# Introductory algebra 12th edition bittinger solutions manual

Information about Introductory algebra 12th edition bittinger solutions manual

Published on January 12, 2018

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1. − − Introductory Algebra 12th Edition Bittinger SOLUTIONS MANUAL Full download: http://testbanklive.com/download/introductory-algebra-12th-edition-bittinger- solutions-manual/ Introductory Algebra 12th Edition Bittinger TEST BANK Full download: http://testbanklive.com/download/introductory-algebra-12th-edition-bittinger- test-bank/ Chapter 2 Solving Equations and Inequalities ExerciseSet2.1 RC2. The correct answer is (c). RC4. The correct answer is (a). 2. t + 17 = 53 35 + 17 ? 53 26. 4 28. 25 30. −16 7 32. 24 10 34. 8.2 1 36. 52 FALSE 4 35 is not a solution. 4. a− 19=17 38. x + 2 = 5 3 6 5 4 9 36 − 19? 17 17 TRUE x = − 6 − 6 = 6 3 36 is a solution. x = − 2 3 5 6. 8y=−72 8(−9) ? −72 40. y − 4 = 6 10 9 y = + −72 −9is a solution. 8. y = 6 TRUE 12 12 19 y = 12 1 3 8 49 ? 6 8 42. − 8 + y = − 4 6 1 y = − 8 + 8 1 6 FALSE 8 49is not a solution. 10. 9x + 5 = 86

2. = 4 −42 9 · 9 + 5 ? 86 5 y = − 8 44. 2.7 46. 16 48. −10.6 86 9is a solution. TRUE 50. 5 1 2 + x 12. 6(y− 2)=18 4 3 3 8 5 6(−5 − 2) ? 18 12 − 4 12 = x 15 8 46(−7) FALSE 12 − 4 12 = x 7 = x −5 is not a solution. 14. 7 16. 34 18. −23 20. −31 22. 23 24. −11 12 3 52. 136 8 54. −5.2 56. 172.72 58. 65t miles 60. x + x = x 2x = x x = 0

3. − − 18 Chapter 2:Solving Equations and Inequalities 62. x + 4 = 5 + x 4 = 5 30. 2 y = 5 4 − 15 5 2 5 4 No solution 2 · 5 y = 2 · 20 − 15 64. |x| + 6 = 19 |x| + 6 − 6 = 19 − 6 |x| = 13 x represents a number whose distance from 0 is 13. Thus y = − 30 2 y = − 3 x = −13 or x = 13. The solutions are −13and 13. 3 32. − 8 15 x = − 16 8 3 8 15 Exercise Set 2.2 − 3 · − 8 x = − 3 · 120 − 16 RC2. The correct answer is (d). RC4. The correct answer is (b). 2. 17 4. 9 6. 7 34. 20 36. −2 38. 8 x = 48 5 x = 2 9 8. −53 40. − 7 y = 12.06 10. 47 7 − 9 · 9 − 7 y 7 = − 9 · (12.06) 84.42 12. −7 14. −7 16. 8 18. −30 y = − 9 y = −9.38 42. −x = 16 8 −x = 8 ( 16) 20. −88 22. 4 x = 16 5 5 4 5 8 8 · − −x = −128 −1 · (−x) = −1 · (−128) x = 128 4 · 5 x = 4 · 16 x = 20 24. 3 x = 12 8 44. −3 · m −3 m −3 = 10 = −3 · 10 8 3 8 m = −30 − 3 − 8 x = − 3 · 12 x = −32 46. −x + 5 26. −x = 9 6 −x = 54 x = −54

4. 2 48. −32y 50. 2 − 5(x + 5) = 2 − 5x − 25 = −5x − 23 52. −2a − 4(5a − 1) = −2a − 20a + 4 = −22a + 4 1 28. 1 = y 8 − 5 54. · b · 10 m2 , or 5b m2 5 8 = −y 5 − 8 = y 56. All real numbers 58. 4|x| = 48 |x| = 12 The distance of x from 0 is 12. Thus, x = 12 or x = −12. 60. 5

5. 2 Exercise Set 2.3 19 62. a+ 1 c 64. To “undo” the last step, divide 22.5 by 0.3. 22.5÷ 0.3 = 75 Now divide 75 by 0.3. 24. −5y − 7y = 144 −12y = 144 y = −12 26. −10y − 3y = −39 −13y = −39 y = 3 75 ÷ 0.3 = 250 The answer should be 250 not 22.5. Exercise Set 2.3 RC2. The correct answer is (a). RC4. The correct answers are (a) and (e). We would usually multiply by 100. 2. 7x + 6 = 13 7x = 7 x = 1 4. 4y + 10 = 46 4y = 36 y = 9 6. 5y − 2 = 53 5y = 55 y = 11 8. 4x − 19= 5 4x = 24 x = 6 10. 5x + 4 = −41 5x = −45 x = −9 12. −91 = 9t + 8 28. x + 1 x = 10 4 5 4 x = 10 4 x = 5 · 10 x = 8 30. 6.8y − 2.4y = −88 4.4y = −88 y = −20 32. 4x − 6 = 6x −6 = 2x −3 = x 34. 5y − 2 = 28 − y 6y = 30 y = 5 36. 4 − 3x = 6 − 7x 4x = 2 1 x = 2 38. 14 − 6a = −2a + 3 11 = 4a 11 = a 4 40. −7z + 2z − 3z − 7 = 17 −8z − 7 = 17 8z = 24 −99= 9t −11 = t 14. −5x − 7 = 108 −5x = 115 x = −23 − z = −3 42. 5 + 4x − 7 = 4x − 2 − x 4x − 2 = 3x− 2 x = 0 44. 5y − 7 + y = 7y + 21 − 5y 16. 3 x 2 − 24 = 3 −36 6y − 7 = 2y + 21 4y = 28 y = 7 2 x = −12 7 1 3 1 2 3 2 46. 8 x − 4 + 4 x = 16 + x, LCM is 16 3 · 2 x = 3 (−12) x = −8 18. 8x + 3x = 55 11x = 55 x = 5 14x − 4 + 12x = 1 + 16x 26x − 4 = 1 + 16x 10x = 5 1 x = 2 3 5 4 20. 8x + 5x = 104 13x = 104 x = 8 22. 7x + 18x = 125 25x = 125 x = 5 48. − 2 + x = − 6 − 3 , LCM is 6 −9 + 6x = −5 − 8 −9 + 6x = −13 6x = −4 2 x = − 3

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7. 20 Chapter 2:Solving Equations and Inequalities 50. 1 + 4m = 3m 5 , LCM is 2 72. 10 − 3(2x − 1) = 1 2 − 2 10 6x + 3 = 1 1 + 8m = 6m − 5 2m = −6 m = −3 − 13 − 6x = 1 −6x = −12 x = 2 2 9 y 3 52. 1 − 3 y = 5 − 5 + 5 , LCM is 15 15 − 10y = 27 − 3y + 9 15 − 10y = 36 − 3y −7y = 21 y = −3 54. 0.96y − 0.79= 0.21y + 0.46 96y − 79= 21y + 46 75y = 125 125 5 y = 75 = 3 56. 1.7t + 8 − 1.62t = 0.4t − 0.32 + 8 170t + 800 − 162t = 40t − 32 + 800 8t + 800 = 40t + 768 −32t = −32 t = 1 58. 5 y + 3 y = 2 + 1 y, LCM is 16 74. 3(t − 2)= 9(t + 2) 3t − 6 = 9t+ 18 −24 = 6t −4 = t 76. 7(5x − 2)= 6(6x − 1) 35x − 14 = 36x − 6 −8 = x 78. 3 − 7x + 10x − 14 = 9 − 6x + 9x − 20 3x − 11 = 3x − 11 −11 = −11 TRUE All real numbers are solutions. 80. 11x − 6 − 4x+ 1 = 9x − 8 − 2x + 12 7x − 5 = 7x + 4 −5 = 4 FALSE The equation has no solution. 16 8 4 5y + 6y = 32 + 4y 11y = 32 + 4y 7y = 32 32 y = 7 60. 8(3x + 2) = 30 24x + 16 = 30 24x = 14 7 x = 12 62. 9= 3(5x − 2) 9 = 15x − 6 15 = 15x 1 = x 64. 17 − t = −t + 68 17 = 68 FALSE The equation has no solution. 82. 5(t + 3) + 9 = 3(t − 2) + 6 5t + 15 + 9 = 3t − 6 + 6 5t + 24 = 3t 24 = −2t −12 = t 84. 13 − (2c + 2) = 2(c + 2) + 3c 13 − 2c − 2 = 2c + 4 + 3c 11 − 2c = 5c + 4 7 = 7c 1 = c 86. 5[3(7 − t) − 4(8 + 2t)]− 20 = −6[2(6 + 3t) − 4] 5[21 − 3t − 32 − 8t] − 20 = −6[12 + 6t − 4] 5[−11 − 11t] − 20 = −6[8 + 6t] −55 − 55t − 20 = −48 − 36t −75 − 55t = −48 − 36t −27 = 19t 27 − 19 = t 2 2 66. y − 3 = − 3 + y 88. 6(2x − 1) − 12 = 7 + 12(x − 1) 2 2 12x − 6 − 12 = 7 + 12x − 12 − 3 = − 3 TRUE All real numbers are solutions. 68. 5x + 5(4x − 1) = 20 5x + 20x − 5 = 20 25x − 5 = 20 25x = 25 x = 1 70. 6b − (3b + 8) = 16 6b − 3b − 8 = 16 3b − 8 = 16 3b = 24 b = 8 12x − 18 = 12x − 5 −18 = −5 FALSE The equation has no solution. 90. 2 + 14x − 9= 7(2x + 1) − 14 2 + 14x − 9 = 14x + 7 − 14 14x − 7 = 14x − 7 −7 = −7 TRUE All real numbers are solutions.

