Solutions for Beginning And Intermediate Algebra 6th Us Edition by Lial by 6alsm

Linear Equations and Inequalities in One Variable

2.1 The Addition Property of Equality

Classroom Examples, Now Try Exercises

1. Note: When solving equations we will write “Add 5” as a shorthand notation for “Add 5 to each side” and “Subtract 5” as a notation for “Subtract 5 from each side.” 123Given 1212312Add12.

9Combineliketerms. x x x −=− −+=−+ =

We check by substituting 9 for x in the original equation.

Check ? 123Originalequation 9123Let9. 33True x x −=− −=−= −=−

Since a true statement results, { }9 is the solution set.

N1. 134Given 1313413Add13. 17Combineliketerms. x x x −= −+=+ =

We check by substituting 17 for x in the original equation.

Check ? 134Originalequation 17134Let17. 44True x x −= −== =

Since a true statement results, { }17 is the solution set.

2. 4.16.3

4.14.16.34.1Add4.1. 2.2 m m m −=− −+=+ =−

Check 2.2:6.36.3True m =−−=−

This is a shorthand notation for showing that if we substitute 2.2 for m, both sides are equal to 6.3, and hence a true statement results. In practice, this is what you will do, especially if you're using a calculator. The solution set is {2.2}.

N2. 5.77.2

5.75.77.25.7Add5.7. 1.5 t t t −=− −+=−+ =−

Check 1.5:7.27.2True t =−−=−

The solution set is {1.5}.

3. 2216 22161616Subtract16. 38 x x x −=+ −−=+− −=

Check 38:2222True x =−−=−

The solution set is {38}.

N3. 1512 15121212Subtract12. 27 x x x −=+ −−=+− −=

Check 27:1515True x =−−=−

The solution set is {27}.

4. 11912 119111211Subtract 11. 9 zz zzzzz z −= −−=− −=

Check 9:108108True z =−−=−

The solution set is {9}.

N4. 52 52Subtract . 5Combine terms. xx xxxxx x −= −−=− −=

Check 5:1010True x =−−=−

The solution set is {5}.

5. 79 1 22 77977 1Subtract . 22222 1 pp ppppp p += +−=− =

Check 99 1:True 22 p ==

The solution set is {1}.

N5. . 25 4 33 22522 4Subtract 33333 4Combineterms. xx xxxxx x += +−=− =

Check 2020 4:True 33 x ==

The solution set is {4}.

(d) This is an equation because of the equals symbol.

61275 125 17 yy y y −++=− +=− =− The solution set is {17}.

6. Equations that have exactly the same solution sets are equivalent equations. 26 2262Subtract2. 4 x x x += +−=− = So 26 x += and 4 x = are equivalent equations. 105

1010510Subtract10. 5 1()1(5)Multiplyby 1. 5 x x x x x −= −−=− −=− −−=−−− = So 105 x −= and 5 x =− arenot equivalent equations.

Subtract 3 from both sides to get 6, x = so 39 x += and 6 x = are equivalent equations. Subtract 4 from both sides to get 4. x = The second equation is 4, x =− so 48 x += and 4 x =− arenot equivalent equations. The pairs of equations in A and C are equivalent.

7. Equations A () 2 560xx−+= and B () 3 xx = are not linear equations in one variable because they cannot be written in the form . AxBC += Note that in a linear equation the exponent on the variable must be 1.

8. Check by replacing the variable(s) in the original equation with the proposed solution. A true statement will result if the proposed solution is correct.

9. 39 3393 12 x x x −= −+=+ = Check this solution by replacing x with 12 in the original equation. ? 39 1239Let12. 99True x x −= −== = Because the final statement is true, {12} is the solution set.

10. 98 9989 17 x x x −= −+=+ = Check 17 x = ? 1798Let17. 88True x −== = Thus, {17} is the solution set.

11. 1219 12121912 31 x x x −= −+=+ = Check 31 x = ? 311219Let31. 1919True x −== = Thus, {31} is the solution set.

12. 1822 18182218 40 x x x −= −+=+ =

Checking yields a true statement, so {40} is the solution set.

13. 69 6696 3 x x x −=− −+=−+ =−

Checking yields a true statement, so {3} is the solution set.

14. 57 5575 2 x x x −=− −+=−+ =−

Checking yields a true statement, so {2} is the solution set.

15. 812 88128 4 r r r += +−=− =

Checking yields a true statement, so {4} is the solution set.

16. 711 77117 4 x x x += +−=− =

Checking yields a true statement, so {4} is the solution set.

Chapter 2 Linear Equations and Inequalities in One Variable

17. 2819 28281928 9 x x x += +−=− =−

Checking yields a true statement, so {9} is the solution set.

18. 4726 47472647 21 x x x += +−=− =−

Checking yields a true statement, so {21} is the solution set.

19. 11 42 1111 4424 21 44 3 4 x x x x +=− +−=−− =−− =−

Check 311 :True 422 x =−−=−

The solution set is 3 . 4 ⎧⎫ ⎨⎬ ⎩⎭

20. 21 36 2212 3363 14 66 5 6 x x x x +=− +−=−− =−− =−

Check 511 :True 666 x =−−=−

The solution set is 5 6 ⎧⎫ ⎨⎬ ⎩⎭

21. 73 73 7737 10 r r r r +=− +=− +−=−− =−

The solution set is {10}.

22. 84 84 8848 12 k k k k +=− +=− +−=−− =−

The solution set is {12}.

23. 215 2151515 13 p p p =+ −=+− −=

The solution set is {13}.

24. 519 5191919 14 z z z =+ −=+− −=

The solution set is {14}.

25. 414 4141414 10 x x x −=− −+=−+ = The solution set is {10}.

26. 722 7222222 15 x x x −=− −+=−+ = The solution set is {15}.

27. 13 35 1333 3555 59 1515 4 15 x x x x −=− −+=−+ −+= =

Check 4549 :True 15151515 x =−=−

The solution set is 4 15 ⎧ ⎫ ⎨ ⎬ ⎩⎭

28. 12 43 1222 4333 38 1212 5 12 x x x x −=− −+=−+ −+= =

Check 5358 :True 12121212 x =−=−

The solution set is 5 12 ⎧ ⎫ ⎨ ⎬ ⎩⎭

29. 8.42.1

8.48.42.18.4 6.3 x x x −=− −+=−+ =

The solution set is {6.3}.

30. 15.55.1

15.515.55.115.5 10.4 x x x −=− −+=−+ =

The solution set is {10.4}.

31. 12.34.6

12.312.34.612.3 16.9 t t t +=− +−=−− =−

The solution set is {16.9}.

32. 21.513.4

21.521.513.421.5 34.9 x x x +=− +−=−− =−

The solution set is {34.9}.

33. 327 32272Subtract2 17or7 xx xxxxx. xx =+ −=+− ==

Check 7:2121True x ==

The solution set is {7}.

34. 549 54494Subtract4 19or9 xx xxxxx. xx =+ −=+− ==

Check 9:4545True x ==

The solution set is {9}.

35. 1049 104999Subtract9 140 4404Subtract4. 4 xx xxxxx. x x x += +−=− += +−=− =−

Check 4:3636True x =−−=−

The solution set is {4}.

36. 857

85777Subtract7

150 5505Subtract5. 5 tt ttttt. t t t += +−=− += +−=− =−

Check 5:3535True t =−−=−

The solution set is {5}.

37. 839 83898Subtract8. 3 xx xxxxx x −= −−=− −=

Check 3:8(3)39(3)True x =−−−=−

The solution set is {3}.

38. 647 64676Subtract7. 4 xx xxxxx x −= −−=− −=

Check 4:6(4)47(4)True x =−−−=−

The solution set is {4}.

39. 625 62555Subtract5. 20 222Add2. 2 tt ttttt t t t −= −−=− −= −+= =

Check 2:6(2)25(2)True t =−=

The solution set is {2}.

40. 463 46333Subtract 3. 60 666Add 6. 6 zz zzzzz z z z −= −−=− −= −+= =

Check 6:4(6)63(6)True z =−=

The solution set is {6}.

41. 27 6 55 22722 6Subtract. 55555 5 6 5 6 ww wwwww w w −= −−=− −= −=

Check 27 6:(6)6(6)True 55 w =−−−=−

The solution set is {6}.

Chapter 2 Linear Equations and Inequalities in One Variable

42. 29 2 77 22922

2Subtract. 77777 7 2 7 2 zz zzzzz z z −= −−=− −= −=

Check 29 2:(2)2(2)True 77 z =−−−=−

The solution set is {2}.

43. 11 5 22 1111 5 2222 50 5505 5 xx xxxx x x x +=− ++=−+ += +−=− =−

The solution set is {5}.

44. 14 7 55 1444 7 5555 5 70 5 7707 7 xx xxxx x x x +=− ++=−+ += +−=− =−

The solution set is {7}.

45. 5.624.6

5.624.64.64.6 1.020 2202 2 xx xxxx x x x += +−=− += +−=− =−

The solution set is {2}.

46. 9.158.1

9.158.18.18.1 1.050 5505 5 xx xxxx x x x += +−=− += +−=− =−

The solution set is {5}.

47. 1.430.4 1.430.40.40.4 1.030 1.03303 3 xx xxxx x x x −= −−=− −= −+=+ =

The solution set is {3}.

48. 1.960.9 1.960.90.90.9 1.060 1.06606 6 tt tttt t t t −= −−=− −= −+=+ =

The solution set is {6}.

49. 54 5444 0 pp pppp p = −=− =

The solution set is {0}.

50. 87 8777 0 zz zzzz z = −=− =

The solution set is {0}.

51. 3720 70 7707 7 xx x x x +−= += +−=− =−

The solution set is{7}.

52. 5440 40 4404 4 xx x x x +−= += +−=− =−

The solution set is {4}.

53. 3724 372242 74 7747 3 xx xxxx x x x +=+ +−=+− += +−=− =−

The solution set is {3}.

54. 9584 958848 54 5545 1 xx xxxx x x x +=+ +−=+− += +−=− =−

Check 1:44True x =−−=−

The solution set is {1}.

55. 8676 867767 66 6666 0 tt tttt t t t +=+ +−=+− += +−=− =

The solution set is {0}.

56. 139129 1391212912 99 9999 0 tt tttt t t t +=+ +−=+− += +−=− = The solution set is {0}.

57. 4759 475595 79 7797 2 xx xxxx x x x −+=−+ −++=−++ += +−=− =

The solution set is {2}.

58. 63710 6377107 310 33103 7 xx xxxx x x x −+=−+ −++=−++ += +−=− =

The solution set is {7}.

59. 5211 522112Add2. 55115Subtract5. 16 xx xxxxx x x −=−− −+=−−+ +−=−− =−

The solution set is {16}.

60. 3891 389919Add9. 31 3313Subtract3. 4 xx xxxxx x x x −=−− −+=−−+ +=− +−=−− =−

The solution set is {4}.

61. 1.240.24 1.240.20.240.2 1.044 4444 0 yy yyyy y y y −=− −−=−− −=− −+=−+ =

The solution set is {0}.

62. 7.766.76 7.766.76.766.7 1.066 6666 0 rr rrrr r r r −=− −−=−− −=− −+=−+ =

The solution set is {0}.

63. 361022 3422Combine terms. 342222Subtract2. 42 4424 2 xx xx xxxxx x x x +−=− −=− −−=−− −=− −+=−+ = The solution set is {2}.

64. 84871 8471Combine terms. 847717Subtract7. 41 4414 3 xx xx xxxxx x x x +−=− −=− −−=−−

−=− −+=−+ =

The solution set is {3}.

65. 5326412 (526)316 33163 13 ttt t t t ++−=+ +−+= +−=− =

Check 13: 1616True t ==

The solution set is {13}.

66. 4636310 613 66136Add6. 19 xxx x x x −+−=+ −= −+=+ =

Check 19: 1313True x ==

The solution set is {19}.

Chapter 2 Linear Equations and Inequalities in One Variable

67. 6573124 138124 1381212412 84 8848 4 xxx xx xxxx x x x +++=+ +=+ +−=+− += +−=− =−

Check 4:4444True x =−−=−

The solution set is {4}.

