Test bank and solution manual of elementary algebra 4e (1)

INSTRUCTOR’S SOLUTIONS MANUAL GEX PUBLISHING SERVICES E LEMENTARY A LGEBRA FOURTH EDITION Tom Carson Franklin Classical School Bill E Jordan Seminole State College of Florida Boston Columbus Indianapolis New York San Francisco Upper Saddle River Amsterdam Cape Town Dubai London Madrid Milan Munich Paris Montreal Toronto Delhi Mexico City São Paulo Sydney Hong Kong Seoul Singapore Taipei Tokyo
The author and publisher of this book have used their best efforts in preparing this book These efforts include the development, research, and testing of the theories and programs to determine their effectiveness The author and publisher make no warranty of any kind, expressed or implied, with regard to these programs or the documentation contained in this book The author and publisher shall not be liable in any event for incidental or consequential damages in connection with, or arising out of, the furnishing, performance, or use of these programs Reproduced by Pearson from electronic files supplied by the author Copyright © 2015, 2011, 2007,2004 Pearson Education, Inc Publishing as Pearson, 75 Arlington Street, Boston, MA 02116 All rights reserved No part of this publication may be reproduced, stored in a retrieval system, or transmitted, in any form or by any means, electronic, mechanical, photocopying, recording, or otherwise, without the prior written permission of the publisher Printed in the United States of America ISBN-13: 978-0-321-91226-8 ISBN-10: 0-321-91226-8 www.pearsonhighered.com
CONTENTS Chapter Foundations of Algebra Chapter Solving Linear Equations and Inequalities 14 Chapter Graphing Linear Equations and Inequalities 53 Chapter Systems of Linear Equations and Inequalities 80 Chapter Polynomials 96 Chapter Factoring .113 Chapter Rational Expressions and Equations 132 Chapter Roots and Radicals .162 Chapter Quadratic Equations .180 Chapter of the way 10 between and 8, so we divide the space between and into 10 equal divisions and place a dot on the 4th mark to the right of 32 The number 7.4 is located 0.4 = Foundations of Algebra Exercise Set 1.1 {q, r, s, t, u, v, w, x, y, z} {Alaska, Hawaii} {2, 4, 6, 8, …} {16, 18, 20, 22, …} 34 First divide the number line between −7 and −8 into tenths The number −7.62 falls between −7.6 and −7.7 on the number line Subdivide this section into hundredths and place a dot on the 2nd mark to the left of −7.6 10 {–2, –1, 0} 12 Rational because and are integers 14 Rational because −12 is an integer and all integers are rational numbers 16 Irrational because