Solution Manual for Finite Mathematics with Applications In the Management, Natural, and Social Sciences, 12th Edition

INSTRUCTOR’S SOLUTIONS MANUAL SAL SCIANDRA Niagara County Community College M ATHEMATICS WITH A PPLICATIONS AND F INITE M ATHEMATICS WITH A PPLICATIONS IN THE M ANAGEMENT , N ATURAL , AND S OCIAL S CIENCES TWELFTH EDITION Margaret L. Lial American River College Thomas Hungerford Saint Louis University John Holcomb Cleveland State University Bernadette Mullins Birmingham-Southern College 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 © 2019, 2015, 2011 Pearson Education, Inc. Publishing as Pearson, 330 Hudson Street, NY NY 10013 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: 978-0-13-477640-8 ISBN-10: 0-13-477640-2 Contents Chapter 1 Algebra and Equations 1 Chapter 2 Graphs, Lines, and Inequalities 47 Chapter 3 Functions and Graphs 94 Chapter 4 Exponential and Logarithmic Functions 160 Chapter 5 Mathematics of Finance 198 Chapter 6 Systems of Linear Equations and Matrices 234 Chapter 7 Linear Programming 316 Chapter 8 Sets and Probability 422 Chapter 9 Counting, Probability Distributions, and Further Topics in Probability 453 Chapter 10 Introduction to Statistics 490 Chapter 11 Differential Calculus 521 Chapter 12 Applications of the Derivative 603 Chapter 13 Integral Calculus 677 Chapter 14 Multivariate Calculus 751 Chapter 1: Algebra and Equations Section 1.1 The Real Numbers 1. True. This statement is true, since every integer can be written as the ratio of the integer and 1. 5 For example, 5  . 1 For Exercises 13–16, let p = –2, q = 3 and r = –5. 13. −3 ( p + 5q ) = −3 [−2 + 5(3) ] = −3 [−2 + 15] = −3 (13) = −39 14. 2 ( q − r ) = 2 (3 + 5 ) = 2 (8 ) = 16 2. False. For example, 5 is a real number, and 10 5 which is not an irrational number. 2 15. 3. Answers vary with the calculator, but 2, 508, 429, 787 is the best. 798, 458, 000 qr 3 + (−5) −2 = = = −2 q  p 3 + (−2) 1 16. 3q 3(3) 9 9 = = = 3 p − 2r 3( −2) − 2( −5) −6 + 10 4 4. 0  (7)  ( 7)  0 This illustrates the commutative property of addition. 5. 6(t  4)  6t  6  4 This illustrates the distributive property. 6. 3 + (–3) = (–3) + 3 This illustrates the commutative property of addition. 7. (–5) + 0 = –5 This illustrates the identity property of addition. 8. (4)( 41 )  1 This illustrates the multiplicative inverse property. 9. 8 + (12 + 6) = (8 + 12) + 6 This illustrates the associative property of addition. 10. 1(20)  20 This illustrates the identity property of multiplication. 11. Answers vary. One possible answer: The sum of a number and its additive inverse is the additive identity. The product of a number and its multiplicative inverse is the multiplicative identity. 17. Let r = 3.8. APR  12r  12(3.8)  45.6% 18. Let r = 0.8. APR  12r  12(0.8)  9.6% 19. Let APR = 11. APR  12r 11  12r 11 r 12 r  .9167% 20. Let APR = 13.2. APR  12r 13.2  12r 13.2 r 12 r  1.1% 21. 3  4  5  5  3  20  5  17  5  12 22. 8  (4) 2  (12) Take powers first. 8 – 16 – (–12) Then add and subtract in order from left to right. 8 – 16 + 12 = –8 + 12 = 4 23. (4  5)  6  6  1  6  6  6  6  0 12. Answers vary. One possible answer: When using the commutative property, the order of the addends or multipliers are changed, while the grouping of the addends or multipliers is changed when using the associative property. Copyright © 2019 Pearson Education, Inc. 1 2 CHAPTER 1 ALGEBRA AND EQUATIONS 24. 2(3  7)  4(8) 4(3)  (3)( 2) Work above and below fraction bar. Do multiplications and work inside parentheses. 2(4)  32 8  32 24     4 12  6 12  6 6 25. 8  4 2  (12) Take powers first. 8 – 16 – (–12) Then add and subtract in order from left to right. 8 – 16 + 12 = –8 + 12 = 4   26. (3  5)   2  3 2  13    Take powers first. –(3 – 5) – [2 – (9 – 13)] Work inside brackets and parentheses. – (–2) – [2 – (–4)] = 2 – [2 + 4] = 2 – 6 = –4 27. 2(3)  ( 32)  2  16   64  1 Work above and below fraction bar. Take roots. 2(3)  ( 32)  ( 24) 30. 34. y is less than or equal to –5. y  5 35. z is at most 7.5. z  7.5 36. w is negative. w0 37. 6  2 38. 3 4  .75 39. 3.14   40. 1 3  .33 42. b + c = a 43. c < a –20 6 2  3 25 6 2  13 Take powers and roots. 36  3(5) 36  15 21   3 7 36  13 49 2040 189 , , 523 37 187 , 2.9884, 63 27, 46. [–1, 10] This represents all real numbers between –1 and 10, including –1 and 10. Draw brackets at –1 and 10 and a heavy line between them. 4587 , 6.735, 691 85 ,  , 10, 31. 12 is less than 18.5. 12 < 18.5 385 117 47 47. 2, 3 All real numbers x such that –2 –2 Start at –2 and draw a heavy line to the right. Use a parenthesis at –2 since it is not part of the graph. 50. (–∞, –2] This represents all real numbers less than or equal to –2. Draw a bracket at –2 and a heavy line to the left. 51. 9  12  9  (12)  3    4  15  19 54.  6  12  4   (6)  16  6  (16)  22 6 6 66 60. 3  5 3( 5) 3  –5 15 35 15 15  15 61. 3  5 35 2 35 2 2 2  2 62. 5  1 5  1 4 5 1 4 6 64. When b ≥ c, b – c is positive. So b  c  b  c . Answers will vary for exercises 65–67. Sample answers are given. example, let a = 1 and b = –1. Then, a  b  1  ( 1)  0  0 , but 4 4 4 a  b  1  (1)  1  1  2 . 66. Yes, if a and b are any two real numbers, it is always true that a  b  b  a . In general, 44 3  10 7 7 7 7 77 10 6 65. No, it is not always true that a  b  a  b . For 56.  4 58. 6  (4) 6 63. When a < 7, a – 7 is negative. So a  7  (a  7)  7  a . 53.  4  1  14  (4)  15 57. 10  3 28 46 52. 8  4  8  (4)  4 55. 5 5 5 __ 5 55 59. 2  8 3 a – b = –(b – a). When we take the absolute value of each side, we get a  b  (b  a )  b  a . 67. 2  b  2  b only when b = 0. Then each side 4  6 10 10 10 10  10 of the equation is equal to 2. If b is any other value, subtracting it from 2 and adding it to 2 will produce two different values. 