oduct pr oper oper opertytytytyty if the product of two factors is zero, oduct proper oduct pr oper then at least one of the factors must itself be zero Glossaryyyyy Glossar Glossar Glossar Glossar 439439439439439 Formula Sheet Axioms of the Real Number System For any real numbers a, b, and c, the following properties hold: PrPrPrPrProper ty Name ty Name oper oper ty Name operty Name ty Name oper Closure Property: Identity Property: Inverse Property: Commutative Property: Associative Property: Distributive Property of Multiplication over Addition: AdAdAdAdAddition dition dition dition dition a + b is a real number + (–a) = 0 = –a + b) + c = a + (b + c) a(b + c) = ab + ac and (b + c)a = ba + ca Multiplicationtiontiontiontion Multiplica Multiplica Multiplica Multiplica a × b is a real number –1 = 1 = a–ab)c = a(bc) PrPrPrPrProper Equality Equality ties of ties of oper oper Equality ties of Equality operties of Equality ties of oper If a = b, then a + c = b + c. If a = b, then ac = bc. If a = b, then a – c = b – c. If a = , then b = a c b c . Absolute Value Order of Operations = x ⎧ x if ⎪⎪⎪⎪ ⎪⎪⎪⎪ x ⎨ 0 if − ⎩ x if x x > = < 0 0 0 Perform operations in the following order: 1. Anything in gggggrrrrrouping symbols ouping symbols ouping symbols ouping symbols — working from the innermost grouping ouping symbols symbols to the outermost. Exponents Exponents Exponents. 2. Exponents Exponents visions tions and dididididivisions visions tions tions Multiplica Multiplica 3. Multiplica visions, working from left to right. Multiplications tions visions Multiplica actions actions subtr subtr ditions ditions 4. AdAdAdAdAdditions actions, again from left to right. subtractions ditions and subtr actions subtr ditions Using Roots ab = ⋅ a b = a b a b Rules of Exponents x m n mn x 1 = x x x0 1 Fractions Adding and subtracting fractions with the same denominator:: Adding and subtracting fractions wi tth different denominators ad + bc bd a b c − = d ad − bc bd Mult iiplying fractions: a b c ⋅ = d ac bd Dividing fractions: a b 440440440440440 FFFFFororororormmmmmulasulasulasulasulas ÷ = ⋅⋅ =d a b c d c ad bc Applications Formulas InInInInInvvvvvestments estments estments estments estments The return (I) earned in one year when p is invested at an interest rate of r (expressed as a fraction): I = pr The total return (I) earned in one year when p1 is invested at an interest rate of r1 and p2 is invested at an interest rate of r2: I = p1r1 + p2r2 Mixtureseseseses Mixtur Mixtur Mixtur Mixtur concentration = amount of substance total volume percent of an ingredient = amount of ingredient total amount ×1100 The concentration (c) and total volume (v) of a mixture are given by cv = c1v1 + c2v2 where c1 and v1 describe the first ingredient and c2 and v2 the second. Speed and WWWWWororororork Rk Rk Rk Rk Raaaaatetetetete Speed and Speed and Speed and Speed and speed = distance time work rate = work completed time taken Graphs Slope of a line Inequalities \|x\| < m means –m < x < m \|x – c\| < m means c – m < x < c + m Special Products of Binomials Point-slope form: y – y1 = m(x – x1) Slope-intercept form: y = mx + b Standard form of a linear equation: Ax + By = C Two lines with slopes m1 and m2 are: parallel if m1 = m2 perpendicular if m1 × m2 = –1 Quadratics Basic form of a quadratic equation: ax2 + bx + c = 0 The quadratic formula: − ± b = x 4 ac −2 b 2 a Completing the square: 2 x + + bx ⎛ ⎜⎜⎜ ⎝ b 2 2 ax + + bx ⎛ ⎜⎜⎜ ⎝ 1 a b 2 ⎞ 2 ⎟⎟⎟ ⎠ ⎛ ⎞ 2 ⎟⎟⎟ = + ⎜⎜⎜ x ⎝ ⎠ b 2 ⎛ ⎞ 2 ⎟⎟⎟ = aa x ⎜⎜⎜ ⎝ ⎠ + ⎞ 2 ⎟⎟⎟ ⎠ b 2 a (a + b)2 = a2 + 2ab + b2 (a – b)2 = a2 – 2ab + b2 (a + b)(a – b) = a2 – b2 Inequality Inequality ties of ties of oper PrPrPrPrProper oper Inequality ties of Inequality operties of Inequality ties of oper For any real numbers a, b, and c: If a < b, then a + c < b + c. If a < b, then a – c < b – c. For any real numbers a, b, and c: If a < b and c > 0, then ac < bc. If a < b and c > 0 , then a c < b c . The discriminant: 2 2 22 b b b − − − 4 ac 4 ac 4 ac > ⇒ 0 = ⇒ 0 < ⇒ 0 2 distinct real roots 1 real double root 0 real roots ⎛ ⎜⎜⎜⎜ The vertex of y = ax2 + bx + c If (x – a)(x – b) = 0, then x = a or x = b. ⎞ ⎟⎟⎟⎟ ⎠ FFFFFororororormmmmmulasulasulasulasulas 441441441441441 Index A abscissa 164 absolute values 19, 21, 128, equations 128-132 inequalities 158-160 addends 14, 21 addition 14, 15, 21, 27, 28 of fractions 52, 53 of polynomials 266, 267, 272 property of equality 76, 77 property of inequalities 147, 148 of rational expressions 403-409 additive identity 16 inverse 16, 266 age-related tasks 99, 251 algebraic expressions 9, 10 annual interest 108, 109 applications 89, 90 of fractional equations 417, 418 of inequalities 150-152 of polynomial division 294, 295 of quadratic equations 359, 360, 379-385 of systems of equations 249-251, 253, 255, 256, 258, 259 associative properties of addition and multiplication 30, 34, 35 axioms 34, 35 B bases 37, 38 binary operations 14, 15, 30 binomials 263 factors 310, 311 multiplication of 276, 281 products of 297-300 squaring of 342 braces 2 as grouping symbols 27 brackets as grouping symbols 27 C canceling common factors 46, 391, 392 fractions 284, 288, 289 closure of sets under addition 14, 15, 21, 22, 34 under multiplication 14, 22, 34 coefficients 12 fractional coefficients 81, 82, 84, 85 coin tasks 92, 93 442442442442442 Indexxxxx Inde Inde Inde Inde combining like terms 67, 68 common denominator 403, 404 common difference 95, 96 common factors 46, 47, 50, 51, 288, 289, 302, 303, 305-308, 313, 314, 322, 323, 329, 330, 391, 392 commutative properties of addition and multiplication 24, 25, 30, 34, 35 completing the square 342-345, 347-353, 368-370 to derive the quadratic formula 355, 356 compound inequalities 155, 156, 158 conjunctions 155, 158 consecutive integer tasks 95, 96, 97, 150, 151 coordinate plane 164, 165 quadrants of 167-169 coordinates 164, 165, 179-181 coordinate of a point on a number line 18 D decimal coefficients 86, 87 degree of a polynomial 264, 278 denominator 46, 288, 388 common denominator 403, 406-409 dependent systems of linear equations 239, 240 difference 24 difference of two squares 298, 325, 326 discriminant 372-378 disjunctions 156, 159, 160 distance 103-106 distributive laws 31 distributive property 34, 35, 67, 68, 70, 71, 276, 281 use in combining like terms 67 use in getting rid of grouping symbols 70, 84 dividend 291 division 24-26 of fractions 50, 51 of polynomials by monomials 283, 284 of polynomials by polynomials 285-292, 294, 295 property of equality 76-79 property of inequalities 139, 140, 142-145 property of square roots 43, 44 by rational expressions 397-401 by zero 17, 26, 33, 388 divisor 291, 303 domain of a relation 420-434 double root 375 E economic tasks 383-385 elements of sets 2 elimination method for solving systems of equations 244-248, 249-251, 255, 256, 258, 259 empty set 3 endpoints of intervals 137, 138, 155, 156, 158-160 equality in equations 74 of functions 431-433 properties of reflexive 14 symmetric 14 transitive 14 of sets 3 equations 74-76, 171, 194 absolute value equations 128-133 equivalent equations 76-79, 86 fractional equations 412-415, 417, 418 linear equations 74-78, 81, 82, 84-87, 89, 90, 175, 176, 229, 230, 249-250 systems of 229, 230 proof of an equation 74 quadratic equations 333-336, 338-341 setting up equations 95-97, 99-101, 103-106 solving equations 76, 77 of straight lines 170, 171, 173-175, 194, 195 205-207 equivalent equations 76-79, 86, 87 fractions / rational expressions 46-48, 390-392, 406-410 inequalities 139, 140, 142-145 evaluating algebraic expressions 13 exponents 28, 37, 38, 73, 274, 275 fractional 38, 40 negative 286, 287 rules of 37, 38, 274, 275, 283-287 expressions algebraic 13 numeric 9 quadratic 310, 311, 313-317, 318-320, 325-328 F factoring 288, 289 291, 292, 294, 295, 318-320, 322, 323, 333-336 monomials 302-303 polynomials 305-308, 310, 311, 313-317, 318-320, 322, 323, 325-330 polynomials using long division 291, 292 prime factorization 53, 54, 81 quadratic expressions and equations 294, 295, 310-320, 325-328, 333-336, 338-341 third-degree polynomials 322, 323 factors 14, 22, 302, 303 common factors 288, 289, 302, 303, 305-308, 313, 314, 322, 323, 329, 330, 390-392 fractional coefficients 81, 82, 84, 85 fractional equations 412-415, 417, 418 fractional powers 38, 40 fractions (see also rational expressions) 46, 47, 49, 50-54, 81, 82, 84, 85 equivalent 46, 47 functions 424-427, 429-433 equality of 431, 432 function notation 429, 430 G gradient (slope) 189-192, 194, 195, H half-open intervals 138 horizontal lines 173, 174 I identity 16 of addition 16 of multiplication 17 properties 30, 31 inconsistent systems of linear equations 237, 238 inequalities 137, 138 absolute value 158, 159, 160 applications of 139, 140, 142, 143 compound inequalities 155, 156 equivalent inequalities 150-153 graphing on a coordinate plane 212-215, 217-220, 222-225 graphing on a number line 137-140, 155, 156, 158-160 multistep inequali
ties 147-148 properties of addition 139, 140, 147, 148 division 143, 145, 147, 148 multiplication 142, 144 subtraction 140 regions defined by 212-215, 217-220 input-output tables 422 inputs of binary operations 15 integer problems 153 197-199, 201-203, 205-207, 209-210 consecutive integer tasks 95-97, 150, graph of a number 18 graphing method for solving systems of equations 229, 230, 234 graphing inequalities on a coordinate plane 212-215, 217-220 inequalities on a number line 137-140, 155, 156, 170, 171, 173, 174 quadratic functions 362, 363, 365, 366, 368, 370 relations 420 straight lines 158-160, 175, 176, 186, 187 in the form Ax + By = C 177, 178, 186, 187 systems of equations 229, 230, 237-240 systems of linear inequalities 222, 223 gravity 379-381 greatest common factor (GCF) 46, 47, 305-308, 391 grouping like terms 67, 68 grouping symbols 27, 70 151 integers 5 interest 108-111 interpreting results 379-381 intersection of sets 7 intervals 137-140, 155-160, 158-160 endpoints of 137-140, 158-160 inverse 16, 17 operations 77-79 properties 34 under addition 17, 266 under multiplication 17, 285 investment tasks 108-111 irrational numbers 6 isolating a variable 76-79 L least common multiple (LCM) 53, 81, 82, 84, 85, 407-410, 412-415 like terms 67-68, 263, 264, 266, 267, 278 grouping like terms 68 limitations on solutions 412 linear equations 74-78, 81, 82, 84-87, 89, 90, 175, 176, 229, 230, 249-250 systems of 229, 230 lines curves (parabolas) 362, 364 of symmetry 362, 368-370 straight lines 170, 171, 173-184, 186, 187, 189-192, 194, 195, 197-199, 201-203 long division 291-292, 294, 295 lowest terms 46, 47 M mapping diagrams 421 mathematical proofs 34, 35, 55-57 minor square root 41 mixture tasks 108-111, 113, 114, 116-119, 255, 256 money tasks 92, 93 monomials 263 dividing polynomials by 283, 284 factoring 302, 303 motion tasks 103-106, 379-381 multiplication 14, 15, 28, 29 of binomials 297-300 of fractions 49-51 of parentheses 148, 267, 268, 276-279, 318 of polynomials 267, 268, 276-281 property of –1 71 property of equality 78, 79 property of inequalities 143, 144 property of square roots 43, 44 by rational expressions 394-396 by zero 32, 33 multiplicative identity 16, 17 inverse 17 multistep inequalities 147, 148 N natural numbers 5 negative exponents 286, 287 of a number 32, 33 numbers 18, 19 as real numbers 5 multiplying and dividing inequalities by negative numbers 144 terms 71 nested grouping symbols 28 nonlinear equations 175 null set 3 number line 18-20 to show inequalities 137-140, 155, 156, 158-160 number system rules of 14-22, 24-35 numerator 46, 47, 288, 289, 388 numeric expressions 9 Indexxxxx Inde Inde Inde Inde 443443443443443 O one as the multiplicative identity 16, 17 open intervals 137 opposite of a number 16 of a polynomial 266 order of operations 27-29, 399, 400 ordered pairs 164, 175, 179, 420, 421, 424 ordinate 164 origin 164 P parabolas 362, 363 parallel lines 197-199, 209 parentheses 148 as grouping symbols 27 percent mixture problems 108-111, 113, 114, 116-119, 255, 256 perfect square trinomials 327, 328, 342-345, 251-253, 353 perpendicular lines 201-203, 210 physical problems 379-381, 383-385 plane 164, 165, 167-169 quadrants of the coordinate plane 167 plotting points 170, 177, 186 point-slope form of equation of a line 189-192, 194, 195, 198, 203 points of intersection 229 points on a line 170, 175, 176, 179, 180 points on a number line 18 polynomials 263 adding 266-268 degree of 264, 278 dividing polynomials by monomials 283, 284 prime factors 81, 302 principal square root 40-42 products 22 of polynomials 276-279 of powers 38 special products of binomials 297-300 proof of an equation 55-57, 74, 75 properties of equality 14, 55-57, 76-79 properties of inequalities 139, 140, 142-145 Q quadrants of the coordinate plane 167 quadratic equations 333 graphing 362, 363, 365, 366, 368-370, 381 quadratic expressions 310, 311 factoring 310, 311, 313-316, 325-330 in two variables 318-320 solving by completing the square 338-341 factoring 335, 336 taking square roots 351-353 using the quadratic formula 355-357, 359, 360, 368-370, 372-378 quotients 25, 26 of powers 38 R radicals 40 radicands 40 range of a relation 420-422, 424-427, 429-433 rates 121, 122, 124-126 regions defined by inequalities 212-214, 217-220 relations 420-422, 424, 425 remainders after division 43, 288, 289, 291, 292 removing fractions from equations 81, 82, 84, 85 removing decimals from equations 86, 87 return on investment 108 “rise over run” formula 189-192, 197, 198 roots 372-378 of quadratic equations 76, 362, 363, 365, 366 square roots 40-44, 338-341 rules of exponents 37, 38, 73, 274, 275, 283, 284, 286 rules of the number system 14-22, 24-35 run 189 S second-degree polynomials 318-320 sequences of integers 95-97 sets 2-6, 420-422, 424-426, 430-433 sign of a number 19 signs of coordinates in different quadrants 167 simplifying algebraic expressions 13, 67, 68, 70, 71, 73-75, 147, 148, 150-153, 263, 264, 266-272, 274, 275, 281, 283-287 inequalities 148 numeric expressions 13 rational expressions 390-392 slope 189-192, 194, 197-199, 201-203, 205-207, 209, 210 slope-intercept form of equation of a line 205-207, 209, 210 dividing polynomials by polynomials rate problems 103-106, 258, 259 solution intervals 137-140, 158-160 288, 289, 291, 292 factoring 305-308, 310, 311, 313-320, 322, 323, 325-330, 333-336 factoring using long division 291, 292 multiplying 276-281 opposite of 266 roots of 372-378 simplifying 264 subtracting 269-271 positive numbers 18 as real numbers 5 postulates 34 powers (exponents) 28, 37, 38, 73, 274, 275, 283, 284 fractional 38, 40 negative 286, 287 rules of 37, 38, 274, 275, 283-287 prime factorization 53, 81 444444444444444 Indexxxxx Inde Inde Inde Inde rational expressions solution sets 155, 160, 174, 212-215 adding and subtracting 403-410 dividing by 397-401 equations with 412-415, 417, 418 equivalent 390-392, 406-408 multiplying by 394-396, 397-401 undefined 388, 389 rational numbers 6 real-life tasks using inequalities 150-153 using quadratic equations 379-381, 383-385 using systems of equations 388-392 real numbers 5, 14-22, 24-35, 433 reciprocals 17, 25, 26, 32, 33, 201, 202, 210, 397-401 of polynomials 285-287 reflexive property of equality 14 solving absolute value inequalities 158-160 compound inequalities 155, 156, 158 conjunctions 155, 156 equations 76 fractional equations 412-415 inequalities 139, 140, 142-145, 147, 148, 150-153 quadratic equations 333-336, 338-341, 351-353, 355-357 speed 103-106 square roots 40-42 properties of 43, 44 square root method for solving quadratics 338-341 stacking method for multiplying polynomials 278, 279 W whole numbers 5 word problems 89, 90, 92, 93, 95-97, 99-101, 103-106, 108-111, 113, 114, 116-119, 121, 122, 124-126, 150-153, 249-251, 253, 255, 256, 258, 259 work-related tasks 121, 122, 124-126 work rates 122, 124-126 X x-axis 164-169 x-intercepts 182, 183, 362, 363, 365, 366, 372-378 Y y-axis 164, 165 y-intercepts 183, 184, 205-207, 362, 363, 365, 366, 370 Z zero 16, 32, 33 as the additive identity 16 division by 17, 26, 32, 33, 388 as a real number 5 zero product 33, 334-336 straight lines 170, 171, 173-184, 186, 187, 189-192, 194, 195, 197-199, 201-203, 205-207 subsets 3 subsets of the real numbers 5, 6 substitution method for solving systems of equations 232, 233, 235, 238, 240, 242, 249, 250, 253 substitution principle (or property) 15 subtraction 24, 25 of fractions 52, of polynomials 269-272 property of equality 76-78 property of inequalities 140, 147-148 of rational expressions 403-409 sum (of two numbers) 21 symmetric property of equality 14 systems of equations 229, 230, 244-248, 249-251, 253, 255, 256, 258, 259 applications of 249-251, 253, 255, 256, 258, 259 dependent systems 239, 240 inconsistent systems 237, 238 solving by elimination 244-248 solving by graphing 229, 230, 237, 239, 240 solving by substitution 232, 233, 238, 240 T terms in algebraic expressions 67, 68 third-degree polynomials 322, 323 time-related problems 103-106 transitive property of equality 14 trial-and-error method for factoring quadratics 314-317 trinomials 263, 327, 328, 342-344 U undefined rational expressions 388, 389 union of sets 7, 8 universal set 2 unknown quantities 9, 76 V values of coefficients 12, 13 of expressions 9, 10 variables 10, 67, 75, 76, 175 verifying points on a line 175, 176 vertex of a parabola 362, 363, 365, 366, 368-370 vertical line test for functions 426, 427 vertical lines 173, 198 Indexxxxx Inde Inde Inde Inde 445445445445445