π cannot be written as a ratio of integers 18 Rational because −0.8 can be expressed as − , 10 the ratio of two integers 36 = because is units from on a number line 38 −8 = because −8 is units from on a number line 20 Rational because 0.13 can be expressed as the 13 fraction , the ratio of two integers 99 40 −4.5 = 4.5 because −4.5 is 4.5 units from on a number line 22 False There are real numbers that are not rational (irrational numbers) 42 24 False There are real numbers that are not natural numbers, such as 0, –2, , 0.6 , and π 26 True 1 is located of the way between 2 and 6, so we divide the space between and into equal divisions and place a dot on the 1st mark to the right of 28 The number 3 3 = because is units from on a 5 5 number line 44 −67.8 = 67.8 because −67.8 is 67.8 units from on a number line 46 < because is farther to the left on a number line than 48 −6 < because −6 is farther to the left on a number line than 50 −19 < −7 because −19 is farther to the left on a number line than −7 52 > −5 because is farther to the right on a number line than −5 2 is located of the way between 5 and −1 , so we divide the space between and −1 into equal divisions and place a dot on the 2nd mark to the left of 30 The number − 54 2.63 < 3.75 because 2.63 is farther to the left on a number line than 3.75 56 −3.5 < −3.1 because −3.5 is farther to the left on a number line than −3.1 Copyright © 2015 Pearson Education, Inc 2 Chapter Foundations of Algebra 5 > because is farther to the right on 6 a number line than 58 60 −4.1 = 4.1 because the absolute value of −4.1 is equal to 4.1 62 −10.4 > 3.2 because the absolute value of −10.4 is equal to 10.4, which is farther to the right on a number line than 3.2 12 ? ⋅ 10 = ⇒ = 16 ⋅ 16 The missing number is 10 14 2⋅3 = ⇒ = ? ⋅ 15 The missing number is 15 16 ? 6÷2 = ⇒ = 8÷ The missing number is 18 27 27 ÷ = ⇒ = 30 ? 30 ÷ 10 The missing number is 10 64 −0.59 = 0.59 because the absolute value of −0.59 and the absolute value of 0.59 are both equal to 0.59 2 < because is farther to the left on 9 a number line than the absolute value of , which is equal to 66 68 −10 > −8 because the absolute value of −10 is 10, the absolute value of −8 is 8, and 10 is farther to the right on a number line than 70 −5.36 < 5.76 because the absolute value of −5.36 is 5.36, the absolute value of 5.76 is 5.76, and 5.36 is farther to the left on a number line than 5.76 > − because the absolute value of 11 11 9 7 − is , the absolute value of − is , and 11 11 11 11 is farther to the right on a number line than 11 11 72 − 74 −12.6, −9.6,1, −1.3 , −2 , 2.9 1 76 −4 , −2 , −2, −0.13, 0.1 ,1.02, −1.06 Exercise Set 1.2 20 8 10 16 20 The LCD of and 11 is 77 ⋅11 55 ⋅ 21 = and = ⋅11 77 11 ⋅ 77 22 The LCD of and 12 is 24 ⋅ 15 ⋅ 14 = and = ⋅ 24 12 ⋅ 24 24 The LCD of 20 and 15 is 60 9⋅3 27 7⋅4 28 − =− and − =− 20 ⋅ 60 15 ⋅ 60 26 The LCD of 21 and 14 is 42 13 ⋅ 26 9⋅3 27 − =− and − =− 21 ⋅ 42 14 ⋅ 42 28 33 = ⋅11 30 42 = ⋅ 21 = ⋅ ⋅ 32 48 = ⋅ 24 = 2⋅8⋅3 = 2⋅ 2⋅ 4⋅3 = 2⋅ 2⋅ 2⋅ 2⋅3 34 810 = ⋅ 405 = ⋅ 81 ⋅ = 2⋅9⋅9⋅5 = ⋅ 3⋅ 3⋅ 3⋅ 3⋅ 36 48 ⋅ ⋅ ⋅ ⋅ = = 84 ⋅ ⋅ ⋅7 38 42 ⋅ ⋅ = = 91 ⋅13 13 Copyright © 2015 Pearson Education, Inc Instructor’s Solutions Manual 40 − 30 ⋅ ⋅5 =− =− 54 ⋅ ⋅3⋅3 64 ⋅ ⋅2 = = 60 ⋅ ⋅ ⋅ 15 42 − 24 ⋅2⋅2⋅ =− =− 162 ⋅ 3⋅3⋅3⋅ 27 66 2⋅2 = = 12 ⋅ ⋅ 3 44 Incorrect is not a factor of the numerator 46 Incorrect The prime factorization of 108 should be ⋅ ⋅ ⋅ ⋅ 48 If 130 of the 250 calories come from fat, the fraction of calories in a serving that comes from 130 fat is 250 130 ⋅ ⋅13 13 = = 250 ⋅ ⋅ ⋅ 25 50 If 120 square feet of the 1830 square feet are used as a home office, the fraction of her home that is 120 used as an office is 1830 120 ⋅2⋅2⋅ ⋅ = = 1830 ⋅ ⋅ ⋅ 61 61 52 There are ⋅ 24 = 168 hours in one week 50 ⋅5⋅5 25 = = 168 ⋅ ⋅ ⋅ ⋅ 84 25 Carla spends of her week sleeping 84 54 50 + 40 + 18 + = 112 hours for the listed activities The non-listed activities take 168 − 112 = 56 hours 56 2⋅2⋅2⋅7 = = 168 ⋅ ⋅ ⋅ ⋅ of her week away from all of the listed activities Carla spends 56 310 ⋅ ⋅ 31 31 = = 1000 ⋅ ⋅ ⋅ ⋅ ⋅ 100 Exercise Set 1.3 Commutative Property of Addition because the order of the addends is changed Additive identity because the sum of a number and is that number Additive inverse because the sum of these opposites is Associative Property of Addition because the grouping is changed 10 Commutative Property of Addition because the order of the addends is changed 12 Additive inverse because the sum of the opposites −4.6 and 4.6 is 14 15 + = 22 16 −5 + (−7 ) = −12 18 −5 + 16 = 11 20 −17 + = −9 22 29 + ( −7 ) = 22 26 690 69 = 1000 100 60 a) 2008 26 ⋅13 13 b) = = 1000 ⋅ ⋅ ⋅ ⋅ ⋅ 500 ⋅3 = = 159 ⋅ 53 53 70 + 12 + = 24 atoms total 12 + = 18 not-carbon atoms 18 ⋅ ⋅3 = = 24 ⋅ ⋅ ⋅ 24 −16 + 13 = −3 58 1000 − 310 = 690 non-victims; 62 68 47 Republicans + Independents = 49 Not 49 of the Senate was not Democrat Democrats; 100 9+5 + = 16 16 16 14 = 16 ⋅7 = ⋅2⋅2⋅2 = Copyright © 2015 Pearson Education, Inc 4 Chapter Foundations of Algebra ⎛ ⎞ −3 + ( −1) 28 − + ⎜ − ⎟ = ⎝ 5⎠ =− 30 − −9 + + = 14 14 14 =− 14 ⋅3 =− ⋅7 =− 32 The LCD of and is 1( ) + = + (2) + 8 2+7 = = = 34 The LCD of and 20 is 20 ( 4) ⎛ ⎞ ⎛ ⎞ − + ⎜− ⎟ = − + ⎜− ⎟ ⎝ 