68. For females: | x  63.5 | 8.4 ; for males: | x  68.9 | 9.3 Copyright © 2019 Pearson Education, Inc. 4 CHAPTER 1 ALGEBRA AND EQUATIONS 69. 1; 30062007 6. To multiply 4 3 and 4 5 , add the exponents since the bases are the same. The product of 4 3 and 70. 8; 2008, 2009, 2010, 2011, 2012, 2013, 2014, 2015 3 4 cannot be found in the same way since the bases are different. To evaluate the product, first do the powers, and then multiply the results. 71. 9; 2006, 2008, 2009, 2010, 2011, 2012, 2013, 2014, 2015 72. 4; 2008, 2010, 2013, 2015 7. 4 2  4 3  4 2  3  4 5 73. 4; 2006, 2007, 2009, 2011 8. 74. 8; 2006, 2007, 2009, 2010, 2011, 2012, 2013, 2014, 2015 9. (6) 2  (6) 5  (6) 2  5  (6) 7 75. | 10.6 − 14.9 |=| −4.3 |= 4.3 10. (2 z ) 5  (2 z ) 6  (2 z ) 5  6  (2 z )11 76. | 63.1 − ( −8.0) |=| 71.1 |= 71.1 44  46  4 4  6  410 7 4 47 28 11. 5u    5u   5u    77. | −1.0 − 63.1|=| −64.1 |= 64.1 12. 78. | 10.6 − ( −5.7) |=| 16.3 |= 16.3 4 6 y 3  6 y 5   6 y 3  6 y 20  (6 y ) 23 79. | −5.7 − ( −8.0) |=| 2.3 |= 2.3 80. | −1.0 − ( −5.7) |=| 4.7 |= 4.7 13. degree 4; coefficients: 6.2, –5, 4, –3, 3.7; constant term 3.7. 81. 3; 30062010, 2015, 2016 14. degree 7; coefficients: 6, 4, 0, 0, –1, 0, 1, 0; constant term 0. 82. 7; 2010, 2011, 2012, 2013, 2014, 2015, 2016 83. 6; 2010, 2011, 2012, 2013, 2014, 2015 84. 3; 2014, 2015, 2016 15. Since the highest power of x is 3, the degree is 3. 16. Since the highest power of x is 5, the degree is 5. Section 1.2 Polynomials 17. 6 1. 11.2  1, 973,822.685 2. (6.54)11  936,171,103.1 3x 3  2 x 2  5x  4 x 3  x 2  8x  3 x 3  4 x 3   2 x 2  x 2   (5 x  8 x)   x 3  x 2  13x 6  18  3.     289.0991339  7 18. 2 p 3  5 p  7  4 p 2  8 p  2  2 p 3  4 p 2  (5 p  8 p )  (7  2)  2 p 3  4 p 2  3 p  9 7 5 4.    .0163339967 9 5. 32 is negative, whereas (3) 2 is positive. Both 33 and (3) 3 are negative. Copyright © 2019 Pearson Education, Inc. SECTION 1.2 POLYNOMIALS 19.  4 y 2  3 y  8    2 y 2  6 y  2    4 y 2  3 y  8   2 y 2  6 y  2   8k 4  6k 3  2k 2  12k 3  9k 2  3k  4 y 2  2 y 2  (3 y  6 y )  (8  2)  8k 4  6k 3  7 k 2  3k  27. (6k – 1)(2k + 3)  (6k )(2k  3)  ( 1)(2k  3) 7b 2  2b  5  3b 2  2b  6  7b 2  2b  5   3b 2  2b  6  7b 2  3b 2   2b  2b    5  6  12k 2  18k  2k  3  12k 2  16k  3 28. (8r + 3)(r – 1) Use FOIL.  4b 2  1   8r 2  8r  3r  3    3 2   2 x  2 x  4 x  3   2 x  8 x  1 2x 3  2x 2  4x  3  2x 3  8x 2  1 3 2  8r 2  5r  3 29. (3y + 5)(2y +1) Use FOIL.  2x 3  2x 2  4 x  3  2 x3  8x 2  1  6 y 2  3 y  10 y  5  2 x 3  2 x 3  2 x 2  8 x 2  (4 x)  ( 3  1)  6 y 2  13 y  5      6 x 2  4 x  4 22.    2k 4k 3  3k 2  k  3 4k 3  3k 2  k  6 y 2  3 y  6 21.   4 y 2  3 y  8  2 y 2  6 y  2  20.  26. (2k  3) 4k 3  3k 2  k 30. (5r – 3s)(5r – 4s) 3 y 3  9 y 2  11y  8  4 y 2  10 y  6  3 y 3  9 y 2  11 y  8  4 y 2  10 y  6   3 y 3  9 y 2  4 y 2   ( 11 y  10 y )  (8  6) 3 2  3 y  13 y  21 y  14    (9m) 2m 2   9m  (6m)  ( 9m)( 1)  24. 2a 4a  6a  8   4.34m 2  8.06m  2.38m  4.42  4.34m 2  5.68m  4.42  8a 3  12a 2  16a  25. (3z  5) 4 z 2  2 z  1      (3z ) 4 z 2  2 z  1  (5) 4 z 2  2 z  1 3 2 2  12 z  6 z  3 z  20 z  10 z  5  12 z 3  14 z 2  7 z  5  12 z 3  20 z 2  6 z 2  10 z  3 z  5  12 z 3  14 z 2  7 z  5 32. (.012x – .17)(.3x + .54) = (.012x)(.3x) + (.012x)(.54) + (–.17)(.3x) + (–.17)(.54) 33. (6.2m – 3.4)(.7m + 1.3) 2   18k 2  9kq  2kq  q 2  .0036 x 2  .04452 x  .0918    2a(6a)  2a(8)  2a 4a 31. (9k + q)(2k – q)  .0036 x 2  .00648 x  .051x  .0918  18m 3  54m 2  9m 2  25r 2  35rs  12 s 2  18k 2  7 kq  q 2 23. 9m 2m 2  6m  1    25r 2  20rs  15rs  12 s 2 34. 2p –3[4p – (8p + 1)] = 2p – 3(4p – 8p – 1) = 2p – 3(– 4p – 1) = 2p + 12p + 3 = 14p + 3 35. 5k – [k + (–3 + 5k)] = 5k – [6k – 3] = 5k – 6k + 3 = –k + 3 Copyright © 2019 Pearson Education, Inc.  5 6 CHAPTER 1 ALGEBRA AND EQUATIONS 42. a. 36. (3x  1)( x  2)  (2 x  5) 2     3 x 2  5 x  2  4 x 2  20 x  25 2  2  3 x  5 x  2  4 x  20 x  25  2  3x  4 x 2 b. According to the bar graph, the net earnings in 2015 were $2757 million. Let x = 15. 4.79 x 3 − 122.5 x 2 + 1104 x − 2863   (5x  20 x)  (–2  25) 3 2 = 4.79 (15 ) − 122.5 (15 ) + 1104 (15 ) − 2863   x 2  15 x  27 = 2300.75 37. R = 5 (1000x) = 5000x C = 200,000 + 1800x P = (5000x) – (200,000 + 1800x) = 3200x – 200,000 38. R = 8.50(1000x) = 8500x C = 225,000 + 4200x P = (8500x) – (225,000 + 4200x) = 4300x – 225,000 According to the polynomial, the net earnings in 2015 were approximately $2301 million. 43. a. b. According to the bar graph, the net earnings in 2012 were $1384 million. Let x = 12. 4.79 x 3 − 122.5 x 2 + 1104 x − 2863 3 2 = 4.79 (12 ) − 122.5 (12 ) + 1104 (12 ) − 2863 39. R =9.75(1000x) = 9750x C  260, 000  ( 3 x 2  3480 x  325) = 1022.12 2  3x  3480 x  259, 675 According to the polynomial, the net earnings in 2012 were approximately $1022 million. P  (9750 x)  ( 3 x 2  3480 x  259, 675)  3 x 2  6270 x  259, 675 44. a. 40. R = 23.50(1000x) = 23,500x C  145, 000  ( 4.2 x 2  3220 x  425)  4.2 x 2  3220 x  144,575 P  (23,500 x)  ( 4.2 x 2  3220 x  144,575)  4.2 x 2  20, 280 x  144,575. 41. a. b. According to the bar graph, the net earnings in 2007 were $673 million. Let x = 7. 4.79 x 3 − 122.5 x 2 + 1104 x − 2863 3 2 = 4.79 ( 7 ) − 122.5 (7 ) + 1104 (7 ) − 2863 = 505.47 According to the polynomial, the net earnings in 2007 were approximately $505 million. b. According to the bar graph, the net earnings in 2013 were $8 million. Let x = 13. 4.79 x 3 − 122.5 x 2 + 1104 x − 2863 3 2 = 4.79 (13 ) − 122.5 (13 ) + 1104 (13 ) − 2863 = 1310.13 According to the polynomial, the net earnings in 2013 were approximately $1,310 million. 45. Let x = 17. 