lationwithheadpain.MaterialsandmethodsThestudyenrolled94females,diagnosedasmigrainewithoutaurafollowingtheInternationalClassiﬁcationofHeadacheDisorders[5],whoweresubsequentlyexaminedattheWomen’sHeadacheCentre,DepartmentofGynae-cologyandObstetricsofTurinUniversity.Theywereallincludedinthestudyduringamigraineattackprovidedthatitstartednomorethan4hpreviously.Accordingtoapredeterminedcomputer-maderandomizationlist,theeli-giblepatientswererandomlyandblindlyassignedtothefollowingtwogroups:groupA(n=46)(averageage35.93years,range15–60),groupB(n=48)(averageage33.2years,range16–58).Beforeenrollment,eachpatientwasaskedtogiveaninformedconsenttoparticipationinthestudy.MigraineintensitywasmeasuredbymeansofaVASbeforeapplyingNCT(T0).IngroupA,aspeciﬁcalgometerexertingamaximumpressureof250g(SEDATELEC,France)waschosentoidentifythetenderpointswithPain–PressureTest(PPT).Everytenderpointlocatedwithintheidentiﬁedareabythepilotstudy(Fig.1,areaM)wastestedwithNCTfor10sstartingfromtheauricle,thatwasipsilateral,tothesideofprevalentcephalicpain.Ifthetestwaspositiveandthereductionwasatleast25%inrespecttobasis,asemi-permanentneedle(ASPSEDATELEC,France)wasinsertedafter1min.Onthecontrary,ifpaindidnotlessenafter1min,afurthertenderpointwaschallengedinthesameareaandsoon.Whenpatientsbecameawareofaninitialdecreaseinthepaininallthezonesoftheheadaffected,theywereinvitedtouseaspeciﬁcdiarycardtoscoretheintensityofthepainwithaVASatthefollowingintervals:after10min(T1),after30min(T2),after60min(T3),after120min(T4),andafter24h(T5).IngroupB,thelowerbranchoftheanthelixwasrepeatedlytestedwiththealgometerforabout30stoensureitwasnotsensitive.OnboththeFrenchandChineseauricularmaps,thisareacorrespondstotherepresentationofthesciaticnerve(Fig.1,areaS)andisspeciﬁcallyusedtotreatsciaticpain.Fourneedleswereinsertedinthisarea,twoforeachear.Inallpatients,theearacupuncturewasalwaysper-formedbyanexperiencedacupuncturist.TheanalysisofthediariescollectingVASdatawasconductedbyanimpartialoperatorwhodidnotknowthegroupeachpatientwasin.TheaveragevaluesofVASingroupAandBwerecalculatedatthedifferenttimesofthestudy,andastatis-ticalevaluationofthedifferencesbetweenthevaluesobtainedinT0,T1,T2,T3andT4inthetwogroupsstudiedwasperformedusingananalysisofvariance(ANOVA)forrepeatedmeasuresfollowedbymultiplettestofBonferronitoidentifythesourceofvariance.Moreover,toevaluatethedifferencebetweengroupBandgroupA,attestforunpaireddatawasalwaysper-formedforeachlevelofthevariable‘‘time’’.Inthecaseofproportions,aChisquaretestwasapplied.AllanalyseswereperformedusingtheStatisticalPackagefortheSocialSciences(SPSS)softwareprogram.Allvaluesgiveninthefollowingtextarereportedasarithmeticmean(±SEM).ResultsOnly89patientsoutoftheentiregroupof94(43ingroupA,46ingroupB)completedtheexperiment.Fourpatientswithdrewfromthestudy,becausetheyexperiencedanunbearableexacerbationofpainintheperiodprecedingthelastcontrolat24h(twofromgroupAandtwofromgroupB)andwereexcludedfromthestatisticalanalysissincetheyrequestedtheremovaloftheneedles.OnepatientfromgroupAdidnotgiveherconsenttotheimplantofthesemi-permanentneedles.IngroupA,themeannumberofFig.1Theappropriatearea(M)versustheinappropriatearea(S)usedinthetreatmentofmigraineattacksS174NeurolSci(2011)32(Suppl1):S173–S175123 16 CHAPTER 1. DATA COLLECTION 1.2 Data basics You collect data on dozens of questions from all of the students at your school. How would you organize all of this data? Eﬀective presentation and description of data is a ﬁrst step in most analyses. This section introduces one structure for organizing data as well as some terminology that will be used throughout this book. We use loan data from Lending Club and county data from the US Census Bureau to motivate and illustrate this section’s learning objectives. Learning objectives 1. Identify the individuals and the variables of a study. 2. Identify variables as categorical or numerical. Identify numerical variables as discrete or con- tinuous. 3. Understand what it means for two variables to be associated. 1.2.1 Observations, variables, and data matrices Figure 1.3 displays rows 1, 2, 3, and 50 of a data set for 50 randomly sampled loans oﬀered through Lending Club, which is a peer-to-peer lending company. These observations will be referred to as the loan50 data set. Each row in the table represents a single loan. The formal name for a row is a case or observational unit. The columns represent characteristics, called variables, for each of the loans. For example, the ﬁrst row represents a loan of $7,500 with an interest rate of 7.34%, where the borrower is based in Maryland (MD) and has an income of $70,000. GUIDED PRACTICE 1.2 What is the grade of the ﬁrst loan in Figure 1.3? And what is the home ownership status of the borrower for that ﬁrst loan? For these Guided Practice questions, you can check your answer in the footnote.5 In practice, it is especially important to ask clarifying questions to ensure important aspects of the data are understood. For instance, it is always important to be sure we know what each variable means and the units of measurement. Descriptions of the loan50 variables are given in Figure 1.4. loan amount 7500 25000 14500 ... 3000 interest rate 7.34 9.43 6.08 ... 7.96 term 36 60 36 ... 36 grade A B A ... A state MD OH MO ... CA total income 70000 254000 80000 ... 34000 homeownership rent mortgage mortgage ... rent 1 2 3 ... 50 Figure 1.3: Four rows from the loan50 data matrix. 5The loan’s grade is A, and the borrower rents their residence. 1.2. DATA BASICS 17 variable loan amount interest rate term grade state total income homeownership description Amount of the loan received, in US dollars. Interest rate on the loan, in an annual percentage. The length of the loan, which is always set as a whole number of months. Loan grade, which takes values A through G and represents the quality of the loan and its likelihood of being repaid. US state where the borrower resides. Borrower’s total income, including any second income, in US dollars. Indicates whether the person owns, owns but has a mortgage, or rents. Figure 1.4: Variables and their descriptions for the loan50 data set. The data in Figure 1.3 represent a data matrix, which is a convenient and common way to organize data, especially if collecting data in a spreadsheet. Each row of a data matrix corresponds to a unique case (observational unit), and each column corresponds to a variable. When recording data, use a data matrix unless you have a very good reason to use a diﬀerent structure. This structure allows new cases to be added as rows or new variables as new columns. GUIDED PRACTICE 1.3 The grades for assignments, quizzes, and exams in a course are often recorded in a gradebook that takes the form of a data matrix. How might you organize grade data using a data matrix?6 GUIDED PRACTICE 1.4 We consider data for 3,142 counties in the United States, which includes each county’s name, the state in which it is located, its population in 2017, how its population changed from 2010 to 2017, poverty rate, and six additional characteristics. How might these data be organized in a data matrix?7 The data described in Guided Practice 1.4 represents the county data set, which is shown as a data matrix in Figure 1.5. These data come from the US Census, with much of the data coming from the US Census Bureau’s American Community Survey (ACS). Unlike the Decennial Census, which takes place every 10 years and attempts to collect basic demographic data from every resident of the US, the ACS is an ongoing survey that is sent to approximately 3.5 million households per year. As stated by the ACS website, these data help communities “plan for hospitals and schools, support school lunch programs, improve emergency services, build bridges, and inform businesses looking to add jobs and expand to new markets, and more.”8 A small subset of the variables from the ACS are summarized in Figure 1.6. 6There are multiple strategies that can be followed. One common strategy is to have each student represented by a row, and then add a column for each assignment, quiz, or exam. Under this setup, it is easy to review a single line to understand a student’s grade history. There should also be columns to include student information, such as one column to list student names. 7Each county may be viewed as a case, and there are eleven pieces of information recorded for each case. A table with 3,142 rows and 11 columns could hold these data, where each row represents a county and each column represents a particular piece of information. 8https://www.census.gov/programs-surveys/acs/about.html 18 CHAPTER 1. DATA COLLECTION ... ... ... ... ... ... ... - ... ... ... ... ... .2. DATA BASICS 19 1.2.2 Types of variables Examine the unemp rate, pop, state, and median edu variables in the county data set. Each of these variables is inherently diﬀerent from the other three, yet some share certain characteristics. First consider unemp rate, which is said to be a numerical variable since it can take a wide range of numerical values, and it is sensible to add, subtract, or take averages with those values. On the other hand, we would not classify a variable reporting telephone area codes as numerical since the average, sum, and diﬀerence of area codes doesn’t have any clear meaning. The pop variable is also numerical, although it seems to be a little diﬀerent than unemp rate. This variable of the population count can only take whole non-negative numbers (0, 1, 2, ...). For this reason, the population variable is said to be discrete since it can only take numerical values with jumps. On the other hand, the unemployment rate variable is said to be continuous. The variable state can take up to 51 values after accounting for Washington, DC: AL, AK, ..., and WY. Because the responses themselves are categories, state is called a categorical variable, and the possible values are called the variable’s levels. Finally, consider the median edu variable, which describes the median education level of
county residents and takes values below hs, hs diploma, some college, or bachelors in each county. This variable seems to be a hybrid: it is a categorical variable but the levels have a natural ordering. A variable with these properties is called an ordinal variable, while a regular categorical variable without this type of special ordering is called a nominal variable. To simplify analyses, any ordinal variable in this book will be treated as a nominal (unordered) categorical variable. Figure 1.7: Breakdown of variables into their respective types. EXAMPLE 1.5 Data were collected about students in a statistics course. Three variables were recorded for each student: number of siblings, student height, and whether the student had previously taken a statistics course. Classify each of the variables as continuous numerical, discrete numerical, or categorical. The number of siblings and student height represent numerical variables. Because the number of siblings is a count, it is discrete. Height varies continuously, so it is a continuous numerical variable. The last variable classiﬁes students into two categories – those who have and those who have not taken a statistics course – which makes this variable categorical. GUIDED PRACTICE 1.6 An experiment is evaluating the eﬀectiveness of a new drug in treating migraines. A group variable is used to indicate the experiment group for each patient: treatment or control. The num migraines variable represents the number of migraines the patient experienced during a 3-month period. Classify each variable as either numerical or categorical.9 9The group variable can take just one of two group names, making it categorical. The num migraines variable describes a count of the number of migraines, which is an outcome where basic arithmetic is sensible, which means this is a numerical outcome; more speciﬁcally, since it represents a count, num migraines is a discrete numerical variable. all variablesnumericalcategoricalcontinuousdiscretenominal(unordered categorical)ordinal(ordered categorical) 20 CHAPTER 1. DATA COLLECTION 1.2.3 Relationships between variables Many analyses are motivated by a researcher looking for a relationship between two or more variables. A social scientist may like to answer some of the following questions: (1) If homeownership is lower than the national average in one county, will the percent of multi-unit structures in that county tend to be above or below the national average? (2) Does a higher than average increase in county population tend to correspond to counties with higher or lower median household incomes? (3) How useful a predictor is median education level for the median household income for US counties? To answer these questions, data must be collected, such as the county data set shown in Figure 1.5. Examining summary statistics could provide insights for each of the three questions about counties. Additionally, graphs can be used to visually explore the data. Scatterplots are one type of graph used to study the relationship between two numerical variables. Figure 1.8 compares the variables homeownership and multi unit, which is the percent of units in multi-unit structures (e.g. apartments, condos). Each point on the plot represents a single county. For instance, the highlighted dot corresponds to County 413 in the county data set: Chattahoochee County, Georgia, which has 39.4% of units in multi-unit structures and a homeownership rate of 31.3%. The scatterplot suggests a relationship between the two variables: counties with a higher rate of multi-units tend to have lower homeownership rates. We might brainstorm as to why this relationship exists and investigate the ideas to determine which are the most reasonable explanations. Figure 1.8: A scatterplot of homeownership versus the percent of units that are in multi-unit structures for US counties. The highlighted dot represents Chattahoochee County, Georgia, which has a multi-unit rate of 39.4% and a homeownership rate of 31.3%. Explore this scatterplot and dozens of other scatterplots using American Community Survey data on Tableau Public . The multi-unit and homeownership rates are said to be associated because the plot shows a discernible pattern. When two variables show some connection with one another, they are called associated variables. Associated variables can also be called dependent variables and vice-versa. Homeownership Rate020406080100020406080100lPercent of Units in Multi−Unit Structures 1.2. DATA BASICS 21 Figure 1.9: A scatterplot showing pop change against median hh income. Owsley County of Kentucky, is highlighted, which lost 3.63% of its population from 2010 to 2017 and had median household income of $22,736. Explore this scatterplot and dozens of other scatterplots using American Community Survey data on Tableau Public . GUIDED PRACTICE 1.7 Examine the variables in the loan50 data set, which are described in Figure 1.4 on page 17. Create two questions about possible relationships between variables in loan50 that are of interest to you.10 EXAMPLE 1.8 This example examines the relationship between a county’s population change from 2010 to 2017 and median household income, which is visualized as a scatterplot in Figure 1.9. Are these variables associated? The larger the median household income for a county, the higher the population growth observed for the county. While this trend isn’t true for every county, the trend in the plot is evident. Since there is some relationship between the variables, they are associated. Because there is a downward trend in Figure 1.8 – counties with more units in multi-unit structures are associated with lower homeownership – these variables are said to be negatively associated. A positive association is shown in the relationship between the median hh income and pop change in Figure 1.9, where counties with higher median household income tend to have higher rates of population growth. If two variables are not associated, then they are said to be independent. That is, two variables are independent if there is no evident relationship between the two. ASSOCIATED OR INDEPENDENT, NOT BOTH A pair of variables is either related in some way (associated) or not (independent). No pair of variables is both associated and independent. 10Two example questions: (1) What is the relationship between loan amount and total income? (2) If someone’s income is above the average, will their interest rate tend to be above or below the average? $0$20k$40k$60k$80k$100k$120k−10%0%10%20%30%Median Household IncomePopulation Changeover 7 Yearsl 22 CHAPTER 1. DATA COLLECTION Section summary • Researchers often summarize data in a table, where the rows correspond to individuals or cases and the columns correspond to the variables, the values of which are recorded for each individual. • Variables can be numerical (measured on a numerical scale) or categorical (taking on levels, such as low/medium/high). Numerical variables can be continuous, where all values within a range are possible, or discrete, where only speciﬁc values, usually integer values, are possible. • When there exists a relationship between two variables, the variables are said to be associated or dependent. If the variables are not associated, they are said to be independent. 1.2. DATA BASICS Exercises 23 1.3 Air pollution and birth outcomes, study components. Researchers collected data to examine the relationship between air pollutants and preterm births in Southern California. During the study air pollution levels were measured by air quality monitoring stations. Speciﬁcally, levels of carbon monoxide were recorded in parts per million, nitrogen dioxide and ozone in parts per hundred million, and coarse particulate matter (PM10) in µg/m3. Length of gestation data were collected on 143,196 births between the years 1989 and 1993, and air pollution exposure during gestation was calculated for each birth. The analysis suggested that increased ambient PM10 and, to a lesser degree, CO concentrations may be associated with the occurrence of preterm births.11 (a) Identify the main research question of the study. (b) Who are the subjects in this study, and how many are included? (c) What are the variables in the study? Identify each variable as numerical or categorical. If numerical, state whether the variable is discrete or continuous. If categorical, state whether the variable is ordinal. 1.4 Buteyko method, study components. The Buteyko method is a shallow breathing technique developed by Konstantin Buteyko, a Russian doctor, in 1952. Anecdotal evidence suggests that the Buteyko method can reduce asthma symptoms and improve quality of life. In a scientiﬁc study to determine the eﬀectiveness of this method, researchers recruited 600 asthma patients aged 18-69 who relied on medication for asthma treatment. These patients were randomnly split into two research groups: one practiced the Buteyko method and the other did not. Patients were scored on quality of life, activity, asthma symptoms, and medication reduction on a scale from 0 to 10. On average, the participants in the Buteyko group experienced a signiﬁcant reduction in asthma symptoms and an improvement in quality of life.12 (a) Identify the main research question of the study. (b) Who are the subjects in this study, and how many are included? (c) What are the variables in the study? Identify each variable as numerical or categorical. If numerical, state whether the variable is discrete or continuous. If categorical, state whether the variable is ordinal. 1.5 Cheaters, study components. Researchers studying the relationship between honesty, age and selfcontrol conducted an experiment on 160 children between the ages of 5 and 15. Participants reported their age, sex, and whether they were an only child or not. The researchers asked each child to toss a fair coin in private and to record the outcome (white or black) on a paper sheet, and said they would only