20 ⎠ ( 4) ⎝ 20 ⎠ ⎛ ⎞ =− + ⎜− ⎟ 20 ⎝ 20 ⎠ 11 =− 20 36 The LCD of 16 and 12 is 48 (3) ( 4) − + =− + 16 12 16 (3) 12 ( 4) 15 12 + 48 48 −15 + 12 = 48 =− 48 =− ⋅16 =− 16 =− 42 −7.8 + ( −9.16) = −16.96 44 −31 + −54 = −31 + 54 = 23 46 −0.6 + −9.1 = 0.6 + 9.1 = 9.7 48 The LCD of and is 20 4 − + = + 5 = (4) (4) + (5 ) (5 ) 16 15 + 20 20 31 = 20 = 50 −7 because + (−7 ) = 52 because −6 + = 54 because −9 + = 56 6 because − + =0 17 17 17 58 –2.8 because 2.8 + ( −2.8) = 60 −b because b + ( −b ) = 62 a a a because − + = b b b 64 − ( −15) = 15 66 − ( − (−1)) = − (1) = −1 68 − 10 = −10 70 − −5 = − (5) = −5 72 − 20 = + ( −20) = −12 74 −7 − 15 = −7 + ( −15) = −22 76 − ( −7 ) = + = 13 78 −13 − ( −6) = −13 + = −7 80 − ⎛ 3⎞ 3 − ⎜− ⎟ = − + ⎝ 4⎠ 4 =0 38 0.06 + 0.17 = 0.23 40 −15.81 + 4.28 = −11.53 Copyright © 2015 Pearson Education, Inc Instructor’s Solutions Manual 104 −256.5 − (−273.15) ; 82 The LCD of and is 24 ⎛ 5⎞ − ⎜− ⎟ = + ⎝ 6⎠ = (3) (3) + −256.5 − (−273.15) = −256.5 + 273.15 = 16.65 ( 4) 106 a) 21.0 – 18.8 (4) b) 21.0 – 18.8 = 2.2 20 + 24 24 29 = 24 = c) The positive difference indicates that the mean composite score in 2010 was greater than the score in 1986 108 $94,207 – $67,790 = $26,417 84 The LCD of and is ⎛ 1⎞ 1 − − ⎜− ⎟ = − + ⎝ 3⎠ =− 1(3) (3) + 110 Masters; $111,149 – $94,207 = $16,942 1( ) Puzzle Problem (2) =− + 6 =− 86 8.1 − 4.76 = 3.34 Exercise Set 1.4 88 0.107 − 5.802 = 0.107 + ( −5.802) Distributive Property of Multiplication over addition Multiplicative Identity because the product of a number and is the number Multiplicative Property of because the product of a number and is Commutative Property of Multiplication because the order of the factors is different = −5.695 90 −7.1 − ( −2.3) = −7.1 + 2.3 = −4.8 92 − −9 − −12 = − (9) − (12) = −9 + ( −12) = −21 10 Associative Property of Multiplication because the grouping of factors is different 94 4.6 − −7.3 = 4.6 − 7.3 = 4.6 + (−7.3) 12 Commutative Property of Multiplication because the order of the factors is different = −2.7 96 24,572.88 + 1284.56 + (−1545.75) + (−2700) + ( −865.45) + (−21,580.50) = −$834.26, which indicates a loss 98 31, 672.88 + 32, 284.56 + 124.75 + 2400 + ( −6545.75) + ( −1200) + ( −165.45) + ( −10,800) = $47,770.99 100 29.15 − 28.83 = 29.15 + (−28.83) = $0.32 102 2887.98 − (−14.35) = 2887.98 + 14.35 = $2902.33 14 (−7 ) = −28 16 (−8)(5) = −40 18 (12)( −4) = −48 20 (−4)(−3) = 12 22 (−8)(−12) = 96 ⎛ 20 ⎞ 2⋅2 2⋅2⋅ 16 24 − ⋅ ⎜ ⎟ = − ⋅ =− ⎝ ⎠ 3 ⎛ 5⎞⎛ 6⎞ ⋅ 26 ⎜ − ⎟ ⎜ − ⎟ = =1 ⎝ 6⎠⎝ 5⎠ ⋅ Copyright © 2015 Pearson Education, Inc 6 Chapter Foundations of Algebra ⎛ ⋅7 ⎞ ⎛ ⎞ ⎛ 21 ⎞ 28 ⎜ ⎟ ⎜ − ⎟ = ⋅ ⎜− =− ⎟ ⎝ ⎠ ⎝ 26 ⎠ ⋅ ⎝ ⋅13 ⎠ 39 62 −48 =8 −6 30 (−2.5) = −20 64 32 −7.1(−0.5) = 3.55 =0 66 −21 ÷ is undefined 34 8.1(−2.75) = −22.275 68 ÷ is indeterminate 36 −4 (5)( −3) = −20 (−3) = 60 70 −8 ÷ 38 (7 )( −8) = 21( −8) = −168 40 (−5)(−3)(−2) = (15)(−2) = −30 72 − 42 −5 (3)(−4)( −2) = −15 (−4)( −2) = 60 ( −2) = −120 44 (−2)(−4)(−30)(−1) = (8)(−30)(−1) = ( −240)(−1) = 240 46 (−1)(−1)(4)(−5)(−3) = (1)(4)(−5)(−3) = ( −5)( −3) = −20 (−3) −8 = ⋅ 32 =− 4 ÷ =− ⋅ 5 = −1 ⎛ 3⎞ ⎛ 2⎞ 74 − ÷ ⎜ − ⎟ = − ⋅ ⎜ − ⎟ ⎝ ⎠ 3 ⎝ 3⎠ = 76 ⎛ 35 ⎞ ⎛ 24 ⎞ ÷ ⎜− ⎟ = ⋅ ⎜− ⎟ 15 ⎝ 24 ⎠ 15 ⎝ 35 ⎠ = = 60 48 20 is the multiplicative inverse of because 20 20 ⋅ = 20 is the multiplicative inverse of − because ⎛ 7⎞ − ⋅ ⎜− ⎟ = ⎝ 6⎠ 50 − 52 is the multiplicative inverse of 17 because 17 17 ⋅ = 17 =− 58 −12 ÷ (−4) = 60 75 = −25 −3 25 78 8.1 ÷ 0.6 = 13.5 80 −10.65 ÷ (−7.1) = 1.5 82 19 ÷ ( −0.06) = −316.6 51 84 25 ÷ = ⋅ 2 51 = = 12 The 12th fret should be placed 12 54 –1 is the multiplicative inverse of −1 because −1 ⋅ (−1) = 56 42 ÷ ( −7 ) = −6 ⎛ 2⋅2⋅2⋅ ⎞ ⋅ − ⋅ ⎜⎝ ⋅ ⎟⎠ saddle or nut 86 (−858) = −$572 ⎛ 3⎞ 88 ⎜ − ⎟ = −$1 ⎝ 8⎠ 90 70.4 (−9.8) = −689.92 N Copyright © 2015 Pearson Education, Inc in from the Instructor’s Solutions Manual 92 94 −2080 ≈ 64.6 slugs −32.2 30 ±7 −15 ÷ (−8) = 1.875 Ω 34 ±13 32 No real-number square root exists 400 = (−6.5) r 96 400 (−6.5)2 =r 9.47Ω ≈ r 38 36 = 42 0.01 = 0.1 44 −25 is not a real number 46 Exercise Set 1.5 Base: 9; Exponent: 4; “nine to the fourth power” Base: –8; Exponent: 2; “negative eight squared” Base: 3; Exponent: 8; “additive inverse of three to the eighth power” 25 = ⋅ ⋅ ⋅ ⋅ = 32 10 (−2)4 = (−2)(−2)(−2)(−2) = 16 14 (−3) 48 40 = (−3)(−3)( −3)( −3)(−3) = −243 16 −35 = −3 ⋅ ⋅ ⋅ ⋅ = −243 18 − ( −3) = − (−3)( −3)( −3) 9 = 100 100 = 10 48 = 16 = 50 ⋅ − = 24 − = 19 54 + ÷ = + = 11 56 −3 ⋅ − ⋅ = −12 − 14 = −26 58 − 32 = − = −1 = − (−27 ) = 27 60 16 − ( −2) = 16 − ( 4) 20 − ( −1) = − (−1)(−1)(−1)(−1) = 16 − 20 = −4 = − (1) 62 32 − 18 ÷ (6 − 3) = 32 − 18 ÷ ⋅ = −1 = − 18 ÷ ⋅ ⎛ 2⎞ ⎛ 2⎞⎛ 2⎞ 22 ⎜ − ⎟ = ⎜ − ⎟ ⎜ − ⎟ = ⎝ 7⎠ ⎝ ⎠ ⎝ ⎠ 49 = − 6⋅3 = − 18 ⎛ 1⎞ ⎛ 1⎞⎛ 1⎞⎛ 1⎞⎛ 1⎞⎛ 1⎞ 24 ⎜ − ⎟ = ⎜ − ⎟ ⎜ − ⎟ ⎜ − ⎟ ⎜ − ⎟ ⎜ − ⎟ ⎝ 3⎠ ⎝ 3⎠ ⎝ 3⎠ ⎝ 3⎠ ⎝ 3⎠ ⎝ 3⎠ =− 243 26 = −9 64 12 − ( −2) − 64 ÷ ⋅ = 12 − ( −8) − 64 ÷ ⋅ = 12 − ( −16) − 16 ⋅ (0.3)4 = (0.3)(0.3)(0.3)(0.3) = 0.0081 28 289 = 17 52 18 ÷ + = + = 12 12 −24 = −2 ⋅ ⋅ ⋅ = −16 36 ±15 (−0.2)4 = (−0.2)(−0.2)(−0.2)(−0.2) = 0.0016 Copyright © 2015 Pearson Education, Inc = 12 + 16 − 32 = 28 − 32 = −4 66 Chapter Foundations of Algebra (−3)3 − 16 − (7 − 2) = (−3)3 − 16 − (5) = −27 − 16 − (5) = 36 + 18 ÷ (−3)( −2) = + 18 ÷ ( −3)( −2) = −27 − 