4.79 x 3 − 122.5 x 2 + 1104 x − 2863 3 2 = 4.79 (17 ) − 122.5 (17 ) + 1104 (17 ) − 2863 = 4035.77 According to the polynomial, the net earnings in 2017 will be approximately $4036 million. Copyright © 2019 Pearson Education, Inc. SECTION 1.2 POLYNOMIALS 46. Let x = 18. 3 2 4.79 x − 122.5 x + 1104 x − 2863 3 2 = 4.79 (18 ) − 122.5 (18 ) + 1104 (18 ) − 2863 = 5254.28 According to the polynomial, the net earnings in 2018 will be approximately $5254 million. 47. Let x = 19. 4.79 x 3 − 122.5 x 2 + 1104 x − 2863 3 2 = 4.79 (19 ) − 122.5 (19 ) + 1104 (19 ) − 2863 = 6745.11 According to the polynomial, the net earnings in 2019 will be approximately $6745 million. 48. The figures for 2013 – 2015 seem high, but plausible. To see how accurate these conclusions are, search Starbucks.com for later annual reports. For exercises 49–52, we use the polynomial 9.5 x3  401.6 x 2  6122 x  25,598. 49. Let x = 10. 9.5(10)3  401.6(10) 2  6122(10)  25,598  4962 Thus, the costs were approximately $4962 million in 2010. The statement is false. 50. Let x = 15. 9.5(15)3  401.6(15) 2  6122(15)  25,598  7934.5 Thus, the costs were approximately $7934.5 million in 2015. The statement is true. 51. Let x = 12. 9.5(12)3  401.6(12) 2  6122(12)  25,598  6451.6 Let x = 15. 9.5(15)3  401.6(15) 2  6122(15)  25,598  7934.5 Thus, the costs were $6451.6 million in 2012 and $7934.5 million in 2015. The statement is false. 52. Let x = 11. 9.5(11)3  401.6(11) 2  6122(11)  25,598  5794.9 Let x = 16. 9.5(16)3  401.6(16) 2  6122(16)  25,598  8456.4 Thus, the costs were $5794.9 million in 2011 and $8456.4 million in 2016. The statement is true. For exercises 53–58, we use the polynomial −72.85 x3 + 2082 x 2 − 16,532 x + 59,357. 53. Let x = 7. −72.85 x3 + 2082 x 2 − 16,532 x + 59,357 = −72.85(7)3 + 2082(7) 2 − 16,532(7) + 59,357 = 20, 663.45 Thus, the profit for PepsiCo Inc in 2007 was approximately $20,663 million. 54. Let x = 10. −72.85 x3 + 2082 x 2 − 16,532 x + 59,357 = −72.85(10)3 + 2082(10) 2 − 16,532(10) + 59,357 = 29,387 Thus, the profit for PepsiCo Inc in 2010 was $29,387 million. 55. Let x = 12. −72.85 x3 + 2082 x 2 − 16,532 x + 59,357 = −72.85(12)3 + 2082(12) 2 − 16,532(12) + 59,357 = 34,896.2 Thus, the profit for PepsiCo Inc in 2012 was approximately $34,896 million. 56. Let x = 15. −72.85 x3 + 2082 x 2 − 16,532 x + 59,357 = −72.85(15)3 + 2082(15) 2 − 16,532(15) + 59,357 = 33,958.25 Thus, the profit for PepsiCo Inc in 2015 was approximately $33,958 million. Copyright © 2019 Pearson Education, Inc. 7 8 CHAPTER 1 ALGEBRA AND EQUATIONS 57. Let x = 13. −72.85 x3 + 2082 x 2 − 16,532 x + 59,357 = −72.85(13)3 + 2082(13) 2 − 16,532(13) + 59,357 = 36, 247.55 60. P  7.2 x 2  5005 x  230, 000  0 Set the profit function equal to 0 and solve for x. 8 x 2  4450 x  215, 000  0 By the quadratic formula, x ≈ 45 or x ≈ –601. Since x represents a positive number, x = 45. Thus, the profit for PepsiCo Inc in 2013 was approximately $36,248 million. Let x = 9. −72.85 x3 + 2082 x 2 − 16,532 x + 59,357 = −72.85(9)3 + 2082(9) 2 − 16,532(9) + 59,357 = 26,103.35 Thus, the profit for PepsiCo Inc in 2009 was approximately $26,104 million. Therefore, the profit was higher in 2013. 58. By comparing the answers to problems 55 and 56, the profit was higher in 2012. In order for the company to make a profit, Therefore, between 40,000 and 45,000 calculators must be sold for the company to make a profit. 61. Let x = 100 (in thousands) 7.2(100) 2  5005(100)  230, 000  342,500 59. P  7.2 x 2  5005 x  230, 000 . Here is part of the screen capture. d. The profit for selling 100,000 calculators is $342,500. 62. Let x = 150 (in thousands) 7.2(150) 2  5005(150)  230, 000  682, 750 d. The profit for selling 150,000 calculators is $682,750. For 25,000, the loss will be $100,375; Section 1.3 Factoring 1. 12 x 2  24 x  12 x  x  12 x  2  12 x( x  2) 2. 5 y  65 xy  5 y (1)  5 y (13 x)  5 y (1  13 x) For 60,000, there profit will be $96,220.    r  r  5r  1 3. r 3  5r 2  r  r r 2  r 5r   r 1 2    t t 2  3t  8 4. t 3  3t 2  8t  t t 2  t (3t )  t (8) There is a loss at the beginning because of large fixed costs. When more items are made, these costs become a smaller portion of the total costs. Copyright © 2019 Pearson Education, Inc. SECTION 1.3 FACTORING 10. 3( x  6) 2  6( x  6) 4  3( x  6) 2 (1)  3( x  6) 2  2( x  6) 2  5. 6 z 3  12 z 2  18 z    6 z  z 2  2 z  3  6 z z 2  6 z (2 z )  6 z (3)  3( x  6) 2 1  2( x  6) 2     2  3  x  6 2 x  24 x  73  3  x  6 1  2 x 2  24 x  72 2 6. 5 x 3  55 x 2  10 x    5 x  x 2  11x  2 2  5 x x 2  5 x(11x)  5 x(2) 7. 3(2 y  1) 2  7(2 y  1) 3  (2 y  1) 2 (3)  (2 y  1) 2  7(2 y  1)  (2 y  1) 2 [3  7(2 y  1)] 11. x 2  5 x  4  ( x  1)( x  4) 12. u 2  7u  6  (u  1)(u  6)  (2 y  1) 2 (3  14 y  7) 13. x 2  7 x  12  ( x  3)( x  4)  (2 y  1) 2 (14 y  4)  2  2 y  1 7 y  2 2 14. y 2  8 y  12  ( y  2)( y  6) 8. (3x  7) 5  4(3 x  7) 3  (3x  7) 3 (3 x  7) 2  (3 x  7) 3 (4)  (3 x  7) 3 (3 x  7) 2  4    (3x  7) 3 9 x 2  42 x  45  (3x  7) 3 9 x 2  42 x  49  4 9. 3( x  5) 4  ( x  5) 6 4  2  3  x  6 1  2 x 2  12 x  36    4  ( x  5)  3  ( x  5) ( x  5) 2 15. x 2  x  6   x  3 x  2 16. x 2  4 x  5   x  5 x  1 17. x 2  2 x  3   x  3 x  1 18. y 2  y  12   y  4 y  3 19. x 2  3 x  4   x  1 x  4  ( x  5) 4 3   x  5    20. u 2  2u  8  u  2u  4  ( x  5) 4 3  x 2  10 x  25 21. z 2  9 z  14   z  2 z  7  2    ( x  5) 4  x 2  10 x  28 22. w2  6w  16   w  2 w  8 23. z 2  10 z  24  ( z  4)( z  6) 24. r 2  16r  60  ( r  6)( r  10) Copyright © 2019 Pearson Education, Inc. 9 10 CHAPTER 1 ALGEBRA AND EQUATIONS  25. 2 x 2  9 x  4  (2 x  1)( x  4) 46. 8k 2  16k  10  2 4k 2  8k  5   2(2k  1)(2k  5) 26. 3w 2  8w  4  (3w  2)( w  2) 47. 4u 2  12u  9  2u  3 2 2 27. 15 p  23 p  4  (3 p  4)(5 p  1) 48. 9 p 2  16  3 p   42  3 p  43 p  4  2 28. 