reward children who report white.13 (a) Identify the main research question of the study. (b) Who are the subjects in this study, and how many are included? (c) The study’s ﬁndings can be summarized as follows: ”Half the students were explicitly told not to cheat and the others were not given any explicit instructions. In the no instruction group probability of cheating was found to be uniform across groups based on child’s characteristics. In the group that was explicitly told to not cheat, girls were less likely to cheat, and while rate of cheating didn’t vary by age for boys, it decreased with age for girls.” How many variables were recorded for each subject in the study in order to conclude these ﬁndings? State the variables and their types. 11B. Ritz et al. “Eﬀect of air pollution on preterm birth among children born in Southern California between 1989 and 1993”. In: Epidemiology 11.5 (2000), pp. 502–511. 12J. McGowan. “Health Education: Does the Buteyko Institute Method make a diﬀerence?” In: Thorax 58 (2003). 13Alessandro Bucciol and Marco Piovesan. “Luck or cheating? A ﬁeld experiment on honesty with children”. In: Journal of Economic Psychology 32.1 (2011), pp. 73–78. 24 CHAPTER 1. DATA COLLECTION 1.6 Stealers, study components. In a study of the relationship between socio-economic class and unethical behavior, 129 University of California undergraduates at Berkeley were asked to identify themselves as having low or high social-class by comparing themselves to others with the most (least) money, most (least) education, and most (least) respected jobs. They were also presented with a jar of individually wrapped candies and informed that the candies were for children in a nearby laboratory, but that they could take some if they wanted. After completing some unrelated tasks, participants reported the number of candies they had taken.14 (a) Identify the main research question of the study. (b) Who are the subjects in this study, and how many are included? (c) The study found that students who were identiﬁed as upper-class took more candy than others. How many variables were recorded for each subject in the study in order to conclude these ﬁndings? State the variables and their types. 1.7 Migraine and acupuncture, Part II. Exercise 1.1 introduced a study exploring whether acupuncture had any eﬀect on migraines. Researchers conducted a randomized controlled study where patients were randomly assigned to one of two groups: treatment or control. The patients in the treatment group received acupuncture that was speciﬁcally designed to treat migraines. The patients in the control group received placebo acupuncture (needle insertion at non-acupoint locations). 24 hours after patients received acupuncture, they were asked if they were pain free. What are the explanatory and response variables in this study? 1.8 Sinusitis and antibiotics, Part II. Exercise 1.2 introduced a study exploring the eﬀect of antibiotic treatment for acute sinusitis. Study participants either received either a 10-day course of an antibiotic (treatment) or a placebo similar in appearance and taste (control). At the end of the 10-day period, patients were asked if they experienced improvement in symptoms. What are the explanatory and response variables in this study? 1.9 Fisher’s irises. Sir Ronald Aylmer Fisher was an English statistician, evolutionary biologist, and geneticist who worked on a data set that contained sepal length and width, and petal length and width from three species of iris ﬂowers (setosa, versicolor and virginica). There were 50 ﬂowers from each species in the data set.15 (a) How many cases were included in the data? (b) How many numerical variables are included in the data? Indicate what they are, and if they are continuous or discrete. (c) How many categorical variables are included in the data, and what are they? List the corresponding levels (categories). Photo by Ryan Claussen (http://ﬂic.kr/p/6QTcuX) CC BY-SA 2.0 license 1.10 Smoking habits of UK residents. A survey was conducted to study the smoking habits of UK residents. Below is a data matrix displaying a portion of the data collected in this survey. Note that “£” stands for British Pounds Sterling, “cig” stands for cigarettes, and “N/A” refers to a missing component of the data.16 1 2 3 . . . 1691 sex Female Male Male . . . Male marital age Single 42 44 Single 53 Married . . . 40 . . . Single grossIncome Under £2,600 £10,400 to £15,600 Above £36,400 . . . £2,600 to £5,200 smoke Yes No Yes . . . Yes amtWeekends 12 cig/day N/A 6 cig/day . . . 8 cig/day amtWeekdays 12 cig/day N/A 6 cig/day . . . 8 cig/day (a) What does each row of the data matrix represent? (b) How many participants were included in the survey? (c) Indicate whether each variable in the study is numerical or categorical. If numerical, identify as contin- uous or discrete. If categorical, indicate if the variable is ordinal. 14P.K. Piﬀ et al. “Higher social class predicts increased unethical behavior”. In: Proceedings of the National Academy of Sciences (2012). 15R.A Fisher. “The Use of Multiple Measurements in Taxonomic Problems”. In: Annals of Eugenics 7 (1936), pp. 179–188. 16National STEM Centre, Large Datasets from stats4schools. 1.2. DATA BASICS 25 1.11 US Airports. The visualization below shows the geographical distribution of airports in the contiguous United States and Washington, DC. This visualization was constructed based on a dataset where each observation is an airport.17 (a) List the variables used in creating this visualization. (b) Indicate whether each variable in the study is numerical or categorical. If numerical, identify as contin- uous or discrete. If categorical, indicate if the variable is ordinal. 1.12 UN Votes. The visualization below shows voting patterns the United States, Canada, and Mexico in the United Nations General Assembly on a variety of issues. Speciﬁcally, for a given year between 1946 and 2015, it displays the percentage of roll calls in which the country voted yes for each issue. This visualization was constructed based on a dataset where each observation is a country/year pair.18 (a) List the variables used in creating this visualization. (b) Indicate whether each variable in the study is numerical or categorical. If numerical, identify as contin- uous or discrete. If categorical, indicate if the variable is ordinal. 17Federal Aviation Administration, www.faa.gov/airports/airport safety/airportdata 5010. 18David Robinson. unvotes: United Nations General Assembly Voting Data. R package version 0.2.0. 2017. url: https://CRAN.R-project.org/package=unvotes. 26 CHAPTER 1. DATA COLLECTION 1.3 Overview of data collection principles How do researchers collect data? Why are the results of some studies more reliable than others? The way a researcher collects data depends upon the research goals. In this section, we look at diﬀerent methods of collecting data and consider the types of conclusions that can be drawn from those methods. Learning objectives 1. Distinguish between the population and a sample and between the parameter and a statistic. 2. Know when to summarize a data set using a mean versus a proportion. 3. Understand why anecdotal evidence is unreliable. 4. Identify the four main types of data collection: census, sample survey, experiment, and obser- vation study. 5. Classify a study as observational or experimental, and determine when a study’s results can be generalized to the population and when a causal relationship can be drawn. 1.3.1 Populations and samples Consider the following three research questions: 1. What is the average mercury content in swordﬁsh in the Atlantic Ocean? 2. Over the last 5 years, what is the average time to complete a degree for Duke undergrads? 3. Does a new drug reduce the number of deaths in patients with severe heart disease? Each research question refers to a target population. In the ﬁrst question, the target population is all swordﬁsh in the Atlantic ocean, and each ﬁsh represents a case. Often times, it is too expensive to collect data for every case in a population. Instead, a sample is taken. A sample represents a subset of the cases and is often a small fraction of the population. For instance, 60 swordﬁsh (or some other number) in the population might be selected, and this sample data may be used to provide an estimate of the population average and answer the research question. GUIDED PRACTICE 1.9 For the second and third questions above, identify the target population and what represents an individual case.19 19(2) Notice that this question is only relevant to students who complete their degree; the average cannot be computed using a student who never ﬁnished her degree. Thus, only Duke undergrads who have graduated in the last ﬁve years are part of the population of interest. Each such student would represent an individual case. (3) A person with severe heart disease represents a case. The population includes all people with severe heart disease. 1.3. OVERVIEW OF DATA COLLECTION PRINCIPLES 27 We collect a sample of data to better understand the characteristics of a population. A variable is a characteristic we measure for each individual or case. The overall quantity of interest may be the mean, median, proportion, or some other summary of a population. These population values are called parameters. We estimate the value of a parameter by taking a sample and computing a numerical summary called a statistic based on that sample. Note that the two p’s (population, parameter) go together and the two s’s (sample, statistic) go together. EXAMPLE 1.10 Earlier we asked the question: what is the average mercury content in swordﬁsh in the Atlantic Ocean? Identify the variable to be measured and the parameter and statistic of interest. The variable is the level of mercury content in swordﬁsh in the Atlantic Ocean. It will be measured for each individual swordﬁsh. The parameter of interest is the average mercury content
in all swordﬁsh in the Atlantic Ocean. If we take a sample of 50 swordﬁsh from the Atlantic Ocean, the average mercury content among just those 50 swordﬁsh will be the statistic. Two statistics we will study are the mean (also called the average) and proportion. When we are discussing a population, we label the mean as µ (the Greek letter, mu), while we label the sample mean as ¯x (read as x-bar ). When we are discussing a proportion in the context of a population, we use the label p, while the sample proportion has a label of ˆp (read as p-hat). Generally, we use ¯x to estimate the population mean, µ. Likewise, we use the sample proportion ˆp to estimate the population proportion, p. EXAMPLE 1.11 Is µ a parameter or statistic? What about ˆp? µ is a parameter because it refers to the average of the entire population. ˆp is a statistic because it is calculated from a sample. EXAMPLE 1.12 For the second question regarding time to complete a degree for a Duke undergraduate, is the variable numerical or categorical? What is the parameter of interest? The characteristic that we record on each individual is the number of years until graduation, which is a numerical variable. The parameter of interest is the average time to degree for all Duke undergraduates, and we use µ to describe this quantity. GUIDED PRACTICE 1.13 The third question asked whether a new drug reduces deaths in patients with severe heart disease. Is the variable numerical or categorical? Describe the statistic that should be calculated in this study.20 If these topics are still a bit unclear, don’t worry. We’ll cover them in greater detail in the next chapter. 20The variable is whether or not a patient with severe heart disease dies within the time frame of the study. This is categorical because it will be a yes or a no. The statistic that should be recorded is the proportion of patients that die within the time frame of the study, and we would use ˆp to denote this quantity. 28 CHAPTER 1. DATA COLLECTION Figure 1.10: In February 2010, some media pundits cited one large snow storm as valid evidence against global warming. As comedian Jon Stewart pointed out, “It’s one storm, in one region, of one country.” —————————– February 10th, 2010. 1.3.2 Anecdotal evidence Consider the following possible responses to the three research questions: 1. A man on the news got mercury poisoning from eating swordﬁsh, so the average mercury concentration in swordﬁsh must be dangerously high. 2. I met two students who took more than 7 years to graduate from Duke, so it must take longer to graduate at Duke than at many other colleges. 3. My friend’s dad had a heart attack and died after they gave him a new heart disease drug, so the drug must not work. Each conclusion is based on data. However, there are two problems. First, the data only represent one or two cases. Second, and more importantly, it is unclear whether these cases are actually representative of the population. Data collected in this haphazard fashion are called anecdotal evidence. ANECDOTAL EVIDENCE Be careful of making inferences based on anecdotal evidence. Such evidence may be true and veriﬁable, but it may only represent extraordinary cases. The majority of cases and the average case may in fact be very diﬀerent. Anecdotal evidence typically is composed of unusual cases that we recall based on their striking characteristics. For instance, we may vividly remember the time when our friend bought a lottery ticket and won $250 but forget most the times she bought one and lost. Instead of focusing on the most unusual cases, we should examine a representative sample of many cases. 1.3. OVERVIEW OF DATA COLLECTION PRINCIPLES 29 1.3.3 Explanatory and response variables When we ask questions about the relationship between two variables, we sometimes also want to determine if the change in one variable causes a change in the other. Consider the following rephrasing of an earlier question about the county data set: If there is an increase in the median household income in a county, does this drive an increase in its population? In this question, we are asking whether one variable aﬀects another. If this is our underlying belief, then median household income is the explanatory variable and the population change is the response variable in the hypothesized relationship.21 EXPLANATORY AND RESPONSE VARIABLES When we suspect one variable might causally aﬀect another, we label the ﬁrst variable the explanatory variable and the second the response variable. For many pairs of variables, there is no hypothesized relationship, and these labels would not be applied to either variable in such cases. ASSOCIATION DOES NOT IMPLY CAUSATION Labeling variables as explanatory and response does not guarantee the relationship between the two is actually causal, even if there is an association identiﬁed between the two variables. We use these labels only to keep track of which variable we suspect aﬀects the other. In many cases, the relationship is complex or unknown. It may be unclear whether variable A explains variable B or whether variable B explains variable A. For example, it is now known that a particular protein called REST is much depleted in people suﬀering from Alzheimer’s disease. While this raises hopes of a possible approach for treating Alzheimer’s, it is still unknown whether the lack of the protein causes brain deterioration, whether brain deterioration causes depletion in the REST protein, or whether some third variable causes both brain deterioration and REST depletion. That is, we do not know if the lack of the protein is an explanatory variable or a response variable. Perhaps it is both.22 21Sometimes the explanatory variable is called the independent variable and the response variable is called the dependent variable. However, this becomes confusing since a pair of variables might be independent or dependent, so we avoid this language. 22nytimes.com/2014/03/20/health/fetal-gene-may-protect-brain-from-alzheimers-study-ﬁnds.html might affectexplanatoryvariableresponsevariable 30 CHAPTER 1. DATA COLLECTION 1.3.4 Observational studies versus experiments There are two primary types of data collection: observational studies and experiments. Researchers perform an observational study when they collect data without interfering with how the data arise. For instance, researchers may collect information via surveys, review medical or company records, or follow a cohort of many similar individuals to study why certain diseases might develop. In each of these situations, researchers merely observe or take measurements of things that arise naturally. When researchers want to investigate the possibility of a causal connection, they conduct an experiment. For all experiments, the researchers must impose a treatment. For most studies there will be both an explanatory and a response variable. For instance, we may suspect administering a drug will reduce mortality in heart attack patients over the following year. To check if there really is a causal connection between the explanatory variable and the response, researchers will collect a sample of individuals and split them into groups. The individuals in each group are assigned a treatment. When individuals are randomly assigned to a group, the experiment is called a randomized experiment. For example, each heart attack patient in the drug trial could be randomly assigned into one of two groups: the ﬁrst group receives a placebo (fake treatment) and the second group receives the drug. See the case study in Section 1.1 for another example of an experiment, though that study did not employ a placebo. EXAMPLE 1.14 Suppose that a researcher is interested in the average tip customers at a particular restaurant give. Should she carry out an observational study or an experiment? In addressing this question, we ask, “Will the researcher be imposing any treatment?” Because there is no treatment or interference that would be applicable here, it will be an observational study. Additionally, one consideration the researcher should be aware of is that, if customers know their tips are being recorded, it could change their behavior, making the results of the study inaccurate. ASSOCIATION = CAUSATION In general, association does not imply causation, and causation can only be inferred from a randomized experiment. Section summary • The population is the entire group that the researchers are interested in. Because it is usually too costly to gather the data for the entire population, researchers will collect data from a sample, representing a subset of the population. • A parameter is a true quantity for the entire population, while a statistic is what is calculated from the sample. A parameter is about a population and a statistic is about a sample. Remember: p goes with p and s goes with s. • Two common summary quantities are mean (for numerical variables) and proportion (for categorical variables). • Finding a good estimate for a population parameter requires a random sample; do not gener- alize from anecdotal evidence. • There are two primary types of data collection: observational studies and experiments. In an experiment, researchers impose a treatment to look for a causal relationship between the treatment and the response. In an observational study, researchers simply collect data without imposing any treatment. • Remember: Correlation is not causation! In other words, an association between two variables does not imply that one causes the other. Proving a causal relationship requires a well-designed experiment. 1.3. OVERVIEW OF DATA COLLECTION PRINCIPLES 31 Exercises 1.13 Air pollution and birth outcomes, scope of inference. Exercise 1.3 introduces a study where researchers collected data to examine the relationship between air pollutants and preterm births in Southern California. During the study air pollution levels were measured by air quality monitoring stations. Length of gestation