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9. −− − Exercise Set 2.4 21 92. 0.9(2x + 8) = 20 − (x + 5) 1.8x + 7.2 = 20 − x − 5 18x + 72 = 200 − 10x − 50 18x + 72 = 150 − 10x 10. d = t 55 12. y = x m 94. −75.14 28x = 78 78 x = 28 39 x = 14 14. z − 21 = t 2 16. y + 3 = x 18. t − 6 = s 20. y − A = x 96. 8y − 88x + 8 = 8(y − 11x + 1) 98. 3x + 2[4 − 5(2x − 1)] = 3x + 2[4 − 10x + 5] = 3x + 2[9 − 10x] = 3x + 18 − 20x = −17x + 18 100. 256÷ 64 ÷ 42 = 256÷ 64 ÷ 16 = 4 ÷ 16 1 = , or 0.25 4 22. y = 10 − x y − 10 = −x −y + 10 = x, or 10 − y = x 24. y = q − x y − q = −x −y + q = x, or q − y = x 102. 1 (8y + 4) 4 17 = 1 − 2 (4y − 8) 26. y = x , or 2 1 − 2 x 2y + 1 − 17 = −2y + 4 2y − 16 = −2y + 4 4y = 20 y = 5 104. 5(3x + 2) = 75 15x + 10 = 75 15x = 65 28. y = Ax B 30. W = mt − b W + b = mt W + b = t m 32. y = bx − c 65 13 1 x = 15 = 3 , or 4 3 y + c = bx y+c = x b ExerciseSet2.4 RC2. y = 1 x w 4 1 w = 4 x − y The correct answer is (c). 34. d = rt d = r t 36. A = πr2 A = r2 π RC4. z = w + 4 z − 4 = w The correct answer is (a). 2. B = 30 · 1800 = 54, 000 Btu’s 4. N = 72 − 7 = 49 − 7 = 42 games 6. a) A = 6s2 = 6 · 32 = 6 · 9 = 54 in2 b) A = s2 , or 1 A = s2 38. A = 1 bh 2 2A = bh 2A = b h 40. A = a + b + c 3 3A = a + b + c 3A − a − b = c 6 6 42. S = rx + s 8. a) P = I · V = 12 · 115 = 1380 watts b) I = P ; V = P S − s = rx S− s

10. = x V I r Copyright c 2015 Pearson Education, Inc.

11. 4 − − 22 Chapter 2:Solving Equations and Inequalities 44. Q = p− q 2 2Q = p − q 2Q + q = p 46. I = Prt I = P rt 48. Ax + By = c By = c − Ax y = c− Ax B 50. P = ab c Pc = ab c = ab P 52. 4a − 8b − 5(5a − 4b) = 4a − 8b − 25a + 20b = −21a + 12b 2. True; seepage85 in the text. 3. True; seepage90 in the text. 4. False; see page 102 in the text. 5. x+ 5 = −3 x+ 5 − 5 = −3− 5 x+ 0 = −8 x = −8 6. −6x = 42 6x = 42 −6 −6 1 · x = −7 x = −7 7. 5y + z = t 5y + z − z = t − z 2 5 2 5 4 5 9 3 5y = t − z 54. − 3 − 6 = − 3 + − 6 = − 6 + − 6 = − 6 = − 2 5y = t− z 5 1 5 3 2 1 5 5 56. − 12 + 4 = − 12 + 12 = − 12 = − 6 y = t− z 5 1 1 5 25 10 25 15 3 58. −2 2 + 6 = 2 + 4 = 4 + 4 = = 3 4 4 8. x + 5 = 11 60. 10x + 4 = 3x − 2 + x 10x + 4 = 4x − 2 6x = −6 x = −1 62. 5a = 3(6 − 3a) 5a = 18 − 9a 14a = 18 9 x = 6 The solution is 6. 9. x + 9 = −3 x+ 9 − 9 = −3− 9 x = −12 The solution is −12. 10. 8 = t + 1 a = 7 64. P = 4m + 7mn P = m(4 + 7n) P = m 4 + 7n 66. Not necessarily; 6 = 2 · 2 + 2 · 1, but 2 · 6, or 12, can be expressed as 2 · 5 + 2 · 1. 68. D = 1 E + F D(E + F) = 1 1 8 − 1 = t + 1 − 1 7 = t The solution is 7. 11. −7 = y + 3 −7 − 3 = y + 3 − 3 −10 = y The solution is −10. 12. x − 6 = 14 x − 6 + 6 = 14 + 6E + F = F = D 1 D − E, or 1− DE D x = 20 The solution is 20. 13. y − 7 = −2 Chapter2Mid-ChapterReview 3 1. The solution of 3 − x = 4x is 5 ; the solution of 5x = −3 3

12. is − 5 . The equations have diﬀerent solutions, so they are not equivalent. The given statement is false. y − 7 + 7 = −2 + 7 y = 5 The solution is 5. Copyright c 2015 Pearson Education, Inc.

13. · − 1 Chapter 2 Mid-Chapter Review 23 3 3 t 14. − 2 + z = − 4 22. − 5 = 3 3 3 3 3 1 − 2 + z + 2 = − 4 + 2 − 5 · t = 3 3 6 1 z = − 4 + 4 3 z = 4 −5 − 5 · t = −5 · 3 t = −15 3 The solution is . 4 15. −3.3 = −1.9 + t The solution is −15. 23. 3 x = 9 4 8 4 3 4 9−3.3 + 1.9 = −1.9 + t + 1.9 −1.4 = t 3 · 4 x = 3 − 8 /3 3 The solution is −1.4. 16. 7x = 42 7x = 42 x = − · · 3/ · 2 · 4/ 3 x = − 2 3 7 7 The solution is − 2 . x = 6 The solution is 6. 17. 17 = −t 17 = −1 · t 24. 3x + 2 = 5 3x+ 2 − 2 = 5 − 2 3x = 3 3x 3 17 = 1· t 3 = 3 −1 −1 −17 = t The solution is −17. x = 1 The solution is 1. 25. 5x+ 4 = −11 18. 6x = −54 5x + 4 4 = 11 4 6x = 54 − − − 6 6 5x = −15 x = −9 5x = −15 The solution is −9. 5 5 19. −5y = −85 x = −3 The solution is −3. −5y = 85 −5 −5 26. 6x − 7 = 2 y = 17 The solution is 17. x 6x − 7 + 7 = 2 + 7 6x = 9 6x = 9 20. 7 = 3 6 6 9 3 3/ x = = 7 · x = 3 1 7 · 7 x = 7 · 3 x = 21 The solution is 21. 21. 2 x = 12 3 3 2 3 2 · 3 x = 2 · 12

14. 1 6 2 · 3 x = 2 3 The solution is . 2 27. −4x − 9 = −5 −4x − 9 + 9 = −5 + 9 −4x = 4 −4x = 4 3· 2/· 6 −4 −4 x = x = −1 · x = 18 The solution is 18. The solution is −1. Copyright c 2015 Pearson Education, Inc.

15. 24 Chapter 2:Solving Equations and Inequalities 28. 6x + 5x = 33 11x = 33 33. 4 + 1 t = 1 5 6 10 4 1 1 11x = 33 30 + t 5 6 = 30 · 10 Clearing fractions 11 11 x = 3 The solution is 3. 29. −3y − 4y = 49 −7y = 49 −7y = 49 4 1 30 30 · 5 + 30 · 6 t = 10 24 + 5t = 3 24 + 5t − 24 = 3 − 24 5t = −21 5t = 21 5 5 −7 −7 21y = −7 The solution is −7. 30. 3x − 4 = 12 − x 3x − 4 + x = 12 − x + x 4x − 4 = 12 4x − 4 + 4 = 12 + 4 4x = 16 4x = 16 t = − 5 21 The solution is − 5 . 34. 0.21n − 1.05 = 2.1 − 0.14n 100(0.21n − 1.05) = 100(2.1 − 0.14n) Clearing decimals 100(0.21n) − 100(1.05) = 100(2.1) − 100(0.14n) 21n − 105 = 210 − 14n 4 4 21n − 105 + 14n = 210 − 14n + 14n x = 4 The solution is 4. 31. 5 − 6x = 9 − 8x 5 − 6x + 8x = 9 − 8x + 8x 5 + 2x = 9 35n − 105 = 210 35n − 105 + 105 = 210 + 105 35n = 315 35n = 315 35 35 5 + 2x− 5 = 9 − 5 2x = 4 2x = 4 The solution is 9. 35. 5(3y − 1)= −35 n = 9 2 2 15y − 5 = −35 x = 2 The solution is 2. 15y − 5 + 5 = −35 + 5 15y = −30 3 3 15y = −30 32. 4y − 2 = 4 + 2y 15 15 3 4 4y − 2 3 3 = 4 + 2y 4 3 Clearing fractions y = −2 The solution is −2. 4 · 4y − 4 · 2 = 4 · 4 + 4 · 2y 16y − 6 = 3 + 8y 16y − 6 − 8y = 3 + 8y − 8y 8y − 6 = 3 8y − 6 + 6 = 3 + 6 8y = 9 36. 7 − 2(5x + 3) = 1 7 − 10x − 6 = 1 1 − 10x = 1 1 − 10x − 1 = 1 − 1 −10x = 0 −10x = 0 10 −10 8y = 9 8 8 9 The solution is . 8 − x = 0 The solution is 0. 37. −8 + t = t − 8

16. −8 + t − t = t − 8 − t −8 = −8 We have an equation that is true for all real numbers. Thus, all real numbers are solutions.

17. Exercise Set 2.5 25 38. z + 12 = −12 + z z + 12 − z = −12 + z − z 46. M = x + y + z 2 x + y + z 12 = −12 We have a false equation. There are no solutions. 39. 4(3x + 2) = 5(2x − 1) 12x + 8 = 10x − 5 12x + 8 − 10x = 10x − 5 − 10x 2x+ 8 = −5 2x+ 8 − 8 = −5− 8 2x = −13 2 · M = 2 2 2M = x + y + z 2M − x − z = x + y + z − x − z 2M − x − z = y 47. Equivalent expressions have the same value for all possi- ble replacements for the variable(s). Equivalent equations have the same solution(s). 48. The equations are not equivalent because they do not have the same solutions. Although 5 is a solution of both equa- 2x = 13 2 2 tions, −5 is a solution of x2 = 25 but not of x = 5. 13 x = − 2 13 The solution is − 2 . 40. 8x − 6 − 2x = 3(2x − 4) + 6 6x − 6 = 6x − 12 + 6 6x − 6 = 6x− 6 6x − 6 − 6x = 6x − 6 − 6x −6 = −6 We have an equation that is true for all real numbers. Thus, all real numbers are solutions. 41. A = 4b A = 4b 49. For an equation x + a = b, add the opposite of a (or sub- tract a) on both sides of the equation. 1 50. It appears that the student added on the right side of 3 1 the equation rather than subtracting . 3 51. For an equation ax = b, multiply by 1/a (or divide by a) on both sides of the equation. 52. Answers may vary. A walker who knows how far and how long she walks each day wants to know her average speed eachday. Exercise Set 2.5 4 4 A = b 4 42. y = x − 1.5 y + 1.5 = x − 1.5 + 1.5 y + 1.5 = x 43. n = s − m n − s = s − m − s n − s = −m −1(n − s) = −1(−m) −n + s = m, or s − n = m 44. 4t = 9w 4t = 9w RC2. The correct answer is (b). RC4. The correct answer is (a). RC6. The correct answer is (c). 2. Solve: p · 76 = 19 p = 0.25 = 25% 4. Solve: 20.4 = 24% · a 85 = a 6. Solve: a = 50% · 50 a = 25 8. Solve: 57 = p · 300 0.19= p 19% = p 10. Solve: 7 = 175% b 4 4 4 = b · 9w t = 4 45. B = at − c B + c = at − c + c B + c = at B+c = at 12. Solve: 16 = p · 40 p = 0.4 = 40% 14. Solve: p · 150 = 39 p = 0.26 = 26% 16. Solve: a = 1% · 1, 000, 000 a a a = 10, 000 B+c = t a 18. Solve: p · 60 = 75 p = 1.25 = 125%