68. 4381112 124112 1241111211 42 4424 2 xxx xx xxxx x x x +++=+ +=+ +−=+− += +−=− =−

Check 2:2020True x =−−=−

The solution set is {2}.

69. 5.24.67.10.94.6

1.94.60.94.6

1.94.60.90.94.60.9

1.04.64.6

1.04.64.64.64.6 0 0 qqq qq qqqq q q q q −−=−− −−=−− −−+=−−+ −−=− −−+=−+ −= =

Check 0:4.64.6True q =−=−

The solution set is {0}.

70. 4.02.79.64.62.7

2.75.64.62.7

2.75.65.64.62.75.6 2.72.7 2.72.72.72.7 0 xxx xx xxxx x x x +−=−+ −=−+ −+=−++ =+ −=+− =

Check 0: 2.72.7True x ==

The solution set is {0}.

71. 51222 73575 5142 7357 5214222Add. 7735777 714 Combine terms. 735 11411 1Subtract. 33533 125 LCD15 1515 7 15 xx xx xxxxx x x x x +=−+ +=− ++=−+ += +−=− =−= =

Check : 722 True 1533 x ==

The solution set is 7 15 ⎧ ⎫ ⎨ ⎬ ⎩⎭

72. To solve the equation, follow the simplification steps below. 63411 74576 632415 LCD30 7430730 63291 Add. 74307 63129111Add. 74730777 7329 Combine terms. 7430 332933 1Add. 443044 5845 LCD60 6060 103 60 ss ss ss sssss s s s s −=−+ −=−+= −=− −+=−+ −= −+=+ =+= = Check 103101101 :True 60140140 s ==

The solution set is 103 . 60 ⎧ ⎫ ⎨ ⎬ ⎩⎭

73. (56)(34)10 563410Distributive prop. 310Combineterms. 33103Subtract3. 7 yy yy y y y +−+= +−−= += +−=− =

Check 7: 1010True y ==

The solution set is { } 7.

74. (83)(17)6 83176 26 2262 4 rr rr r r r +−+= +−−= += +−=− =

Check 4: 66True r ==

The solution set is {4}.

75. 2(5)(9)3 21093 13 1131 4 pp pp p p p +−+=− +−−=− +=− +−=−− =−

Check 4:33True p =−−=−

The solution set is { } 4.

76. 4(6)(83)5 424835 165 1616516 21 kk kk k k k +−+=− +−−=− +=− +−=−− =−

Check 21:55True k =−−=−

The solution set is { } 21.

77. 6(21)(137)0 1261370 130 1313013 13 bb bb b b b −++−= −−+−= −= −+=+ =

Check 13: 00True b ==

The solution set is {13}.

78. 5(33)(161)0 15151610 160 1616016 16 ww ww w w w −−++= −+++= += +−=− =−

Check 16: 00True w =−=

The solution set is {16}.

79. 10(21)19(1) 20101919 201019191919 1019 10101910 29 29 xx xx xxxx x x x x −+=−+ −+=−− −++=−−+ −+=− −+−=−− −=− =

Check 29:570570True x =−=−

The solution set is {29}.

80. 2(32)5(3) 64515 6455155 415 44154 11 11 rr rr rrrr r r r r −+=−− −+=−+ −++=−++ −= −−=− −= =−

Check 11: 7070True r =−=

The solution set is {11}.

81. 2(82)3(27)2(42)0 164621840 180 1818018 18 ppp ppp p p p −+−−−+= −−−+−−= −= −+=+ =

Check 18: 00True p ==

The solution set is {18}.

82. 5(12)4(3)7(3)0 5101242170 140 140 14 zzz zzz z zzz z −−+−−+= −++−−−= −−= −−+=+ −=

Check 14:00True z =−=

The solution set is {14}.

Chapter 2 Linear Equations and Inequalities in One Variable

83. 4(71)3(25)4(35)6 28461512206 186 1818618 12 xxx xxx x x x −+−−+=− −+−−−=−

Check 12:66True x =−=−

The solution set is {12}.

84. 9(23)4(53)5(4)3 182720122053 673

Check 64:33True m =−=−

The solution set is {64}.

85. Answers will vary. One example is 68 x −=−

86. Answers will vary. One example is 1 1. 2 x +=

87. “Three times a number is 17 more than twice the number.” 3217 322172 17 xx xxxx x =+ −=+− = The number is 17 and { }17 is the solution set.

88. “One added to three times a number is three less than four times the number.” 1343 133433 13 1333 4 xx xxxx x x x +=− +−=−− =− +=−+ =

The number is 4 and { }4 is the solution set.

89. “If six times a number is subtracted from seven times the number, the result is 9. ”

769 9 xx x −=− =−

The number is –9 and { }9 is the solution set.

90. “If five times a number is added to three times the number, the result is the sum of seven times the number and 9.”

5379 879 87797 9 xxx xx xxxx x +=+ =+ −=+− =

The number is 9 and { }9 is the solution set.

2.2 The Multiplication Property of Equality

Classroom Examples, Now Try Exercises 1. 1575 1575 Divideby15. 1515 5 x x x = = =

Check 5:7575True x ==

The solution set is { }5 N1. 880 880 Divideby8. 88 10 x x x = = =

Check 10:8080True x ==

The solution set is { } 10. 2. 820 820

Divide by 8. 88 205

Writeinlowestterms. 82 x x x =− =− =−=−

Check 5 :2020True 2 x =−−=−

The solution set is 5 2 ⎧ ⎫ ⎨ ⎬ ⎩⎭

N2. 1024 1024

Divideby10. 1010 2412

Writeinlowestterms. 105 x x x =− =− =−=−

Check 12 :2424True 5 x =−−=−

The solution set is 12 5 ⎧ ⎫ ⎨ ⎬ ⎩⎭

3. 5.040.7

5.040.7

Divideby0.7. 0.70.7

7.2 x x x =− =− =−

Check 7.2:5.045.04True x =−=

The solution set is { } 7.2.

N3. 7.021.3

⋅=⋅ =−

Check 28:1616True t =−−=−

7.021.3 Divideby1.3. 1.31.3

5.4 x x x =− =− =−

Check 5.4:7.027.02True x =−=

The solution set is { } 5.4.

4. 6 4 1 6 4 1 44(6)Multiplyby4. 4 24 x x x x =− =− ⋅=− =−

Check 24:66True x =−−=−

The solution set is { } 24.

N4. 7 5 1 7 5 1 55(7)Multiplyby5. 5 35 x x x p =− =− ⋅=− =−

Check 35:77True p =−−=−

The solution set is { } 35. 5. 2 12 3 3233 (12)Multiplyby. 2322 312 1 21 18 t t t t −=− ⎛⎞ −−=−−−

⋅=−⋅ =

Check 18:1212True t =−=−

The solution set is { } 18. N5. () 4 16 7 7477 16Multiplyby. 4744 716 1 41 28 z z t t =− ⎛⎞ =− ⎜⎟ ⎝⎠

The solution set is { } 28. 6. 7 171 (1)(1)(1)(7) 17 7 p ppp p p p −=− −⋅=−−=−⋅

⋅= =

Check 7:77True p =−=−

The solution set is { } 7.

N6. 9 191 (1)(1)(1)(9)Multiplyby1. 19 9 x xxx x x x −=− −⋅=−−=−⋅

⋅= =

Check 9:99True x =−=−

The solution set is { } 9.

7. 4920

520 Divideby5. 55 4 rr r r r −= −= =− =−

520Combineterms.

Check 4:2020True r =−=

The solution set is { } 4. N7. 9621

321 Divideby3. 33 7 nn n n n −= = = =

321Combineterms.

Check 7:2121True n ==

The solution set is { } 7.

Exercises

1. (a) multiplication property of equality; to get x alone on the left side of the equation, multiply each side by 1 3 (or divide each side by 3).

(b) addition property of equality; to get x alone on the left side of the equation, add 3 (or subtract 3) on each side.

(c) multiplication property of equality; to get x alone on the left side of the equation, multiply each side by 1 (or divide each side by 1 ).

(d) addition property of equality; to get x alone on the right side of the equation, add 6 (or subtract 6) on each side.

2. Choice C doesn’t require the use of the multiplicative property of equality. After the equation is simplified, the variable x is alone on the left side.

547 7 xx x −= =

3. Choice B; to find the solution of 3 , 4 x −=− multiply (or divide) each side by 1, or use the rule “If , xa−= then .xa =− ”

4. Choice A; to find the solution of 24, x −=− multiply (or divide) each side of the equation by 1, or use the rule “If , xa−= then .xa =− ”

5. To get just x on the left side, multiply both sides of the equation by the reciprocal of 4 , 5 which is 5 4

6. To get just x on the left side, multiply both sides of the equation by the reciprocal of 2 , 3 which is 3 2

7. This equation is equivalent to 1 5. 10 x = To get just x on the left side, multiply both sides of the equation by the reciprocal of 1 , 10 which is 10.

8. This equation is equivalent to 1 10. 100 x = To get just x on the left side, multiply both sides of the equation by the reciprocal of 1 , 100 which is 100.

9. To get just x on the left side, multiply both sides of the equation by the reciprocal of 9 , 2 which is 2 9

10. To get just x on the left side, multiply both sides of the equation by the reciprocal of 8 , 3 which is 3 8

11. This equation is equivalent to 10.75. x −= To get just x on the left side, multiply both sides of the equation by the reciprocal of 1, which is 1.

12. This equation is equivalent 10.48. x −= To get just x on the left side, multiply both sides of the equation by the reciprocal of 1, which is 1.

13. To get just x on the left side, divide both sides of the equation by the coefficient of x, which is 6.

14. To get just x on the left side, divide both sides of the equation by the coefficient of x, which is 7.

15. To get just x on the left side, divide both sides of the equation by the coefficient of x, which is 4.

16. To get just x on the left side, divide both sides of the equation by the coefficient of x, which is 13.

17. To get just x on the left side, divide both sides of the equation by the coefficient of x, which is 0.12.

18. To get just x on the left side, divide both sides of the equation by the coefficient of x, which is 0.21.

19. This equation is equivalent to 125. x −= To get just x on the left side, divide both sides of the equation by the coefficient of x, which is 1

20. This equation is equivalent to 150. x −= To get just x on the left side, divide both sides of the equation by the coefficient of x, which is 1.

21. 636 636

Divide by 6. 66 16 6 x x x x = = = =

Check 6:3636True x ==

The solution set is { } 6.

22. 864 864

Divideby8. 88 8 x x x = = =

Check 8:6464True x ==

The solution set is {8}.

23. 215 215 Divideby2. 22 15 2 m m m = = =

Check 15 :1515True 2 m ==

The solution set is 15 2 ⎧⎫ ⎨⎬ ⎩⎭

24. 310 310

Divideby3. 33 10 3 m m m = = =

Check 10 :1010True 3 m ==

The solution set is 10 3 ⎧⎫ ⎨⎬ ⎩⎭

25. 420 420 Divideby4. 44 5 x x x =− = =−

Check 5:2020True x =−−=−

The solution set is {5}.

26. 560 560

Divideby5. 55 12 x x x =− = =−

Check 12:6060True x =−−=−

The solution set is {12}.

27. 728 728

Divideby7. 77 4 x x x −= =− =−

Check 4:2828True x =−=

The solution set is {4}.

28. 936 936

Divideby9. 99 4 x x x −= =− =−

Check 4:3636True x =−=

The solution set is {4}.

29. 1036 1036

Divideby10. 1010 3618

Lowestterms 105 t t t =− = =−=−

Check 18 :3636True 5 t =−−=−

The solution set is 18 , 5 ⎧ ⎫ ⎨ ⎬ ⎩⎭ or {3.6}.

30. 1054 1054

Divideby10. 1010 5427

Lowestterms 105 s s s =− = =−=−

Check 27 :5454True 5 s =−−=−

The solution set is 27 , 5 ⎧ ⎫ ⎨ ⎬ ⎩⎭ or {5.4}.