8 x 2  14 x  3  (4 x  1)(2 x  3) 49. 25 p 2  10 p  4 This polynomial cannot be factored 29. 4 z 2  16 z  15  (2 z  5)(2 z  3) 30. 12 y 2  29 y  15  (3 y  5)(4 y  3) 50. 10 x 2  17 x  3  5 x  12 x  3 31. 6 x 2  5 x  4  (2 x  1)(3 x  4) 51. 4r 2  9v 2  2r  3v 2r  3v  32. 12 z 2  z  1  (4 z  1)(3 z  1) 52. x 2  3 xy  28 y 2   x  7 y  x  4 y  33. 10 y 2  21 y  10  (5 y  2)(2 y  5) 53. x 2  4 xy  4 y 2   x  2 y  34. 15u 2  4u  4  (5u  2)(3u  2) 2  54. 16u 2  12u  18  2 8u 2  6u  9   2  4u  32u  3 35. 6 x 2  5 x  4  (2 x  1)(3 x  4) 36. 12 y 2  7 y  10  (3 y  2)(4 y  5) 55. 3a 2  13a  30  (3a  5)( a  6) . 37. 3a 2  2a  5  3a  5a  1 56. 3k 2  2k  8  (3k  4)(k  2)  38. 6a 2  48a  120  6 a 2  8a  20  57. 21m 2  13mn  2n 2  (7 m  2n)(3m  n)  6(a  10)(a  2) 58. 81 y 2  100  (9 y  10)(9 y  10) 39. x 2  81  x 2  (9) 2  ( x  9)( x  9) 59. y 2  4 yz  21z 2  ( y  7 z )( y  3 z ) 40. x 2  17 xy  72 y 2  ( x  8 y )( x  9 y ) . 60. 49a 2  9 This polynomial cannot be factored. 41. 9 p 2  12 p  4  (3 p ) 2  2(3 p)(2)  2 2 61. 121x 2  64  (11x  8)(11x  8)  (3 p  2) 2 62. 4 z 2  56 zy  196 y 2 42. 3r 2  r  2  (3r  2)( r  1) .   4 z 2  14 zy  49 y 2 43. r 2  3rt  10t 2  (r  2t )(r  5t ) .  4  z 2  2( z )(7 y )  (7 y ) 2   4( z  7 y ) 2 44. 2a 2  ab  6b 2  (2a  3b)(a  2b). 2 2 2 45. m  8mn  16n  (m)  2(m)(4n)  (4n)  ( m  4n) 2   63. a 3  64  a 3  (4) 3  (a  4) a 2  4a  16 2  64. b 3  216  b 3  6 3  (b  6) b 2  6b  36 Copyright © 2019 Pearson Education, Inc.   SECTION 1.3 FACTORING      2 x  32 x  3  x 2  9 76. 4 x 4  27 x 2  81  4 x 2  9 x 2  9 65. 8r 3  27 s 3 3  (2r )  (3s) 3 2 2  (2r  3s) 2r   2r 3s   3s      3 66. 1000 p  27 q    (2r  3s) 4r 2  6rs  9 s 2      y2    x 2  y 2  x 4  x 2 y 2  y 4    x  y  x  y   x 2  xy  y 2    x 2  xy  y 2  3   79. x 8  8 x 2  x 2  x 6  8  x 2   x 2   2 3    2 2 4 2  x  x  2 x  2 x  4 78. x 6  y 6  x 2  67. 64m 3  125  (4m) 3  (5) 3 2 2  (4m  5) 4m   4m 5  5         2a  3b 2a  3b  4a 2  9b 2  (10 p  3q ) 100 p 2  30 pq  9q 2  (4m  5) 16m 2  20m  25  77. 16a 4  81b 4  4a 2  9b 2 4a 2  9b 2 3  (10 p ) 3  (3q ) 3  68. 216 y 3  343 3 3     2 2   x 3  x 3    2 3     3 3 3 3  x  x  2  x  2 3   x 3  x  2  x 2  2 x  4   x  2  x 2  2 x  4 80. x 9  64 x 3  x 3 x 6  64  x 3 x 6  2 6  (6 y ) 3  (7) 3   (6 y  7) 36 y 2  42 y  49  69. 1000 y 3  z 3  (10 y ) 3  ( z ) 3  (10 y  z ) (10 y ) 2  (10 y )( z )  ( z ) 2    (10 y  z ) 100 y 2  10 yz  z 2     (5 p  2q ) 25 p 2  10 pq  4q 2    71. x 4  5 x 2  6  x 2  2 x 2  3  2 2  2 72. y  7 y  10  y  2 y  5      3  2 x 2  1 x 2  1  3  2 x 2  1 ( x  1)( x  1) 6 x 4  3x 2  3  3 2 x 4  x 2  1     74. z 4  3 z 2  4  z 2  4 z 2  1     z  2 z  2 z 2  1    82. The sum of two squares can be factored when the terms have a common factor. An example is (3x) 2  32  9 x 2  9  9( x 2  1) 75. x 4  x 2  12  x 2  4 x 2  3   correct complete factorization because 3x 2  3 contains a common factor of 3. This common factor should be factored out as the first step. This will reveal a difference of two squares, which requires further factorization. The correct factorization is 73. b 4  b 2  b 2 b 2  1  b 2 b  1b  1   81. 6 x 4  3x 2  3  2 x 2  1 3 x 2  3 is not the 70. 125 p 3  8q 3  (5 p ) 3  (2q ) 3 4 11    x  2 x  2 x 2  3 Copyright © 2019 Pearson Education, Inc. 12 CHAPTER 1 ALGEBRA AND EQUATIONS 83. ( x  2) 3  ( x  2)( x  2) 2 9. 2  ( x  2)( x  4 x  4) 3 2 3 2 3 y 2  12 y 9 y3 2  x  4x  2x  8x  4 x  8   x  6 x  12 x  8, which is not equal to x 3  8 . The correct factorization is x 3  8  ( x  2)( x 2  2 x  4).  10. 15k 2  45k 9k 2 2 ( x  3)( x  2)  x  2 x  3 x  6  x  x  6 11. 12. Section 1.4 Rational Expressions 2 m m6 r 2  r  12 y4 3y2 15k ( k  3) 3k  3k 3k  5(k  3)  3k  3k 5(k  3)  3k m 2  4m  4 r2  r  6   3y 3y 2  2 84. Factoring and multiplication are inverse operations. If we factor a polynomial and then multiply the factors, we get the original polynomial back. For example, we can factor x 2  x  6 to get ( x  3)( x  2) . Then if we multiply the factors, we get 3 y ( y  4)   (m  2)(m  2) m  2  (m  3)(m  2) m  3 (r  3)(r  2) r  2  (r  4)(r  3) r  4 13. x 2  2 x  3  x  3 x  1 x  3    x  1 x  1 x  1 x2  1 14.  z  2  z  2 z2  4z  4  2 z   2 z  2 z  2 z 4 5 7p 15. 3a 2 8 3a 2  8 3  3  3 64 2a 16a 64  2a 3y2 4 16. 2u 2 10u 3 2u 2  10u 3 5    4 4 9u 18 8u 8u  9u 5. 5m  15 5( m  3) 5   4m  12 4( m  3) 4 17. 7 x 14 x3 7 x 66 y 7 x  66 y 3 y      11 66 y 11 14 x3 11  14 x3 x 2 6. 10 z  5 5(2 x  1) 1   20 z  10 5(2 x  1)  2 2 18. 6 x 2 y 21xy 6 x 2 y y y     2x y 2 x 21xy 7 7. 4( w  3) 4  ( w  3)( w  6) w  6 19. 2a  b 15 (2a  b)  15 15 5     3c 4(2a  b) (2a  b)  12c 12c 4c 8. 6( x  2) 6 6  or  x4 ( x  4)( x  2) x  4 20. 4( x  2) 3w 2 4 w 2 ( x  2)  3 3w    w 8( x  2) 4 w( x  2)  2 2 8x 2 x  8x x   1. 56 x 7  8 x 7 2. 3. 4. 27 m 27 m 81m 27 m  3m  3 25 p 2 5  5 p2 35 p 7 p 5p  3  2  2 18 y 4 6 y 2  3y 2 24 y 6y2  4  2  2 1 3m 2 Copyright © 2019 Pearson Education, Inc. SECTION 1.4 RATIONAL EXPRESIONS 21. 22. 23. 24. 25. 15 p  3 10 p  2 15 p  3 3    6 3 6 10 p  2 3(5 p  1)  3  3  2  2  (5 p  1) 3(5 p  1)  3 3   3(5 p  1)  2  2 4 27. 2k  8 3k  12 2k  8 3    6 3 6 3k  12 2(k  4) 3   6 3(k  4) 6(k  4) 6 1    18(k  4) 18 3 28. 9 y  18 3 y  6 9( y  2) 3( y  2)    6 y  12 15 y  30 6( y  2) 15( y  2) 27( y  2)( y  2) 27 3    90( y  2)( y  2) 90 10 12r  24 6r  12 12(r  2) 6( r  2)    36r  36 8r  8 36( r  1) 8( r  1) 3(r  2) r2   3(r  1) 4( r  1) r  2 4(r  1) 4    3(r  1) 3( r  2) 9 4a  12 4a  12 a 2  a  20 a2  9  2   2a  10 a  a  20 2a  10 a2  9 4(a  3) (a  5)(a  4)   2(a  5) (a  3)(a  3) 4(a  3)(a  5)(a  4)  2(a  5)(a  3)(a  3) 2(a  4)  a3 12r  16  26. 