18. 28. Solve: 8 = 2% · b 400 = b 30. Solve: a = 7.0% · 8909 a ≈ 624 million 32. Solve: a = 4.4% · 8909 a ≈ 392 million 34. Solve: a = 0.5% · 8909 a ≈ 45 million 36. Solve: a = 54.8% · \$2360 a ≈ \$1293 million 38. Solve: a = 7.9% · \$5000 · 26 Chapter 2:Solving Equations and Inequalities 20. Any number is 100% of itself, so 70 is 100% of 70. We could also do this exercise as follows: Solve: p · 70 = 70 p = 1 = 100% 22. Solve: 54 = 24% · b 225 = b 54. Decrease: 4381− 962 = 3419 To ﬁnd the percent decrease, solve: 3419= p · 4381 0.780 ≈ p 78.0% ≈ p 56. 5x − 21 24. Solve: a = 40% · 2 18b 3· 6/· /b 3 a = 0.8 58. − 12b = − 2 · /b − 2 26. Solve: 40 = 2% · b 2000 = b 60. [3(x + 4) − 6] − [8 + 2(x − 5)] = [3x + 12 − 6] − [8 + 2x − 10] = 3x+ 6 − [2x − 2] = 3x + 6 − 2x + 2 = x + 8 62. Note: 4 ft 8 in.= 56 in. Solve: 56 = 84.4% · b 66 ≈ b Dana’s ﬁnal adult height will be about 66 in., or 5 ft 6 in. Exercise Set 2.6 a = \$395 40. Solve: 43 = p · 116 0.37 ≈ p 37% ≈ p 42. a) Solve: a = 20% · \$75 a = \$15 b) \$75+ \$15 = \$90 44. a) Solve: \$6.75 = 18% · b \$37.50 = b b) \$37.50 + \$6.75= \$44.25 46. Solve: 46.2 = 44% · b b = 105billion pieces of junk mail 48. Increase, in millions of dollars: 1070 − 950 = 120 To ﬁnd the percent of increase, solve: 120 = p · 950 0.126 ≈ p 12.6% ≈ p 50. Decrease: 301− 273 = 28 To ﬁnd the percent decrease, solve: 28 = p · 301 RC2. Translate the problem to an equation. RC4. Check the answer in the original problem. 2. Solve: c − 89= 60 c = 149calories 4. Solve: x + (x + 2) = 72 x = 35 If x = 35, then x + 2 = 37. The lengths of the pieces are 35 in. and 37 in. 6. Solve: 3a + 72, 000 = 876, 000 a = \$268, 000 8. Solve: m + 59= 385 m = 326 ft 10. Solve: x + (x + 1) = 547 x = 273 If x = 273, then x + 1 = 274. The numbers are 273 and 274. 12. Solve: a + (a + 1) + (a + 2) = 108 a = 35 Whitney, Wesley, and Wanda’s ages are 35, 36, and 37, respectively. 0.093 ≈ p 9.3% ≈ p 52. Increase: 764, 495 − 582, 996 = 181, 499 14. Solve: x + 1 (x + 1) + 2(x + 2) 2 − 7 = 2101 x = 601

19. To ﬁnd the percent increase, solve: 181, 499 = p · 582, 996 0.311 ≈ p 31.1% ≈ p

20. − − If x = 601, then x + 1 = 602 and x + 2 = 603. The integers are 601, 602, and 603. Exercise Set 2.7 27 16. Solve: 2(w + 2) + 2w = 10 3 1 w = 2 , or 1 2 1 1 1 54. Let d = the number of dimes. Then 2d = the number of quarters, and d + 10 = the number of nickels. The value of d dimes is 0.10d. If w = 1 , then w + 2 = 1 + 2 = 3 . The value of 2d quarters is 0.25(2d). 2 2 2 1 1 The value of d + 10 nickels is 0.05(d + 10). The length is 3 2 in. and the width is 1 2 in. Solve: 0.10d + 0.25(2d) + 0.05(d + 10) = 20 18. Solve: p − 15%p = 33.15 p = \$39 20. Solve: p + 6.5%p = 117.15 p = \$110 22. Solve: 84.45 + 0.55m = 250 m = 301mi 24. Solve: x + 4x + (x + 4x) − 45 = 180 x = 22.5 If x = 22.5, then 4x = 90 and (x + 4x) − 45 = 67.5. The measures of the angles are 22.5◦ , 90◦ , and67.5◦ . 26. Solve: x + 3x + (x − 15) = 180 x = 39 If x = 39, then 3x = 117 and x − 15 = 24. The measures of the angles are 39◦ , 117◦ , and24◦ . 28. Solve: a + 0.06a = 6996 a = \$6600 d = 30 If d = 30,then2d = 60 and d + 10 = 40. Susanne got 60 quarters, 30 dimes, and 40 nickels. Exercise Set 2.7 RC2. 3x − 5 ≤ −x + 1 4x ≤ 6 We see that 3x − 5 ≤ −x + 1 and 2x ≤ 6 are not equivalent. RC4. 2 − t > −3t + 4 2t > 2 We see that 2 − t > −3t + 4 and 2t > 2 are equivalent. 2. a) Yes, b) yes, c) yes, d) yes, e) no 4. a) No, b) yes, c) yes, d) no, e) yes x 0 30. Solve: b + 0.1b = 7194 b = \$6540 6. 5 4 3 2 1 0 1 2 3 4 5 32. Solve: 1.80 + 2.20m = 26 m = 11 mi 8. 5 4 3 2 1 y 5 0 1 2 3 4 5 34. Solve: c + 20%c = \$24.90 10. x 2 c = \$20.75 5 4 3 2 1 0 1 2 3 4 5 36. Solve: t+2t+27 = 34 3 12. 5 x 2 t = 25 5 4 3 2 1 0 1 2 3 4 5 If t = \$25, then 2t = \$50. The prices of the other two shirts were \$25 and \$50. 14. 5 x 0 38. Solve: 2x + 85 = 3 x 4 x = −68 5 4 3 2 1 16. x+ 5 > 2 x > −3 0 1 2 3 4 5 4 3 32 15 17 {x|x > −3} 40. − 5 + 8 = − 40 + 40 = − 40 4 3 42. − 5 ÷ 8 4 8 32 = 5 · 3 = 15

21. 5 4 3 2 1 0 1 2 3 4 5 44. −25.6 ÷ (−16) = 1.6 46. −25.6 − (−16) = −25.6 + 16 = −9.6 48. (−0.02) ÷ (−0.2) = 0.1 50. c + (4 + d) 52. Solve: 3 · 7 + 3m = 78 m = 19 The student answered 19multiple-choice questions cor- rectly. 18. x + 8 ≤ −11 x ≤ −19 {x|x ≤ −19} 19 40 30 20 10 0 20. y − 9 > −15 y > −6 {y|y > −6} 10 20 30 40 Copyright c 2015 Pearson Education, Inc.

22. 36 28 Chapter 2:Solving Equations and Inequalities 22. 2x + 4 > x + 7 x > 3 46. 1 −4x ≤ 9 1 {x|x > 3} x ≥ − 36 1 24. 3x + 18 ≤ 2x + 16 x ≤ −2 x|x ≥ − 7 {x|x ≤ −2} 48. − 9 > 63x 26. 9x − 8 < 8x − 9 x < −1 {x|x < −1} 28. −8 + p > 10 1 63 7 − 9 > x 1 − 81 > x p > 18 {p|p > 18} 1 x x < − 81 1 5 50. 3 + 4y < 35 30. x − 3 < 6 7 x ≤ 6 7 x x ≤ 6 4y < 32 y < 8 {y|y < 8} 52. 5y − 9 ≤ 21 5y ≤ 30 32. x + 1 > 1 y ≤ 6 8 2 {y|y ≤ 6} 3 x > 8 54. 8y − 6 < −54 8y < −48 x x > 3 8 y < −6 34. 8x ≥ 32 x ≥ 4 {x|x ≥ 4} 5 4 3 2 1 36. −16x > −64 x < 4 {x|x < 4} 0 1 2 3 4 5 {y|y < −6} 56. 48 > 13 − 7y 35 > −7y −5 < y {y|y > −5} 58. 15x + 5 − 14x ≤ 9 x + 5 ≤ 9 x ≤ 4 5 4 3 2 1 0 1 2 3 4 5 60. {x|x ≤ 4} 8 < 9x + 8 8x 3 38. 3x < −4 4 x < − 3 4 x x < − 3 − − − −8 < x + 5 −13 < x {x|x > −13} 62. 9 − 8y > 5 − 7y + 2 9 − 8y > 7 − 7y 40. −3x ≤ 15 x ≥ −5 {x|x ≥ −5} 42. −7x < −21 x > 3 {x|x > 3} 44. −5y > −23 23 y < 5

23. 2 > y {y|y < 2} 6 4 . 6 − 18x ≤ 4 − 12x − 5x 6 − 18x ≤ 4 − 17x 2 ≤ x {x|x ≥ 2} 66. 18 − 6y − 4y < 63 + 5y 18 − 10y < 63 + 5y −45 < 15y y y < 23 −3 < y {y|y > −3} 5 Copyright c 2015 Pearson Education, Inc.

24. − Exercise Set 2.8 29 68. 0.96y − 0.79 ≤ 0.21y+ 0.46 96y − 79 ≤ 21y + 46 75y ≤ 125 5 82. 0.4(2x + 8) ≥ 20 − (x + 5) 0.8x + 3.2 ≥ 20 − x − 5 0.8x + 3.2 ≥ 15 − x 8x + 32 ≥ 150− 10x y ≤ 3 5 18x ≥ 118 118 y y ≤ 3 x ≥ 18 59 , or 70. 2 + x < 4 x ≥ 9 3 5 15 59 15 2 + x < 15 4 3 5 · 15 x x ≥ 9 10 + 3x < 4 3x < −6 84. 1 − 3 y ≥ 5 − 5 + 5 x < −2 {x|x < −2} 2 15 1 − 3 y 9 y 3 ≥ 15 5 − 5 + 5 72. 3x 7 15 15 − 10y ≥ 27 − 3y + 9 15 − 10y ≥ 36 − 3y 4 − 8 ≥ − −21 ≥ 7y 8 3x 7 −3 ≥ y 4 − 8 ≥ 8(−15) {y|y ≤ −3} 6x − 7 ≥ −120 6x ≥ −113 86. 8.12 − 9.23 = 8.12 + (−9.23) = −1.11 113 x ≥ − 6 113 3 1 3 88. − 4 − 8 = 4 + 1 6 − 8 = − 8 + 1 7 − 8 = − 8 x x ≥ − 6 90. 10 ÷ 2 · 5 − 32 + (−5)2 = 10 ÷ 2 · 5 − 9 + 25 = 5 · 5 − 9 + 25 74. 4(2y − 3) > 28 8y − 12 > 28 8y > 40 y > 5 {y|y > 5} 76. 3(5 + 3m)− 8 ≤ 88 15 + 9m − 8 ≤ 88 9m + 7 ≤ 88 9m ≤ 81 m ≤ 9 = 25 − 9 + 25 = 16 + 25 = 41 92. 9(3 + 5x)− 4(7 + 2x) = 27 + 45x − 28 − 8x = −1+ 37x, or 37x − 1 94. The solutions of |x| < 3 are all points whose distance from 0 is less than 3. This is equivalent to −3 < x < 3. The graph is as follows: x 3 {m|m ≤ 9} 78. 7(5y − 2) > 6(6y − 1) 35y − 14 > 36y − 6 −8 > y 5 4 3 2 1 96. x+ 4 > 3 + x 4 > 3 0 1 2 3 4 5 {y|y < −8} 80. 5(x + 3) + 9 ≤ 3(x − 2) + 6 5x + 15 + 9 ≤ 3x − 6 + 6 5x + 24 ≤ 3x 2x ≤ −24

25. All real numbers are solutions. Exercise Set 2.8 RC2. q ≤ r x ≤ −12 {x|x ≤ −12} RC4. r ≥ q, so we choose q ≤ r. RC6. q ≥ r, so we choose r ≤ q. 2. n ≥ 5 4. 75 < p < 100 6. a ≤ 180 8. A > 40 L Copyright c 2015 Pearson Education, Inc.