31. 672 672

Divideby 6. 66 12 x x x −=− =− =

Check 12:7272True x =−=−

The solution set is {12}.

32. 464 464

Divideby4. 44 16 x x x −=− =− =

Check 16:6464True x =−=−

The solution set is {16}.

Chapter 2 Linear Equations and Inequalities in One Variable

33. 40 40 Divideby4. 44

0 r r r = = =

Check 0:00True r ==

The solution set is {0}.

39. 0.28 0.28 0.20.2 40 t t t = = =

Check 40:88True t ==

The solution set is { } 40.

34. 70 70 Divideby7. 77

0 x x x = = =

Check 0:00True x ==

The solution set is {0}.

35. 12 1()112Multiplyby 1. 12 x x x −= −⋅−=−⋅− =−

Check 12:1212True x =−=

The solution set is {12}.

36. 14 1()114Multiplyby 1. 14 t t t −= −⋅−=−⋅− =−

Check 14:1414True t =−=

The solution set is { } 14.

37. 3 4 3 1()1 4 3 4 x x x −=−

Check 333 :True 444 x =−=−

The solution set is 3 4 ⎧⎫ ⎨⎬ ⎩⎭

38. 1 2 1 1()1 2 1 2 x x x −=− ⎛⎞ −⋅−=−⋅−⎜⎟ ⎝⎠ =

Check 111 :True 222 x =−=−

The solution set is 1 2 ⎧⎫ ⎨⎬ ⎩⎭

40. 0.918 0.918 0.90.9 20 x x x = = =

Check 20:1818True x ==

The solution set is { } 20.

41. 0.39 0.39 0.30.3 30 x x x −= = =−

Check 30:99True x =−=

The solution set is { } 30.

42. 0.520 0.520

Divide by0.5. 0.50.5 40 x x x −= =− =−

Check 40:2020True x =−=

The solution set is { } 40.

43. 0.61.44

0.61.44 Divide by0.6. 0.60.6

2.4 x x x =− = =−

Check 2.4:1.441.44True x =−−=−

The solution set is { } 2.4.

44. 0.82.96

0.82.96 Divide by0.8. 0.80.8

3.7 x x x =− = =−

Check 3.7:2.962.96True x =−−=−

The solution set is { } 3.7.

45. 9.12.6 9.12.6 Divide by 2.6. 2.62.6

3.5 x x x −=− =− =

Check 3.5:9.19.1True x =−=−

The solution set is { } 3.5.

46. 7.24.5

7.24.5 Divide by 4.5. 4.54.5 1.6 x x x −=− =− = Check 1.6:7.27.2True x =−=−

The solution set is { } 1.6.

47. 2.125.62

2.125.62 Divide by 2.1. 2.12.1 12.2 m m m −= =− =−

Check 12.2:25.6225.62True m =−=

The solution set is { } 12.2.

52. 15 5 1 15 5 1 5515 5 75 x x x x = = ⋅=⋅ =

Check 75: 15=15 True x =

The solution set is { } 75.

48. 3.932.76

3.932.76 3.93.9 8.4 x x x −= = =−

Check 8.4:32.7632.76True x =−=

The solution set is { } 8.4.

49. 1 12 4 1 44(12)Multiply by4. 4 148 48 x x x x =− ⋅=− =− =−

Check 48:1212True x =−−=−

The solution set is { } 48.

50. 1 3 5 1 55(3)Multiplyby5. 5 15 p p p =− ⋅=− =− Check 15:33True p =−−=−

The solution set is { } 15.

51. 12 6 1 12 6 1 6612 6 72 z z z z = = ⋅=⋅ = Check 72:1212True z ==

The solution set is { } 72.

53. 5 7 1 5 7 1 77(5) 7 35 x x x x =− =− ⎛⎞ =−

Check 35:55True x =−−=−

The solution set is { } 35.

54. 3 8 1 3 8 1 88(3) 8 24 r r r r =− =− ⎛⎞ =− ⎜⎟ ⎝⎠ =−

Check 24:33True r =−−=−

The solution set is { } 24.

55. 2 4 7 7 727 (4) Multiply by . 272 2 14 p p p = ⎛⎞ = ⎜⎟ ⎝⎠ =

Check 14:44True p ==

The solution set is { } 14.

56. () 3 9 8 8 838 9 Multiply by . 383 3 24 x x x = ⎛⎞⎛⎞⎛⎞ = ⎜⎟⎜⎟⎜⎟ ⎝⎠⎝⎠⎝⎠ = Check 24:99 True x ==

The solution set is { } 24.

57. 5 15 6 6 656 (15)Multiply by . 565 5 18 t t t −=−

−−=−−−

⎝⎠ = Check 18:1515True t =−=−

The solution set is { } 18.

58. 3 21 4 4 434 (21) Multiply by . 343 3 28 z z z −=−

= Check 28:2121True z =−=−

The solution set is { } 28.

59. 73 95 9 9793 Multiply by . 7975 7 27 35 x x x −=

−−=−⋅−

Check 2733 True 3555 : x ==

The solution set is 27 . 35 ⎧⎫ ⎨⎬ ⎩⎭

60. 54 69 6 6564 Multiply by . 5659 5 2348 53315 x x x −= ⎛⎞⎛⎞⎛⎞⎛⎞ −−=−−

Check 844 :True 1599 x =−=

The solution set is 8 . 15 ⎧⎫ ⎨⎬

61. 4321 721 721 77 3 xx x x x += = = =

Check 3:2121True x ==

The solution set is { } 3.

62. 83121 11121 11121 1111 11 xx x x x += = = = Check 11:121121True x ==

The solution set is { } 11.

63. 6810 210 210 22 5 rr r r r −= −= = =−

Check 5:1010True r =−=

The solution set is { } 5. 64. 3724 424 424 44 6 pp p p p −= −= = =− Check 6:24=24True p =−

The solution set is { } 6.

Check 20: 862True x =−=

The solution set is { } 20.

25 4 39 65 4 99 1 4 9 1 994 9 36 xx xx x x x −= −= = ⋅=⋅ = Check 36:24204True x =−=

The solution set is { } 36.

67. 76463 963 963 99 7 mmm m m m +−= = = =

Check 7:6363True m ==

The solution set is { } 7.

68. 92768 468 468 44 17 rrr r r r +−= = = =

Check 17: 6868True r ==

The solution set is { } 17.

69. 6470 90 90 99 0 xxx x x x −+−= −= = =

Check 0:00True x ==

The solution set is { } 0.

70. 5480 90 90 99 0 xxx x x x −+−= −= = =

Check 0:00True x ==

The solution set is { } 0.

71. 843 53 53 55 3 5 www w w w −+=− =− = =−

Check 3 :33True 5 w =−−=−

The solution set is 3 5 ⎧⎫ ⎨⎬ ⎩⎭

72. 934 74 74 77 4 7 xxx x x x −+=− =− =− =−

Check 4 :44True 7 x =−−=−

The solution set is { } 4 7

73. 111 3 3412 111

3Distributive property 3412 431 3LCD12 121212 1 3Lowest terms 6 1 66(3)Multiply by 6. 6 18 xxx x x x x x −+= ⎛⎞−+=⎜⎟ ⎝⎠ ⎛⎞−+==⎜⎟ ⎝⎠ = ⎛⎞ = ⎜⎟ ⎝⎠ =

Check 18:64.51.53True x =−+=

The solution set is { } 18.

74. 21 1 18 20 510 211 18Distributive prop. 51020 821 18LCD20 202020 9 18Add fractions. 20 2092020 (18)Multiply by . 92099 40 xxx x x x x x +−= ⎛⎞+−=⎜⎟ ⎝⎠ ⎛⎞+−==⎜⎟ ⎝⎠ = ⎛⎞ = ⎜⎟ ⎝⎠ =

Check 40:164218True x =+−=

The solution set is { } 40.

75. 0.90.50.13 0.53Combine terms. 0.53 Divideby0.5. 0.50.5 6 www w w w −+=− =− = =−

Check 6:33True w =−−=−

The solution set is { } 6.

76. 0.50.60.31

0.21Simplify.

0.21 Divide by 0.2. 0.20.2 5 xxx x x x −+=− =− = =−

Check 5:11True x =−−=− The solution set is { } 5.

77. Answers will vary. One example is 3 6. 2 x =−

78. Answers will vary. One example is 10017. x =

79. “When a number is multiplied by 4, the result is 6.” 46 46 44 3 2 x x x = = = The number is 3 2 and 3 2 ⎧ ⎫ ⎨ ⎬ ⎩⎭ is the solution set.

80. “When a number is multiplied by4, the result is 10.” 410 410 44 105 42 x x x −= = =−=−

The number is 5 2 and 5 2 ⎧⎫ ⎨⎬ ⎩⎭ is the solution set.

81. “When a number is divided by 5, the result is 2.” 2 5 1 (5)(5)(2) 5 10 x x x = ⎛⎞ −−=− ⎜⎟ ⎝⎠ =−

The number is 10 and { }10 is the solution set.

82. “If twice a number is divided by 5, the result is 4.” 2 4 5 5254 2521 10 x x x = ⎛⎞⎛⎞⎛⎞⎛⎞ = ⎜⎟⎜⎟⎜⎟⎜⎟

= The number is 10 and { }10 is the solution set.

2.3 More on Solving Linear Equations

Classroom Examples, Now Try Exercises

1. Step1 (not necessary)

Step2 5419

544194Subtract4. 515Combineterms. p p p −+= −+−=− −=

Step3 515

Divide by5. 55 3 p p =− =−

Step4

Check 3:15419True p =−+=

The solution set is { } 3.

N1. Step1 (not necessary)

Step2

723

72737Subtract7. 210Combineterms. m m m +=− +−=−− =−

Step3

210

Divide by2. 22 5 m m = =−

Step4

Check 5: 7103True m =−−=−

The solution set is { } 5.

2. Step1 (not necessary)

Step2 5825 588258Add 8. 5105Combine terms. 551055Add5. 1010Combine terms. xx xxxxx x x x −=− −+=−+ =− +=−+ = Step3 1010 Divideby10. 1010 1 x x = = Step4

Check 1: 5825True x =−=−

The solution set is { } 1.

N2. Step1 (not necessary)

Step2 2349 232492Subtract . 329Combine terms. 39299Add 9. 122Combine terms. qq qqqq2q q q q +=− +−=−− =− +=−+ = Step3 2 12 Divide by 2. 2 2 6 q q = = Step4

Check 6: 123249True q =+=− The solution set is { } 6.

3. Step1 113(1)516 1133516 Distributive property 314516 Combine terms. xx xx xx ++=+ ++=+ +=+ Step2 3141451614 Subtract 14. 352 35525 Subtract 5. 22 xx xx xxxxx x +−=+− =+ −=+− −= Step3 22 Divideby2. 22 1 x x =− =−

Step4

Check 1: 1111True x =−=

The solution set is { } 1.

N3. Step1 3(6)577 318577Distributiveproperty 21877Combineterms. zzz zzz zz −−=−+ −−=−+ −−=−+ Step2

218187718Add 18. 2725 277257Add 7. 525 zz zz zzzzz z −−+=−++ −=−+ −+=−++ = Step3 525 Divide by 5. 55 5 z z = = Step4

Check 5: 2828True z =−=−

The solution set is { } 5.

4. Step1 4(7)9 479 Distributive property 379 xx xx x −+= −−= −= Step2

37797 Add 7. 316 x x −+=+ = Step3 316 Divideby3. 33 16 3 x x = = Step4

Check 16 :99True 3 x == The solution set is 16 . 3 ⎧ ⎫ ⎨ ⎬ ⎩⎭

N4. Step1 5(9)4 594 Distributive property 494 xxx xxx xx −+=− −−=− −=− Step2

49949 Add 9. 45 45 Subtract . 35 xx xx xxxxx x −+=−+ =+ −=+− = Step3 35 Divide by3. 33 5 3 x x = =

Step4

Check 577 : True 333 x =−=−

The solution set is 5 . 3 ⎧⎫

5. Step1 23(26)4(1)8 2618448 Distributive property 41844 zz zz zz −+=+− −−=+−

Step2

4184444 Add 4. 184 18444 Subtract 4. 220 zz zz zzzzz z

Step3 220 Divideby22. 2222 0 z z =− = Step4

Check ? 0: 23(2)4(1)8 44 True z =−=− −=−

The solution set is { } 0.