data were collected on 143,196 births between the years 1989 and 1993, and air pollution exposure during gestation was calculated for each birth. (a) Identify the population of interest and the sample in this study. (b) Comment on whether or not the results of the study can be generalized to the population, and if the ﬁndings of the study can be used to establish causal relationships. 1.14 Cheaters, scope of inference. Exercise 1.5 introduces a study where researchers studying the relationship between honesty, age, and self-control conducted an experiment on 160 children between the ages of 5 and 15. The researchers asked each child to toss a fair coin in private and to record the outcome (white or black) on a paper sheet, and said they would only reward children who report white. Half the students were explicitly told not to cheat and the others were not given any explicit instructions. Diﬀerences were observed in the cheating rates in the instruction and no instruction groups, as well as some diﬀerences across children’s characteristics within each group. (a) Identify the population of interest and the sample in this study. (b) Comment on whether or not the results of the study can be generalized to the population, and if the ﬁndings of the study can be used to establish causal relationships. 1.15 Buteyko method, scope of inference. Exercise 1.4 introduces a study on using the Buteyko shallow breathing technique to reduce asthma symptoms and improve quality of life. As part of this study 600 asthma patients aged 18-69 who relied on medication for asthma treatment were recruited and randomly assigned to two groups: one practiced the Buteyko method and the other did not. Those in the Buteyko group experienced, on average, a signiﬁcant reduction in asthma symptoms and an improvement in quality of life. (a) Identify the population of interest and the sample in this study. (b) Comment on whether or not the results of the study can be generalized to the population, and if the ﬁndings of the study can be used to establish causal relationships. 1.16 Stealers, scope of inference. Exercise 1.6 introduces a study on the relationship between socioeconomic class and unethical behavior. As part of this study 129 University of California Berkeley undergraduates were asked to identify themselves as having low or high social-class by comparing themselves to others with the most (least) money, most (least) education, and most (least) respected jobs. They were also presented with a jar of individually wrapped candies and informed that the candies were for children in a nearby laboratory, but that they could take some if they wanted. After completing some unrelated tasks, participants reported the number of candies they had taken. It was found that those who were identiﬁed as upper-class took more candy than others. (a) Identify the population of interest and the sample in this study. (b) Comment on whether or not the results of the study can be generalized to the population, and if the ﬁndings of the study can be used to establish causal relationships. 32 CHAPTER 1. DATA COLLECTION 1.17 Relaxing after work. The General Social Survey asked the question, “After an average work day, about how many hours do you have to relax or pursue activities that you enjoy?” to a random sample of 1,155 Americans. The average relaxing time was found to be 1.65 hours. Determine which of the following is an observation, a variable, a sample statistic (value calculated based on the observed sample), or a population parameter. (a) An American in the sample. (b) Number of hours spent relaxing after an average work day. (c) 1.65. (d) Average number of hours all Americans spend relaxing after an average work day. 1.18 Cats on YouTube. Suppose you want to estimate the percentage of videos on YouTube that are cat videos. It is impossible for you to watch all videos on YouTube so you use a random video picker to select 1000 videos for you. You ﬁnd that 2% of these videos are cat videos. Determine which of the following is an observation, a variable, a sample statistic (value calculated based on the observed sample), or a population parameter. (a) Percentage of all videos on YouTube that are cat videos. (b) 2%. (c) A video in your sample. (d) Whether or not a video is a cat video. 1.4. OBSERVATIONAL STUDIES AND SAMPLING STRATEGIES 33 1.4 Observational studies and sampling strategies You have probably read or heard claims from many studies and polls. A background in statistical reasoning will help you assess the validity of such claims. Some of the big questions we address in this section include: • If a study ﬁnds a relationship between two variables, such as eating chocolate and positive health outcomes, is it reasonable to conclude eating chocolate improves health outcomes? • How do opinion polls work? How do research organizations collect the data, and what types of bias should we look out for? Learning objectives 1. Identify possible confounding factors in a study and explain, in context, how they could con- found. 2. Distinguish among and describe a convenience sample, a volunteer sample, and a random sample. 3. Identify and describe the eﬀects of diﬀerent types of bias in sample surveys, including under- coverage, non-response, and response bias. 4. Identify and describe how to implement diﬀerent random sampling methods, including simple, systematic, stratiﬁed, and cluster. 5. Recognize the beneﬁts and drawbacks of choosing one sampling method over another. 6. Understand when it is valid to generalize and to what population that generalization can be made. 1.4.1 Observational studies Generally, data in observational studies are collected only by monitoring what occurs, while experiments require the primary explanatory variable in a study be assigned for each subject by the researchers. Making causal conclusions based on experiments is often reasonable. However, making the same causal conclusions based on observational data is treacherous and is not recommended. Observational studies are generally only suﬃcient to show associations. GUIDED PRACTICE 1.15 Suppose an observational study tracked sunscreen use and skin cancer, and it was found people who use sunscreen are more likely to get skin cancer than people who do not use sunscreen. Does this mean sunscreen causes skin cancer?23 Some previous research tells us that using sunscreen actually reduces skin cancer risk, so maybe there is another variable that can explain this hypothetical association between sunscreen usage and skin cancer. One important piece of information that is absent is sun exposure. Sun exposure is what is called a confounding variable (also called a lurking variable, confounding factor, or a confounder). 23No. See the paragraph following the exercise for an explanation. 34 CHAPTER 1. DATA COLLECTION CONFOUNDING VARIABLE A confounding variable is a variable that is associated with both the explanatory and response variables. Because of the confounding variable’s association with both variables, we do not know if the response is due to the explanatory variable or due to the confounding variable. Sun exposure is a confounding factor because it is associated with both the use of sunscreen and the development of skin cancer. People who are out in the sun all day are more likely to use sunscreen, and people who are out in the sun all day are more likely to get skin cancer. Research shows us the development of skin cancer is due to the sun exposure. The variables of sunscreen usage and sun exposure are confounded, and without this research, we would have no way of knowing which one was the true cause of skin cancer. EXAMPLE 1.16 In a study that followed 1,169 non-diabetic adults who had been hospitalized for a ﬁrst heart attack, the people that reported eating chocolate had increased survival rate over the next 8 years than those that reported not eating chocolate. Also, those who ate more chocolate tended to live longer on average. The researchers controlled for several confounding factors, such as age, physical activity, smoking, and many other factors. Can we conclude that the consumption of chocolate caused the people to live longer? This is an observational study, not a controlled randomized experiment. Even though the researchers controlled for many possible variables, there may still be other confounding factors. (Can you think of any that weren’t mentioned?) While it is possible that the chocolate had an eﬀect, this study cannot prove that chocolate increased the survival rate of patients. EXAMPLE 1.17 The authors who conducted the study did warn in the article that additional studies would be necessary to determine whether the correlation between chocolate consumption and survival translates to any causal relationship. That is, they acknowledged that there may be confounding factors. One possible confounding factor not considered was mental health. In context, explain what it would mean for mental health to be a confounding factor in this study. Mental health would be a confounding factor if, for example, people with better mental health tended to eat more chocolate, and those with better mental health also were less likely to die within the 8 year study period. Notice that if better mental health were not associated with eating more chocolate, it would not be considered a confounding factor since it wouldn’t explain the observed associated between eating chocolate and having a better survival rate. If better mental health were associated only with eating chocolate and not with a better survival rate, then it would also not be confounding for the same reason. Only if a variable that is associated with both the explanatory variable of interest (chocolate) and the outcome variable in the study (survival during the 8 year study period) can it be considered a confounding factor. While one method to justify making causal conclusions from observational studies is to exhaust the se
arch for confounding variables, there is no guarantee that all confounding variables can be examined or measured. In the same way, the county data set is an observational study with confounding variables, and its data cannot be used to make causal conclusions. sun exposureuse sunscreenskin cancer? 1.4. OBSERVATIONAL STUDIES AND SAMPLING STRATEGIES 35 GUIDED PRACTICE 1.18 Figure 1.8 shows a negative association between the homeownership rate and the percentage of multi-unit structures in a county. However, it is unreasonable to conclude that there is a causal relationship between the two variables. Suggest one or more other variables that might explain the relationship visible in Figure 1.8.24 Observational studies come in two forms: prospective and retrospective studies. A prospective study identiﬁes individuals and collects information as events unfold. For instance, medical researchers may identify and follow a group of similar individuals over many years to assess the possible inﬂuences of behavior on cancer risk. One example of such a study is The Nurses’ Health Study, started in 1976 and expanded in 1989. This prospective study recruits registered nurses and then collects data from them using questionnaires. Retrospective studies collect data after events have taken place, e.g. researchers may review past events in medical records. Some data sets, such as county, may contain both prospectively- and retrospectively-collected variables. Local governments prospectively collect some variables as events unfolded (e.g. retails sales) while the federal government retrospectively collected others during the 2010 census (e.g. county population counts). 1.4.2 Sampling from a population We might try to estimate the time to graduation for Duke undergraduates in the last 5 years by collecting a sample of students. All graduates in the last 5 years represent the population, and graduates who are selected for review are collectively called the sample. The goal is to use information from the sample to generalize or make an inference to the population. In order to be able to generalize, we must randomly select a sample from the population of interest. The most basic type of random selection is equivalent to how raﬄes are conducted. For example, in selecting graduates, we could write each graduate’s name on a raﬄe ticket and draw 100 tickets. The selected names would represent a random sample of 100 graduates. Figure 1.11: In this graphic, ﬁve graduates are randomly selected from the population to be included in the sample. Why pick a sample randomly? Why not just pick a sample by hand? Consider the following scenario. 24Answers will vary. Population density may be important. If a county is very dense, then this may require a larger fraction of residents to live in multi-unit structures. Additionally, the high density may contribute to increases in property value, making homeownership infeasible for many residents. all graduatessample 36 CHAPTER 1. DATA COLLECTION Figure 1.12: Instead of sampling from all graduates equally, a nutrition major might inadvertently pick graduates with health-related majors disproportionately often. EXAMPLE 1.19 Suppose we ask a student who happens to be majoring in nutrition to select several graduates for the study. What kind of students do you think she might select? Do you think her sample would be representative of all graduates? Perhaps she would pick a disproportionate number of graduates from health-related ﬁelds. Or perhaps her selection would be well-representative of the population. When selecting samples by hand, we run the risk of picking a biased sample, even if that bias is unintentional or diﬃcult to discern. If the student majoring in nutrition picked a disproportionate number of graduates from healthrelated ﬁelds, this would introduce undercoverage bias into the sample. Undercoverage bias occurs when some individuals of the population are inherently less likely to be included in the sample than others, making the sample not representative of the population. In the example, this bias creates a problem because a degree in health-related ﬁelds might take more or less time to complete than a degree in other ﬁelds. Suppose that it takes longer. Since graduates from other ﬁelds would be less likely to be in the sample, the undercoverage bias would cause her to overestimate the parameter. Sampling randomly resolves the problem of undercoverage bias, if the sample is randomly selected from the entire population of interest. If the sample is randomly selected from only a subset of the population, say, only graduates from health-related ﬁelds, then the sample will not be representative of the population of interest. Generalizations can only be made to the population from which the sample is randomly selected. The most basic random sample is called a simple random sample, which is equivalent to using a raﬄe to select cases. This means that each case in the population has an equal chance of being included and there is no implied connection between the cases in the sample. A common downfall is a convenience sample, where individuals who are easily accessible are more likely to be included in the sample. For instance, if a political survey is done by stopping people walking in the Bronx, this will not represent all of New York City. It is often diﬃcult to discern what sub-population a convenience sample represents. Figure 1.13: Due to the possibility of non-response, surveys studies may only reach a certain group within the population. It is diﬃcult, and often times impossible, to completely ﬁx this problem. all graduatessamplegraduates fromhealth−related fieldspopulation of interestsamplepopulation actuallysampled 1.4. OBSERVATIONAL STUDIES AND SAMPLING STRATEGIES 37 Similarly, a volunteer sample is one in which people’s responses are solicited and those who choose to participate, respond. This is a problem because those who choose to participate may tend to have diﬀerent opinions than the rest of the population, resulting in a biased sample. GUIDED PRACTICE 1.20 We can easily access ratings for products, sellers, and companies through websites. These ratings are based only on those people who go out of their way to provide a rating. If 50% of online reviews for a product are negative, do you think this means that 50% of buyers are dissatisﬁed with the product?25 The act of taking a random sample helps minimize bias; however, bias can crop up in other ways. Even when people are picked at random, e.g. for surveys, caution must be exercised if the non-response is high. For instance, if only 30% of the people randomly sampled for a survey actually respond, then it is unclear whether the results are representative of the entire population. This non-response bias can skew results. Even if a sample has no undercoverage bias and no non-response bias, there is an additional type of bias that often crops up and undermines the validity of results, known as response bias. Response bias refers to a broad range of factors that inﬂuence how a person responds, such as question wording, question order, and inﬂuence of the interviewer. This type of bias can be present even when we collect data from an entire population in what is called a census. Because response bias is often subtle, one must pay careful attention to how questions were asked when attempting to draw conclusions from the data. EXAMPLE 1.21 Suppose a high school student wants to investigate the student body’s opinions on the food in the cafeteria. Let’s assume that she manages to survey every student in the school. How might response bias arise in this context? There are many possible correct answers to this question. For example, students might respond diﬀerently depending upon who asks the question, such as a school friend or someone who works in the cafeteria. The wording of the question could introduce response bias. Students would likely respond diﬀerently if asked “Do you like the food in the cafeteria?” versus “The food in the cafeteria is pretty bad, don’t you think?” WATCH OUT FOR BIAS Undercoverage bias, non-response bias, and response bias can still exist within a random sample. Always determine how a sample was chosen, ask what proportion of people failed to respond, and critically examine the wording of the questions. When there is no bias in a sample, increasing the sample size tends to increase the precision and reliability of the estimate. When a sample is biased, it may be impossible to decipher helpful information from the data, even if the sample is very large. 