26. 30 Chapter 2:Solving Equations and Inequalities 3 1 10. T ≤ −2◦ 12. x < 5 14. n ≥ 18 48. Solve: 3 + w > 22 4 2 w > 26 1 The puppy’s weight will exceed 22 lb 26 weeks after its 2 16. c ≤ \$4857.95 18. c ≥ \$3.19 1 weight is 3 lb. 50. Solve: h + (h + 3) > 27 h > 12 20. 2 n − 5 > 17 73 + 75 + 89 + 91 + s George worked more than 12 hr, and Joan worked more than 15 hr. 22. Solve: 5 ≥ 85 52. −22 The solution set is {s|s ≥ 97}. 9 54. 8x + 3x = 66 24. Solve: 5 C + 32 > 98.6 11x = 66 The solution set is {C |C > 37◦ }. 26. Solve: −0.028t + 20.8 < 19.0 2 t > 64 7 x = 6 56. 9x − 1 + 11x − 18 = 3x − 15 + 4 + 17x 20x − 19= 20x − 11 2 We have 1920+ 64 7 2 = 1984 7, so the solution set is −19= −11 The equation has no solution. {Y |Y ≥ 1985}, where Y represents the year. 28. Solve: 80 + 16n ≤ 750 n ≤ 41.875 At most,41 people can attend the banquet. 30. Solve: 53 + L ≤ 165 The solution set is {L|L ≤ 112 in.}. 58. Solve: a = 10% · 310 a = 31 60. Solve: 80 = p · 96 0.833 = p 83.3% = p, or 1 83 3 % = p 32. Solve: 0.45 + 0.25h ≥ 2.20 h ≥ 7 Laura parks for at least 7 half hours,or for at least 3.5 hr. 62. Solve: 4 + 2.5(h − 1) > 16.5 h > 6 hr 34. Solve: 45 + 30t > 150 t > 3.5 hr 36. Solve: 5 + 7 + 8 +c 7 4 ≥ 64. Solve: 14 < 4 + 2.50(h − 1) < 24 5 hr < h < 9hr Chapter 2 Vocabulary Reinforcement c ≥ 8 credits 38. Solve: b + (b − 2) + (b + 3) > 19 b > 6 cm 40. Solve: 16l ≥ 264. l ≥ 16.5 yd 42. Solve: c > 0.8(21, 000) c > \$16, 800 44. Solve: 5 ≤ 0.75r (See Exercise 43.) 2 r ≥ 6 3 g 1. Any replacement for the variable that makes an equation true is called a solution of the equation. 2. The addition principle for equations states that for any real numbers a, b, and c, a = b is equivalent to a+c = b+c. 3. The multiplicationprinciple states that for any real num- bers a, b, and c, a = b is equivalent to a · c = b · c. 4. An inequality is a number sentence with <, ≤, >, or ≥ as its verb. 5. Equations with the same solution are called equivalent equations.

27. · ·46. Solve: 1 8 h 2 ≤ 12 h ≤ 3 ft Chapter2ConceptReinforcement 1. True; see page 102 in the text. Copyright c 2015 Pearson Education, Inc.

28. > − Chapter 2 Summary and Review:Review Exercises 31 2. True; for any numbern, n ≥ n is true because n = n is true. 3. False; the solution set of 2x − 7 ≤ 11 is {x|x ≤ 9}; the solution set of x < 2 is {x|x < 2}. The inequalities do not have the same solution set, so they are not equivalent. 4. True; if x > y, then −1·x < −1·y (reversing the inequality symbol), or −x < −y. Chapter2StudyGuide 1. 4(x − 3) = 6(x + 2) 4x − 12 = 6x + 12 6. Graph: x ≤ −1 The solutions of x ≤ −1are all numbers less than or equal to −1. We shade all points to the left of −1and use a closed circle at −1to indicate that −1 is part of the graph. x 1 1 0 7. 6y + 5 > 3y − 7 6y + 5 − 3y > 3y − 7 − 3y 3y + 5 > −7 3y + 5 − 5 > −7 − 5 3y > −12 3y −12 4x − 12 − 6x = 6x + 12 − 6x 3 3 −2x − 12 = 12 y > −4 The solution set is y y > 4 . −2x − 12 + 12 = 12 + 12 −2x = 24 −2x = 24 { | − } Chapter 2 Review Exercises −2 −2 x = −12 The solution is −12. 1. x+ 5 = −17 x+ 5 − 5 = −17− 5 x = −22 2. 4 + 3y − 7 = 3 + 3(y − 2) 4 + 3y − 7 = 3 + 3y − 6 The solution is −22. 3y − 3 = −3 + 3y 3y − 3 − 3y = −3 + 3y − 3y −3 = −3 Every real number is a solution of the equation −3 = −3, so all real numbers are solutions of the original equation. 3. 4(x − 3) + 7 = −5 + 4x + 10 4x − 12 + 7 = −5 + 4x + 10 4x − 5 = 5 + 4x 4x − 5 − 4x = 5 + 4x − 4x −5 = 5 We get a false equation, so the original equation has no solution. 4. A = 1 bh 2. n − 7 = −6 n − 7 + 7 = −6 + 7 n = 1 The solution is 1. 3. x − 11 = 14 x − 11 + 11 = 14 + 11 x = 25 The solution is 25. 4. y − 0.9 = 9.09 y − 0.9 + 0.9 = 9.09+ 0.9 y = 9.99 The solution is 9.99. 5. 2 1 2 − 3 x = − 6 1 2 · A = 2 · 2 bh 3 − 2 · 2 3 1 − 3 x = 2 · − 6 2A = bh 2A = bh 1 · x = 3/· 1 2 · 2 · 3/ h h 1 x = 2A = b h 5. Graph: x > 1 The solutions of x > 1 are all numbers greater than 1. We shade all points to the right of 1 and use an open circle at 1 to indicate that 1 is not part of the graph. 4 1 The solution is . 4 6. −8x = −56 8x = 56 −8 −8

29. x 1 6 5 4 3 2 1 0 1 2 3 4 5 6 x = 7 The solution is 7. Copyright c 2015 Pearson Education, Inc.

30. 3 − 32 Chapter 2:Solving Equations and Inequalities x 7. − 4 = 48 1 4 · 4 · (−x) = 4 · 48 13. 7x − 6 = 25x 7x − 6 − 7x = 25x − 7x −6 = 18x −x = 192 −1 · (−1 · x) = −1 · 192 x = −192 The solution is −192. 8. 15x = −35 15x = −35 6 18x = 18 18 6/ · 1 = x− 3 · 6/ 1 − 3 = x 1 The solution is − . 15 15 5/ · 7 3 14. 14y = 23y − 17 − 10 x = − · 7 14y = 23y − 27 Collecting like terms x = − 3 7 The solution is − 3 . 14y − 23y = 23y − 27 − 23y −9y = −27 −9y = −27 −9 −9 9. 4 y = 5 3 − 16 y = 3 5 4 5 3 The solution is 3. 4 · 5 y = 4 · 15 y = − 64 − 16 15 15. 0.22y − 0.6 = 0.12y + 3 − 0.8y 0.22y − 0.6 = −0.68y + 3 Collecting like terms 0.22y − 0.6 + 0.68y= −0.68y + 3 + 0.68y The solution is − 64 . 10. 5 − x = 13 0.9y − 0.6 = 3 0.9y − 0.6 + 0.6 = 3 + 0.6 0.9y = 3.6 5 − x − 5 = 13 − 5 0.9y 3.6 −x = 8 −1 · (−1 · x) = −1 · 8 x = −8 The solution is −8. = 0.9 0.9 y = 4 The solution is 4. 11. 1 x 5 = 3 16. 1 x 4 1 8 x = 3 − 1 16 x 4 − 8 8 2 1 1 x 1 5 5 3 5 4 x − 8 + 8 = 8 + 8 8 − 8 x = 3 − 16 x 1 1 x = 3 1 8 − 16 x 4 x = 1 1 1 x + x = 3 1 1 x + x 1 8 16 − 16 16 4 · 4 x = 4 · 1 2 1 x + x = 3 x = 4 The solution is 4. 12. 5t + 9 = 3t − 1 5t+ 9 − 3t = 3t − 1 − 3t 16 16 3 16 16 3 3 · 16 x = 3 16 x = 3 · 3 16· 3/ 2t + 9 = −1

31. 2t+ 9 − 9 = −1− 9 2t = −10 2t = 10 2 2 t = −5 The solution is −5. x = 1· x = 16 The solution is 16. 17. 14y + 17 + 7y = 9 + 21y + 8 21y + 17 = 21y + 17 21y + 17 − 21y = 21y + 17 − 21y 17 = 17 TRUE All real numbers are solutions. Copyright c 2015 Pearson Education, Inc.