2 9r  6r  24 4r  12 6(r  3) 4(3r  4) 2(3r  4)    2  r 4( 3) 3 3r  2r  8 3r 2  2r  8   2(3r  4) 2   (3r  4)(r  2) r  2  k 2  k  6 k 2  3k  4  k 2  k  12 k 2  2k  3 (k  3)(k  2) (k  4)(k  1)   (k  4)(k  3) (k  3)(k  1) (k  3)(k  2)(k  4)(k  1) k  2   (k  4)(k  3)(k  3)(k  1) k  3 n2  n  6  n2  9 n 2  2n  8 n 2  7 n  12 (n  3)(n  2) (n  3)(n  3)   (n  4)(n  2) (n  3)(n  4) n3 n3 n3 n4 n4      n4 n4 n4 n3 n4 Answers will vary for exercises 29 and 30. Sample answers are given. 29. To find the least common denominator for two fractions, factor each denominator into prime factors, multiply all unique prime factors raising each factor to the highest frequency it occurred. 30. To add three rational expressions, first factor each denominator completely. Then, find the lowest common denominator and rewrite each expression with that denominator. Next, add the numerators and place over the common denominator. Finally, simplify the resulting expression and write it in lowest terms. 31. The common denominator is 35z. 2 1 25 1 7 10 7 3       7 z 5 z 7 z  5 5 z  7 35 z 35 z 35 z 32. The common denominator is 12z. 4 5 44 53 16 15 1       3z 4 z 3 z  4 4 z  3 12 z 12 z 12 z 33. 6r  18 13  34. r  2 r  2 (r  2)  (r  2)   3 3 3 r 2r 2 4   3 3 3 y  1 3 y  1 (3 y  1)  (3 y  1) 2 1     8 8 8 8 4 35. The common denominator is 5x. 4 1 4  5 1  x 20 x 20  x       5x x 5 x  5 5  x 5x 5x Copyright © 2019 Pearson Education, Inc. 14 CHAPTER 1 ALGEBRA AND EQUATIONS 36. The common denominator is 4r. 6 3 6  4 3  r 24 3r      r 4 r  4 4  r 4 r 4r 24  3r 3(8  r )   4r 4r 37. The common denominator is m(m – 1). 1 2 m 1 ( m  1)  2    m  1 m m  (m  1) (m  1)  m m 2(m  1)   m(m  1) m(m  1) m  2(m  1) m  2m  2   m(m  1) m(m  1) 3m  2  m(m  1) 38. The common denominator is (y + 2)y. 8 3 8y 3( y  2) 8 y  3( y  2)     y  2 y ( y  2) y y ( y  2) ( y  2) y 8y  3y  6 5y  6 5y  6 or   ( y  2) y ( y  2) y y ( y  2) 39. The common denominator is 5(b + 2). 7 2 75 2    b  2 5 b  2 b  2  5 5 b  2 35  2 37   5 b  2 5 b  2 40. The common denominator is 3(k + 1). 4 3 4 33    3  k  1 k  1 3 k  1 3 k  1 49 13   3  k  1 3  k  1 41. The common denominator is 20(k – 2). 2 5 8 25    5(k  2) 4(k  2) 20( k  2) 20( k  2) 8  25 33   20(k  2) 20( k  2) 43. First factor the denominators in order to find the common denominator. x 2  4 x  3   x  3 x  1 x 2  x  6   x  3 x  2 The common denominator is (x – 3)(x–1)(x + 2). 2 5  2 2 x  4x  3 x  x  6 2 5   ( x  3)( x  1) ( x  3)( x  2) 2( x  2) 5( x  1)   ( x  3)( x  1)( x  2) ( x  3)( x  2)( x  1) 2( x  2)  5( x  1) 2 x  4  5x  5   ( x  3)( x  2)( x  1) ( x  3)( x  1)( x  2) 7x  1  ( x  3)( x  1)( x  2) 44. First factor the denominators in order to find the common denominator. m 2  3m  10   m  5 m  2 m 2  m  20   m  5 m  4  The common denominator is (m  5)(m  2)(m  4) . 3 7  m 2  3m  10 m 2  m  20 3 7   (m  5)(m  2) (m  5)(m  4) 3(m  4) 7(m  2)   (m  5)(m  2)(m  4) (m  5)( m  4)(m  2) 3m  12  7 m  14 10m  26   (m  5)(m  2)(m  4) (m  5)( m  2)( m  4) 45. First factor the denominators in order to find the common denominator. y 2  7 y  12   y  3 y  4 y 2  5 y  6   y  3 y  2 42. The common denominator is 6(p + 4). 11 5 22 5    3( p  4) 6( p  4) 6( p  4) 6( p  4) 22  5 17   6( p  4) 6( p  4) Copyright © 2019 Pearson Education, Inc. SECTION 1.4 RATIONAL EXPRESIONS The common denominator is (y + 4)(y + 3)(y + 2). 2y y  y 2  7 y  12 y 2  5 y  6 2y y   ( y  4)( y  3) ( y  3)( y  2) 2 y ( y  2) y ( y  4)   ( y  4)( y  3)( y  2) ( y  4)( y  3)( y  2)  2 y ( y  2)  y ( y  4) 2 y 2  4 y  y 2  4 y  ( y  4)( y  3)( y  2) ( y  4)( y  3)( y  2)  y2 ( y  4)( y  3)( y  2) r 2  2r  8   r  4 r  2 The common denominator is (r  8)(r  2)(r  4) . 3r r  2 2 r  10r  16 r  2r  8 3r r   (r  8)(r  2) (r  4)(r  2) 3r (r  8)  r (r  4)   (r  8)(r  2)(r  4) (r  4)( r  2)( r  8)   (r  8)(r  2)(r  4) 1  1x xh h The common denominator in the numerator is x(x + h) 1  1x xh h 50. 46. First factor the denominators in order to find the common denominator. r 2  10r  16   r  8 r  2  r 2  4r  3r 2  24r 49. h The common denominator of the numerator is ( x  h) 2 x 2 . 1  12 ( x  h) 2 x h 1 x 48.    x 1  x  1x   x  1 x 1  1x  x  1  x  1x  x  1 x 1  1x 2  2y     x  ( x  h)  2 2 ( x  h) x  1 h x 2  x 2  2 xh  h 2 2 2 ( x  h) x 2 xh  h 2 ( x  h) 2 x 2 h 2 x  h  1 h h(2 x  h) ( x  h) 2 x 2 h ( x  h) 2 x 2 a. The probability that a dart will land in the  x2 shaded region is . 4 x2 b.  x2 2   4 4x 52. The radius of the dartboard is x + 2x + 3x = 6x, so the area of the dartboard is  6 x   36 x 2 . 2 2  2y Multiply both numerator and denominator by the common denominator, y. The area of the shaded region is  x 2 . a.    2 y  2  2( y  1)  y  1 y  2   2 y  2 2( y  1) y  1 2  2y y 2  2y 2 y 2 y  2 ( x  h) 2 x2 2 2  ( x  h) x ( x  h) 2 x 2  h 1 2 2 51. The length of each side of the dartboard is 2x, so the area of the dartboard is 4 x 2 . The area of the shaded region is  x 2 . Multiply both numerator and denominator of this complex fraction by the common denominator, x.  1 h x( x  h) 1  12 ( x  h) 2 x (r  8)( r  2)( r  4) 1  1x 1  1x 1  1x x xh x( x  h)   h h h h h 1  h  x ( x  h) x( x  h) h 1 1  or  x ( x  h) x( x  h) 2 47. x  ( x  h) x( x  h)    r 2  4r  3r 2  24r 4r  20r  (r  8)(r  2)(r  4) 15 b. The probability that a dart will land in the  x2 . shaded region is 36 x 2  x2 1  2 36 36 x Copyright © 2019 Pearson Education, Inc. 16 CHAPTER 1 ALGEBRA AND EQUATIONS 53. The length of each side of the dartboard is 5x, so the area of the dartboard is 25 x 2 . The area of the shaded region is x 2 . 59. .072 45  .744 45  1.2  3.84 45  2 Let x = 20. Then 2 The probability that a dart will land in the x2 shaded region is . 25 x 2 a. b. x2 1  2 25 25 x 2 .505 ( 20 ) − 4.587 ( 20 ) + 27.6 ≈ 6.6 20 + 1 The cost of an ad will not reach $7 million in 2020. 