25Answers will vary. From our own anecdotal experiences, we believe people tend to rant more about products that fell below expectations than rave about those that perform as expected. For this reason, we suspect there is a negative bias in product ratings on sites like Amazon. However, since our experiences may not be representative, we also keep an open mind. 38 CHAPTER 1. DATA COLLECTION GUIDED PRACTICE 1.22 A researcher sends out questionnaires to 50 randomly selected households in a particular town asking whether or not they support the addition of a traﬃc light in their neighborhood. Because only 20% of the questionnaires are returned, she decides to mail questionnaires to 50 more randomly selected households in the same neighborhood. Comment on the usefulness of this approach.26 1.4.3 Simple, systematic, stratiﬁed, cluster, and multistage sampling Almost all statistical methods for observational data rely on a sample being random and unbiased. When a sample is collected in a biased way, these statistical methods will not generally produce reliable information about the population. The idea of a simple random sample was introduced in the last section. Here we provide a more technical treatment of this method and introduce four new random sampling methods: systematic, stratiﬁed, cluster, and multistage.27 Figure 1.14 provides a graphical representation of simp
le versus systematic sampling while Figure 1.15 provides a graphical representation of stratiﬁed, cluster, and multistage sampling. Simple random sampling is probably the most intuitive form of random sampling. Consider the salaries of Major League Baseball (MLB) players, where each player is a member of one of the league’s 30 teams. For the 2019 season, N, the population size or total number of players, is 750. To take a simple random sample of n = 120 of these baseball players and their salaries, we could number each player from 1 to 750. Then we could randomly select 120 numbers between 1 and 750 (without replacement) using a random number generator or random digit table. The players with the selected numbers would comprise our sample. Two properties are always true in a simple random sample: 1. Each case in the population has an equal chance of being included in the sample. 2. Each group of n cases has an equal chance of making up the sample. The statistical methods in this book focus on data collected using simple random sampling. Note that Property 2 – that each group of n cases has an equal chance making up the sample – is not true for the remaining four sampling techniques. As you read each one, consider why. Though less common than simple random sampling, systematic sampling is sometimes used when there exists a convenient list of all of the individuals of the population. Suppose we have a roster with the names of all the MLB players from the 2019 season. To take a systematic random sample, number them from 1 to 750. Select one random number between 1 and 750 and let that player be the ﬁrst individual in the sample. Then, depending on the desired sample size, select every 10th number or 20th number, for example, to arrive at the sample.28 If there are no patterns in the salaries based on the numbering then this could be a reasonable method. EXAMPLE 1.23 A systematic sample is not the same as a simple random sample. Provide an example of a sample that can come from a simple random sample but not from a systematic random sample. Answers can vary. If we take a sample of size 3, then it is possible that we could sample players numbered 1, 2, and 3 in a simple random sample. Such a sample would be impossible from a systematic sample. Property 2 of simple random samples does not hold for other types of random samples. 26The researcher should be concerned about non-response bias, and sampling more people will not eliminate this issue. The same type of people that did not respond to the ﬁrst survey are likely not going to respond to the second survey. Instead, she should make an eﬀort to reach out to the households from the original sample that did not respond and solicit their feedback, possibly by going door-to-door. 27Multistage sampling is not part of the AP syllabus. 28If we want a sample of size n = 150, it would make sense to select every 5th player since 750/150 = 5. Suppose we randomly select the number 741. Then player 741, 746, 1, 6, 11, · · · , 731, and 736 would make up the sample. 1.4. OBSERVATIONAL STUDIES AND SAMPLING STRATEGIES 39 Figure 1.14: Examples of simple random sampling and systematic sampling. In the top panel, simple random sampling was used to randomly select 18 cases. In the lower panel, systematic random sampling was used to select every 7th individual. Index●●●●●●●●●●●●●●●●●●●●●●●●●●●●●●●●●●●●●●●●●●●●●●●●●●●●●●●●●●●●●●●●●●●●●●●●●●●●●●●●●●●●●●●●●●●●●●●●●●●●●●●●●●●●●●●●●●●●●●●●●●●●●●●●●●●●●●●●●●●●●●●●●●●●●●●●●●●●●●●●●●●●●●●●●●●●●●●●●●●●●●●●●●●●●●●●●●●●●●●●●●●●●●●●●●●●●●●●●●●●●●●●●●●●●●●●●●●●●●●●●●●●●●●●●●●●●●●●●●●●●●●●●●●●●●●●●●●●●●●●●●●●●●●●●●●●●●●●●●●●●●●●●●●●●●●●●●●●●●●●●●●●●●●●●●●●●●●●●●●●● 40 CHAPTER 1. DATA COLLECTION Sometimes there is a variable that is known to be associated with the quantity we want to estimate. In this case, a stratiﬁed random sample might be selected. Stratiﬁed sampling is a divide-and-conquer sampling strategy. The population is divided into groups called strata. The strata are chosen so that similar cases are grouped together and a sampling method, usually simple random sampling, is employed to select a certain number or a certain proportion of the whole within each stratum. In the baseball salary example, the 30 teams could represent the strata; some teams have a lot more money (we’re looking at you, Yankees). EXAMPLE 1.24 For this baseball example, brieﬂy explain how to select a stratiﬁed random sample of size n = 120. Each team can serve as a stratum, and we could take a simple random sample of 4 players from each of the 30 teams, yielding a sample of 120 players. Stratiﬁed sampling is inherently diﬀerent than simple random sampling. For example, the stratiﬁed sampling approach described would make it impossible for the entire Yankees team to be included in the sample. EXAMPLE 1.25 Stratiﬁed sampling is especially useful when the cases in each stratum are very similar with respect to the outcome of interest. Why is it good for cases within each stratum to be very similar? We should get a more stable estimate for the subpopulation in a stratum if the cases are very similar. These improved estimates for each subpopulation will help us build a reliable estimate for the full population. For example, in a simple random sample, it is possible that just by random chance we could end up with proportionally too many Yankees players in our sample, thus overestimating the true average salary of all MLB players. A stratiﬁed random sample can assure proportional representation from each team. Next, let’s consider a sampling technique that randomly selects groups of people. Cluster sampling is much like simple random sampling, but instead of randomly selecting individuals, we randomly select groups or clusters. Unlike stratiﬁed sampling, cluster sampling is most helpful when there is a lot of case-to-case variability within a cluster but the clusters themselves don’t look very diﬀerent from one another. That is, we expect individual strata to be homogeneous (self-similar), while we expect individual clusters to be heterogeneous (diverse) with respect to the variable of interest. Sometimes cluster sampling can be a more economical random sampling technique than the alternatives. For example, if neighborhoods represented clusters, this sampling method works best when each neighborhood is very diverse. Because each neighborhood itself encompasses diversity, a cluster sample can reduce the time and cost associated with data collection, because the interviewer would need only go to some of the neighborhoods rather than to all parts of a city, in order to collect a useful sample. Multistage sampling, also called multistage cluster sampling, is a two (or more) step strategy. The ﬁrst step is to take a cluster sample, as described above. Then, instead of including all of the individuals in these clusters in our sample, a second sampling method, usually simple random sampling, is employed within each of the selected clusters. In the neighborhood example, we could ﬁrst randomly select some number of neighborhoods and then take a simple random sample from just those selected neighborhoods. As seen in Figure 1.15, stratiﬁed sampling requires observations to be sampled from every stratum. Multistage sampling selects observations only from those clusters that were randomly selected in the ﬁrst step. It is also possible to have more than two steps in multistage sampling. Each cluster may be naturally divided into subclusters. For example, each neighborhood could be divided into streets. To take a three-stage sample, we could ﬁrst select some number of clusters (neighborhoods), and then, within the selected clusters, select some number of subclusters (streets). Finally, we could select some number of individuals from each of the selected streets. 1.4. OBSERVATIONAL STUDIES AND SAMPLING STRATEGIES 41 Figure 1.15: Examples of stratiﬁed, cluster, and multistage sampling. In the top panel, stratiﬁed sampling was used: cases were grouped into strata, and then simple random sampling was employed within each stratum. In the middle panel, cluster sampling was used, where data were binned into nine cluster and three clusters were randomly selected. In the bottom panel, multistage sampling was used. Data were binned into the nine clusters, three of the cluster were randomly selected, and then six cases were randomly sampled in each of the three selected clusters. IndexlllllllllllllllllllllllllllllllllllllllllllllllllllllllllllllllllllllllllllllllllllllllllllllllllllllllllllllllllllllllllllllllllllllllllllllllllllllllllllllllllllllllllllStratum 1Stratum 2Stratum 3Stratum 4Stratum 5Stratum 6IndexlllllllllllllllllllllllllllllllllllllllllllllllllllllllllllllllllllllllllllllllllllllllllllllllllllllllllllllllllllllllllllllllllllllllllllllllllllllllllllllllllllllllllllllllllllllllllllllllllllllllllllllllCluster 1Cluster 2Cluster 3Cluster 4Cluster 5Cluster 6Cluster 7Cluster 8Cluster 9lllllllllllllllllllllllllllllllllllllllllllllllllllllllllllllllllllllllllllllllllllllllllllllllllllllllllllllllllllllllllllllllllllllllllllllllllllllllllllllllllllCluster 1Cluster 2Cluster 3Cluster 4Cluster 5Cluster 6Cluster 7Cluster 8Cluster 9 42 CHAPTER 1. DATA COLLECTION EXAMPLE 1.26 Suppose we are interested in estimating the proportion of students at a certain school that have part-time jobs. It is believed that older students are more likely to work than younger students. What sampling method should be employed? Describe how to collect such a sample to get a sample size of 60. Because grade level aﬀects the likelihood of having a part-time job, we should take a stratiﬁed random sample. To do this, we can take a simple random sample of 15 students from each grade. This will give us equal representation from each grade. Note: in a simple random sample, just by random chance we might get too many students who are older or younger, which could make the estimate too high or too low. Also, there are no well-deﬁned
clusters in this example. We wouldn’t want to use the grades as clusters and sample everyone from a couple of the grades. This would create too large a sample and would not give us the nice representation from each grade aﬀorded by the stratiﬁed random sample. EXAMPLE 1.27 Suppose we are interested in estimating the malaria rate in a densely tropical portion of rural Indonesia. We learn that there are 30 villages in that part of the Indonesian jungle, each more or less similar to the next. Our goal is to test 150 individuals for malaria. What sampling method should be employed? A simple random sample would likely draw individuals from all 30 villages, which could make data collection extremely expensive. Stratiﬁed sampling would be a challenge since it is unclear how we would build strata of similar individuals. However, multistage cluster sampling seems like a very good idea. First, we might randomly select half the villages, then randomly select 10 people from each. This would probably reduce our data collection costs substantially in comparison to a simple random sample and would still give us reliable information. ADVANCED SAMPLING TECHNIQUES REQUIRE ADVANCED METHODS The methods of inference covered in this book generally only apply to simple random samples. More advanced analysis techniques are required for systematic, stratiﬁed, cluster, and multistage random sampling. 1.4. OBSERVATIONAL STUDIES AND SAMPLING STRATEGIES 43 Section summary • In an observational study, one must always consider the existence of confounding factors. A confounding factor is a “spoiler variable” that could explain an observed relationship between the explanatory variable and the response. Remember: For a variable to be confounding it must be associated with both the explanatory variable and the response variable. • When taking a sample from a population, avoid convenience samples and volunteer sam- ples, which likely introduce bias. Instead, use a random sampling method. • Generalizations from a sample can be made to a population only if the sample is random. Furthermore, the generalization can be made only to the population from which the sample was randomly selected, not to a larger or diﬀerent population. • Random sampling from the entire population of interest avoids the problem of undercoverage bias. However, response bias and non-response bias can be present in any type of sample, random or not. • In a simple random sample, every individual as well as every group of individuals has the same probability of being in the sample. A common way to select a simple random sample is to number each individual of the population from 1 to N. Using a random digit table or a random number generator, numbers are randomly selected without replacement and the corresponding individuals become part of the sample. • A systematic random sample involves choosing from of a population using a random starting point, and then selecting members according to a ﬁxed, periodic interval (such as every 10th member). • A stratiﬁed random sample involves randomly sampling from every strata, where the strata should correspond to a variable thought to be associated with the variable of interest. This ensures that the sample will have appropriate representation from each of the diﬀerent strata and reduces variability in the sample estimates. • A cluster random sample involves randomly selecting a set of clusters, or groups, and then collecting data on all individuals in the selected clusters. This can be useful when sampling clusters is more convenient and less expensive than sampling individuals, and it is an eﬀective strategy when each cluster is approximately representative of the population. • Remember: Individual strata should be homogeneous (self-similar), while individual clusters should be heterogeneous (diverse). For example, if smoking is correlated with what is being estimated, let one stratum be all smokers and the other be all non-smokers, then randomly select an appropriate number of individuals from each strata. Alternately, if age is correlated with the variable being estimated, one could randomly select a subset of clusters, where each cluster has mixed age groups. 44 CHAPTER 1. DATA COLLECTION Exercises 1.19 Course satisfaction across sections. A large college class has 160 students. All 160 students attend the lectures together, but the students are divided into 4 groups, each of 40 students, for lab sections administered by diﬀerent teaching assistants. The professor wants to conduct a survey about how satisﬁed the students are with the course, and he believes that the lab section a student is in might aﬀect the student’s overall satisfaction with the course. (a) What type of study is this? (b) Suggest a sampling strategy for carrying out this study. 