N5. Step1

244(72)4(1) 2428844Distributive property 4844 tt tt tt −−=−

484444Add 4. 84 8444Subtract 4. 40 tt tt ttttt

Step3 40 Divide by4. 44 0 t t

Step4

Check ? 0:244(7)4(1) 44True t =−=− −=−

The solution set is { } 0.

6. 3102414 210210 2101021010Subtract 10. 22 2222Subtract 2 00True xxx xx xx xx xxxxx. −+=−+ +=+ +−=+− =

The variable x has “disappeared,” and a true statement has resulted. The original equation is an identity. This means that for every real number value of x, the equation is true. Thus, the solution set is {all real numbers}.

N6. 3(7)2521 321321 3212132121 Subtract 21. 33 3333Add 3. 00True xxx xx xx xx xxxxx −−=−+ −+=−+ −+−=−+− −=− −+=−+ =

7. 386(1)3

38663Distributive prop. 3836Combine terms. 383363Subtract 3. 86False

The variable x has “disappeared,” and a false statement has resulted. This means that for every real number value of x, the equation is false. Thus, the equation has no solution and its solution set is the empty set, or null set, symbolized ∅

N7. 41234(3) 4123412Distr. prop. 412415Combine. 41244154Add 4. 1215False

8. Step1 1531 31242 xx −=+

The LCD of all the fractions in the equation is 12, so multiply each side by 12 to clear the fractions.

The solution set is { } 7. N8. Step1 153 6 284 xxx+=−

The LCD of all the fractions in the equation is 8, so multiply each side by 8 to clear the fractions.

Step2 966486Subtract6. 348 xxxxx x −=−− =−

Step3 348 Divide by 3. 33 16 x x = =−

Step4

Check ? 16: 810126 1818True x =−−−=−− −=−

The solution set is { } 16. 9. Step1 () () () ()() () () ()() 2 3 12 312 43 12 1231122 43 1 12312124 4 338124 398824 5124 xx xx xx xx xx x +−+=− ⎡⎤+−+=−⎢⎥

⎡⎤ ⎡⎤ +−+=−⎢⎥ ⎣⎦ ⎣⎦ +−+=− +−−=− −+=−

Step2 511241 Subtract 1. 525 Combine like terms x x. −+−=−− −=−

Step3

525 Divide by 5. 55 5 x x =− = Step4

Check 5:242True x =−=−

The solution set is { } 5.

N9. Step1 21 32(2)(34)4 21 6(2)(34)6(4) 32 21 6(2)6(34)24 32 4(2)3(34)24 4891224 5424 xx xx xx xx xx x +−+=− ⎡⎤+−+=−⎢⎥ ⎣⎦ ⎡⎤⎡⎤+−+=−⎢⎥⎢⎥ ⎣⎦⎣⎦ +−+=− +−−=− −−=−

Step2 544244Add 4. 520 Combine like terms. x x −−+=−+ −=−

Step3

520 Divide by5. 55 4 x x =− =

Step4

Check 4: 484 True x =−=−

The solution set is { } 4.

Chapter 2 Linear Equations and Inequalities in One Variable

10. Step1

0.5(23)4.50.1(7) xx −=−+

To clear decimals, multiply both sides by 10. [][] 100.5(23)104.50.1(7)

5(23)451(7)

1015457 101538 xx xx xx xx −=−+

Step2 101538 101438 1010143810 1428 xxxx x

Step3 1428 1414 2 x x

Check ? 2: 0.5(8)4.50.1(5) 44 True x =−=− =

The solution set is { } 2.

N10. Step1

0.05(13)0.20.08(30) tt −−=

To clear decimals, multiply both sides by 100. [][] 1000.05(13)0.21000.08(30)

5(13)208(30)

65520240 6525240

Step2 65256524065 25175 t t

−= Step3 25175 2525 7 t t = =− Step4

Check ? 7:11.42.4 2.42.4True t =−+=− =

The solution set is { } 7.

11. To find the other number, you would divide 36 by x, so an expression for the other number is 36 . x

N11. First, suppose that the sum of two numbers is 18, and one of the numbers is 10. How would you find the other number? You would subtract 10 from 18. Instead of using 10 as one of the numbers, use m. This gives us the expression 18 m for the other number.

Exercises

1. Use the addition property of equality to subtract 8 from each side.

2. Combine like terms on the left.

3. Clear the parentheses by using the distributive property.

4. Use the multiplication property of equality to multiply each side by 4 . 3

5. Clear fractions by multiplying by the LCD, 6.

6. Clear the decimals by multiplying each side by 10.

7. (a) 66 = (The original equation is a(n) identity.) This goes with choice B, {all real numbers}.

(b) 0 x = (The original equation is a(n) conditional.) This goes with choice A, {0}.

8. ()() 1000.0310310330, xxx −=−=−⎡⎤ ⎣⎦ so choice D is correct.

9. 3214

322142 Subtract2.

312 Divide by3. 33 4 x x x x x += +−=− = = = Check 4: 12214True x =+=

312 Combine like terms.

The solution set is { } 4.

10. 4327

433273Subtract3.

424 Divideby4. 44 6 x x x x x += +−=− = = = Check 6: 24327True x =+=

424Combine like terms.

The solution set is { } 6.

11. 5421

544214Add 4.

525 Divide by 5. 55 5 z z z z z −−= −−+=+ −= =− =−

525Combine like terms.

Check 5: 25421True z =−−=

The solution set is { } 5.

12. 7410

744104Add 4.

Divide by 7. 77 2 w w w w w −−= −−+=+ −= =−

714Combine like terms. 714

Check 2: 14410True w =−−=

The solution set is { } 2.

13. 452

2.3 More on Solving Linear Equations

15. 29411

2911 Subtract 4. 22 Subtract 9. 1 Divide by2. xx xx x x +=+ −+= −= =−−

Check 1: 77True x =−=

The solution set is { } 1.

16. 7892

282 Subtract9. 210 Subtract8. 5 Divide by2. pp pp p p +=− −+=− −=− =−

Check 5: 4343True p ==

The solution set is { } 5.

17. For this equation, Step 1 is not needed.

52 Divide by 2. 22 5 2 pp ppppp p p p −= −−=− −=− =− =

45424Subtract 4. 52Combine like terms.

Step2 588738 Subtract 8. 531 53313 Subtract 3. 21 mm mm mmmmm m +−=+− =− −=−− =−

Step3 21 22 1 2 m m = =−

Step4

Check 5 : 1055True 2 p =−=

The solution set is 5 2 ⎧⎫ ⎨⎬ ⎩⎭

14. 623

62636Subtract 6. 23Combine like terms.

23=Divide by3. 33 2 3 qq qqqqq q q q −= −−=−

Check 2 : 422True 3 q =−=

The solution set is 2 3 ⎧⎫ ⎨⎬ ⎩⎭

Substitute 1 2 for m in the original equation. ? ? 5873 111 5873 Let . 222 53 87 22 111 True 22 mm m +=+ ⎛⎞⎛⎞ −+=+−=− ⎜⎟⎜⎟ ⎝⎠⎝⎠ ⎛⎞−+=+−⎜⎟ ⎝⎠ = The solution set is 1 2 ⎧ ⎫ ⎨ ⎬ ⎩⎭

Chapter 2 Linear Equations and Inequalities in One Variable

18. 426

426 Subtract .

326

32262 Subtract 2. 38 38 Divide by3. 33 8 3 rr rrrrr r r r r r +=− +−=−− +=− +−=−− =− =− =−

Check 8326818 :True 33333 r =−−=−+−

The solution set is { }8 . 3

19. 125107

12571077 Add7. 5510 555105 Add 5. 515 515 Divide by5. 55 3 xx xxxxx x x x x x −−=− −−+=−+ −−= −−+=+ −= =− =−

Check 3: 3651021True x =−−=+

The solution set is { } 3.

20. 163138 16381388 Add8. 8313

833133 Add 3. 816 816 Divide by8. 88 2 ww wwwww w w w w w −−=− −−+=−+ −−= −−+=+ −= =− =−

Check 2: 3231316True w =−−=+

The solution set is { } 2.

21. 125115

22. 41513

4121 Combine like terms. 211 Add 2. 22 Add 1. 1 Divide by 2. xxx xx xx x x −−=−++ −−=−+ −−= −= =−−

Check 1: 33True x =−=

The solution set is { } 1. 23. 75252

2242Combine liketerms. 222Subtract2. 02Subtract2. 0Divideby2. rrrr rr rr r r −+=+− +=+ =+ = =

Check 0:22True r ==

The solution set is {0}.

24. 946763 5646Combine terms. 66Subtract4. 0Subtract6. pppp pp pp p −+=+− +=+ += = Check 0:66True p ==

The solution set is {0}.

25. 3(42)530

126530Distributive prop. 17630Combine like terms. 18630Add1. 1824Subtract6. 244 Divideby18. 183 xxx xxx xx xx x x ++=− ++=− +=− += = ==

Check 48686 :True 333 x ==

The solution set is 4 . 3 ⎧ ⎫ ⎨ ⎬ ⎩⎭

26. 5(23)4225

125105 Combine like terms.

255 Subtract 10. 210 Add 5. 5 Divide by 2. hhh hh hh h h −=+− −=+ −= = =

Check 5: 5555True h ==

The solution set is { } 5.

10154225Distributive prop. 615225Combine terms. 6210Subtract15. 410Subtract2. 105 Divideby4. 42 mmm mmm mm mm mm m +−=+ +−=+ +=+ =+ = ==

Check 5 :3030True 2 m ==

The solution set is 5 2 ⎧ ⎫ ⎨ ⎬ ⎩⎭

27. 273(51)

27351Distributiveproperty 2752Combine liketerms. 372Add5. 35Subtract7. 5 3 pp pp pp pp p p −+=−+ −+=−− −+=−+ += =− =−

Check 53131 :True 333 p =−=

The solution set is 5 3 ⎧⎫ ⎨⎬ ⎩⎭

28. 493(2)

4932Distributiveproperty 495Combine liketerms. 595Add1. 54Subtract9. 4 Divideby5. 5 xx xx xx xx x x +=−− +=−+ +=−+ += =− =−

Check 42929 :True 555 x =−=

The solution set is 4 5 ⎧⎫ ⎨⎬ ⎩⎭

29. 115(2)65 1151065 61065 6106656 105 xxx xxx xx xxxx −+=+ −−=+ −=+ −−=+− −=

Since 105−= is a false statement, the equation has no solution set, symbolized by ∅

30. 64(1)24 64424 2424 44Subtract2. xxx xxx xx x −+=+ −−=+ −=+ −=

Since 44−= is a false statement, the equation has no solution set, symbolized by ∅

31. 6(35)2(1010) 18302020 182010Subtract30. 210Subtract20. 5Divideby2. ww ww ww ww w +=+ +=+ =− −=− =−

Check 5:120120True w ==

The solution set is {5}.

2.3 More on Solving Linear Equations

32. 4(21)6(3) 84618Distributive property 14418Add6. 1414Add4. 1Divideby14. xx xx xx x x −=−+ −=−− −=− =− =−

Check 1:1212True x =−−=−

The solution set is {1}.

33. (42)(35)3 1(42)1(35)3 42353 33 0 0 xx xx xx x x x −+−−−= −+−−−= −−++= −+= −= =

Check 0:33True x ==

The solution set is {0}.

34. (65)(58)3 1(65)1(58)3 65583 33 0 0 kk kk kk k k k −−−−+=− −−−−+=− −++−=− −−=− −= =

Check 0:33True k =−=−

The solution set is {0}.

35. 3(24)6(2) 612612 1212Subtract6. 00Add12. xx xx x −=− −=− −=− =

The variable has “disappeared.” Since the resulting statement is a true one, any real number is a solution. We indicate the solution set as {all real numbers}.