60. Let x = 45. Then .072 45  .744 45  1.2  3.84 45  2 Let x = 22. Then 2 54. The length of each side of the dartboard is 3x, so the area of the dartboard is 9 x 2 . The area of the shaded region is 12 x 2 . 2 .505 ( 22 ) − 4.587 ( 22 ) + 27.6 ≈ 7.4 22 + 1 The cost of an ad will not reach $8 million in 2022. The probability that a dart will land in the 1 2 x x2 . shaded region is 2 2  9x 18 x 2 a. b. Let x = 45. Then 61. x2 1  2 18 18 x Let x = 11. Then 2 55. Average cost = total cost C divided by the number of calculators produced. 7.2 x 2  6995 x  230, 000 1000 x 56. Let x = 20 (in thousands). 7.2(20) 2  6995(20)  230, 000  $18.35 1000(20) Let x = 50 (in thousands). 7.2(50) 2  6995(50)  230, 000  $11.24 1000(50) Let x = 125 (in thousands). 7.2(125) 2  6995(125)  230, 000  $7.94 1000(125) .049 (11) + 2.40 (11) + .83 ≈ 2.55 11 + 2 The hourly insurance cost in 2011 was $2.55. 62. Let x = 17. Then 2 .049 (17 ) + 2.40 (17 ) + .83 ≈ 2.94 17 + 2 The hourly insurance cost in 2017 is $2.94. 63. Let x = 20. Then 2 .049 ( 20 ) + 2.40 ( 20 ) + .83 ≈ 3.11 20 + 2 The hourly insurance cost in 2020 will be $3.11. The annual cost will be 3.11(2100) = $6531. 64. Let x = 23. Then 2 .049 ( 23 ) + 2.40 ( 23 ) + .83 Let x = 13. Then 57. 2 .505 (13 ) − 4.587 (13 ) + 27.6 ≈ 3.8 13 + 1 The ad cost approximately $3.8 million in 2013 58. ≈ 3.28 23 + 2 The hourly insurance cost in 2023 will be $3.28. The annual cost will be 3.28(2100) = $6888. No; the annual cost will not reach $10,000 by 2023. Let x = 16. Then 2 .505 (16 ) − 4.587 (16 ) + 27.6 ≈ 4.9 16 + 1 The ad cost approximately $4.9 million in 2016. Copyright © 2019 Pearson Education, Inc. SECTION 1.5 EXPONENTS AND RADICALS Section 1.5 Exponents and Radicals 1. 75 73  7 5  3  7 2  49 2. 614 8   6  1, 679, 616 6 6 3. 4c   4 c  16c 4. 2 x  4  24 x 4  16 x 4 2 2 2 a  17.  3  b  3 2 6 5v   125v 6  125v 2 2 3 16v 4 1  1 7 10. 10 3  1 10 3 16 1 7  1 1000 1 1 11. 6 5   5   7776 6 12. 13.   x  4   y  3  1  x  4 1   y 3  1 x  12  4  22. 12 5 2  121 2 1 y3 6     6 2  36 1 4 15.   3 2 9 3    4 16   498.83 5  23. 64 2 3   641 3   (4) 2  16  2    3  24. 64 3 2    641 2    83  512   4  8  25.    27  4 3 4  271 3  3 4 81 3   13      4  2 16 2  8   27  26.    64  1 3  64     27  4 2 2 6 27. 1 14.   6 2 1 21. (5.71)1 4  (5.71) .25  1.55 Use a calculator. 3u  2u   27u 4u   108u 9. 7 1  1 20. 81 3  2 because 2 3  8 . 3 2v 4  b3  b3    a a  19. 491 2  7 because 7 2  49 .  5 53 125 6.    3 3  3 3  xy  x y x y 8. 1 2 1 , but 16 2 1 1 (2) 4   4 16 (2) 5 7.  y2  y4    2 x  x  18. 2 4   2 2 5 32 2 5.    5  5 x x x 2 3 2  x  16.  2  y  28.  4 3  7 4  7 3  7 1  1 7 5 3 42 4 53  2 7 4 7 3 13 29. 4 3  4 6  4 3 30. 9 9  910  91  9 31. 410  4 6 4 4  410  4 6  4 4  48 Copyright © 2019 Pearson Education, Inc. 17 18 32. 33. 34. 35. CHAPTER 1 ALGEBRA AND EQUATIONS 5 4  5 6 5 1 z6  z2 z5 k6 k9 k 12    21 2 p1 2  21 3   p 3   25 6 p 3 2 z8  z  z 85  z 3 5 k 15  k 12 42.  k 15 12  k 3 34   3 2 3 2 3 2 4 4 x p 5 x 2 10 x    44. 3x 3 2 2 x  3 2  x 3 2  3x 3 2  2 x  3 2  3x 3 2  x 3 2  6 x 0  3x 6 2  6  3x 3 x2   y 2  45. 13 4 23  3x 2 3 y 2 1  x 2 3  y 4 3 3x 2 3 y 2 1 1  2 4 3  2 3 3y 3y 25 x10 5 q 5r 3   q 5r 3  q 5  r13  qr 3 38. 2 y 2 z 2   2 3  y 2   z 2  1 3 3 2 3 6 6 y z  c1 2   d 3  c 3 2 d 3 2    46.  14 3 c 3  d 1 4  c 3 4 d 3 4  3 3 z6 23 y 6  3 1 3  c (3 2)  (3 4) d (3 2)  (3 4)  c 3 4d 3 4 8y6 2 2 2 7a 2 5b 3 2 7 2 a 2 53 2 b 3 2 a 2  47.   5a 3 2 7b 4 53 2 a 3 2 7 4 b 4 7 2 b 4  3 2   1  1  1   23  3   2   4     p  5 8 12 z6 2 p   5 p   2  p  (5)  p   1   2 3  p 3   2   p 4  5  2 2 p  48. 25 p 7 4 x 1 2 x3 2 y 2 4  4 2  x 6  34  x 12  1296 x 18  xy  3 2 a1 2 49b 5 2 4 x 1 2  xy 1 2 x3 2 y 2  41 2 x1 2 x1 2 y1 2 x3 2 y 2  2 xy1 2 x 3 2 y 2  2 x 1 2 y  3 2 2  12 32 x y 4 1 x 3   3x 3  2 2 4 4   4 1    x 3   3   x 3  2 34 3   p 9 37. 1 3 32  2 p  5 p5 3 31 p 7 1    53 4   k 1 3   53 2 k 2 43. p 2 3 2 p1 3  5 p  p 2 3 2 p1 3  p 2 3 (5 p ) 5x  5 2  x   5 2 x 6 36.   40. 32 3 3 3 p 7 5k 2   5k 1 3   5 2 4 k 3k 1 4  59 4 k 13 4    31 p 6  311 p 6 (7)  32 p1  39. 13  21 2 p1 2  21 3  p1 31 p 2 x 13 41. (2 p)1 2  2 p 3  5 4  5 6  51  53 1296 49.   x1 2 x 2 3  x 4 3  x1 2 x 2 3  x1 2 x 4 3  x 7 6  x11 6 x18   50. x1 2 3 x 3 2  2 x 1 2  3 x1 2 x 3 2  2 x1 2 x 1 2  3x 2  2 Copyright © 2019 Pearson Education, Inc. SECTION 1.5 EXPONENTS AND RADICALS 51. This is a difference of two squares.  x  y  x  y    x    y  12 12 12 12 2 12 12 2  x y 52. Use FOIL.  x1 3  y1 2 2 x1 3  y 3 2  x 13 13  2x x 13 3 2 y  2x y 8 96  8 8  12  8 8 12  8 4  3 70.  8 4 3  8  2 3  16 3 71. 50  72  5 2  6 2   2 72 75  192  5 3  8 3  13 3 73. 5 20  45  2 80 13 12 y 12 32 y  52 5  3 5  24 5  2 x 2 3  2 x1 3 y1 2  x1 3 y 3 2  y 2 13 53. (3 x)  10 5  3 5  8 5  15 5 3  3 x , (f) 74.  3  2 3  2   3   2 2  3  4  1 75.  5  2  5  2    5    2  54. 3×1 3  3 3 x , (b) 1 1 55. (3 x) 1 3  3 , (h) 13 3x (3x) 56. 3x 1 3 3 3  1 3  3 , (d) x x 57. (3x )1 3  3 3x , (g) 58. 3x 1 3  59. (3x ) 1 3 3 x 13 63. 4 625  625 64. 7 128  ( 128)1 7  2 3 1 2 3 3 1 2    1  2 1  2 1  2 (1) 2  2 2 78. 3 3 79. 80. 81  9  729  9 67. 81  4  77 68. 49  16  33 69. 5 15  75  25  3  25 3  5 3    2 1 5 4    1  5  1  5  1 2 1 2 9  3 9  3 3  3 27  9 3  3 3  3    2 3 3 3 3 3 3 32  3    3   3 1  2     2 2 1 5 2 1 5    1 5 1 5 1 5 1 5  5  3 1 2  64  641 6  2 14  1 2 1  3 1  2  3  3 2 63  7  3 7  7  3  7  21 65. 66. 77. 125  1251 3  5 6 4  3 4  4 . A correct statement would be 3 4  3 4  3 16 .  60. 3×1 3  3 3 x , (e) 2 3 1   , (c) (3 x)1 3 3 3x 62. 2 76. 1 3 2  52  3 3  3 , (a) x 61. 19 24  6 3 24  6 3   4 3 93 6    3 1 3  2 3 1 3 1 3  2    3 4 32 32 32 3  2 3  3  2 1 3    1  3 1 1 Copyright © 2019 Pearson Education, Inc. 20 81. 