1.20 Housing proposal across dorms. On a large college campus ﬁrst-year students and sophomores live in dorms located on the eastern part of the campus and juniors and seniors live in dorms located on the western part of the campus. Suppose you want to collect student opinions on a new housing structure the college administration is proposing and you want to make sure your survey equally represents opinions from students from all years. (a) What type of study is this? (b) Suggest a sampling strategy for carrying out this study. 1.21 Internet use and life expectancy. The following scatterplot was created as part of a study evaluating the relationship between estimated life expectancy at birth (as of 2014) and percentage of internet users (as of 2009) in 208 countries for which such data were available.29 (a) Describe the relationship between life expectancy and percentage of internet users. (b) What type of study is this? (c) State a possible confounding variable that might explain this relationship and describe its potential eﬀect. 1.22 Stressed out, Part I. A study that surveyed a random sample of otherwise healthy high school students found that they are more likely to get muscle cramps when they are stressed. The study also noted that students drink more coﬀee and sleep less when they are stressed. (a) What type of study is this? (b) Can this study be used to conclude a causal relationship between increased stress and muscle cramps? (c) State possible confounding variables that might explain the observed relationship between increased stress and muscle cramps. 1.23 Evaluate sampling methods. A university wants to determine what fraction of its undergraduate student body support a new $25 annual fee to improve the student union. For each proposed method below, indicate whether the method is reasonable or not. (a) Survey a simple random sample of 500 students. (b) Stratify students by their ﬁeld of study, then sample 10% of students from each stratum. (c) Cluster students by their ages (e.g. 18 years old in one cluster, 19 years old in one cluster, etc.), then randomly sample three clusters and survey all students in those clusters. 1.24 Random digit dialing. The Gallup Poll uses a procedure called random digit dialing, which creates phone numbers based on a list of all area codes in America in conjunction with the associated number of residential households in each area code. Give a possible reason the Gallup Poll chooses to use random digit dialing instead of picking phone numbers from the phone book. 29CIA Factbook, Country Comparisons, 2014. Percent Internet UsersLife Expectancy at Birth0%20%40%60%80%100%5060708090 1.4. OBSERVATIONAL STUDIES AND SAMPLING STRATEGIES 45 1.25 Haters are gonna hate, study conﬁrms. A study published in the Journal of Personality and Social Psychology asked a group of 200 randomly sampled men and women to evaluate how they felt about various subjects, such as camping, health care, architecture, taxidermy, crossword puzzles, and Japan in order to measure their attitude towards mostly independent stimuli. Then, they presented the participants with information about a new product: a microwave oven. This microwave oven does not exist, but the participants didn’t know this, and were given three positive and three negative fake reviews. People who reacted positively to the subjects on the dispositional attitude measurement also tended to react positively to the microwave oven, and those who reacted negatively tended to react negatively to it. Researchers concluded that “some people tend to like things, whereas others tend to dislike things, and a more thorough understanding of this tendency will lead to a more thorough understanding of the psychology of attitudes.”30 (a) What are the cases? (b) What is (are) the response variable(s) in this study? (c) What is (are) the explanatory variable(s) in this study? (d) Does the study employ random sampling? (e) Is this an observational study or an experiment? Explain your reasoning. (f) Can we establish a causal link between the explanatory and response variables? (g) Can the results of the study be generalized to the population at large? 1.26 Family size. Suppose we want to estimate household size, where a “household” is deﬁned as people living together in the same dwelling, and sharing living accommodations. If we select students at random at an elementary school and ask them what their family size is, will this be a good measure of household size? Or will our average be biased? If so, will it overestimate or underestimate the true value? 1.27 Sampling strategies. A statistics student who is curious about the relationship between the amount of time students spend on social networking sites and their performance at school decides to conduct a survey. Various research strategies for collecting data are described below. In each, name the sampling method proposed and any bias you might expect. (a) He randomly samples 40 students from the study’s population, gives them the survey, asks them to ﬁll it out and bring it back the next day. (b) He gives out the survey only to his friends, making sure each one of them ﬁlls out the
survey. (c) He posts a link to an online survey on Facebook and asks his friends to ﬁll out the survey. (d) He randomly samples 5 classes and asks a random sample of students from those classes to ﬁll out the survey. 1.28 Reading the paper. Below are excerpts from two articles published in the NY Times: (a) An article titled Risks: Smokers Found More Prone to Dementia states the following:31 “Researchers analyzed data from 23,123 health plan members who participated in a voluntary exam and health behavior survey from 1978 to 1985, when they were 50-60 years old. 23 years later, about 25% of the group had dementia, including 1,136 with Alzheimer’s disease and 416 with vascular dementia. After adjusting for other factors, the researchers concluded that pack-a-day smokers were 37% more likely than nonsmokers to develop dementia, and the risks went up with increased smoking; 44% for one to two packs a day; and twice the risk for more than two packs.” Based on this study, can we conclude that smoking causes dementia later in life? Explain your reasoning. (b) Another article titled The School Bully Is Sleepy states the following:32 “The University of Michigan study, collected survey data from parents on each child’s sleep habits and asked both parents and teachers to assess behavioral concerns. About a third of the students studied were identiﬁed by parents or teachers as having problems with disruptive behavior or bullying. The researchers found that children who had behavioral issues and those who were identiﬁed as bullies were twice as likely to have shown symptoms of sleep disorders.” A friend of yours who read the article says, “The study shows that sleep disorders lead to bullying in school children.” Is this statement justiﬁed? If not, how best can you describe the conclusion that can be drawn from this study? 30Justin Hepler and Dolores Albarrac´ın. “Attitudes without objects - Evidence for a dispositional attitude, its measurement, and its consequences”. In: Journal of personality and social psychology 104.6 (2013), p. 1060. 31R.C. Rabin. “Risks: Smokers Found More Prone to Dementia”. In: New York Times (2010). 32T. Parker-Pope. “The School Bully Is Sleepy”. In: New York Times (2011). 46 CHAPTER 1. DATA COLLECTION 1.5 Experiments You would like to determine if drinking a cup of tea each morning will cause students to perform better on tests. What are diﬀerent ways you could design an experiment to answer this question? What are possible sources of bias, and how would you try to minimize them? The goal of an experiment is to be able to draw a causal conclusion about the eﬀect of a treatment – in this case, drinking tea. If the design is poor, a causal conclusion cannot be drawn, even if you observe an association between drinking tea and performing better on tests. This is why it is crucial to start with a well-designed experiment. Learning objectives 1. Identify the subjects/experimental units, treatments, and response variable in an experiment. 2. Identify the three main principles of experiment design and explain their purpose: direct control, randomization, and replication. 3. Explain placebo eﬀect and describe when and how to implement a single-blind and a double- blind experiment. 4. Identify and describe how to implement the following three experimental designs: completely randomized design, blocked design, and matched pairs design. 5. Explain the purpose of random assignment or randomization in each of the three experimental designs. 6. Explain how to randomize treatments in a completely randomized design using technology or a table of random digits (make sure this is explained). 7. Explain when it is reasonable to draw a causal conclusion about the eﬀect of a treatment. 8. Identify the number of factors in experiment, the number of levels for each factor and the total number of treatments. 1.5.1 Reducing bias in human experiments In the last section we investigated observational studies and sampling strategies. While these are eﬀective tools for answering certain research questions, often times researchers want to measure the eﬀect of a treatment. In this case, they must carry out an experiment. Just as randomization is essential in sampling in order to avoid selection bias, randomization is essential in the context of experiments to determine which subjects will receive which treatments. If the researcher chooses which patients are in the treatment and control groups, she may unintentionally place sicker patients in the treatment group, biasing the experiment against the treatment. Randomized experiments are essential for investigating cause and eﬀect relationships, but they do not ensure an unbiased perspective in all cases. Human studies are perfect examples where bias can unintentionally arise. Here we reconsider a study where a new drug was used to treat heart attack patients. In particular, researchers wanted to know if the drug reduced deaths in patients. These researchers designed a randomized experiment because they wanted to draw causal conclusions about the drug’s eﬀect. Study volunteers33 were randomly placed into two study groups. One group, the treatment group, received the drug. The other group, called the control group, In an experiment, the explanatory variable is also called a did not receive any drug treatment. It has two levels: yes and no, thus it factor. Here the factor is receiving the drug treatment. 33Human subjects are often called patients, volunteers, or study participants. 1.5. EXPERIMENTS 47 is categorical. The response variable is whether or not patients died within the time frame of the study. It is also categorical. Put yourself in the place of a person in the study. If you are in the treatment group, you are given a fancy new drug that you anticipate will help you. On the other hand, a person in the other group doesn’t receive the drug and sits idly, hoping her participation doesn’t increase her risk of death. These perspectives suggest there are actually two eﬀects: the one of interest is the eﬀectiveness of the drug, and the second is an emotional eﬀect that is diﬃcult to quantify. Researchers aren’t usually interested in the emotional eﬀect, which might bias the study. To circumvent this problem, researchers do not want patients to know which group they are in. When researchers keep the patients uninformed about their treatment, the study is said to be blind or single-blind. But there is one problem: if a patient doesn’t receive a treatment, she will know she is in the control group. The solution to this problem is to give fake treatments to patients in the control group. A fake treatment is called a placebo, and an eﬀective placebo is the key to making a study truly blind. A classic example of a placebo is a sugar pill that is made to look like the actual treatment pill. Often times, a placebo results in a slight but real improvement in patients. This eﬀect has been dubbed the placebo eﬀect. The patients are not the only ones who should be blinded: doctors and researchers can accidentally bias a study. When a doctor knows a patient has been given the real treatment, she might inadvertently give that patient more attention or care than a patient that she knows is on the placebo. To guard against this bias, which again has been found to have a measurable eﬀect in some instances, most modern studies employ a double-blind setup where researchers who interact with subjects and are responsible for measuring the response variable are, just like the subjects, unaware of who is or is not receiving the treatment.34 GUIDED PRACTICE 1.28 Look back to the study in Section 1.1 where researchers were testing whether stents were eﬀective at reducing strokes in at-risk patients. Is this an experiment? Was the study blinded? Was it double-blinded?35 1.5.2 Principles of experimental design Well-conducted experiments are built on three main principles. Direct Control. Researchers assign treatments to cases, and they do their best to control any other diﬀerences in the groups. They want the groups to be as identical as possible except for the treatment, so that at the end of the experiment any diﬀerence in response between the groups can be attributed to the treatment and not to some other confounding or lurking variable. For example, when patients take a drug in pill form, some patients take the pill with only a sip of water while others may have it with an entire glass of water. To control for the eﬀect of water consumption, a doctor may ask all patients to drink a 12 ounce glass of water with the pill. Direct control refers to variables that the researcher can control, or make the same. A researcher can directly control the appearance of the treatment, the time of day it is taken, etc. She cannot directly control variables such as gender or age. To control for these other types of variables, she might consider blocking, which is described in Section 1.5.3. Randomization. Researchers randomize patients into treatment groups to account for variables that cannot be controlled. For example, some patients may be more susceptible to a disease than others due to their dietary habits. Randomizing patients into the treatment or control group helps even out the eﬀects of such diﬀerences, and it also prevents accidental bias from entering the study. 34There are always some researchers involved in the study who do know which patients are receiving which treatment. However, they do not interact with the study’s patients and do not tell the blinded health care professionals who is receiving which treatment. 35The researchers assigned the patients into their treatment groups, so this study was an experiment. However, the patients could distinguish what treatment they received, so this study was not blind. The study could not be double-blind since it was not blind. 48 CHAPTER 1. DATA COLLECTION Replication. The more cases researchers observe, the more accurately they can estimate the eﬀect of
the explanatory variable on the response. In an experiment with six subjects, even if there is randomization, it is quite possible for the three healthiest people to be in the same treatment group. In a randomized experiment with 100 people, it is virtually impossible for the healthiest 50 people to end up in the same treatment group. In a single study, we replicate by imposing the treatment on a suﬃciently large number of subjects or experimental units. A group of scientists may also replicate an entire study to verify an earlier ﬁnding. However, each study should ensure a suﬃciently large number of subjects because, in many cases, there is no opportunity or funding to carry out the entire experiment again. It is important to incorporate these design principles into any experiment. If they are lacking, the inference methods presented in the following chapters will not be applicable and their results may not be trustworthy. In the next section we will consider three types of experimental design. 1.5.3 Completely randomized, blocked, and matched pairs design A completely randomized experiment is one in which the subjects or experimental units are randomly assigned to each group in the experiment. Suppose we have three treatments, one of which may be a placebo, and 300 subjects. To carry out a completely randomized design, we could randomly assign each subject a unique number from 1 to 300, then subjects with numbers 1-100 would get treatment 1, subjects 101-200 would get treatment 2, and subjects 201- 300 would get treatment 3. Note that this method of randomly allocating subjects to treatments in not equivalent to taking a simple random sample. Here we are not sampling a subset of a population; we are randomly splitting subjects into groups. While it might be ideal for the subjects to be a random sample of the population of interest, that is rarely the case. Subjects must volunteer to be part of an experiment. However, because randomization is incorporated in the splitting of the groups, we can still use statistical techniques to check for a causal connection, though the precise population for which the conclusion applies may be unclear. For example, if an experiment to determine the most eﬀective means to encourage individuals to vote is carried out only on college students, we may not be able to generalize the conclusions of the experiment to all adults in the population. Researchers sometimes know or suspect that another variable, other than the treatment, inﬂuences the response. Under these circumstances, they may carry out a blocked experiment. In this design, they ﬁrst group individuals into blocks based on the identiﬁed variable and then randomize subjects within each block to the treatment groups. This strategy is referred to as blocking. For instance, if we are looking at the eﬀect of a drug on heart attacks, we might ﬁrst split patients in the study into low-risk and high-risk blocks. Then we can randomly assign half the patients from each block to the control group and the other half to the treatment group, as shown in Figure 1.16. At the end of the experiment, we would incorporate this blocking into the analysis. By blocking by risk of patient, we control for this possible confounding factor. Additionally, by randomizing subjects to treatments within each block, we attempt to even out the eﬀect of variables that we cannot block or directly control. EXAMPLE 1.29 An experiment will be conducted to compare the eﬀectiveness of two methods for quitting smoking. Identify a variable that the researcher might wish to use for blocking and describe how she would carry out a blocked experiment. The researcher should choose the variable that is most likely to inﬂuence the response variable whether or not a smoker will quit. A reasonable variable, therefore, would be the number of years that the smoker has been smoking. The subjects could be separated into three blocks based on number of years of smoking and each block randomly divided into the two treatment groups. 