36. 3(64)2(69) 18121218 1818Add12. xx xx x −=−+ −=−+ = Since 1818 = is a true statement, the solution set is {all real numbers}.

37. 6(41)12(23) 2462436 636Subtract24. xx xx x −=+ −=+ −=

The variable has "disappeared," and the resulting equation is false. Therefore, the equation has no solution set, symbolized by .∅

38. 6(28)4(36) 12481224 4824Subtract12. xx xx x +=− +=− =−

Since 4824 =− is a false statement, the equation has no solution set, symbolized by ∅

39. The least common denominator of all the fractions in the equation is 10. 315 1010 5102 315 10101010 5102 61025 51025 525

Check 55 5:True 22 t ==

The solution set is {5}.

40. The least common denominator of all the fractions in the equation is 14, so multiply both sides by 14 and solve for r. 2117 14214 722 4287119 247119 17119 119 7

Check 7:1212True r ==

The solution set is {7}.

41. The least common denominator of all the fractions in the equation is 12, so multiply both sides by 12 and solve for x 12512315 436 946010 56010 605 605 55 12 xxx xxx xx x x x

Check 12:94510True x =−+=

The solution set is {12}.

42. The least common denominator of all the fractions in the equation is 15, so multiply both sides by 15 and solve for x 15215122 535 310306 7306 30 xxx xxx xx x

Check 30:620212True x =−−+−=

The solution set is {30}.

43. The least common denominator of all the fractions in the equation is 35, so multiply both sides by 35 and solve for x. () () () ()() 11 35324352 75 5327470 151072870 81870 888 888 88 11 xx xx xx x x x x

Check 11:532True x =−=

The solution set is {11}.

44. The least common denominator of all the fractions in the equation is 12, so multiply both sides by 12 and solve for x. () ()() ()() 11 12313123 46 3312336 932636 11336 1133 1133 1111 3 xx xx xx x x x x

Check 3:213True x =+=

The solution set is {3}.

45. The LCD of all the fractions is 4.

Check 0:44True x ==

The solution set is {0}.

46. The least common denominator of all the fractions in the equation is 9, so multiply both sides by 9 and solve for p. () ()()

48. The least common denominator is 12. () () () () () 51 1211280 64 51 121211280 64 101212380 22123240 1912240 19228 228 12 19 qqq qqq qqq qq q q q

Check 12:2323True q =−=

The solution set is {12}.

49. To clear the equation of decimals, we multiply by 100.

100(0.753.2)100(0.550.5) 753205550 7532050555050 12532055 12532032055320 125375 125375 125125 3 xx xx xxxx x x x x x −=− −=− −+=−+ −= −+=+ = = =

Check 3:0.950.95True x =−=−

The solution set is {3}.

Check 0:33True p ==

The solution set is {0}.

47. The least common denominator of all the fractions in the equation is 6, so multiply both sides by 6 and solve for k.

50. 100(1.350.6)100(1.652.1) 13560165210 13560135165210135 6016575 6016516575165 22575 22575 7575 3 xx xx xxxx x x x x x −=+ −=+ −−=+− −=+ −−=+− −= = =−

Check 3:4.654.65True x =−−=−

The solution set is { } 3.

Check 1111 18:True 22 k =−=−

The solution set is {18}.

51. Solve the equation for t 0.80.1521.35

100(0.80.15)100(21.35) 8015200135 80158020013580 15120135 15135120135135 150120 150120 120120 5 4 tt tt

Check 5 :1.151.15True 4 t ==

The solution set is 5 4 ⎧⎫

52. To eliminate the decimal in 3.4, we need to multiply both sides by 10. But to eliminate the decimal in –0.12 and 0.84, we need to multiply by 100, so we choose 100. [] 0.123.40.845

100[0.123.4]1000.845 1234084500 12340128450012 34084512 340848451284 256512 256512 512512 1 2 pp pp pp pppp p p p p

:3.343.34True

The solution set is 1 2

53. To eliminate the decimal in 0.2 and 0.1, we need to multiply both sides by 10. But to eliminate the decimal in 0.05, we need to multiply by 100, so we choose 100. ()()

1000.2600.051000.160 206051060 1200560010 12005600 5600 600 120 5 xx xx xx x x x

Check 120:1818True x ==

The solution set is {120}.

()() ()() ()() 0.3300.150.230 1000.3300.151000.230 3030152030 9001560020 9005600 5300 60 xx xx xx xx x x x

Check 60:1818True x ==

The solution set is {60}.

55. 1.000.05(12)0.10(63) xx+−=

To clear the equation of decimals, we multiply both sides by 100. [][] 1001.000.05(12)1000.10(63)

1005(12)10(63) 100605630

9560630 95570 570 6 95 xx xx xx x x x +−= +−= +−=

Check 6:6.36.3True x ==

The solution set is {6}.

56. [][] 0.920.98(12)0.96(12)

1000.920.98(12)1000.96(12)

9298(12)96(12)

921176981152 611761152 624 24 4 6 xx xx xx xx x x x +−= +−=

Check 4:11.5211.52True x ==

The solution set is {4}.

60. 9(35)12(31)51 2745361251 27453663 45963 189 2 kk kk kk k k k −=−− −=−− −=− −=− = =

Check 2:99True k ==

The solution set is {2}.

57. 0.6(10,000)0.80.72(10,000) 60(10,000)8072(10,000)

600,00080720,00072

600,0008720,000 8120,000 120,000 8 15,000 xx xx xx x x x x +=+ +=+ +=+ += = = =

Check 15,000:18,00018,000True x ==

The solution set is {15,000}.

61. 244(72)4(1) 2428844 4844 484444 444 40 0 tt tt tt tttt t t t −−=− −+=− −+=− −+−=−− −+=− = =

Check 0:44True t =−=−

The solution set is {0}.

58. 0.2(5000)0.30.25(5000) 20(5000)3025(5000)

100,00030125,00025

5100,000125,000 525,000 25,000 5000 5 xx xx xx x x x +=+ +=+ +=+ += = ==

Check 5000:25002500 True x ==

The solution set is {5000}.

59. 10(21)8(21)14 201016814 20101622 41022 432 8 xx xx xx x x x −=++ −=++ −=+ −= = =

Check 8:150150True x ==

The solution set is {8}.

62. 82(2)4(1) 84244 4244 422442 424 20 0 xx xx xx xxxx x x x −−=+ −+=+ +=+ +−=+− =+ = =

Check 0:44True x ==

The solution set is {0}.

63. 4(8)2(26)20 43241220 432432 44 00 xx xx xx xx +=++ +=++ +=+ = = Since 00 = is a true statement, the solution set is {all real numbers}.

64. 4(3)2(28)4 4124164 412412 1212 xx xx xx +=+− +=+− +=+ = Since 1212 = is a true statement, the solution set is {all real numbers}.

65. To clear fractions, multiply both sides by the LCD, which is 4. 13

4(2)(4)4(5) 24 13 4(2)4(4)420 24

2(2)3(4)420 24312420 516420 1620 4 xxx xxx xxx xxx xx x x ⎡⎤+++=+

Check 4:99True x ==

The solution set is {4}.

66. To clear fractions, multiply both sides by the LCD, which is 6. 11

6(3)(6)6(3) 36 11

6(3)6(6)6(3) 36

2(3)1(6)6(3) 266618 3618 318 18 6 3 xxx xxx xxx xxx xx x x ⎡⎤++−=+

Check 6:33True x =−−=−

The solution set is {6}.

67. To eliminate the decimals, multiply both sides by 10.

[] 100.1(80)0.210(14)

1(80)2140 802140 380140 360 20 xx xx xx x x x ++= ++= ++= += = =

Check 20:1414True x ==

The solution set is {20}.

68. To clear the decimals, multiply both sides by 10.

3(15)4(25)250 3454100250 7145250 7105 15 xx xx x x x +++= +++= += = =

Check 15:2525True x ==

The solution set is {15}.

69. 9(1)32(31)8 993628 6966 96 vvv vvv vv +−=+− +−=+− +=− =−

Because 96 =− is a false statement, the equation has no solution set, symbolized by ∅

70. 8(3)46(21)10 824412610 1224124 244 ttt ttt tt −+=+− −+=+− −=− −=−

Because 244−=− is a false statement, the equation has no solution set, symbolized by ∅

71. The sum of q and the other number is 11. To find the other number, you would subtract q from 11, so an expression for the other number is 11. q

72. The sum of r and the other number is 34. To find the other number, you would subtract r from 34, so an expression for the other number is 34. r

73. The product of x and the other number is 9. To find the other number, you would divide 9 by x, so an expression for the other number is 9 x

74. The product of m and the other number is 6. To find the other number, you would divide 6 by m, so an expression for the other number is 6 m

75. If a baseball player gets 65 hits in one season, and h of the hits are in one game, then 65 h of the hits came in the rest of the games.

76. If a hockey player scores 42 goals in one season, and n of the goals are in one game, then 42 n of the goals came in the rest of the games.

77. If Monica is x years old now, then 15 years from now she will be 15 x + years old. Five years ago, she was 5 x years old.

78. If Chandler is y years old now, then four years ago he was 4 y years old. Eleven years from now, he will be 11 y + years old.

79. Since the value of each quarter is 25 cents, the value of r quarters is 25r cents.

80. Since the value of each dime is 10 cents, the value of y dimes is 10y cents.

81. Since each bill is worth 5 dollars, the number of bills is . 5 t

82. Since each bill is worth 10 dollars, the number of bills is . 10 v

83. Since each adult ticket costs x dollars, the cost of 3 adult tickets is 3x. Since each child's ticket costs y dollars, the cost of 2 children's tickets is 2y. Therefore, the total cost is 32xy + (dollars).

84. Since each adult ticket costs p dollars, the cost of 4 adult tickets is 4p. Since each child's ticket costs q dollars, the cost of 6 children's tickets is 6q. Therefore, the total cost is 46pq + (dollars).

Summary Exercises Applying Methods for Solving Linear Equations

1. This is an equation since it has an equals sign. 23 5Subtract2. x x +=− =− Check 5:33True x =−−=− The solution set is {5}.

2. This is an expression since it does not have an equals sign.

4638714 ppp −+−=−

3. This is an expression since it does not have an equals sign.

(1)(32)132 32 mmmm m −−−+=−+−− =−−

4. This is an equation since it has an equals sign. 69123 3912 321 7 qq q q q −=+ −= = = Check 7:3333True q == The solution set is {7}.

5. This is an equation since it has an equals sign. ()5933 5939Distributive property 299Subtract3. 20Add9. 0Divideby2. xx xx xx x x −=− −=− −=− = = Check 0:99True x =−=− The solution set is {0}.

6. This is an equation since it has an equals sign. To clear fractions, multiply both sides by the LCD, which is 12. 21 12812 34 8963 5960 596 96 5 xx xx x x x ⎛⎞⎛⎞ +=

+= += =− =−

Check 962424 :True 555 x =−−=−

The solution set is 96 . 5 ⎧ ⎫ ⎨ ⎬ ⎩⎭

7. This is an expression since it does not have an equals sign. 26(1)4(2)10 2664810 106 zz zz z −+−−− =−−−+− =−−

8. This is an equation since it has an equals sign. 7(2)2(2) 71424 81424 6144 618 3 ppp ppp pp p p p −+=+ −+=+ −=+ −= = = Check 3:1010True p == The solution set is {3}.

9. This is an expression since it does not have an equals sign. 1212 (10)5 2323 34 5 66 1 5 6 xxxx xx x +−=+− =+− =−+

Chapter 2 Linear Equations and Inequalities in One Variable

10. This is an expression since it does not have an equals sign.

4(2)3(21)4863 211 kkkk k −++−=−−+− =−

11. 614 14 Divideby6. 6 7 3 z z −=− =− =

Check 7 :1414True 3 z =−=−

The solution set is 7 . 3 ⎧⎫ ⎨⎬ ⎩⎭

12. 2816

28Subtract8. 4Divideby2. m m m += = =

Check 4:1616True m ==

The solution set is {4}.