32. Chapter 2 Summary and Review:Review Exercises 33 18. 4(x + 3) = 36 4x + 12 = 36 25. Since 4 ≤ 4 is true, 4 is a solution. 26. y + 2 1 4x + 12 − 12 = 36 − 12 3 ≥ 6 4x = 24 4x = 24 2 2 y + 3 − 3 1 2 ≥ 6 − 3 1 4 4 4 y ≥ − x = 6 6 6 3The solution is 6. 19. 3(5x − 7) = −66 15x − 21 = −66 y ≥ − 6 1 y ≥ − 2 1 15x − 21 + 21 = −66 + 21 15x = −45 15x = −45 The solution set is 27. 9x ≥ 63 y y ≥ − 2 . 15 15 x = −3 The solution is −3. 20. 8(x − 2) − 5(x + 4) = 20 + x 8x − 16 − 5x − 20 = 20 + x 3x − 36 = 20 + x 3x − 36 − x = 20 + x − x 2x − 36 = 20 2x − 36 + 36 = 20 + 36 2x = 56 2x = 56 9x 63 9 ≥ 9 x ≥ 7 The solution set is {x|x ≥ 7}. 28. 2 + 6y > 14 2 + 6y − 2 > 14 − 2 6y > 12 6y 12 6 > 6 y > 2 The solution set is {y|y > 2}. 2 2 29. 7 − 3y ≥ 27 + 2y The solution is 28. x = 28 7 − 3y − 2y ≥ 27 + 2y − 2y 7 − 5y ≥ 27 21. −5x + 3(x + 8) = 16 −5x + 3x + 24 = 16 7 − 5y − 7 ≥ 27 − 7 −5y ≥ 20 −2x + 24 = 16 −2x + 24 − 24 = 16 − 24 −5y −5 ≤ y 20 Reversing the inequality symbol −5 −2x = −8 −2x = −8 −2 −2 x = 4 The solution is 4. 22. 6(x − 2) − 16 = 3(2x − 5) + 11 6x − 12 − 16 = 6x − 15 + 11 6x − 28 = 6x− 4 6x − 28 − 6x = 6x − 4 − 6x ≤ −4 The solution set is {y|y ≤ −4}. 30. 3x+ 5 < 2x − 6 3x+ 5 − 2x < 2x − 6 − 2x x+ 5 < −6 x+ 5 − 5 < −6 − 5 x < −11 The solution set is {x|x < −11}. 31. −4y < 28 −28 = −4 False 4y 28 > Reversing the inequality symbol There is no solution. −4 −4 23. Since −3≤ 4 is true, −3is a solution. y > −7 The solution set is y y > 7 .

33. 24. Since 7 ≤ 4 is false, 7 is not a solution. { | − } Copyright c 2015 Pearson Education, Inc.

34. − ≤ 4 4 3 34 Chapter 2:Solving Equations and Inequalities 32. 4 − 8x < 13 + 3x 4 − 8x − 3x < 13 + 3x − 3x 4 − 11x < 13 4 − 11x − 4 < 13 − 4 38. V = 1 Bh 3 1 3 · V = 3 · 3 Bh 3V = Bh −11x < 9 11x 9 > Reversing the inequality symbol 3V = Bh h h −11 −11 3V = B 9 h x > − 11 The solution set is 9 x x > − 11 . 39. A = a + b 2 a + b 33. 4x 1 3 1 1 1 2 · A = 2 · 2 2A = a + b 2A − b = a + b − b − · (−4x) ≥ − · Reversing the inequality symbol 2A − b = a 1 x ≥ − 12 The solution set is 1 x x ≥ − . 40. y = mx + b y − b = mx + b − b y − b = mx 12 34. 4x − 6 < x + 3 4x − 6 − x < x+ 3 − x 3x − 6 < 3 3x − 6 + 6 < 3 + 6 3x < 9 3x 9 3 < 3 x < 3 The solution set is {x|x < 3}. The graph is as follows: x 3 y− b = mx m m y− b = x m 41. Familiarize. Let w = the width, in miles. Then w + 90 = the length. Recall that the perimeter P of a rectangle with length l and width w is given by P = 2l + 2w. Translate. Substitute 1280 for P and w + 90 for l in the formula above. P = 2l + 2w 1280 = 2(w + 90) + 2w Solve. We solve the equation. 0 3 1280 = 2(w + 90) + 2w 35. In order to be a solution of −2 < x ≤ 5, a number must be a solution of both −2< x and x ≤ 5. The solution set is graphed as follows: 2 x 5 1280 = 2w + 180 + 2w 1280 = 4w + 180 1280 − 180 = 4w + 180 − 180 1100 = 4w 2 0 5 36. The solutions of y > 0 are those numbers greater than 0. 1100 4 = 4w 4 The graph is as follows: y 0 0 37. C = πd C = πd 275 = w If w = 275, then w + 90 = 275 + 90 = 365. Check. The length, 365mi, is 90 mi morethan the width, 275mi. The perimeter is 2 · 365 mi +2 · 275 mi = 730 mi + 550 mi = 1280mi. The answer checks. State. The length is 365 mi, and the width is 275 mi. π π 42. Familiarize. Let x = the number on the ﬁrst marker. C = d π Then x + 1 = the number on the second marker. Translate. First number plus second number is 691.

35. ↓ ↓ ↓ ↓ ↓ x + (x + 1) = 691

36. Chapter 2 Summary and Review:Review Exercises 35 Solve. We solve the equation. x + (x + 1) = 691 2x + 1 = 691 2x+ 1 − 1 = 691− 1 2x = 690 2x = 690 Solve. We solve the equation. x + (x + 50) + (2x − 10) = 180 4x + 40 = 180 4x + 40 − 40 = 180 − 40 4x = 140 4x = 140 2 2 4 4 x = 345 If x = 345, then x + 1 = 345 + 1 = 346. Check. 345 and 346 are consecutive integers and 345+ 346 = 691. The answer checks. State. The numbers on the markers are 345 and 346. 43. Familiarize. Let c = the cost of the entertainment center in February. Translate. x = 35 If x = 35, then x + 50 = 35 + 50 = 85 and 2x − 10 = 2 · 35 − 10 = 70 − 10 = 60. Check. The measure of the second angle is 50◦ more than the measure of the ﬁrst angle, and the measure of the third angle is 10◦ less than twice the measure of the ﬁrst angle. The sum of the measure is 35◦ + 85◦ + 60◦ = 180◦ . The answer checks. State. The measures of the angles are 35◦ , 85◦ , and60◦ . Cost in February plus \$332 is Cost in June 46. Translate. ↓ ↓ ↓ ↓ ↓ c + 332 = 2449 Solve. We solve the equation. c + 332 = 2449 c + 332 − 332 = 2449 − 332 c = 2117 Check. \$2117 + \$332 = \$2449, so the answer checks. State. The entertainment center cost \$2117in February. 44. Familiarize. Let a = the number of subscriptions Ty sold. Translate. What number is 20% of 75? ↓ ↓ ↓ ↓ ↓ a = 20% · 75 Solve. We convert 20% to decimal notation and multiply. a = 20% · 75 a = 0.2 · 75 a = 15 Thus, 15 is 20% of 75. 47. Translate. 15 is what percent of 80? Commission per subscription times number sold is Total commission ↓ ↓ ↓ ↓ ↓ 15 = p · 80 Solve. We solve the equation. ↓ ↓ ↓ ↓ ↓ 15 = p · 80 4 × a = 108 Solve. We solve the equation. 4 · a = 108 4· a = 108 4 4 a = 27 Check. \$4 · 27 = \$108, so the answer checks. State. Ty sold27 magazine subscriptions. 45. Familiarize. Let x = the measure of the ﬁrst angle. Then x + 50 = the measure of the second angle, and 2x − 10 = the measure of the third angle. Recall that the sum of measures of the angles of a triangle is 180◦ . Translate. 15 = pM· 80 80 80 0.1875 = p 18.75% = p Thus, 15 is 18.75% of 80. 48. Translate. 18 is 3% of w9hat number? ↓ ↓ ↓ ↓ ↓ 18 = 3% · b Solve. We solve the equation. 18 = 3% · b 18 = 0.03 · b Measure of ﬁrst angle measure of + second angle measure of + third angle is 180◦ . 18 = 0.03 0.03· b 0.03 600 = b ↓ ↓ ↓ ↓ ↓ ↓ ↓ x + (x + 50) + (2x − 10) = 180 Thus, 18 is 3% of 600.

37. 49. We subtract to ﬁnd the increase. 164, 440 − 87, 872 = 76, 568

38. ↓ ↓ ↓ ↓ ↓ 28.9 = p · 102.4 36 Chapter 2:Solving Equations and Inequalities The increase is 76,568. Now we ﬁnd the percent increase. 76, 568 is what percent of 87, 872? ↓ ↓ ↓ ↓ ↓ 76, 568 = p · 87, 872 Solve. We solve the equation. s + 8% · s = 78, 300 s + 0.08s = 78, 300 1.08s = 78, 300 1.08s = 78, 300 We divide by 87,872 on both sides and then convert to 1.08 1.08 percent notation. 76, 568 = p · 87, 872 76, 568 = p· 87, 872 s = 72, 500 Check. 8% of \$72, 500 = 0.08· \$72, 500 = \$5800 and \$72, 500 + \$5800 = \$78,300, so the answer checks. 87, 872 0.871 ≈ p 87.1% ≈ p 87, 872 State. The previous salary was \$72,500. 53. Familiarize. Let a = the amount the organization actu- ally owes. This is the price of the supplies without sales The percent increase is about 87.1%. 50. We subtract to ﬁnd the decrease, in billions. tax added. Then the incorrect amount is a + 5% of a, or a + 0.05a, or 1.05a. Translate. 102.4 − 73.5 = 28.9 Now we ﬁnd the percent decrease. Incorrect amount is \$145.90. 28.9is what percent of 102.4? ↓ ↓ ↓ 1.05a = 145.90 Solve. We solve the equation. 1.05a = 145.90 We divide by 102.4on both sides and then convert to per- cent notation. 1.05a = 1.05 145.90 1.05 28.9 = p · 102.4 a ≈ 138.95 Check. 5% of \$138.95 is 0.05 · \$138.95 ≈ \$6.95, and 28.9 102.4 = p· 102.4 102.4 \$138.95 + \$6.95= \$145.90, so the answer checks. 0.282 ≈ p 28.2% ≈ p The percent decrease is about 28.2%. State. The organization actually owes \$138.95. 54. Familiarize. Let s represent the score on the next test. Translate. 51. Familiarize. Let p = the price beforethe reduction. The average score is at least 80. Translate. 71 + 7 ↓ 2 + 86 + s ↓ ↓ Price beforereduction minus 30% of price is \$154. ↓ ↓ ↓ ↓ ↓ ↓ ↓ Solve. 5 + 8 5 ≥ 80 p − 30% · p = 154 Solve. We solve the equation. p − 30% · p = 154 p − 0.3p = 154 0.7p = 154 0.7p = 154 71 +75 +82 +86+s 5 ≥ 80 71 + 75 + 82 + 86 + s 5 ≥ 5 · 80 5 71 + 75 + 82 + 86 + s ≥ 400 314 + s ≥ 400 s ≥ 86 0.7 0.7 Check. As a partial check we show that the average is at p = 220 least 80 when the next test score is 86.