1.5. EXPERIMENTS 49 Figure 1.16: Blocking using a variable depicting patient risk. Patients are ﬁrst divided into low-risk and high-risk blocks, then each block is evenly separated into the treatment groups using randomization. This strategy ensures an equal representation of patients in each treatment group from both the low-risk and high-risk categories. Numbered patientscreateblocksLow−risk patientsHigh−risk patientsrandomlysplit in halfrandomlysplit in halfControlTreatmentl1l2l3l4l5l6l7l8l9l10l11l12l13l14l15l16l17l18l19l20l21l22l23l24l25l26l27l28l29l30l31l32l33l34l35l36l37l38l39l40l41l42l43l44l45l46l47l48l49l50l51l52l53l54l2l5l6l8l13l16l17l21l23l29l33l34l36l37l39l41l45l46l47l50l53l54l1l3l4l7l9l10l11l12l14l15l18l19l20l22l24l25l26l27l28l30l31l32l35l38l40l42l43l44l48l49l51l52l1l2l3l4l5l6l7l8l9l10l11l12l13l14l15l16l17l18l19l20l21l22l23l24l25l26l27l28l29l30l31l32l33l34l35l36l37l38l39l40l41l42l43l44l45l46l47l48l49l50l51l52l53l54 50 CHAPTER 1. DATA COLLECTION Even in a blocked experiment with randomization, other variables that aﬀect the response can be distributed unevenly among the treatment groups, thus biasing the experiment in one direction. In a matched pairs A third type of design, known as matched pairs addresses this problem. experiment, pairs of people are matched on as many variables as possible, so that the comparison happens between very similar cases. This is actually a special type of blocked experiment, where the blocks are of size two. An alternate form of matched pairs involves each subject receiving both treatments. Randomization can be incorporated by randomly selecting half the subjects to receive treatment 1 ﬁrst, followed by treatment 2, while the other half receives treatment 2 ﬁrst, followed by treatment. GUIDED PRACTICE 1.30 How and why should randomization be incorporated into a matched pairs design?36 GUIDED PRACTICE 1.31 Matched pairs sometimes involves each subject receiving both treatments at the same time. For example, if a hand lotion was being tested, half of the subjects could be randomly assigned to put Lotion A on the left hand and Lotion B on the right hand, while the other half of the subjects would put Lotion B on the left hand and Lotion A on the right hand. Why would this be a better design than a completely randomized experiment in which half of the subjects put Lotion A on both hands and the other half put Lotion B on both hands?37 Because it is essential to identify the type of data collection method used when choosing an appropriate inference procedure, we will revisit sampling techniques and experiment design in the subsequent chapters on inference. 1.5.4 Testing more than one variable at a time Some experiments study more than one factor (explanatory variable) at a time, and each of these factors may have two or more levels (possible values). For example, suppose a researcher plans to investigate how the type and volume of music aﬀect a person’s performance on a particular video game. Because these two factors, type and volume, could interact in interesting ways, we do not want to test one factor at a time. Instead, we want to do an experiment in which we test all combinations of the factors. Let’s say that volume has two levels (soft and loud) and that type has three levels (dance, classical, and punk). Then, we would want to have experiment groups for each of the six (2 x 3 = 6) combinations: soft dance, soft classical, soft punk, loud dance, loud classical, loud punk. Each combination is a treatment. Therefore, this experiment will have 2 factors and 6 treatments. To replicate each treatment 10 times, one would need to play the game 60 times. GUIDED PRACTICE 1.32 A researcher wants to compare the eﬀectiveness of four diﬀerent drugs. She also wants to test each of the drugs at two doses: low and high. Describe the factors, levels, and treatments of this experiment.38 As the number of factors and levels increases, the number of treatments become large and the analysis of the resulting data becomes more complex, requiring the use of advanced statistical methods. We will investigate only one factor at a time in this book. 36Assume that all subjects received treatment 1 ﬁrst, followed by treatment 2. If the variable being measured happens to increase naturally over the course of time, it would appear as though treatment 2 had a greater eﬀect than it really did. 37The dryness of people’s skins varies from person to person, but probably less so from one person’s right hand to left hand. With the matched pairs design, we are able control for this variability by comparing each person’s right hand to her left hand, rather than comparing some people’s hands to other people’s hands (as you would in a completely randomized experiment). 38There are two factors: type of drug, which has four levels, and dose, which has 2 levels. There will be 4 x 2 = 8 treatments: drug 1 at low dose, drug 1 at high dose, drug 2 at low dose, and so on. 1.5. EXPERIMENTS Section summary 51 • In an experiment, researchers impose a treatment to test its eﬀects. In order for observed diﬀerences in the response to be attributed to the treatment and not to some other factor, it is important to make the treatment groups and the conditions for the treatment groups as similar as possible. • Researchers use direct control, ensuring that variables that are within their power to modify (such as drug dosage or testing conditions) are made the same for each treatment group. • Researchers randomly assign subjects to the treatment groups so that the eﬀects of uncontrolled and potentially confounding variables are evened out among the treatment groups. • Replication, or imposing the treatments on many subjects, gives more data and decreases the likelihood that the treatment groups diﬀer on some characteristic due to chance alone (i.e. in spite of the randomization). • An ideal experiment is randomized, controlled, and double-blind. • A completely rando
mized experiment involves randomly assigning the subjects to the diﬀerent treatment groups. To do this, ﬁrst number the subjects from 1 to N. Then, randomly choose some of those numbers and assign the corresponding subjects to a treatment group. Do this in such a way that the treatment group sizes are balanced, unless there exists a good reason to make one treatment group larger than another. • In a blocked experiment, subjects are ﬁrst separated by a variable thought to aﬀect the response variable. Then, within each block, subjects are randomly assigned to the treatment groups as described above, allowing the researcher to compare like to like within each block. • When feasible, a matched-pairs experiment is ideal, because it allows for the best comparison of like to like. A matched-pairs experiment can be carried out on pairs of subjects that are meaningfully paired, such as twins, or it can involve all subjects receiving both treatments, allowing subjects to be compared to themselves. • A treatment is also called a factor or explanatory variable. Each treatment/factor can have multiple levels, such as yes/no or low/medium/high. When an experiment includes many factors, multiplying the number of levels of the factors together gives the total number of treatment groups. • In an experiment, blocking, randomization, and direct control are used to control for confound- ing factors. 52 CHAPTER 1. DATA COLLECTION Exercises 1.29 Light and exam performance. A study is designed to test the eﬀect of light level on exam performance of students. The researcher believes that light levels might have diﬀerent eﬀects on males and females, so wants to make sure both are equally represented in each treatment. The treatments are ﬂuorescent overhead lighting, yellow overhead lighting, no overhead lighting (only desk lamps). (a) What is the response variable? (b) What is the explanatory variable? What are its levels? (c) What is the blocking variable? What are its levels? 1.30 Vitamin supplements. To assess the eﬀectiveness of taking large doses of vitamin C in reducing the duration of the common cold, researchers recruited 400 healthy volunteers from staﬀ and students at a university. A quarter of the patients were assigned a placebo, and the rest were evenly divided between 1g Vitamin C, 3g Vitamin C, or 3g Vitamin C plus additives to be taken at onset of a cold for the following two days. All tablets had identical appearance and packaging. The nurses who handed the prescribed pills to the patients knew which patient received which treatment, but the researchers assessing the patients when they were sick did not. No signiﬁcant diﬀerences were observed in any measure of cold duration or severity between the four groups, and the placebo group had the shortest duration of symptoms.39 (a) Was this an experiment or an observational study? Why? (b) What are the explanatory and response variables in this study? (c) Were the patients blinded to their treatment? (d) Was this study double-blind? (e) Participants are ultimately able to choose whether or not to use the pills prescribed to them. We might expect that not all of them will adhere and take their pills. Does this introduce a confounding variable to the study? Explain your reasoning. 1.31 Light, noise, and exam performance. A study is designed to test the eﬀect of light level and noise level on exam performance of students. The researcher believes that light and noise levels might have diﬀerent eﬀects on graduate and undergraduate students, so wants to make sure both are equally represented in each treatment. The light treatments considered are ﬂuorescent overhead lighting, yellow overhead lighting, no overhead lighting (only desk lamps). The noise treatments considered are no noise, construction noise, and human chatter noise. (a) What is the response variable? (b) How many factors are considered in this study? Identify them, and describe their levels. (c) What is the role of the program type (graduate versus undergraduate) variable in this study? 1.32 Music and learning. You would like to conduct an experiment in class to see if students learn better if they study without any music, with music that has no lyrics (instrumental), or with music that has lyrics. Brieﬂy outline a design for this study. 1.33 Soda preference. You would like to conduct an experiment in class to see if your classmates prefer the taste of regular Coke or Diet Coke. Brieﬂy outline a design for this study. 1.34 Exercise and mental health. A researcher is interested in the eﬀects of exercise on mental health and he proposes the following study: Use stratiﬁed random sampling to ensure representative proportions of 18-30, 31-40 and 41- 55 year olds from the population. Next, randomly assign half the subjects from each age group to exercise twice a week, and instruct the rest not to exercise. Conduct a mental health exam at the beginning and at the end of the study, and compare the results. (a) What type of study is this? (b) What are the treatment and control groups in this study? (c) Does this study make use of blocking? If so, what is the blocking variable? (d) Does this study make use of blinding? (e) Comment on whether or not the results of the study can be used to establish a causal relationship between exercise and mental health, and indicate whether or not the conclusions can be generalized to the population at large. (f) Suppose you are given the task of determining if this proposed study should get funding. Would you have any reservations about the study proposal? 39C. Audera et al. “Mega-dose vitamin C in treatment of the common cold: a randomised controlled trial”. In: Medical Journal of Australia 175.7 (2001), pp. 359–362. 1.5. EXPERIMENTS 53 Chapter highlights Chapter 1 focused on various ways that researchers collect data. The key concepts are the diﬀerence between a sample and an experiment and the role that randomization plays in each. • Researchers take a random sample in order to draw an inference to the larger population from which they sampled. When examining observational data, even if the individuals were randomly sampled, a correlation does not imply a causal link. • In an experiment, researchers impose a treatment and use random assignment in order to draw causal conclusions about the eﬀects of the treatment. While often implied, inferences to a larger population may not be valid if the subjects were not also randomly sampled from that population. Related to this are some important distinctions regarding terminology. The terms stratifying and blocking cannot be used interchangeably. Likewise, taking a simple random sample is diﬀerent than randomly assigning individuals to treatment groups. • Stratifying vs Blocking. Stratifying is used when sampling, where the purpose is to sample a subgroup from each stratum in order to arrive at a better estimate for the parameter of interest. Blocking is used in an experiment to separate subjects into blocks and then compare responses within those blocks. All subjects in a block are used in the experiment, not just a sample of them. • Random sampling vs Random assignment. Random sampling refers to sampling a subset of a population for the purpose of inference to that population. Random assignment is used in an experiment to separate subjects into groups for the purpose of comparison between those groups. When randomization is not employed, as in an observational study, neither inferences nor causal conclusions can be drawn. Always be mindful of possible confounding factors when interpreting the results of observation studies. 54 CHAPTER 1. DATA COLLECTION Chapter exercises 1.35 Pet names. The city of Seattle, WA has an open data portal that includes pets registered in the city. For each registered pet, we have information on the pet’s name and species. The following visualization plots the proportion of dogs with a given name versus the proportion of cats with the same name. The 20 most common cat and dog names are displayed. The diagonal line on the plot is the x = y line; if a name appeared on this line, the name’s popularity would be exactly the same for dogs and cats. (a) Are these data collected as part of an experiment or an observational study? (b) What is the most common dog name? What is the most common cat name? (c) What names are more common for cats than dogs? (d) Is the relationship between the two variables positive or negative? What does this mean in context of the data? 1.36 Stressed out, Part II. In a study evaluating the relationship between stress and muscle cramps, half the subjects are randomly assigned to be exposed to increased stress by being placed into an elevator that falls rapidly and stops abruptly and the other half are left at no or baseline stress. (a) What type of study is this? (b) Can this study be used to conclude a causal relationship between increased stress and muscle cramps? 1.37 Chia seeds and weight loss. Chia Pets – those terra-cotta ﬁgurines that sprout fuzzy green hair – made the chia plant a household name. But chia has gained an entirely new reputation as a diet supplement. In one 2009 study, a team of researchers recruited 38 men and divided them randomly into two groups: treatment or control. They also recruited 38 women, and they randomly placed half of these participants into the treatment group and the other half into the control group. One group was given 25 grams of chia seeds twice a day, and the other was given a placebo. The subjects volunteered to be a part of the study. After 12 weeks, the scientists found no signiﬁcant diﬀerence between the groups in appetite or weight loss.40 (a) What type of study is this? (b) What are the experimental and control treatments in this study? (c) Has blocking been used in this study? If so, what is the blocking variable? (d) Has blinding been used in this study? (e) Comment on whether or not we can make a causal statement, and indicate whether or not we ca