16 − 25 = −43 − 25 = −68 = + ( −6)(−2) = + 12 68 18 ÷ (−6 + 3)( + 1) = 18 ÷ (−3)(5) = − (5 ) = 18 80 − ⎡⎣3 − (9 + 3)⎤⎦ + 64 = −30 = − (3 − 12) + 64 70 −15.54 ÷ 3.7 + (−2) + 49 = − ( −9) + 64 = − ( −9) + = −15.54 ÷ 3.7 + 16 + = −4.2 + 16 + = 11.8 + = 18.8 = + 72 + = 84 72 16.3 + 2.8 ⎡⎣(8 + ) ÷ − ⎦ ( = 25 − 22 ⎡⎣9 − (−5)⎤⎦ + 34 ) = 25 − 22 (9 + 5) + 34 = 16.3 + 2.8 (15 ÷ − 16) = 25 − 22 (14) + 34 = 16.3 + 2.8 (3 − 16) = − (14) + 81 = 16.3 + 2.8 ( −13) = − 56 + 81 = −51 + 81 = 30 = 16.3 + ( −36.4) = −20.1 74 −2 − 15 + 52 − 32 = −2 −6 + 52 − 32 = −2 (6) + 52 − 32 = −2 (6) + 25 − = −12 + 25 − =4 76 ⎛ 2⎞ ⎛ 2⎞ ÷ ⎜ − ⎟ + ⎜ − ⎟ (5)(−14) ⎝ 3⎠ ⎝ 7⎠ = 2⋅ ⎛ ⎞ ⎛ ⎞ ⎛ ⎞ ⎛ 2⋅ ⎞ ⋅ ⎜− ⎟ + ⎜− ⎟ ⎜ ⎟ ⎜− ⎝ ⎠ ⎝ ⎠⎝1⎠⎝ ⎠⎟ 20 + 80 =− + 4 75 = = 18 =− 83 − 58 − 22 ⎡⎣9 − (3 − 8)⎤⎦ + 34 82 2⎤ = 16.3 + 2.8 15 ÷ − 42 100 − 64 + 18 ÷ ( −3)( −2) 78 ⎛ 16 ⎞ ⎛ ⎞ ⎛3 2⎞ 84 ⎜ − ⎟ ÷ −⎜ ⎟÷⎜ ⎟ ⎝4 3⎠ 81 ⎝ 27 ⎠ ⎝ ⎠ 8⎞ ⎛ 16 ⎞ ⎛ ⎞ ⎛9 = ⎜ − ⎟÷ −⎜ ⎟÷⎜ ⎟ ⎝ 12 12 ⎠ 81 ⎝ 27 ⎠ ⎝ ⎠ ⎛ 16 ⎞ ⎛ ⎞ ⎛1⎞ = ⎜ ⎟÷ −⎜ ⎟÷⎜ ⎟ ⎝ 12 ⎠ 81 ⎝ 27 ⎠ ⎝ ⎠ ⎛ ⎞ ⎛ 16 ⎞ ⎛ ⎞ = ⎜ ⎟÷ −⎜ ⎟÷⎜ ⎟ ⎝ 12 ⎠ ⎝ 27 ⎠ ⎝ ⎠ = 1 16 ÷ − ÷ 12 27 = 16 ⋅ ⋅ − 12 27 4 − 12 3 16 = − 12 12 13 =− 12 = Copyright © 2015 Pearson Education, Inc Instructor’s Solutions Manual 86 (−18) ÷ ⎛⎜⎝ ⎞⎟⎠ − + 16 92 ⎛3⎞ = ( −18) ÷ ⎜ ⎟ − 25 ⎝2⎠ ⎡⎣ 24 − (6 − 2)⎤⎦ −33 + 42 + −8 24 = −8 = −3 15 ⋅ −5 31 94 = −10 − = −15 ( 62 − + 25 ) + 20 − (2 + 4) = ⎛ 5⎞ = 18 ⋅ ⎜ − ⎟ ÷ (−3) + + ( 4) ⎝ 6⎠ ⎛ 5⎞ = 18 ⋅ ⎜ − ⎟ ÷ (−3) + 12 ⎝ 6⎠ = −15 ÷ (−3) + (12) = + (12) = + 24 = 29 ( −3) + − 11 53 − (6 − 12) = = −18 + − 11 53 − (−6) −11 − 11 125 − (−6) 11 − 11 = 125 + 12 = 137 =0 62 − (4 + 32) + 20 − 62 62 − (36) + 20 − 36 36 − (36) 24 − 36 36 − 108 = −12 −72 = −12 =6 ⎛ 5⎞ = 18 ⋅ ⎜ − ⎟ ÷ (−3) + + ⎝ 6⎠ ⎛ ⎞ = 318 ⋅ ⎜ − ⎟ ÷ (−3) + (12) ⎝ 61 ⎠ = = ⎛ 5⎞ 88 18 ⋅ ⎜ − ⎟ ÷ (−3) + + (7 − 3) ⎝ 6⎠ 90 −11 + 3 (8) = ⎛3⎞ = −15 ÷ ⎜ ⎟ − ⎝2⎠ −27 + 16 + 3 (24 − 16) = ⎛3⎞ = ( −18) ÷ ⎜ ⎟ − ⎝2⎠ =− ⎡⎣ 24 − (4)⎤⎦ = 96 ( − 9) + − 100 − 36 = ( −5) + 23 − 64 −25 + = 8−8 −24 = Because the divisor is 0, the answer is undefined 98 Distributive Property The parentheses were not simplified first 100 Commutative Property of Addition The addition was not performed from left to right 102 Mistake: Subtracted before multiplying Correct: 19 − (10 − 8) = 19 − ⋅ = 19 − 12 =7 104 Mistake: Treated −34 as (−3) Correct: −34 + 20 ÷ − (16 − 24) = −34 + 20 ÷ − (−8) = −81 + 20 ÷ − ( −8) = −81 + + = −69 Copyright © 2015 Pearson Education, Inc 10 Chapter Foundations of Algebra 28 0.81 + ( x + 0.3) 106 Since the instructor drops one quiz, the 4, there is a total of quizzes Add the quiz scores and divide by + + + + + + + 62 = = 7.75 8 26 −8 − (m − n ) 