39. Check. 30% of \$220 is 0.3· \$220 = \$66 and \$220− \$66 = \$154, so the answer checks. 71 + 75 + 82 + 86 + 86 = 5 400 = 80 5 State. The price before the reduction was \$220. 52. Familiarize. Let s = the previous salary. Translate. State. The lowest grade Noah can get on the next test and have an average test score of 80 is 86. 55. Familiarize. Let w represent the width of the rectangle, in cm. The perimeter is given by P = 2l +2w, or 2·43+2w, Previous salary plus 8% of previous salary is \$78,300. or 86 + 2w. Translate. ↓ ↓ ↓ ↓ ↓ ↓ ↓ s + 8% · s = 78, 300 The perimeter is greater than 120 cm . ↓ ↓ ↓ 86 + 2w > 120

40. Chapter 2 Test 37 Solve. 86 + 2w > 120 2w > 34 w > 17 Check. We check to see if the solution seems reasonable. When w = 16 cm, P = 2 · 43 + 2 · 16, or 118cm. When w = 17 cm, P = 2 · 43 + 2 · 17, or 120cm. When w = 18 cm, P = 2 · 43 + 2 · 18, or 122cm. It appearsthat the solutionis correct. State. The solution set is {w|w > 17 cm}. 56. 4(3x − 5) + 6 = 8 + x 12x − 20 + 6 = 8 + x 12x − 14 = 8 + x 12x − 14 − x = 8 + x − x 11x − 14 = 8 11x − 14 + 14 = 8 + 14 11x = 22 11x = 22 11 11 x = 2 The solution is 2. This is between 1 and 5, so the correct answer is C. 57. 3x + 4y = P 3x + 4y − 3x = P − 3x 4y = P − 3x 4y = P− 3x Chapter 2 Discussion and Writing Exercises 1. The end result is the same either way. If s is the original salary, the newsalary after a 5% raise followed by an 8% raise is 1.08(1.05s). If the raises occur in the opposite order, the new salary is 1.05(1.08s). By the commutative and associate laws of multiplication, we see that these are equal. However, it would be better to receive the 8% raise ﬁrst, because this increaseyields a higher salary initially than a 5% raise. 2. No; Erin paid 75% of the original price and was oﬀered credit for 125% of this amount, not to be used on sale items. Now 125% of 75% is 93.75%, so Erin wouldhave a credit of 93.75%of the original price. Since this credit can be applied only to nonsale items, she has less purchasing power than if the amount she paid were refunded and she could spend it on sale items. 3. The inequalities are equivalent by the multiplication prin- ciple for inequalities. If we multiply both sides of one in- equality by −1, the other inequality results. 4. For any pair of numbers, their relative position on the number line is reversed when both are multiplied by the same negative number. For example, −3 is to the left of 5 on the number line (−3 < 5), but 12 is to the right of −20. That is, −3(−4) > 5(−4). 5. Answers may vary. Fran is more than 3 years older than Todd. 6. Let n represent “a number.” Then “ﬁve morethan a num- 4 4 ber” translates to n + 5, or 5 + n, and “ﬁve is more than a number” translates to 5 > n. y = P− 3x 4 Answer A is correct. 58. 2|x| + 4 = 50 2|x| = 46 |x| = 23 The solutions are the numbers whose distance from 0 is 23. Those numbers are −23and 23. Chapter 2 Test 1. x + 7 = 15 x+ 7 − 7 = 15 − 7 x = 8 The solution is 8. 59. |3x| = 60 The solutions are the values of x for which the distance of 2. t − 9 = 17 3 · x from 0 is 60. Then we have: 3x = −60 or 3x = 60 x = −20 or x = 20 The solutions are −20and 20. 60. y = 2a − ab + 3 y − 3 = 2a − ab y − 3 = a(2 − b) y− 3 = a 2 − b t − 9 + 9 = 17 + 9 t = 26 The solution is 26. 3. 3x = −18 3x = 18 3 3 x = −6 The solution is −6.

41. − − · 1 38 Chapter 2:Solving Equations and Inequalities 4. 4 x = 28 7 7 4 7 10. −3x − 6(x − 4) = 9 −3x − 6x + 24 = 9 − 4 · − 7 x = − 4 · (−28) 7· 4/· 7 −9x + 24 = 9 −9x + 24 − 24 = 9 − 24 x = · −9x = −15 x = 49 The solution is 49. 5. 3t+ 7 = 2t − 5 3t+ 7 − 2t = 2t − 5 − 2t t + 7 = −5 t + 7 − 7 = −5− 7 t = −12 The solution is −12. −9x = −15 −9 −9 3/ · 5 x = 3/ 3· 5 x = 3 5 The solution is . 3 11. We multiply by 10 to clear the decimals. 0.4p + 0.2 = 4.2p − 7.8 − 0.6p 6. 1 x 3 = 2 10(0.4p + 0.2) = 10(4.2p 7.8 0.6p) 2 − 5 5 − − 1 3 3 2 3 2 x − 5 + 5 = 5 + 5 1 2 x = 1 1 2 · 2 x = 2 · 1 x = 2 The solution is 2. 4p + 2 = 42p − 78 − 6p 4p + 2 = 36p − 78 4p + 2 − 36p = 36p − 78 − 36p −32p + 2 = −78 −32p + 2 − 2 = −78− 2 −32p = −80 −32p = −80 −32 −32 7. 8 − y = 16 8 − y − 8 = 16 − 8 −y = 8 5· ✧16 p = 2 ✧16 5 p = 2 −1 · (−1 · y) = −1 · 8 y = −8 The solution is −8. 5 The solution is . 2 12. 4(3x − 1)+ 11 = 2(6x + 5) − 8 2 3 12x − 4 + 11 = 12x + 10 − 8 8. − 5 + x = − 4 12x + 7 = 12x + 2 2 2 3 2 − 5 + x + 5 = − 4 + 5 15 8 x = − 20 + 20 7 x = − 20 7 The solution is − 20 . 9. 3(x + 2) = 27 3x+ 6 = 27 3x+ 6 − 6 = 27 − 6 3x = 21 3x = 21 12x + 7 − 12x = 12x + 2 − 12x 7 = 2 FALSE There are no solutions. 13. −2 + 7x + 6 = 5x + 4 + 2x 7x + 4 = 7x + 4 7x+ 4 − 7x = 7x + 4 − 7x 4 = 4 TRUE All real numbers are solutions. 14. x + 6 ≤ 2 x+ 6 − 6 ≤ 2 − 6 x ≤ −4 3 3 x = 7 The solution is 7. The solution set is {x|x ≤ −4}.

42. 5 5 4 Chapter 2 Test 39 15. 14x + 9 > 13x − 4 14x + 9 − 13x > 13x − 4 − 13x x+ 9 > −4 x+ 9 − 9 > −4 − 9 x > −13 The solution set is {x|x > −13}. 23. 6x − 3 < x + 2 6x − 3 − x < x+ 2 − x 5x − 3 < 2 5x − 3 + 3 < 2 + 3 5x < 5 5x 5 < 16. 12x ≤ 60 5 5 12x 60 12 ≤ 12 x ≤ 5 The solution set is {x|x ≤ 5}. 17. −2y ≥ 26 x < 1 The solution set is {x|x < 1}. The graph is as follows: x 1 0 1 24. In order to be a solution of the inequality −2 ≤ x ≤ 2, −2y 26 Reversing the inequality symbol a number must be a solution of both −2 ≤ x and x ≤ 2. −2 ≤ −2 y ≤ −13 The solution set is {y|y ≤ −13}. 18. −4y ≤ −32 The solution set is graphed as follows: 2 x 2 2 0 2 25. Translate. −4y 32 −4 ≥ −4 Reversing the inequality symbol W9hat number is 24% of 75? y ≥ 8 The solution set is {y|y ≥ 8}. 1 19. −5x ≥ 4 1 1 1 − · (−5x) ≤ − · Reversing the inequality symbol 1 x ≤ − 20 ↓ ↓ ↓ ↓ ↓ a = 24% · 75 Solve. We convert 24% to decimal notation and multiply. a = 24% · 75 a = 0.24 · 75 a = 18 Thus, 18 is 24% of 75. The solution set is 1 x x ≤ − 20 . 26. Translate. 15.84 is what percent 20. 4 − 6x > 40 4 − 6x − 4 > 40 − 4 −6x > 36 ↓ ↓ ↓ ↓ ↓ 15.84 = p · 96 Solve. 15.84 = p · 96−6x < 36 Reversing the inequality symbol 15.84 p96 −6 −6 = · x < −6 The solution set is {x|x < −6}. 21. 5 − 9x ≥ 19+ 5x 5 − 9x − 5x ≥ 19+ 5x − 5x 5 − 14x ≥ 19 5 − 14x − 5 ≥ 19 − 5 −14x ≥ 14 96 96 0.165 = p 16.5% = p Thus, 15.84 is 16.5% of 96. 27. Translate. 800 is 2% of what number? ↓ ↓ ↓ ↓ ↓ 800 = 2% · b 14x 14 Reversing the inequality symbol Solve. −14 ≤ −14 x ≤ −1 The solution set is {x|x ≤ −1}. 800 = 2% · b 800 = 0.02 · b 22. The solutions of y ≤ 9are shown by shading the point for 800 = 0.02 0.02· b 0.02

43. 9and all points to the left of 9. The closed circle at 9 indicates that 9is part of the graph. y 9 40, 000 = b Thus, 800is 2% of 40,000. 0 4 9

44. 40 Chapter 2:Solving Equations and Inequalities 28. We subtract to ﬁnd the increase. 29.2 − 18.2 = 11 Now we ﬁnd the percent of increase. State. The total cost of raising a child to age 17 is about \$230,556. 31. Familiarize. Let x = the ﬁrst integer. Then x + 1 = the 11 is what percent of 18.2? second and x + 2 = thethird. Translate. ↓ ↓ ↓ ↓ ↓ 11 = p · 18.2 We divide by 18.2 on both sides and then convert to per- cent notation. First integer plus second integer plus third integer is 7530. 11 = p · 18.2 ↓ ↓ ↓ ↓ ↓ ↓ ↓ x + (x + 1) + (x + 2) = 7530 11 18.2 = p· 18.2 18.2 Solve. x + (x + 1) + (x + 2) = 7530 0.604 ≈ p 60.4% ≈ p The percent increase is about 60.4%. 3x + 3 = 7530 3x+ 3 − 3 = 7530 − 3 3x = 7527 29. Familiarize. Let w = the width of the photograph, in cm. Then w + 4 = the length. Recall that the perimeter 3x = 3 7527 3 P of a rectangle with length l and width w is given by P = 2l + 2w. Translate. We substitute 36 for P and w + 4 for l in the formula above. P = 2l + 2w 36 = 2(w + 4) + 2w Solve. We solve the equation. 36 = 2(w + 4) + 2w 36 = 2w + 8 + 2w x = 2509 If x = 2509, then x + 1 = 2510 and x + 2 = 2511. Check. The numbers 2509,2510,and 2511 are consecutive integers and 2509+2510+2511 = 7530. The answer checks. State. The integers are 2509, 2510, and 2511. 32. Familiarize. Let x = the amount originally invested. Us- ing the formula for simple interest, I = Prt, the interest earned in one year will be x · 5% · 1, or 5%x. Translate. 36 = 4w + 8 36 − 8 = 4w + 8 − 8 28 = 4w Amount invested plus interest is amount after 1 year. 28 = 4w 4 4 7 = w If w = 7, then w + 4 = 7 + 4 = 11. Check. The length, 11 cm, is 4 cm more than the width, 7 cm. The perimeter is 2 · 11 cm+ 2 · 7 cm = 22 cm + 14 cm= 36 cm. The answer checks. ↓ ↓ ↓ ↓ ↓ x + 5%x = 924 Solve. We solve the equation. x + 5%x = 924 x + 0.05x = 924 1.05x = 924 1.05x = 924 State. The width is 7 cm, and the length is 11 cm. 1.05 1.05 30. Familiarize. Let t = the total cost of raising a child to age 17. Translate. x = 880 Check. 5% of \$880 is 0.05· \$880 = \$44 and \$880+ \$44 = \$924, so the answer checks. Cost for child care and K-12 education is 18% of Total cost State. \$880 was originally invested. 33. Familiarize. Using the labels on the drawing in the text, we let x = the length of the shorter piece, in meters, and Solve. ↓ ↓ ↓ ↓ ↓ 41, 500 = 18% · t x + 2 = the length of the longer piece. Translate.