n generalize the conclusion to the population at large. 1.38 City council survey. A city council has requested a household survey be conducted in a suburban area of their city. The area is broken into many distinct and unique neighborhoods, some including large homes, some with only apartments, and others a diverse mixture of housing structures. For each part below, identify the sampling methods described, and describe the statistical pros and cons of the method in the city’s context. (a) Randomly sample 200 households from the city. (b) Divide the city into 20 neighborhoods, and sample 10 households from each neighborhood. (c) Divide the city into 20 neighborhoods, randomly sample 3 neighborhoods, and then sample all households from those 3 neighborhoods. (d) Divide the city into 20 neighborhoods, randomly sample 8 neighborhoods, and then randomly sample 50 households from those neighborhoods. (e) Sample the 200 households closest to the city council oﬃces. 40D.C. Nieman et al. “Chia seed does not promote weight loss or alter disease risk factors in overweight adults”. In: Nutrition Research 29.6 (2009), pp. 414–418. 1.5. EXPERIMENTS 55 1.39 Flawed reasoning. Identify the ﬂaw(s) in reasoning in the following scenarios. Explain what the individuals in the study should have done diﬀerently if they wanted to make such strong conclusions. (a) Students at an elementary school are given a questionnaire that they are asked to return after their parents have completed it. One of the questions asked is, “Do you ﬁnd that your work schedule makes it diﬃcult for you to spend time with your kids after school?” Of the parents who replied, 85% said “no”. Based on these results, the school oﬃcials conclude that a great majority of the parents have no diﬃculty spending time with their kids after school. (b) A survey is conducted on a simple random sample of 1,000 women who recently gave birth, asking them about whether or not they smoked during pregnancy. A follow-up survey asking if the children have respiratory problems is conducted 3 years later. However, only 567 of these women are reached at the same address. The researcher reports that these 567 women are representative of all mothers. (c) An orthopedist administers a questionnaire to 30 of his patients who do not have any joint problems and ﬁnds that 20 of them regularly go running. He concludes that running decreases the risk of joint problems. 1.40 Income and education in US counties. The scatterplot below shows the relationship between per capita income (in thousands of dollars) and percent of population with a bachelor’s degree in 3,143 counties in the US in 2010. (a) What are the explanatory and response variables? (b) Describe the relationship between the two variables. Make sure to discuss unusual observations, if any. (c) Can we conclude that having a bachelor’s degree increases one’s income? 1.41 Eat better, feel better? In a public health study on the eﬀects of consumption of fruits and vegetables on psychological well-being in young adults, participants were randomly assigned to three groups: (1) dietas-usual, (2) an ecological momentary intervention involving text message reminders to increase their fruits and vegetable consumption plus a voucher to purchase them, or (3) a fruit and vegetable intervention in which participants were given two additional daily servings of fresh fruits and vegetables to consume on top of their normal diet. Participants were asked to take a nightly survey on their smartphones. Participants were student volunteers at the University of Otago, New Zealand. At the end of the 14-day study, only participants in the third group showed improvements to their psychological well-being across the 14-days relative to the other groups.41 (a) What type of study is this? (b) Identify the explanatory and response variables. (c) Comment on whether the results of the study can be generalized to the population. (d) Comment on whether the results of the study can be used to establish causal relationships. (e) A newspaper article reporting on the study states, “The results of this study provide proof that giving young adults fresh fruits and vegetables to eat can have psychological beneﬁts, even over a brief period of time.” How would you suggest revising this statement so that it can be supported by the study? 41Tamlin S Conner et al. “Let them eat fruit! The eﬀect of fruit and vegetable consumption on psychological well-being in young adults: A randomized controlled trial”. In: PloS one 12.2 (2017), e0171206. Percent with Bachelor's Degree$0$20k$40k$60k0%20%40%60%80%Per Capita Income 56 CHAPTER 1. DATA COLLECTION 1.42 Screens, teens, and psychological well-being. In a study of three nationally representative largescale data sets from Ireland, the United States, and the United Kingdom (n = 17,247), teenagers between the ages of 12 to 15 were asked to keep a diary of their screen time and answer questions about how they felt or acted. The answers to these questions were then used to compute a psychological well-being score. Additional data were collected and included in the analysis, such as each child’s sex and age, and on the mother’s education, ethnicity, psychological distress, and employment. The study concluded that there is little clear-cut evidence that screen time decreases adolescent well-being.42 (a) What type of study is this? (b) Identify the explanatory variables. (c) Identify the response variable. (d) Comment on whether the results of the study can be generalized to the population, and why. (e) Comment on whether the results of the study can be used to establish causal relationships. 1.43 Stanford Open Policing. The Stanford Open Policing project gathers, analyzes, and releases records from traﬃc stops by law enforcement agencies across the United States. Their goal is to help researchers, journalists, and policymakers investigate and improve interactions between police and the public.43 The following is an excerpt from a summary table created based oﬀ of the data collected as part of this project. State County Arizona Apaice County Arizona Apaice County Apaice County Arizona Cochise County Arizona Cochise County Arizona Cochise County Arizona · · · Wood County Wood County Wood County · · · Wisconsin Black Wisconsin Hispanic Wisconsin White Driver’s race Black Hispanic White Black Hispanic White · · · No. of stops per year 266 1008 6322 1169 9453 10826 · · · 16 27 1157 % of stopped cars searched 0.08 0.05 0.02 0.05 0.04 0.02 · · · 0.24 0.04 0.03 drivers arrested 0.02 0.02 0.01 0.01 0.01 0.01 · · · 0.10 0.03 0.03 (a) What variables were collected on each individual traﬃc stop in order to create to the summary table above? (b) State whether each variable is numerical or categorical. If numerical, state whether it is continuous or discrete. If categorical, state whether it is ordinal or not. (c) Suppose we wanted to evaluate whether vehicle search rates are diﬀerent for drivers of diﬀerent races. In this analysis, which variable would be the response variable and which variable would be the explanatory variable? 1.44 Space launches. The following summary table shows the number of space launches in the US by the type of launching agency and the outcome of the launch (success or failure).44 1957 - 1999 2000 - 2018 Failure 13 281 - Success Failure 10 33 5 295 3751 - Success 562 711 65 Private State Startup (a) What variables were collected on each launch in order to create to the summary table above? (b) State whether each variable is numerical or categorical. If numerical, state whether it is continuous or discrete. If categorical, state whether it is ordinal or not. (c) Suppose we wanted to study how the success rate of launches vary between launching agencies and over time. In this analysis, which variable would be the response variable and which variable would be the explanatory variable? 42Amy Orben and AK Baukney-Przybylski. “Screens, Teens and Psychological Well-Being: Evidence from three time-use diary studies”. In: Psychological Science (2018). 43Emma Pierson et al. “A large-scale analysis of racial disparities in police stops across the United States”. In: arXiv preprint arXiv:1706.05678 (2017). 44JSR Launch Vehicle Database, A comprehensive list of suborbital space launches, 2019 Feb 10 Edition. 57 Chapter 2 Summarizing data 2.1 Examining numerical data 2.2 Numerical summaries and box plots 2.3 Normal distribution 2.4 Considering categorical data 2.5 Case study: malaria vaccine (special topic) 58 After collecting data, the next stage in the investigative process is to describe and summarize the data. In this chapter, we will look at ways to summarize numerical and categorical data graphically, numerically, and verbally. While in practice, numerical and graphical summaries are done using computer software, it is helpful to understand how these summaries are created and it is especially important to understand how to interpret and communicate these ﬁndings. For videos, slides, and other resources, please visit www.openintro.org/ahss 2.1. EXAMINING NUMERICAL DATA 59 2.1 Examining numerical data How do we visualize and describe the distribution of household income for counties within the United States? What shape would the distribution have? What other features might be important to notice? In this section, we will explore techniques for summarizing numerical variables. We will apply these techniques using county-level data from the US Census Bureau, which was introduced in Section 1.2, and a new data set email50, that comprises information on a random sample of 50 emails. Learning objectives 1. Use scatterplots to represent bivariate data and to see the relationship between two numerical variables. Describe the direction, form, and strength of the relationship, as well as any unusual observations. 2. Understand what the term distribution means and how to summarize it in a table or a graph. 3. Create univariate displays, including
stem-and-leaf plots, dot plots, and histograms, to visualize the distribution of a numerical variable. Be able to read oﬀ speciﬁc information and summary information from these graphs. 4. Identify the shape of a distribution as approximately symmetric, right skewed, or left skewed. Also, identify whether a distribution is unimodal, bimodal, multimodal, or uniform. 5. Read and interpret a cumulative frequency or cumulative relative frequency histogram. 2.1.1 Scatterplots for paired data Sometimes researchers wish to see the relationship between two variables. When we talk of a relationship or an association between variables, we are interested in how one variable behaves as the other variable increases or decreases. A scatterplot provides a case-by-case view of data that illustrates the relationship between two numerical variables. A scatterplot is shown in Figure 2.1, illustrating the relationship between the number of line breaks (line breaks) and number of characters (num char) in emails for the email50 data set. In any scatterplot, each point represents a single case. Since there are 50 cases in email50, there are 50 points in Figure 2.1. EXAMPLE 2.1 A scatterplot requires bivariate, or paired data. What does paired data mean? We say observations are paired when the two observations correspond to the same case or individual. In unpaired data, there is no such correspondence. In our example the two observations correspond to a particular email. The variable that is suspected to be the response variable is plotted on the vertical (y) axis and the variable that is suspected to be the explanatory variable is plotted on the horizontal (x) axis. In this example, the variables could be switched since either variable could reasonably serve as the explanatory variable or the response variable. 60 CHAPTER 2. SUMMARIZING DATA Figure 2.1: A scatterplot of line breaks versus num char for the email50 data. DRAWING SCATTERPLOTS (1) Decide which variable should go on each axis, and draw and label the two axes. (2) Note the range of each variable, and add tick marks and scales to each axis. (3) Plot the dots as you would on an (x, y) coordinate plane. The association between two variables can be positive or negative, or there can be no association. Positive association means that larger values of the ﬁrst variable are associated with larger values of the second variable. Additionally, the association can follow a linear trend or a curved (nonlinear) trend. EXAMPLE 2.2 What would it mean for two variables to have a negative association? What about no association? Negative association implies that larger values of the ﬁrst variable are associated with smaller values of the second variable. No association implies that the values of the second variable tend to be independent of changes in the ﬁrst variable. EXAMPLE 2.3 Figure 2.2 shows a plot of median household income against the poverty rate for 3,142 counties. What can be said about the relationship between these variables? The relationship is evidently nonlinear, as highlighted by the dashed line. This is diﬀerent from previous scatterplots we’ve seen, which show relationships that do not show much, if any, curvature in the trend. There is also a negative association, as higher rates of poverty tend to be associated with lower median household income. GUIDED PRACTICE 2.4 What do scatterplots reveal about the data, and how are they useful?1 1Answers may vary. Scatterplots are helpful in quickly spotting associations relating variables, whether those associations come in the form of simple trends or whether those relationships are more complex. Number of Lines010203040506070020040060080010001200llllllllllllllllllllllllllllllllllllllllllllllllllNumber of Characters (in thousands) 2.1. EXAMINING NUMERICAL DATA 61 Figure 2.2: A scatterplot of the median household income against the poverty rate for the county data set. A statistical model has also been ﬁt to the data and is shown as a dashed line. Explore dozens of scatterplots using American Community Survey data on Tableau Public . GUIDED PRACTICE 2.5 Describe two variables that would have a horseshoe-shaped association in a scatterplot (∩ or ∪).2 2.1.2 Stem-and-leaf plots and dot plots Sometimes two variables is one too many: only one variable may be of interest. In these cases we want to focus not on the association between two variables, but on the distribution of a single, or univariate, variable. The term distribution refers to the values that a variable takes and the frequency of these values. Here we introduce a new data set, the email50 data set. This data set contains the number of characters in 50 emails. To simplify the data, we will round the numbers and record the values in thousands. Thus, 22105 is recorded as 22. 22 7 1 2 42 0 1 5 9 17 64 10 43 0 29 10 2 0 5 12 6 7 0 3 27 26 5 3 6 10 25 7 25 26 0 11 4 1 11 0 4 14 9 25 1 14 3 1 9 16 Figure 2.3: The number of characters, in thousands, for the data set of 50 emails. 2Consider the case where your vertical axis represents something “good” and your horizontal axis represents something that is only good in moderation. Health and water consumption ﬁt this description: we require some water to survive, but consume too much and it becomes toxic and can kill a person. If health was represented on the vertical axis and water consumption on the horizontal axis, then we would create a ∩ shape. 0%10%20%30%40%50%$0$20k$40k$60k$80k$100k$120kPoverty Rate (Percent)Median Household Income 62 CHAPTER 2. SUMMARIZING DATA Rather than look at the data as a list of numbers, which makes the distribution diﬃcult to discern, we will organize it into a table called a stem-and-leaf plot shown in Figure 2.4. In a stemand-leaf plot, each number is broken into two parts. The ﬁrst part is called the stem and consists of the beginning digit(s). The second part is called the leaf and consists of the ﬁnal digit(s). The stems are written in a column in ascending order, and the leaves that match up with those stems are written on the corresponding row. Figure 2.4 shows a stem-and-leaf plot of the number of characters in 50 emails. The stem represents the ten thousands place and the leaf represents the thousands place. For example, 1 \| 2 corresponds to 12 thousand. When making a stem-and-leaf plot, remember to include a legend that describes what the stem and what the leaf represent. Without this, there is no way of knowing if 1 \| 2 represents 1.2, 12, 120, 1200, etc. 0 \| 00000011111223334455566777999 1 \| 0001124467 2 \| 25556679 3 \| 4 \| 23 5 \| 6 \| 4 Legend: 1 \| 2 = 12,000 Figure 2.4: A stem-and-leaf plot of the number of characters in 50 emails. GUIDED PRACTICE 2.6 There are a lot of numbers on the ﬁrst row of the stem-and-leaf plot. Why is this the case?3 When there are too many numbers on one row or there are only a few stems, we split each row into two halves, with the leaves from 0-4 on the ﬁrst half and the leaves from 5-9 on the second half. The resulting graph is called a split stem-and-leaf plot. Figure 2.5 shows the previous stem-and-leaf redone as a split stem-and-leaf. 0 \| 000000111112233344 0 \| 55566777999 1 \| 00011244 1 \| 67 2 \| 2 2 \| 5556679 3 \| 3 \| 4 \| 23 4 \| 5 \| 5 \| 6 \| 4 Legend: 1 \| 2 = 12,000 Figure 2.5: A split stem-and-leaf. 3There are a lot of numbers on the ﬁrst row because there are a lot of values in the data set less than 10 thousand. 2.1. EXAMINING NUMERICAL DATA 63 GUIDED PRACTICE 2.7 What is the smallest number in the email50 data set? What is the largest?4 Another simple graph for univariate numerical data is a dot plot. A dot plot uses dots to show the frequency, or number of occurrences, of the values in a data set. The higher the stack of dots, the greater the number occurrences there are of the corresponding value. An example using the same data set, number of characters from 50 emails, is shown in Figure 2.6. Figure 2.6: A dot plot of num char for the email50 data set. GUIDED PRACTICE 2.8 Imagine rotating the dot plot 90 degrees clockwise. What do you notice?5 These graphs make it easy to observe important features of the data, such as the location of clusters and presence of gaps. EXAMPLE 2.9 Based on both the stem-and-leaf and dot plot, where are the values clustered and where are the gaps for the email50 data set? There is a large cluster in the 0 to less than 20 thousand range, with a peak around 1 thousand. There are gaps between 30 and 40 thousand and between the two values in the 40 thousands and the largest value of approximately 64 thousand. Additionally, we can easily identify any observations that appear to be unusually distant from the rest of the data. Unusually distant observations are called outliers. Later in this chapter we will provide numerical rules of thumb for identifying outliers. For now, it is suﬃcient to identify them by observing gaps in the graph. In this case, it would be reasonable to classify the emails with character counts of 42 thousand, 43 thousand, and 64 thousand as outliers since they are numerically distant from most of the data. OUTLIERS ARE EXTREME An outlier is an observation that appears extreme relative to the rest of the data. WHY IT IS IMPORTANT TO LOOK FOR OUTLIERS Examination of data for possible outliers serves many useful purposes, including 1. Identifying asymmetry in the distribution. 2. Identifying data collection or entry errors. For instance, we re-examined the email pur- ported to have 64 thousand characters to ensure this value was accurate. 3. Providing insight into interesting properties of the data. 4The smallest number is less than 1 thousand, and the largest is 64 thousand. That is a big range! 5It has a similar shape as the stem-and-leaf plot! The values on the horizontal axis correspond to the stems and the number of dots in each interval correspond the number of leaves needed for each stem. llllllllllllllllllllllllllllllllllllllllllllllllllNumber of Characters (in thousands)010203040506070 64 CHAPTER 2.

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