13. 12.563.75 63.75

Divideby12.5. 12.5 5.1 x x =− = =−

Check 5.1:63.7563.75True x =−−=−

The solution set is {5.1}.

14. 12

12Multiplyby1. x x −=− =−

Check 12:1212True x =−=−

The solution set is {12}.

15. 4 20 5 55 (20)Multiplyby. 44 25 x x =− ⎛⎞ =− ⎜⎟ ⎝⎠ =−

Check 25:2020True x =−−=−

The solution set is {25}.

16. 7512

212Combine like terms. 6Divide by 2. mm m m −=− =− =−

Check 6:1212True m =−−=−

The solution set is {6}.

17. 6 6Multiplyby1. x x −= =−−

Check 6:66True x =−=

The solution set is {6}.

18. 8 2 1 8 2 2(8)Multiplyby2. 16 x x x = −= =−− =−

Check 16:88True x =−=

The solution set is {16}.

19. 42(32)6 4646 66 xx xx +−= +−= = Since 66 = is a true statement, the solution set is {all real numbers}.

20. 16.27.5 23.7Add 16.2. x x −= =

Check 23.7:7.57.5True x ==

The solution set is {23.7}.

21. 7(29)39 72939 5939 530 6 mm mm m m m −−= −+= += = =

Check 6:3939True m ==

The solution set is {6}.

22. () 2432 2432 232 422 40 0 0 4 mm mm mm m m m −+=− −−=− −−=− −−=− −= ==

Check 0:22True m =−=−

The solution set is {0}.

23. 3(4)2(52)29 31210429 2229 7 mm mm m m −−++= −+++= += =

Check 7:2929True m ==

The solution set is {7}.

24. To eliminate the decimals, multiply both sides by 10.

28. To eliminate the decimals, multiply both sides by 10.

[][] 102.313.7101.32.9 231371329 1013729 10108 10.8 xx xx x x x +=+ +=+ += =− =−

Check 10.8:11.1411.14True x =−−=−

The solution set is {10.8}.

321(4)66 3218466 188466 1818 1 xx xx xx x x x −+−=− −+−=− −+−=− −=− = =

[] 100.32.1(4)10(6.6)

Check 1:0.36.36.6True x =−−=−

The solution set is {1}.

25. To eliminate the decimals, multiply both sides by 100.

29. To eliminate the decimals, multiply both sides by 10.

[][] 100.2(50)0.8100.4(50) 2(50)84(50) 10082004 1004200 4100 25 rr rr rr r r r +=+ +=+ +=+ += = =

Check 25:102030True r =+=

The solution set is {25}.

86(9)124 8654124 1454124 1470 5 xx xx xx x x x ++= ++= ++= += = =

[] 1000.080.06(9)100(1.24)

Check 5:0.40.841.24True x =+=

The solution set is {5}.

26. 3(5)125(2) 31512510 514510 1410 mmm mmm mm +−+=+ +−+=+ +=+ = Because 1410 = is a false statement, the equation has no solution set, symbolized by ∅

27. 2593(4)5 2593125 39317 917 ttt ttt tt −+−=−− −+−=−− −=− −=−

Because 917−=− is a false statement, the equation has no solution set, symbolized by ∅

30. 974(32)3 891283 8989 99 rrr rr rr ++=+− +=+− +=+ = Since 99 = is a true statement, the solution set is {all real numbers}.

31. 2(37)(115)2 6141152 52 3 3 xx xx x x x +−+= +−−= −+= −=− =

Check 3:48462True x =−=

The solution set is {3}.

32. To eliminate the decimals, multiply both sides by 10.

[][] 100.6(100)0.4100.5(92) 6(100)45(92) 60064460 6002460 2140 70 xx xx xx x x x −+= −+= −+= −= −=− =

Check 70:182846True x =+=

The solution set is {70}.

33. To clear fractions, multiply both sides by the LCD, which is 4.

Step6

20 plus 8 is 12 times 5 is 60, so 20 is the number.

N1. Step2

Let the x = number.

Check 2:4.531.5True x =−−=−−

The solution set is {2}.

34. To clear fractions, multiply both sides by the LCD, which is 12. 31 12(2)(52)12(2) 43 9(2)4(52)24 91820824 173824 1714 14 17 zz

The solution set is 14 . 17 ⎧⎫

2.4 Applications of Linear Equations

Classroom Examples, Now Try Exercises

1. Step2

Let the x = number.

Step3

Thea number productofincreased 5,andby 8,is60.

560 8 x

Step4

Solve the equation. () 5860

54060Distributive property

5100Subtract 40. 20Divide by 5. x x x x +=− +=− =− =−

Step5 The number is 20

Theproductofincreased 7,andby 3,is63.

763 3 x ↓↓↓↓ ⋅=− +

Step3 () a number

Step4

Solve the equation. () 7363

72163Distributive property

784Subtract 21. 12Divide by 7. x x x x +=− +=− =− =−

Step5

The number is 12

Step6 12 plus 3 is 9, times 7 is 63, so 12 is the number.

2. Step2

Let the x = number. Step3 theproductofthe19less Ifadded9andaresultthanthe 5 istonumber,isnumber.

5919 xx ↓↓↓↓↓ +=−

Step4

Solve the equation. 5919 595195Subtract5. 924 924Subtract. 824 824

Divideby8. 88 3 xx xx xx xxxxx x x x +=− +−=−− =− −=−− =− = =−

Step5

The number is 3

Step6

9 times 3 is 27. 5 added to 27 is 22, which is 19 less than 3, so 3 is the number.

N2. Step2

Let the x = number.

Step3 the7less If 5addedaresultthan3times istonumber,isthenumber.

N3. Step2

Let the x = number of medals Great Britain won.

Let 21the x −= number of medals Germany won.

Step3

5=37 xx

+−

Step4

Solve the equation. 537

55375Subtract5. 312 33123Subtract3.

Step5

The number is 6.

Step6

5 added to 6 is 11. 3 times 6 is 18, and 7 less than 18 is 11, so 6 is the number.

3. Step2

Let the x = number of medals France won. Then 4 x + number of medals Japan won.

Step3

the number ofthe number of Themedalsmedals totalis France wonplusJapan won.

72=+ 4 xx

Step4

Solve the equation. ()472 2472Combine like terms.

268Subtract 4. 34Divide by 2. xx x x x ++= += = =

Step5 France won 34 medals and Japan won 34438 += medals.

Step6

38 is 4 more than 34, and the sum of 38 and 34 is 72.

the number ofthe number of Themedals Great medals totalisBritain wonplusGermany won.

109=+ 21 xx

Step4

Solve the equation. ()21109 221109Combine like terms.

2130Add 21.

65Divide by 2. xx x x x +−= −= = =

Step5

Great Britain won 65 medals and Germany won 652144 −= medals.

Step6

44 is 21 less than 65, and the sum of 65 and 44 is 109.

4. Step2

Let the x = number of orders for muffins. Then 1 the 6 x = number of orders for croissants.

Step3 orders for Theorders for croissants. totalismuffinsplus

Step4

Solve the equation. 1 5611 6 61 56LCD6 66 7 56Combineliketerms. 6 6676 (56)Multiplyby. 7767 48 xxxx xx x x x =+= =+= =

Step5

The number of orders for muffins was 48, so the number of orders for croissants was 1 (48)8. 6 =

Step6

One-sixth of 48 is 8, and the sum of 48 and 8 is 56.

N4. Step2

Let the x = number of orders for chocolate scones.

Then 2 the 3 x = number of orders for bagels.

Step3 orders for

Step4

Solve the equation.

Step5

The number of orders for chocolate scones was 315, so the number of orders for bagels was 2 (315)210. 3 = Step6

Two-thirds of 315 is 210, and the sum of 315 and 210 is 525.

5. Step2

Let the x = number of members. Then 2the x = number of nonmembers. (If each member brought two nonmembers, there would be twice as many nonmembers as members.)

Step3

Numbernumberthe total ofofin membersplusnonmembersisattendance. +227xx ↓↓↓↓↓ =

Step4

Solve the equation. 227 327 327 33 9 xx x x x += = = =

Step5

There were 9 members and 2918 ⋅= nonmembers.

Step6 18 is twice as much as 9, and the sum of 9 and 18 is 27.

N5. Step2

Let the x = number of residents. Then 4the x = number of guests. (If each resident brought four guests, there would be four times as many guests as residents.)

Step3

Numbernumberthe total ofofin residentsplusguestsisattendance. +4175xx ↓↓↓↓↓ =

Step4

Solve the equation. 4175 5175 5175 55 35 xx x x x += = = =

Step5

There were 35 residents and 435140 ⋅= guests.

Step6 140 is four times as much as 35, and the sum of 35 and 140 is 175.

6. Step2

Let the x = length of the middle-sized piece. Then 10the x += length of the longest piece and 5the x −= length of the shortest piece.

Step3 middle-total

Shortestplussizedpluslongestislength.

Step4

Solve the equation. (5)(10)50

N6.

Step5

The middle-sized piece is 15 inches long, the longest piece is 151025 += inches long, and the shortest piece is 15510 −= inches long.

Step6

Since 25 inches is 10 inches longer than 15 inches, 15 inches is 5 inches longer than 10 inches, and 15251050 ++= inches (the length of the pipe), the answers are correct.

Step2

Let the x = time spent practicing free throws. Then 2the x = time spent lifting weights and 2the x += time spent watching game films.

Step3

Freeliftingwatchingtotal throwsplusweightsplusfilmsistime.

Step4

Solve the equation.

Step5

The time spent practicing free throws is 1 hour, the time spent lifting weights is 2(1)2 = hours, and the time spent watching game films is 123 += hours.

Step6

Since 2 hours is twice as much time as 1 hour, 3 hours is 2 more hours than 1 hour, and the sum of the times is 1236 ++= hours (the total time spent), the answers are correct.

7. Step2

Let the x = lesser page number.

Then 1the x += greater page number.

Step3

Because the sum of the page numbers is 569, an equation is (1)569.xx++=

Step4

Solve the equation 21569Combineliketerms.

2568Subtract1.

284Divideby2. x x x += = =

Step5

The lesser page number is 284, and the greater page number is 2841285. +=

Step6

285 is one more than 284, and the sum of 284 and 285 is 569.

N7.

Step2

Let the x = lesser page number. Then 1the x += greater page number.

Step3

Because the sum of the page numbers is 593, an equation is (1)593.xx++=

Step4

Solve the equation 21593Combineliketerms.

2592Subtract1.

296Divideby2. x x x += = =

Step5

The lesser page number is 296, and the greater page number is 2961297. +=

Step6

297 is one more than 296, and the sum of 296 and 297 is 593.

8. Let the x = lesser even integer. Then 2the x += greater even integer. From the given information, we have 6(2)86. xx ⋅++=

Solve the equation. 7286 784 12 x x x += = =

The lesser even integer is 12 and the greater consecutive even integer is 12214. += Six times 12 is 72 and 72 plus 14 is 86.

N8. Let the x = lesser odd integer. Then 2the x += greater odd integer. From the given information, we have 23(2)191. xx ⋅++=

Solve the equation. 236191 56191 5185 37 xx x x x ++= += = =

The lesser odd integer is 37 and the greater consecutive odd integer is 37239. += Two times 37 is 74, three times 39 is 117, and 74 plus 117 is 191.

9. Let the x = degree measure of the angle. Then 90the x −= degree measure of its complement.

Theeighttimes complementtheangle. is

908 xx

Solve the equation. 908

909 10 xx x x −= = = The measure of the angle is 10°.

N9. Let the x = degree measure of the angle. Then 90the x −= degree measure of its complement. Thetwice complementistheangle.

902 xx

Solve the equation. 902 903 30 xx x x −= = = The measure of the angle is 30°.

10. Step2

Let the x = degree measure of the angle. Then 90the x −= degree measure of its complement, and 180 x −= the degree measure of its supplement.

Step3 complementplussupplementequals174 90180174 xx ↓↓↓↓↓ −+−=

Step4

Solve the equation. 2702174

296Subtract270. 48Divideby2. x

Step5

The measure of the angle is 48°.