45. 41, 500 = 18% · t 41, 500 = 0.18t Length of shorter piece plus length of longer piece is 8 m . 41, 500 =0.18t ↓ ↓ ↓ ↓ ↓ 0.18 0.18 x + (x + 2) = 8 230, 556 ≈ t Check. 18% of \$230,556 is about \$41,500, so the answer checks.

46. Chapter 2 Test 41 Solve. We solve the equation. x + (x + 2) = 8 2x + 2 = 8 2x+ 2 − 2 = 8 − 2 State. Jason can spend no more than \$105in the sixth month. The solution set can be expressed as {s|s ≤ \$105}. 36. Familiarize. Let c = the number of copies made. For 3 months, the rental charge is 3 · \$225,or \$675. Expressing 2x = 6 3.2 as \$0.032, the charge for the copies is given by 2x = 6 \$0.032 · c. Translate. 2 2 x = 3 If x = 3, then x + 2 = 3 + 2 = 5. Rental charge plus copy charge is no more than \$4500. Check. One piece is 2 m longer than the other and the sum of the lengths is 3 m+5 m, or 8 m. The answerchecks. State. The lengths of the pieces are 3 m and 5 m. Solve. ↓ ↓ ↓ ↓ ↓ 675 + 0.032c ≤ 4500 34. Familiarize. Let l = the length of the rectangle, in yd. The perimeter is given by P = 2l + 2w, or 2l + 2 · 96, or 2l + 192. Translate. 675 + 0.032c ≤ 4500 0.032c ≤ 3825 c ≤ 119, 531 Check. We check to see if the solution seems reasonable. The perimeter is at least 540 yd . When c = 119, 530, the total cost is \$675 + \$0.032(119, 530), or about \$4499.96. Solve. ↓ ↓ ↓ 2l + 192 ≥ 540 When c = 119, 532, the total cost is \$675 + \$0.032(119, 532), or about \$4500.02. 2l + 192 ≥ 540 2l ≥ 348 l ≥ 174 Check. We check to see if the solution seems reasonable. It appearsthat the solutionis correct. State. No more than 119,531 copies can be made. The solution set can be expressed as {c|c ≤ 119, 531}. 37. A = 2πrh When l = 174 yd, P = 2 · 174 + 2 · 96, or 540 yd. When l = 175 yd, P = 2 · 175 + 2 · 96, or 542 yd. A 2πh A 2πrh = 2πh It appearsthat the solutionis correct. State. For lengths that are at least 174yd, the perimeter will be at least 540 yd. The solution set can be expressed as {l|l ≥ 174 yd}. 35. Familiarize. Let s = the amount Jason spends in the sixth month. Translate. = r 2πh 38. y = 8x + b y − b = 8x + b − b y − b = 8x y− b = 8x 8 8 y− b Average spending is no more than \$95. = x 8 39. We subtract to ﬁnd the increase, in millions. 98 + 89 + 11 ↓ ↓ ↓ 5 + 83 + s Solve. 0 + 8 6 ≤ 95 70.3 − 40.4 = 29.9 Now we ﬁnd the percent increase. 98 + 89 + 110 + 85 + 83 + s 6 ≤ 95 29.9is what percent of 40.4? 98 + 89 + 110 + 85 + 83 + s 6 6 ≤ 6 · 95 ↓ ↓ ↓ ↓ ↓ 29.9 = p · 40.4 We divide by 40.4 on both sides and then convert to per- 98 + 89 + 110 + 85 + 83 + s ≤ 570 465 + s ≤ 570 s ≤ 105 cent notation. 29.9 = p · 40.4 29.9 = p· 40.4

47. Check. As a partial check we show that the average spending is \$95 when Jason spends \$105 in the sixth month. 40.4 0.74 ≈ p 74% ≈ p 40.4 98 + 89 + 110 + 85 + 83 + 105 = 6 570 6 = 95 The percent increase is about 74%. Answer D is correct. Copyright c 2015 Pearson Education, Inc.

48. 4 42 Chapter 2:Solving Equations and Inequalities 40. c = 1 a − d (a − d) · c = a − d · ac − dc = 1 ac − dc − ac = 1 − ac 1 a − d 4. 2w − 4 5. Since −4 is to the right of −6, we have −4 > −6. 6. Since 0 is to the right of −5, we have 0 > −5. 7. Since −8 is to the left of 7, we have −8 < 7. −dc = 1 − ac 2 8. The opposite of is 2 2 − because + − 2 = 0. −dc = 1− ac 5 5 5 5 −c −c 2 5 2 The reciprocal of is because 5 = 1. d = 1− ac 5 2 5 · 2 1− ac −c −1 1− ac 1(1− ac) −1 +ac 9. The distance of 3 from 0 is 3, so |3| = 3. Since = · = −c −1 −c = −1(−c) , or c 3 10. The distance of − from 0 is 3 3 3 , so − = . ac− 1 , we can also express the result as d = ac− 1 . 4 4 4 c 41. 3|w| − 8 = 37 3|w| = 45 |w| = 15 c 11. The distance of 0 from 0 is 0, so |0| = 0. 12. −6.7 + 2.3 One negative number and one positive number. The abso- lute values are 6.7 and 2.3. The diﬀerence of the absolute The solutions are the numbers whose distance from 0 is 15. They are −15 and 15. 42. Familiarize. Let t = the number of tickets given away. values is 6.7 − 2.3 = 4.4. The negative number has the larger absolute value, so the sum is negative. −6.7 + 2.3 = −4.4 Translate. We add the number of tickets given to the ﬁve 1 7 1 7 1 14 15 people. 1 1 1 13. − 6 − 3 = − 6 + − 3 = − 6 + − 6 = − 6 = 3 t + 4 t + 5 t + 8 + 5 = t Solve. 3/ · 5 = 5 − 2 · 3/ − 2 1 1 1 5 4 5· 4 5· 4/ 5 3 t + 4 t + 5 t + 8 + 5 = t 14. − 8 − 3 = 8 3 = = 20 15 12 · 2 · 4/ · 3 6 60 t + 60 t + 60 t + 8 + 5 = t 47 60 t + 13 = t 47 15. (−7)(5)(−6)(−0.5) = −35(3) = −105 16. 81 ÷ (−9) = −9 17. −10.8 ÷ 3.6 = −3 13 = t − 60 t 4 25 4 8 4 8 32 · 13 = 60 47 t − t 18. − 5 ÷ − 8 = − 5 · − 25 = 5 = · 25 125 13 = 60 60 60 13 60 t 60 13 19. 5(3x + 5y + 2z) = 5 · 3x + 5 · 5y + 5 · 2z = 15x + 25y + 10z 13 · 13 = 13 · 60 t 60 = t 1 1 1

49. 20. 4(−3x − 2)= 4(−3x) − 4 · 2 = −12x − 8 21. −6(2y − 4x) = −6 · 2y − (−6)(4x) = −12y − (−24x) = −12y + 24x Check. 3 · 60 = 20, 4 · 60 = 15, 5 · 60 = 12; then 22. 64 + 18x + 24y = 2 32 + 2 9x+ 2 12y = 2(32 + 9x + 12y) 20 + 15 + 12 + 8 + 5 = 60. The answer checks. State. 60 tickets were given away. CumulativeReviewChapters1-2 1. y− x = 12 − 6 = 6 = 2/ · 3 = 3 · · · 23. 16y − 56 = 8 · 2y − 8 · 7 = 8(2y − 7) 24. 5a − 15b + 25 = 5 · a − 5 · 3b + 5 · 5 = 5(a − 3b + 5) 25. 9b + 18y + 6b + 4y = 9b + 6b + 18y + 4y = (9 + 6)b + (18 + 4)y = 15b + 22y 4 2. 3x = y 4 4 3· 5 = 15 4 4 2/ · 2 2 26. 3y + 4 + 6z + 6y = 3y + 6y + 4 + 6z = (3 + 6)y + 4 + 6z = 9y + 4 + 6z 3. x − 3 = 3 − 3 = 0

50. − − = 8 − Cumulative Review Chapters 1 - 2 43 27. −4d − 6a + 3a − 5d + 1 = −4d − 5d − 6a + 3a + 1 = (−4 − 5)d + (−6 + 3)a + 1 38. 3 x = 36 4 4 3 4 = −9d − 3a + 1 28. 3.2x + 2.9y −5.8x−8.1y = 3.2x − 5.8x + 2.9y − 8.1y − 3 − 4 x = − 3 · 36 x = 4· 36 3 = 4· 3/· 12 − 3/ · 1 = (3.2 − 5.8)x + (2.9 − 8.1)y = −2.6x − 5.2y 29. 7 − 2x − (−5x) − 8 = 7 − 2x + 5x − 8 = −1 + 3x x = −48 The solution is −48. 30. −3x − (−x + y) = −3x + x − y = −2x − y 39. 2 x = 5 3 − 20 5 2 5 3 31. −3(x − 2) − 4x = −3x + 6 − 4x = −7x + 6 32. 10 − 2(5 − 4x) = 10 − 10 + 8x = 8x 2 · 5 x = 2 x = − 20 3· 5/ 33. [3(x + 6) − 10] − [5 − 2(x − 8)] = [3x + 18 − 10] − [5 − 2x + 16] = [3x + 8] − [21 − 2x] = 3x+ 8 − 21 + 2x = 5x − 13 − 2 · 4 · 3 x = − 8 3 The solution is − 8 . 40. 5.8x = −35.96 34. x + 1.75 = 6.25 5.8x = 5.8 35.96 5.8 x + 1.75 − 1.75 = 6.25− 1.75 x = 4.5 The solution is 4.5. 35. 5 y = 2 2 5 2 5 2 2 5 · 2 y = 5 · 5 x = −6.2 The solution is −6.2. 41. −4x + 3 = 15 −4x + 3 − 3 = 15 − 3 −4x = 12 −4x = 12 4 −4 −4 y = 25 4 The solution is . 25 36. −2.6 + x = 8.3 −2.6 + x + 2.6 = 8.3 + 2.6 x = 10.9 The solution is 10.9. 37. 4 1 + y 1 x = −3 The solution is −3. 42. −3x + 5 = −8x − 7 −3x + 5 + 8x = −8x − 7 + 8x 5x+ 5 = −7 5x+ 5 − 5 = −7− 5 5x = −12 5x = 12 2 3 5 5 1 1 1 1 12 4 2 +y−4 2 = 8 4 3 2 x = − 5 12 2 y = 8 6 3 − 4 6 The solution is − 5 . 8 y = 7 6 5 y = 3 6 5 Th e sol uti on is 3 6 3 − 4 6 . 2 2 8 = 7 + 1 6 6 = 7 +

51. 8 8 =7 6 6 4 3 . 4 y − 4 + y = 6 y + 2 0 − 4 y 5 y − 4 = 2 y + 2 0 5 y − 4 − 2 y = 2 y + 2 0 − 2 y 3 y − 4 = 2 0 3 y − 4 + 4 = 2 0 + 4 3 y = 2 4 3 y = 2 4

52. 3 3 y = 8 The solution is 8.