82. CHAPTER 1 ALGEBRA AND EQUATIONS 3 2 3 2 3 2   3 2 3 2 3 2 92 7   9  6 2  2 11  6 2 1 7 1 7 1 7   2  3 2  3 1 7 1 7  22 7  3 3 7 6  2  2 7  3  21 .188 86 Let x = 18. Then 6.67 (18 ) ≈ 11.5 The domestic revenue for 2018 will be about $11.5 billion. .188 87. Let x = 20. Then 6.67 ( 20 ) ≈ 11.7 The domestic revenue for 2020 will be about $11.7 billion. .188 88. Let x = 23. Then 6.67 ( 23 ) ≈ 12.0 The domestic revenue for 2023 will be about $12.0 billion. kM f Note that because x represents the number of units to order, the value of x should be rounded to the nearest integer. For exercises 89–92, we use the model revenue = 1.08 x.527 , x ≥ 1, x = 1 corresponds to the first quarter of the year 2013. k = $1, f = $500, M = 100,000 1  100, 000 x  200  14.1 500 The number of units to order is 14. The advertising revenue in the first quarter of 2016 was approximately $4.2 billion. 83. x  a. b. c. k = $3, f = $7, M = 16,700 3  16, 700 x  84.6 7 The number of units to order is 85. k = $1, f = $5, M = 16,800 1  16,800 x  3360  58.0 5 The number of units to order is 58. 84. h  12.3T 1 3 If T = 216, find h. 13 h  12.3(216)  73.8 A height of 73.8 in. corresponds to a threshold weight of 216 lb. For exercises 85–88, we use the model revenue = 6.67 x.188 , x ≥ 5, x = 5 corresponds to 2005. .188 85. Let x = 15. Then 6.67 (15 ) ≈ 11.1 The domestic revenue for 2015 were about $11.1 billion. .527 89. Let x = 13. Then 1.08 (13) .527 90. Let x = 18. Then 1.08 (18 ) ≈ 4.2 ≈ 5.0 The advertising revenue in the second quarter of 2017 was approximately $5.0 billion. .527 91. Let x = 24. Then 1.08 ( 24 ) ≈ 5.8 The advertising revenue in the fourth quarter of 2018 will be approximately $5.8 billion. .527 92. Let x = 27. Then 1.08 ( 27 ) ≈ 6.1 The advertising revenue in the third quarter of 2019 will be approximately $6.1 billion. For exercises 93–96, we use the model Part-time Students = 2.72 x.238 ; x ≥ 10, x = 10 corresponds to 1990. 93. Let x = 29. Then 2.72(29).238 ≈ 6.1 According to the model, there were approximately 6.1 million part-time students attending college or university in 2009. 94. Let x = 34. Then 2.72(34).238 ≈ 6.3 According to the model, there were approximately 6.3 million part-time students attending college or university in 2014. Copyright © 2019 Pearson Education, Inc. SECTION 1.6 FIRST-DEGREE EQUATIONS 5. 2a  1  4(a  1)  7 a  5 2a  1  4a  4  7 a  5 2a  1  11a  9 2a  2a  1  11a  2a  9 1  9a  9 1  9  9a  9  9 10  9a 10 9a 10   a 9 9 9 6. 3  k  2  6  4k  3k  1 3k  6  6  4k  3k  1 3k  12  k  1 3k  12  (  k )  k  1  ( k ) 2k  12  1 2k  12  12  1  12 2k  13 2k 13 13  k 2 2 2 95. Let x = 38. Then 2.72(38).238 ≈ 6.5 According to the model, there will be approximately 6.5 million part-time students attending college or university in 2018 96. Let x = 43. Then 2.72(43).238 ≈ 6.7 According to the model, there will be approximately 6.7 million part-time students attending college or university in 2023 Section 1.6 First-Degree Equations 1. 3 x  8  20 3x  8  8  20  8 3 x  12 1 1 (3x )  (12) 3 3 x4 2. 4 – 5y = 19 Add –4 to both sides. 4  5 y  (4)  19  (4) 5 y  15 7. 2[ x  (3  2 x)  9]  3 x  8 2( x  3  2 x  9)  3 x  8 2( x  6)  3 x  8 2 x  12  3 x  8 12  5 x  8 20  5 x  4  x 1 Multiply both sides by  . 5 1 1  (5 y )   (15) 5 5 y  3 3. 8. 2  4  k  2  3  k  1  14  2k .6k  .3  .5k  .4 .6k  .5k  .3  .5k  .5k  .4 .1k  .3  .4 .1k  .3  .3  .4  .3 .1k  .7 .1k .7  k 7 .1 .1 4. 2.5  5.04m  8.5  .06m 2.5  5.04m  .06m  8.5  .06 m  .06 m 2.5  5.1m  8.5 2.5  5.1m  ( 2.5)  8.5  ( 2.5) 5.1m  6.0 5.1m 6.0  5.1 5.1 6.0 m  m  1.18 5.1 2(4k  8  3k  3)  14  2k 2(k  5)  14  2k 2k  10  14  2k 2k  10  2k  14  2k  2k 10  14  4k 10  14  14  4k  14 24  4k 24 4k   6  k 4 4 Copyright © 2019 Pearson Education, Inc. 21 22 CHAPTER 1 ALGEBRA AND EQUATIONS 9. 3x 4 3  ( x  1)  2  (3x  4) 5 5 10 Multiply both sides by the common denominator, 10.  3x  4 10    10   ( x  1) 5 5 12. 3  (10)(2)  (10)   (3 x  4)  10  2(3x)  8( x  1)  20  3(3 x  4) 6 x  8 x  8  20  9 x  12 2 x  8  32  9 x 2 x  9 x  32  8 7 x  40 1 1 40 (7 x)  (40)  x  7 7 7 10. 11. 4 1 3  ( x  2)   2  x  1   3 2 4 4 8 1 3 x   x2 3 3 2 2 4 19 3  x2 x 3 6 2 4 19 4 3 4 x  x  x2 x 3 6 3 2 3 19 1   x2 6 6 19 1  2 x22 6 6 7 1   x 6 6  7 1 6     6   x  7  x  6 6 5y 2y 8  5 6 3 5 2y  y    6   8  6 5    6   3   5y   2y  6    6(8)  6(5)  6    6   3  5 y  48  30  4 y 9 y  48  30 9 y  78 78 26  y 9 3 13. x 3x 3 1 2 5 Multiply both sides by the common denominator, 10, to eliminate the fractions. x   3x  10   3   10   1 2  5  5 x  30  6 x  10 5 x  30  5 x  6 x  10  5 x 30  x  10 30  10  x  10  10 40  x m 1 6m  5   2 m 12 m 1   6m  5  12m     12m   2 m  12  m 1 (12m)    (12m)    m(6m)  m(5) 2 m 6m 2  12  6m 2  5m 12  5m 1 1 12 (12)  (5m)    m 5 5 5 3k 9k  5 11k  8   2 6 k Multiply both sides by the common denominator, 6k to eliminate the fractions.  3k 9k  5   11k  8   6k  6k      2  k  6  14.   3k   9k  5   11k  8 6k     6k   6k   6k      2  6   k  k  9k 2  k (9k  5)  6(11k )  6(8) 9k 2  9k 2  5k  66k  48 5k  66k  48 5k  66k  66k  48  66k 71k  48 48 48 71k  k 71 71 71 Copyright © 2019 Pearson Education, Inc. SECTION 1.6 FIRST-DEGREE EQUATIONS 15. 4 8 3   0 x  3 2x  5 x  3 4 3 8   0 x  3 x  3 2x  5 7 8  0 x  3 2x  5 Multiply each side by the common denominator, (x – 3)(2x + 5).  7   8  ( x  3)(2 x  5)   ( x  3)(2 x  5)   x  3   2 x  5   ( x  3)(2 x  5)(0) 17. 3 1  2 2m  4 m  2 3 1  2 2(m  2) m  2   3 2(m  2)   2(m  2)   1   2(m  2)   2(m  2)(2)  m  2   1   2(m  2)   2(m  2)(2)  m  2  3  2  4(m  2) 3  2  4m  8 7(2 x  5)  8( x  3)  0 14 x  35  8 x  24  0 6 x  59  0 6 x  59  x   16. 23 3  6  4m  9  4m  m   59 6 5 3 4   2p  3 p  2 2p 3 5 3 5 4 5     2p  3 p  2 2p 3 2p 3 2p 3 3 1   p2 2p 3 Multiply both sides by the common denominator, (2p + 3)(p – 2).  