Step6

The complement of 48° is 904842 °−°=° and the supplement is 18048132. °−°=° The sum of 42° and 132° is 174°.

N10.

Step2

Let the x = degree measure of the angle. Then 90the x −= degree measure of its complement, and 180 x −= the degree measure of its supplement.

Step3 The46lessthan3times supplementequalsitscomplement.

1803(90)46 xx °

Step4

Solve the equation.

1803(90)46 180270346Distributiveprop. 1802243Combineterms. 1802224Add3.

244Subtract180. 22Divideby2.

Step5

The measure of the angle is 22°

Step6

The complement of 22° is 902268, °−°=° and 46° less than 3 times 68° is 3(68)4620446158. °−°=°−°=° The supplement is 18022158. °−°=°

Exercises

1. Choice D, 1 6, 2 is not a reasonable answer in an applied problem that requires finding the number of cars on a dealer's lot, since you cannot have 1 2 of a car. The number of cars must be a whole number.

2. Choice D, 25, is not a reasonable answer since the number of hours a light bulb is on during a day cannot be more than 24.

3. Choice A, 10, is not a reasonable answer since distance cannot be negative.

4. Choice C, 5, is not a reasonable answer since time cannot be negative.

5. Step2

Let the x = number.

Step3

The sum of a number and9is26. 926 x ↓↓↓↓ +=−

Step4

Solve the equation. 926 35 x x +=− =−

Step5

The number is 35.

Step6

The sum of 35 and 9 is 26.

6. Step2

Let the x = number.

Step3

The difference of a number and 11 is31. 1131 x ↓↓↓ −=−

Step4

Solve the equation. 1131 20 x x −=− =−

Step5

The number is 20.

Step6

Check that 20 is the correct answer by substituting this result into the words of the original problem. The difference of 20 and 11 is 31.

7. Step2

Let the x = number.

Step3

Theproductandanumber of8,increasedby6,is104.

8(6)104 x ↓↓↓↓ ⋅+=

Step4

Solve the equation.

8(6)104

848104 856 7 x x x x += += = =

Step5

The number is 7.

Step6

Check that 7 is the correct answer by substituting this result into the words of the original problem. 7 increased by 6 is 13. The product of 8 and 13 is 104, so 7 is the number.

8. Step2

Let the x = number.

Step3

Theproductand3morethan of5,twiceanumber, is85.

(23) 585 x ↓↓↓↓ + ⋅=

Step4

Solve the equation. 101585 1070 7 x x x += = =

Step5

The number is 7.

Step6

Check that 7 is the correct answer by substituting this result into the words of the original problem. If the number is 7, then twice the number is 14, and 3 more than twice the number is 17. The product of 5 and 17 is 85, so 7 is the number.

9. Step2

Let the x = unknown number. Then 52 x + represents “2 is added to five times a number,” and 45 x + represents “5 more than four times the number.”

Step3

5245 xx+=+

Step4

Solve the equation 5245 25 3 xx x x +=+ += =

Step5

The number is 3.

Step6

Check that 3 is the correct answer by substituting this result into the words of the original problem. Two added to five times a number is 25(3)17 += and 5 more than four times the number is 54(3)17. += The values are equal, so the number 3 is the correct answer.

10. Step2

Let the x = unknown number. Then 84 x + represents “four times a number added to 8,” and 53 x + represents “three times the number, added to 5.”

Step3

8453xx+=+

Step4

Solve the equation 8453 85 3 xx x x +=+ += =−

Step5

The number is 3.

Step6

Check that 3 is the correct answer by substituting this result into the words of the original problem. Four times a number is added to 8 is 84(3)4 +−=− and three times the number added to 5 is 53(3)4. +−=− The values are equal, so the number 3 is the correct answer.

11. Step2

Let the x = unknown number. Then 2 x is two subtracted from the number, 3(2) x is triple the difference, and 6 x + is six more than the number.

Step3

3(2)6 xx−=+

Step4

Solve the equation. 366 266 212 6 xx x x x −=+ −= = =

Step5

The number is 6.

Step6

Check that 6 is the correct answer by substituting this result into the words of the original problem. Two subtracted from the number is 624. −= Triple this difference is 3(4)12, = which is equal to 6 more than the number, since 6612. +=

12. Step2

Let the x = unknown number. Then 3 x + is 3 is added to a number, 2(3) x + is this sum is doubled, and 2 x + is 2 more than the number.

Step3

2(3)2 xx+=+

Step4

Solve the equation. 262 62 4 xx x x +=+ += =−

Step5

The number is 4.

Step6

Check that 4 is the correct answer by substituting this result into the words of the original problem. Three added to the number is 1, double this value is 2. Two more than 4 is also 2, so the number is 4.

13. Step2

Let the x = unknown number. Then 3 4 x is 3 4

of the number, and 3 6 4 x + is 6 added to 3 4 of the number. 4 x is 4 less than the number.

Step3 3 64 4 xx+=−

Step4

Solve the equation.

Step5

The number is 40.

Step6

Check that 40 is the correct answer by substituting this result into the words of the original problem. 3 4 of 40 is 30. When 6 is added to 30, the sum is 36, which is 4 less than 40 because 6 is added to 30.

14. Step2

Let the x = unknown number. Then 2 3 x is 2 3 of the number, and 2 10 3 x + is 2 3 of the number added to 10. 5 x + is 5 more than the number.

Step3 2 105 3 xx +=+

Step4

Solve the equation.

Step5

The number is 15.

Step6

Check that 15 is the correct answer by substituting this result into the words of the original problem. 2 3 of 15 is 10. When 10 is added to 10, the sum is 20, which is 5 more than 15 because 5 is added to 10.

15. Step2

Let the x = unknown number. Then 3x is three times the number, 7 x + is 7 more than the number, 2x is twice the number, and 112 x is the difference between 11 and twice the number.

Step3

3(7)112 xxx ++=−−

Step4

Solve the equation 47112 6711 618 3 xx x x x +=−− +=−

Step5

The number is 3.

Step6

Check that 3 is the correct answer by substituting this result into the words of the original problem. The sum of three times a number and 7 more than the number is 3(3)(37)5 −+−+=− and the difference between –11 and twice the number is 112(3)5. −−−=− The values are equal, so the number 3 is the correct answer.

16. Step2

Let the x = unknown number. Then 24 x + is 4 is added to twice the number, 2(24) x + is the sum multiplied by 2, and 34 x + is the number is multiplied by 3 and 4 is added to the product.

Step3

2(24)34 xx+=+

Step4

Solve the equation. 4834 84 4 xx x x +=+ += =−

Step5

The number is 4.

Step6

Check that 4 is the correct answer by substituting this result into the words of the original problem.

Twice the number is 2(4)8. −=− Four added to twice the number is 844.−+=− This sum multiplied by 2 is 2(4)8. −=− The number multiplied by 3 is 3(4)12. −=− Four added to this product is 1248.−+=− Because both results are 8, the answer, 4, checks.

17. Step1

We must find the number of Democrats and the number of Republicans

Step2

Let the x = number of Republicans. Then 4the x −= number of Democrats.

Step3 () Number of number oftotal DemocratsRepublicansequalsmembers.

Step4

Solve the equation. ()4150 24150 2154 77 xx x x x −+= −= = = Step5

There were 77 Republicans and 77473 −= Democrats.

Step6

Check that the numbers found are the correct answers by substituting the result into the words of the original problem. 73 is 4 fewer than 77, and 7773150. +=

18. Step1

We must find the two consecutive integers.

Step2

Let the x = lesser of the two consecutive even integers.

Then 2the x += greater of the two consecutive even integers.

Step3 The lesserthe greater consecutiveconsecutivethe even integereven integeristotal.

Step4

Solve the equation. ()2254 22254 2252 126 xx x x x ++= += = =

Step5

The lesser even integer is 126, and the greater even integer is 1262128. +=

Step6

Check that the numbers found are the correct answers by substituting the result into the words of the original problem. 126 and 128 are consecutive even integers, and 126128254. +=

19. Letthe x = number of drive-in movie screens in Ohio.

Then2thenumber of drive-in movie screens inNew York. x +=

Since the total number of screens was 56, we can write the equation (2)56.xx++=

Solve the equation. 2256 254 27 x x x += = = Since 27,229.xx=+=

There were 27 drive-in movie screens in Ohio and 29 in New York. Since 29 is 2 more than 27 and 272956, += this answer checks.

20. Let the x = number of Cheers viewers. Then8thenumber of *** viewers. xMASH +=

Since the total number of viewers is 92 (all numbers in millions), we can write the equation (8)92.xx++=

Solve the equation. 2892 284 42 x x x += = = Since 42,850.xx=+=

There were 42 million Cheers viewers and 50 million M*A*S*H viewers. Since 425092, += this answer checks.

21. Let the x = number of Republicans. Then 6the x += number of Democrats. Since the total number of Democrats and Republicans was 98, we can write the equation (6)98.xx++=

Solve the equation.

2698

292 46 x x x += = = Since 46,652.xx=+=

There were 46 Republicans and 52 Democrats. Since 465298, += this answer checks.

22. Let the x = number of Democrats. Then 32the x += number of Republicans. Since the total number of Democrats and Republicans was 432, we can write the equation (32)432.xx++=

Solve the equation.

232432

2400

200 x x x += = = Since 200,32232.xx=+=

There were 200 Democrats and 232 Republicans. Since 200232432, += this answer checks.

23. Let revenue x = from ticket sales for Madonna. Then 29the x −= revenue from the ticket sales for Bruce Springsteen.

Since the total revenue from ticket sales was $427 (all numbers in millions), we can write the equation (29)427.xx+−=

Solve the equation.

24. Let the x = number of Toyota Camry sales. Then 27 the x −= number of Honda Accord sales.

Since the total number of sales was 691 (all numbers in thousands), we can write the equation (27)691.xx+−=

Solve the equation.

227691

2718

359 x x x −= = =

Since 359,27332.xx=−=

There were 359 thousand Toyota Camry sales and 332 thousand Honda Accord sales. Since 332 is 27 less than 359 and 359332691, += this answer checks.

25. Let the x = number of games the Heat lost. Then 42the x += number of games the Heat won.

Since the total number of games played was 82, we can write the equation (42)82.xx++=

Solve the equation.

5282

580

16 x x x += = =

Since 16,4266.xx=+=

The Heat won 66 games and lost 16 games. Since 661682, += this answer checks.

26. Let the x = number of games the Indians lost. Then 248the x −= number of games the Indians won.

229427

2456

228 x x x −= = =

Since 228,29199.xx=−=

Madonna took in $228 million and Bruce Springsteen took in $199 million. Since 199 is 29 less than 228 and 228199427, += this answer checks.

Since the total number of games played was 162, we can write the equation (248)162.xx+−=

Solve the equation.

70 x x x −= = =

348162

3210

Since 70,24892.xx=−=

The Indians won 92 games and lost 70 games. Since 9270162, += this answer checks.

27. Let the x = number of mg of vitamin C in a one-cup serving of pineapple juice.

Then 43the x −= number of mg of vitamin C in a one-cup serving of orange juice. Since the total amount of vitamin C in a serving of the two juices is 122 mg, we can write (43)122.xx+−=

Solve the equation.

53122

25 x x x −= = =

5125

Since 25,4397.xx=−=

A one-cup serving of pineapple juice has 25 mg of vitamin C and a one-cup serving of orange juice has 97 mg of vitamin C. Since 97 is 3 less than four times 25 and 2597122, += this answer checks.

28. Let the x = number of calories in a one-cup serving of tomato juice. Then 39the x += number of calories in a onecup serving of pineapple juice. Since the total number of calories in a serving of the two juices is 173, we can write (39)173.xx++=

Solve the equation.

Since 22 168,(168)112. 33 xx===

There were 112 DVDs sold.

30. Letthe x = number of calories burned doing aerobics. Then 2 the 5 x = number of calories burned doing weight training. The total number of calories burned was 371, so 2 371. 5 xx+= Solve the equation.