53. · 5 44 Chapter 2:Solving Equations and Inequalities 44. −3(x − 2) = −15 −3x + 6 = −15 −3x + 6 − 6 = −15− 6 −3x = −21 −3x = −21 48. 0(x + 3) + 4 = 0 0 + 4 = 0 4 = 0 FALSE There is no solution. 49. 3x − 1 < 2x + 1 −3 −3 3x 1 2x < 2x + 1 2x x = 7 The solution is 7. 45. First we will multiply by the least common multiple of all the denominatorsto clear the fractions. 1 5 1 3 x − 6 = 2 + 2x − − − x − 1 < 1 x − 1 + 1 < 1 + 1 x < 2 The solution set is {x|x < 2}. 50. 3y + 7 > 5y + 13 1 6 3 x − 1 5 1 = 6 6 2 5 1 + 2x 3y + 7 − 5y > 5y + 13 − 5y −2y + 7 > 13 6 · 3 x − 6 · 6 = 6 · 2 + 6 · 2x 2x − 5 = 3 + 12x 2x − 5 − 12x = 3 + 12x − 12x −2y + 7 − 7 > 13 − 7 −2y > 6 −2y < 6 Reversing the inequality symbol −10x − 5 = 3 −10x − 5 + 5 = 3 + 5 −10x = 8 −10x = 8 −2 −2 y < −3 The solution set is {y|y < −3}. 51. 5 − y ≤ 2y − 7 −10 −10 8 2/ · 4 5 − y − 2y ≤ 2y − 7 − 2y 5 − 3y ≤ −7 x = − 10 = − 4 x = − 5 5 − 3y − 5 ≤ −7− 5 −3y ≤ −12 4 The solution is − 5 . −3y 12 −3 ≥ −3 Reversing the inequality symbol 46. First we will multiply by 10 to clear the decimals. −3.7x + 6.2 = −7.3x − 5.8 10(−3.7x + 6.2) = 10(−7.3x − 5.8) −37x + 62 = −73x − 58 −37x + 62 + 73x = −73x − 58 + 73x 36x + 62 = −58 36x + 62 − 62 = −58 − 62 36x = −120 36x = −120 y ≥ 4 The solution set is {y|y ≥ 4}. 52. H = 65 − m H − 65 = 65 − m − 65 H − 65 = −m −1(H − 65) = −1 · (−1 · m) −H + 65 = m, or 65 − H = m 53. I = Prt 36 36 10 · ✧12 I = Prt x = − 3 · ✧12 Pr Pr I 10 = t x = − 3 Pr 10 The solution is − 3 . 47. 4(x + 2) = 4(x − 2) + 16 4x+ 8 = 4x − 8 + 16 4x + 8 = 4x + 8

54. 54. Translate. W9hat number is 24% of 105? ↓ ↓ ↓ ↓ ↓ a = 24% · 105 Solve. We convert 24% to decimal notation and multiply. a = 24% · 105 4x+ 8 − 4x = 4x + 8 − 4x 8 = 8 TRUE All real numbers are solutions. a = 0.24 a = 25.2 · 105 Thus, 25.2 is 24% of 105.

55. ↓ ↓ ↓ ↓ ↓ ↓ ↓ p − 25% · p = 18.45 Cumulative Review Chapters 1 - 2 45 55. Translate. 39.6 is what percent of 88? ↓ ↓ ↓ ↓ ↓ 39.6 = p · 88 Solve. We solve the equation. 39.6 = p · 88 Solve. m + (m + 17) = 107 2m + 17 = 107 2m + 17 − 17 = 107 − 17 2m = 90 2m = 90 39.6 = p· 88 2 2 88 88 0.45 = p 45% = p Thus, 39.6 is 45% of 88. 56. Translate. \$163.35 is 45% of w9hat number? m = 45 The exercise asks only for the amount Melinda paid, but we also ﬁnd the amount Susan paid so that we can check the answer. If m = 45,thenm + 17 = 45 + 17 = 62. Check. \$62 is \$17 more than \$45, and \$45 + \$62 = \$107. The answer checks. State. Melinda paid \$45 for her rollerblades. 163.35 = 45% · b Solve. 163.35 = 45% · b 163.35 = 0.45· b 59. Familiarize. Let x = the amount originally invested. Us- ing the formula for simple interest, I = Prt, the interest earned in one year will be x · 8% · 1, or 8%x. Translate. 163.35 = 0.45· b Amount amount 0.45 363 = b 0.45 invested plus interest is after 1 year. Thus, \$163.35 is 45% of \$363. 57. Familiarize. Let p = the price beforethe reduction. Solve. ↓ ↓ ↓ ↓ ↓ x + 8%x = 1134 Translate. Price before reduction minus 25% of price is \$18.45. x + 8%x = 1134 x + 0.08x = 1134 1.08x = 1134 Solve. We solve the equation. 1.08x = 1.08 1134 1.08 p − 25% · p = 18.45 p − 0.25p = 18.45 0.75p = 18.45 0.75p = 18.45 x = 1050 Check. 8% of \$1050 is 0.08·\$1050= \$84 and \$1050+\$84 = \$1134, so the answer checks. State. \$1050 was originally invested. 0.75p 0.75 p = 24.6 Check. 25% of \$24.60 is 0.25·\$24.60 = \$6.15and \$24.60− \$6.15 = \$18.45, so the answer checks. State. The price before the reduction was \$24.60. 60. Familiarize. Let l = the length of the ﬁrst piece of wire, in meters. Then l +3 = the length of the second piece and 4 5 l = the length of the third piece. Translate. 58. Familiarize. Let m = the amount Melinda paid for her rollerblades. Then m + 17 = the amount Susan paid for hers. Length of ﬁrst piece plus length of second piece plus length of third piece is 143 m . Translate. ↓ ↓ ↓ ↓ ↓ ↓ ↓ 4 Amount Melinda paid plus amount Susan paid is \$107. l + (l + 3) + 5 l = 143

56. ↓ ↓ ↓ ↓ ↓ m + (m + 17) = 107

57. · 46 Chapter 2:Solving Equations and Inequalities Solve. 4 l + (l + 3) + 5 l = 143 Translate. Final salary is \$48, 418.24. 5 5 4 ↓ ↓ ↓ 5 l + 5 l + 3 + 5 l = 143 14 Solve. 1.0712s = 48, 418.24 5 l + 3 = 143 1.0712s = 48, 418.24 14 5 l + 3 − 3 = 143 − 3 1.0712s = 1.0712 48, 418.24 1.0712 14 5 l = 140 s = 45, 200 Check. 4% of \$45,200 is 0.04 · \$45, 200 = \$1808 and 5 14 5 \$45, 200 + \$1808 = \$47,008. Then 3% of \$47,008 is 14 · 5 l = 14 · 140 5· 10 · ✧14 0.03 · \$47, 008 = \$1410.24 and \$47,008 + \$1410.24 = \$48, 418.24. The answer checks. l = ✧14 1 l = 50 4 4 If l = 50,thenl + 3 = 50 + 3 = 53 and 5 l = 5 · 50 = 40. Check. The second piece is 3 m longer than the ﬁrst 4 State. At the beginning of the year the salary was \$45,200. 64. First we subtract to ﬁnd the amount of the reduction. 9in. − 6.3 in. = 2.7 in. Translate. piece and the third piece is 5 as long as the ﬁrst. Also, 2.7 in. is what percent of 9in.? 50 m+ 53 m+ 40 m= 143 m. The answer checks. State. The lengths of the pieces are 50 m, 53 m, and 40 m. 61. Familiarize. Let s = Nadia’s score on the fourth test. Solve. ↓ ↓ ↓ ↓ ↓ 2.7 = p · 9 2.7 = p · 9 Translate. The averagescore is9at least 80. 2.7 = 9 p· 9 9 0.3 = p 82 + 7 ↓ ↓ ↓ 8 + s Solve. 6 + 7 4 ≥ 80 82 +76 +78 +s 4 ≥ 80 30% = p The drawing should be reduced 30%. 65. 4|x| − 13 = 3 4|x| = 16 82 + 76 + 78 + s 4 4 ≥ 4 · 80 |x| = 4 82 + 76 + 78 + s ≥ 320 236 + s ≥ 320 s ≥ 84 Check. As a partial check we show that the average is at least 80 when the fourth test score is 84. The solutions are the numbers whosedistance from 0 is 4. They are −4and 4. 66. First we multiply by 28 to clear the fractions. 2 +5x = 11 + 8x+ 3 4 28 7 82 + 76 + 78 + 84 = 4 320 4 = 80 2 + 5x 28 4 11 = 28 28 8x + 3 + 7 State. Scores greater than or equal to 84 will earn Nadia 28(2 + 5x) 11 28(8x + 3) at least a B. The solutionset is {s|s ≥ 84}. 62. −125÷ 25 · 625÷ 5 = −5· 625÷ 5 = −3125 ÷ 5 = −62 5 Answer C is correct.

58. ·63. Familiarize. Let s = the salary at the beginning of the year. After a 4% increase the new salary is s + 4%s, or s + 0.04s, or 1.04s. Then after a 3% cost-of-livingadjust- ment the ﬁnal salary is 1.04s + 3% · 1.04s, or 1.04s + 0.03 · 1.04s, or 1.04s + 0.0312s, or 1.0712s. = 28 + 4 28 7 7(2 + 5x) = 11 + 4(8x + 3) 14 + 35x = 11 + 32x + 12 14 + 35x = 32x + 23 14 + 3x = 23 3x = 9 x = 3 The solution is 3.

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