5 3   (2 p  3)( p  2)   2 p  3 p  2   3    p  2   p  2 2 p  3 18. 8 5  4 3k  9 k  3 Multiply both sides by the common denominator, 3k – 9. 5   8 (3k  9)    (3k  9)4  3k  9 k  3  5   8    3(k  3)    12k  36 (3k  9)   3k  9   k  3  8  15  12k  36 7  12k  36 29 k 29  12k  12 19. 9.06 x  3.59(8 x  5)  12.07 x  .5612 9.06 x  28.72 x  17.95  12.07 x  .5612 9.06 x  28.72 x  12.07 x  17.95  .5612 25.71x  18.5112 18.5112 x  .72 25.71 20. 5.74(3.1  2.7 p)  1.09 p  5.2588 17.794  15.498 p  1.09 p  5.2588 15.498 p  1.09 p  5.2588  17.794 14.408 p  23.0528 23.0528 p  1.6 14.408  1     p  22 p  3  2 p  3  3  2 p  3  1  p  2 6 p  9   p  2 11 5  p  2 5  2 p  33   p  2 4 5 p  10  6 p  9  4 p  8  p  19  4 p  8 11 11  5 p    p 5 9 4 5 p  11  p   Copyright © 2019 Pearson Education, Inc. 24 21. CHAPTER 1 ALGEBRA AND EQUATIONS 2.63r  8.99 3.90r  1.77  r 1.25 2.45 Multiply by the common denominator (1.25)(2.45) to eliminate the fractions. 2.452.63r  8.99  1.253.90r  1.77    2.451.25 r 6.4435r  22.0255  4.875r  2.2125  3.0625 r 1.5685r  19.813  3.0625r 19.813  1.494r 19.813 1.494r   1.494 1.494 r  13.26 22. 8.19m  2.55 8.17 m  9.94   4m 4.34 1.04 1.048.19m  2.55  4.34 8.17 m  9.94   4m 1.044.34  8.5176m  2.652  35.4578m  43.1396  18.0544m 26.9402m  45.7916  18.0544 m 45.7916  44.9946m 45.7916  m  1.02 m 44.9946 23. 4(a  x)  b  a  2 x 4a  4 x  b  a  2 x 4a  b  a  2 x 5a  b  2 x 5a  b 2 x  2 2 5a  b b  5a   x or x  2 2 25. 5(b – x) = 2b + ax First, use the distributive property. 5b – 5x = 2b + ax 5b  2b  ax  5 x 3b  ax  5 x 3b  (a  5) x 3b (a  5) x 3b   x a5 a5 a5 Now use the distributive property on the right. 3b  (a  5) x 3b (a  5) x  a5 a5 3b x a5 26. bx – 2b = 2a – ax Isolate terms with x on the left. bx  ax  2a  2b ax  bx  2a  2b (a  b) x  2(a  b) 2(a  b) x x2 ab 27. PV  k for V 1 1 ( PV )  (k ) P P k V P 28. i  prt for p i p rt 29. 24. (3a  b)  bx  a ( x  2) 3a  b  bx  ax  2a Isolate terms with x on the right. 3a  b  ax  2a  bx 5a  b  ax  bx 5a  b 5a  b  (a  b) x  x ab V  V0  gt for g V  V0  gt V  V0 gt  t t V  V0 g t S  S 0  gt 2  k 30. S  S 0  k  gt 2 S  S0  k t 2  gt 2 Copyright © 2019 Pearson Education, Inc. t 2  S  S0  k t2 g SECTION 1.6 FIRST-DEGREE EQUATIONS 31. 1 ( B  b)h for B 2 1 1 A  Bh  bh 2 2 2 A  Bh  bh Multiply by 2. 2 A  bh  Bh 2 A  bh Bh 1 Multiply by .  h h h 2 A  bh 2 A  b  B h h A 5 32. C  ( F  32) for F 9 9 9 C  F  32  C  32  F 5 5 37. 38. 33. 2h  1  5 2h  1  5 or 2h  1  5 2h  6 or 2h  4 h  3 or h  2 5  10 r 3 5  10 r 3 5  10  r  3 5  10r  30 35  10r 35 7  r 10 2 3 4 2h  1 3 4 2h  1 3  4 2h  1 3  8h  4 7  8h 7 h 8 39. 34. 4m  3  12 4m  3  12 or 4m  3  12 4m  15 or 4m  9 15 9 m or m 4 4 40. 35. 6  2 p  10 6  2 p  10 or 6  2 p  10 2 p  4 or 2 p  16 p  2 or p  8 36. 5 x  7  15 5 x  7  15 or 5 x  7  15 5 x  8 or 5 x  22 8 22 x or x 5 5 41. 42. or or or or or or or or or or 5  10 r 3 5  10 r  3 5  10r  30 25  10r 25 5  r 10 2 3  4 2h  1 3  4 2h  1 3  8h  4 1  8h 1 h 8 5  F  32 9 9 95 5         F  32 5 59 9  F  32  23  F The temperature –5°C = 23°F. 5  5  F  32 9 9 95 15         F  32 5 59 27  F  32  5  F The temperature –15°C = 5°F. 15  5  F  32 9 9 95 22         F  32 5 59 39.6  F  32  71.6  F The temperature 22°C = 71.6°F. 22  5  F  32 9 9 95 36         F  32 5 59 64.8  F  32  96.8  F The temperature 36°C = 96.8°F. 36  Copyright © 2019 Pearson Education, Inc. 25 26 CHAPTER 1 ALGEBRA AND EQUATIONS 43. Let x = 10. y = 1.15 x + 1.62 y = 1.15(10) + 1.62 y = 13.12 Therefore, the gross federal debt in 2010 was $13.12 trillion. 44. Let x = 15. y = 1.15 x + 1.62 y = 1.15(15) + 1.62 y = 18.87 Therefore, the gross federal debt in 2015 was $18.87 trillion. 45. y = 1.15 x + 1.62 Substitute 22.32 for y. 22.32 = 1.15 x + 1.62 20.7 = 1.15 x ⇒ 18 = x Therefore, the federal deficit will be $22.32 trillion in 2018. 46. y = 1.15 x + 1.62 Substitute 24.62 for y. 24.62 = 1.15 x + 1.62 23 = 1.15 x ⇒ 20 = x Therefore, the federal deficit will be $24.628.83 trillion in 2020. 47. y = 1.15 x + 1.62 Substitute 25.77 for y. 25.77 = 1.15 x + 1.62 24.15 = 1.15 x ⇒ 21 = x Therefore, the federal deficit will be $825.77 trillion in 2021. 48. y = 1.15 x + 1.62 Substitute 30.37 for y. 30.37 = 1.15 x + 1.62 28.75 = 1.15 x ⇒ 25 = x Therefore, the federal deficit will be $30.378 trillion in 2025. 49. E = .108x + 1.517 Substitute $2.7052422 in for E. 2.705 = .108 x + 1.517 1.188 = .108 x ⇒ 11 = x The health care expenditures were at $2.7052250 trillion in 2011. 50. E = .108x + 1.517 Substitute $3.029 in for E. 3.029 = .108 x + 1.517 1.512 = .108 x ⇒ 14 = x The health care expenditures were at $3.029 trillion in 2014. 51. E = .108x + 1.517 Substitute $3.461 in for E. 3.461 = .108 x + 1.517 1.944 = .108 x ⇒ 18 = x The health care expenditures will be $3.4612422 trillion in 2018. 52. E = .108x + 1.517 Substitute $3.893 in for E. 3.893 = .108 x + 1.517 2.376 = .108 x ⇒ 22 = x The health care expenditures will be $3.893 trillion in 2022. 53. 114.8 ( x − 2010 ) = 5 y − 3390.5 Substitute 746.98 for y and solve for x. 114.8 ( x − 2010 ) = 5 (746.98 ) − 3390.5 114.8 x − 230, 748 = 344.4 114.8 x = 231, 092.4 x = 2013 The amount of income was $746.98 billion in 2013. 54. 114.8 ( x − 2010 ) = 5 y − 3390.5 Substitute 815.86 for y and solve for x. 114.8 ( x − 2010 ) = 5 (815.86 ) − 3390.5 114.8 x − 230, 748 = 688.8 114.8 x = 231, 436.8 x = 2016 The amount of income was $815.86 billion in 2016. 55. 114.8 ( x − 2010 ) = 5 y − 3390.5 Substitute 907.7 for y and solve for x. 114.8 ( x − 2010 ) = 5 (907.7 ) − 3390.5 114.8 x − 230, 748 = 1148 114.8 x = 231,896 x = 2020 The amount of income will be $907.7 billion in 2020. Copyright © 2019 Pearson Education, Inc.