108 Assume that Lisa will not make lower than 68 and that score will be dropped Add the test scores (268) and subtract from the lowest possible points for an A (4 tests multiplied by a score of 90 = 360 points) 360 – 268 = 92 36 Mistake: Order is incorrect Correct: m − 110 Add the unemployment figures for each month and divide by 12, the number of months in a year ⎛14,937 + 14,542 + 14, 060 + 13, 237 ⎞ ⎜ +13, 421 + 14, 409 + 14, 428 + 14,008 ⎟ ⎜ ⎟ ⎝ +13,520 + 13,102 + 12, 613 + 12, 692 ⎠ 12 164,969 = 12 ≈ 13, 747 thousand people = 13, 747, 000 people 112 Add the ending averages and divide by 5, the number of days ⎛13, 075.35 + 13, 071.72 + 13, 007.47 ⎞ ⎜⎝ +12,969.70 + 12,885.82 ⎟⎠ 65, 010.06 = ≈ 13, 002.01 30 (c − d ) − ( a + b ) 32 ab − x 34 5n − (n + 2) 38 Mistake: Wrote 19 as a dividend instead of a divisor hk Correct: or hk ÷ 19 19 40 l − 42 l 48 t + 54 v2 r 44 2r 46 60 − n 50 π r 52 56 1− πr v2 c2 58 Mistake: Could be translated as (a − 7) Correct: Seven less than two times a 60 Mistake: Could be translated as y + Correct: Four times the sum of y and six 62 Mistake: Could be translated as (m − 3)(m + 2) Correct: m minus the product of three and the sum of m and two 64 The product of one-half the height and the sum of a and b 66 The product of π , the radius squared, and the height Exercise Set 1.6 4n T −6 + y 68 Twice the product of π , the radius, and the sum of the radius and the height m2 10 y − 13 r 12 r ÷ or 14 b3 + 16 x + 18 (n + 4) 20 22 3a + 24 x ÷ y + or 70 The product of a and x squared added to the product of b and x added to c Puzzle Problem a) n + 1, n + b) n + 2, n + c) n + 2, n + (2 − l )3 x +7 y Copyright © 2015 Pearson Education, Inc Instructor’s Solutions Manual 11 14 Let m = −4 , n = −5 Exercise Set 1.7 2m + 2n = ( −4) + ( −5) Let m = 5, n = 8n − (m + 1) = 8(3) − (5 + 1) = (16) + (−5) = 8(3) − (6) = 32 + ( −10) = 24 − 12 = 22 = 12 = 22 Let y = 16 Let x = −2, y = −3, z = − 0.4 ( y − 2) = − 0.4 (5 − 2) −2 x3 y + z = −2 (−2) ( −3) + = − 0.4(3) = −2 (−8)(−3) + = − 1.2 = −48 + = 4.8 = −46 Let n = −1 18 Let h = 16 , k = −3 h + k = −3 16 + n − 8n + = ( −1) − ( −1) + = − ( −1) + = −3 ( 4) + (3) = 1+ +1 = −12 + = 10 = −3 Let r = − 20 Let m = 2, n = ⎛ 1⎞ ⎛ 1⎞ 3r − 9r + = ⎜ − ⎟ − ⎜ − ⎟ + ⎝ 3⎠ ⎝ 3⎠ ⎛1⎞ ⎛ 1⎞ = ⎜ ⎟ − ⎜− ⎟ + ⎝9⎠ ⎝ 3⎠ +3+6 28 =9 = 3 = 10 Let l = −0.4 −6 − (l − 5) = −6 − ( −0.4 − 5) = −6 − ( −5.4) = −6 + 10.8 = 4.8 12 Let m = , n = −2 − 2m − 4n = − (3) − (−2) = − (9) − −8 4m ( 2) = n+4 4+4 ( 4) = 16 = =2 22 Let a = 1, x = 64, y = 36 − a2 x+ y = = − 12 64 + 36 −1 100 = ⋅10 = 30 = 15 = − 18 − = −18 − = −26 Copyright © 2015 Pearson Education, Inc 12 Chapter Foundations of Algebra 24 a) Let a = 1, b = 0.5, c = −4, d = ad − bc = 1(6) − 0.5 ( −4) 7−0 = , which is undefined 0 because the denominator is 32 If y = , we have = 6+2 ⎛ 1⎞ 3 3⎜− ⎟ − − ⎝ 2⎠ 2 = = 34 If y = − , we have , −1 + ⎛ 1⎞ ⎜− ⎟ +1 ⎝ 2⎠ which is undefined because the denominator is =8 , c = 2, d = ⎛1⎞ ad − bc = −3 ⎜ ⎟ − (2) ⎝2⎠ =− − (5 ) ( ) =− − (5) ( 2) b) Let a = −3, b = 36 (b − 5) = ⋅ b − ⋅ = 4b − 20 38 −7 (3 − 2m ) = −7 ⋅ − (−7 ) ⋅ 2m = −21 − (−14m ) 15 16 − 10 10 31 =− 10 =− = −21 + 14m 40 26 a) Let x1 = 2, y1 = 1, x2 = 5, y2 = ( x2 − x1 )2 + ( y2 − y1 )2 = (5 − 2)2 + (7 − 1)2 = 32 + 62 = + 36 = 45 ≈ 6.7 b) Let x1 = −1, y1 = 2, x2 = −7, y2 = −2 = (−7 − (−1)) + (−2 − 2)2 = (−6) + (−4) = −9 x − 10.5 44 –14 50 46 52 − 48 –1 56 5b − 13b = −8b 58 −5 y + 12 y = y 60 −7m − 6m = −13m = 36 + 16 62 −5.1x + 3.4 x = −1.7 x = 52 ≈ 7.2 64 8 = , which is −3 + undefined because the denominator is 28 If x = −3 , we have 30 If a = , we have 42 −1.5 (6 x + ) = −1.5 ⋅ x + ( −1.5) ⋅ 54 6m + 7m = 13m ( x2 − x1 )2 + ( y2 − y1 )2 4⎛ 2⎞ 4 ⎜⎝ −10h + ⎟⎠ = ( −10h ) + ⋅ 5 = −8h + 45 −5 ( ) (4 − 4)(4 − 2) = −20 −20 = , (0)(2) which is undefined If a = , we have −5 ( ) −10 −10 = = , which is − − − 2 ( )( ) ( )( ) (5 ) ( 4) z− z = z− z (5) ( 4) 15 28 z− z 20 20 13 z =− 20 = 66 −15w − 6w − 11w = −21w − 11w = −32w 68 y + + y − = y + y + − undefined Copyright © 2015 Pearson Education, Inc = y2 − Instructor’s Solutions Manual 13 70 −4a + 9b − a + + 2b − = −4a − a + 9b + 2b + − = −5a + 11b − 72 −3h + k − − 8h − 7k + 19 + x = −3h − 8h + k − 7k + x − + 19 = −11h + x + 14 74 0.4t + t − 2.8 − t + 0.9t − = 0.4t − t + t + 0.9t − 2.8 − = −0.6t + 1.9t − 6.8 76 y+4− x+ − y 4 = − x+ y− y+4+ 1( 2) (3) = − x+ y− y+ + (2) 1(3) 3 12 x+ y− y+ + 8 3 3 14 = − x+ y+ =− 78 m − 3n + 14 − m − n − 10 = m − m − 3n − n + 14 − 10 1( 4) (10) = m− m− n − n + 14 − ( 4) 1(10) 10 30 m − m − n − n + 14 − 8 10 10 39 = m− n+9 10 = 80 a) −5n + (8 − 2n ) b) − 7n c) Let n = 0.2 − 7n = − (0.2) = − 1.4 = 6.6 Puzzle Problem F = 2, O = 9, R = 7, T = 8, Y = 6, E = 5, N = 0, S = 3, I = 1, X = 29786 850 + 850 31486 Copyright © 2015 Pearson Education, Inc  ... sum of y and six 62 Mistake: Could be translated as (m − 3)(m + 2) Correct: m minus the product of three and the sum of m and two 64 The product of one-half the height and the sum of a and b... 10 16 20 The LCD of and 11 is 77 ⋅11 55 ⋅ 21 = and = ⋅11 77 11 ⋅ 77 22 The LCD of and 12 is 24 ⋅ 15 ⋅ 14 = and = ⋅ 24 12 ⋅ 24 24 The LCD of 20 and 15 is 60 9⋅3 27 7⋅4 28 − =− and − =− 20 ⋅ 60...
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