41 x x x += = =

49173

4164

Since 41,39132.xx=+=

A one-cup serving of tomato juice has 41 calories and a one-cup serving of pineapple juice has 132 calories. Since 132 is 9 more than three times 41 and 41132173, += this answer checks.

29. Let the x = number of Blu-Ray discs sold. Then 2 the 3 x = number of DVDs sold.

The total number of Blu-Rays and DVDs sold was 280, so 2 280. 3 xx+=

Since 22 265,(265)106. 55 xx===

Weight training burns 106 calories.

31. Let the x = number of kg of onions. Then 6.6the x = number of kg of grilled steak. The total weight of these two ingredients was 617.6 kg, so 6.6617.6.xx+= Solve the equation. 16.6617.6 7.6617.6

Since 617.6617.6 ,6.66.6536.3. 7.67.6 xx

To the nearest tenth of a kilogram, 81.3 kg of onions and 536.3 kg of grilled steak were used to make the taco.

32. Let the x = population of China. Then 0.9the x = population of India. The combined population of the two countries was 2.5 billion, so 0.92.5.xx+= Solve the equation. 10.92.5 1.92.5 2.525 1.919 xx x x += = == Since 252545 1.3,0.90.91.2. 191938 xx

To the nearest tenth of a billion, the population of China was 1.3 billion and the population of India was 1.2 billion.

33. Let the x = value of the 1945 nickel. Then 2the x = value of the 1950 nickel. The total value of the two coins is $24.00, so 224.xx+= Solve the equation.

324 8Divideby8. x x = =

Since 8,22(8)16.xx===

The value of the 1945 Philadelphia nickel is $8.00 and the value of the 1950 Denver nickel is $16.00.

34. Let the x = number of pounds of nickel. Then 3the x = number of pounds of copper. The total number of pounds would be 560, so 3560.xx+= Solve the equation.

4560

560 140 4 x x = ==

Since 140,33(140)420.xx=== To make 560 pounds of five-cent coins, use 420 pounds of copper.

35. Let the x = number of ounces of rye flour. Then 4the x = number of ounces of whole-wheat flour.

The total number of ounces would be 32, so 432.xx+=

Solve the equation.

532

32 6.4 5 x x = ==

Since 6.4,44(6.4)25.6.xx=== To make a loaf of bread weighing 32 oz, use 6.4 oz of rye flour and 25.6 oz of whole-wheat flour.

36. Let the x = number of milligrams of inert ingredients.

Then 9 the x = number of milligrams of active ingredients.

The total number of milligrams would be 250, so 9250.xx+= Solve the equation.

10250

250 25 10 x x = ==

Since 25,99(25)225.xx=== In a single 250-mg caplet, there would be 25 mg of inert ingredients and 225 mg of active ingredients.

37. Let the x = number of tickets booked on United Airlines.

Then 7the x += number of tickets booked on American Airlines, and 24the x += number of tickets booked on Southwest Airlines.

The total number of tickets booked was 55, so (7)(24)55.xxx++++=

Solve the equation.

41155

444

44 11 4 x x x += = ==

Since 11,711718xx=+=+= and 242(11)426. x +=+= He booked 18 tickets on American, 11 tickets on United, and 26 tickets on Southwest.

38. Let the x = number of hours making telephone calls.

Then 0.5the x += number of hours writing emails, and 2 the x = number of hours attending meetings.

The total number of hours is 7.5, so (0.5)27.5.xxx +++=

Solve the equation.

40.57.5

47 7 1.75 4 x x x += = ==

Since 1.75,0.51.750.52.25xx=+=+= and 22(1.75)3.5. x == In her 7.5 hour-day, she spent 1.75 hours making telephone calls, 2.25 hours writing e-mails, and 3.5 hours attending meetings.

39. Let the x = length of the shortest piece. Then 5the x += length of the middle piece, and 9the x += length of the longest piece.

The total length is 59 inches, so (5)(9)59.xxx++++=

Solve the equation.

41. Letthe x = number of silver medals.

Then the x = number of bronze medals, and 17the x += number of gold medals.

The total number of medals earned by the United States was 104, so (17)104.xxx+++=

Solve the equation.

317104

387 29 x x x += = =

Since 29, x = 1746. x +=

The United States earned 46 gold medals, 29 silver medals, and 29 bronze medals. The answer checks since 462929104. ++=

42. Letthe x = number of bronze medals.

Then 4the x += number of silver medals, and 15the x += number of gold medals.

The total number of medals earned by China was 88, so (4)(15)88.xxx++++=

Solve the equation.

31988

369 23 x x x += = =

31459

345 15 x x x += = =

Since 15, 520 xx=+= and + 924. x =

The shortest piece should be 15 inches, the middle piece should be 20 inches, and the longest piece should be 24 inches. The answer checks since 15202459. ++=

40. Let the x = length of the shortest piece. Then 2the x = length of the middle piece, and 8the x += length of the longest piece. The total length is 3 feet, or 36 inches, so (2)(8)36.xxx+++=

Solve the equation.

Since 23,427xx=+= and 1538. x +=

China earned 38 gold medals, 27 silver medals, and 23 bronze medals. The answer checks since 23273888. ++=

43. Let the x = distance of Mercury from the sun. Then 31.2the x += distance of Venus from the sun, and 57the x += distance of Earth from the sun.

Since the total of the distances from these three planets is 196.2 (all distances in millions of miles), we can write the equation ()()31.257196.2.xxx ++++=

Solve the equation.

388.2196.2 3108

4836

428 7 x x x += = =

Since 7, 214xx== and + 815. x =

The shortest piece should be 7 inches, the middle piece should be 14 inches, and the longest piece should be 15 inches. The answer checks since 7141536. ++=

36 x x x += = =

Mercury is 36 million miles from the sun, Venus is 3631.267.2 += million miles from the sun, and Earth is 365793 += million miles from the sun. The answer checks since 3667.293196.2. ++=

44. Let the x = number of satellites of Saturn. Then 5the x += number of satellites of Jupiter, and 35the x −= number of satellites of Uranus.

Since the total number of satellites is 156, (5)(35)156.xxx+++−=

Solve the equation.

330156

3186

186 62 3 x x x −= = ==

Since 62,567xx=+= and 3527. x −=

Uranus has 27 known satellites.

45. Let the x = measure of angles A and B. Then 60the x += measure of angle C

The sum of the measures of the angles of any triangle is 180°, so (60)180.xxx+++=

Solve the equation.

Since 68,169.xx=+=

The lockers have numbers 68 and 69. Since 6869137, += this answer checks.

48. Let the x = number on the first check. Then 1the x += number on the second check. Since the sum of the numbers is 357, we can write the equation (1)357.xx++=

Solve the equation.

21357

2356

356 178 2 x x x += = ==

Since 178,1179.xx=+=

The checkbook check numbers are 178 and 179. Since 178179357, += this answer checks.

49. Because the two pages are back-to-back, they must have page numbers that are consecutive integers.

40 x x x += = =

360180

3120

Angles A and B have measures of 40 degrees, and angle C has a measure of 4060100 += degrees. The answer checks since 4040100180. ++=

46. Let the x = measure of angles B and C

Then 141the x += measure of angle A

The sum of the measures of the angles of any triangle is 180°, so (141)180. xxx+++=

Solve the equation.

Let the x = lesser page number.

Then 1the x += greater page number. (1)203

21203

2202

101 xx x x x ++= += = =

Since 101,1102.xx=+=

The page numbers are 101 and 102. This answer checks since the sum is 203.

3141180

339 13 x x x += = =

Since 13,141154.xx=+=

The measure of angle A is 154°; the measures of angles B and C are both 13°. The answer checks since 1541313180. ++=

47. Let the x = number on the first locker. Then 1the x += number on the next locker. Since the numbers have a sum of 137, we can write the equation (1)137.xx++=

Solve the equation.

21137

2136

136 68 2 x x x += = ==

50. Let the x = first apartment number. Then 1the x += second apartment number. Since the sum of the numbers is 59, we have (1)59.xx++=

Solve the equation.

2159 258 29 x x x += = =

Since 29,130.xx=+=

The apartment numbers are 29 and 30. Since 293059, += this answer checks.

51. Let the x = lesser even integer.

Then 2the x += greater even integer.

“The lesser added to three times the greater gives a sum of 46” can be written as 3(2)46. 3646 4646 440 10 xx xx x x x ++= ++= += = =

Since 10,212.xx=+=

The integers are 10 and 12. This answer checks since 103(12)46. +=

52. Let the x = lesser even integer.

Then 2the x += greater even integer.

“Six times the lesser added to the greater gives a sum of 86” can be written as 6286. xx++= 7286 784 12 x x x += = =

Since 12,214.xx=+=

The integers are 12 and 14. This answer checks since 6 times 12 is 72, plus 14 is 86.

53. Let the x = lesser odd integer.

Then 2the x += greater odd integer.

“59 more than the lesser is 4 times the greater” can be written as () 5942.xx+=+

5948 5938 513 17 xx x x x +=+ =+ = =

Since 17,219.xx=+=

The integers are 17 and 19. This answer checks since 175976 += and 41976. ⋅=

54. Let the x = lesser odd integer.

Then 2the x += greater odd integer.

“Twice the greater is 17 more than the lesser” can be written as 2(2)17. xx+=+ 2417 417 13. xx x x +=+ += =

Since 13,215.xx=+=

The integers are 13 and 15. This answer checks since 2(15)1317. =+

55. Let the x = lesser integer.

Then 1the x += greater integer. 3(1)43 3343 4343 440 10 xx xx x x x ++= ++= += = =

Since 10,111.xx=+=

The integers are 10 and 11. This answer checks since 103(11)43. +=

56. Let the x = lesser integer.

Then 1the x += greater integer. 53(1)59 53359 8359 856 7 xx xx x x x ++= ++= += = =

Since 7,18.xx=+=

The integers are 7 and 8. This answer checks since 5(7)3(8)352459. +=+=

57. Let the x = first even integer.

Then 2the x += second even integer, and 4the x += third even integer. (2)(4)60 3660 354 18 xxx x x x ++++= += = =

Since 18,220xx=+= and 422. x +=

The first even integer is 18. This answer checks since 18202260. ++=

58. Let the x = first odd integer.

Then 2the x += second odd integer, and 4the x += third odd integer. (2)(4)69 3669 363 21 xxx x x x ++++= += = =

Since 21,223xx=+= and 425. x +=

The third odd integer is 25. This answer checks since 21232569. ++=

59. Let the x = first odd integer.

Then 2the x += second odd integer, and 4the x += third odd integer.

63. Let the x = measure of the angle.

Then 180the x −= measure of its supplement. The phrase “supplement is eight times its measure” can be written as 1808. xx−=

2[(4)6][2(2)]23

2(2)2423

24319 419 15 xxx xxx xx x x +−=++− −=++− −=− −=− =

Since 15,217xx=+= and 419. x +=

The integers are 15, 17, and 19.

60. Let the x = first even integer.

Then 2the x += second even integer, and 4the x += third even integer.

24316 416 20 20 xxx xxx xx x x x ++=+− ++=+− +=− −+=− −=− =

(4)3(2)22

43622

Since 20,222xx=+= and 424. x +=

The integers are 20, 22, and 24.

61. Let the x = measure of the angle.

Then 90the x −= measure of its complement.

The phrase “complement is four times its measure” can be written as 904. xx−=

Solve the equation.

905

90 18 5 x x = ==

The measure of the angle is 18°. The complement is 901872, °−°=° which is four times 18°.

62. Let the x = measure of the angle.

Then 90the x −= measure of its complement.

The phrase “complement is five times its measure” can be written as 905. xx−=

Solve the equation.

906

90 15 6 x x = ==

The measure of the angle is 15°. The complement is 901575, °−°=° which is five times 15°.

Solve the equation.

1809

180 20 9 x x = ==

The measure of the angle is 20°. The supplement is 18020160, °−°=° which is eight times 20°.

64. Let the x = measure of the angle. Then 180the x −= measure of its supplement. The phrase “supplement is three times its measure” can be written as 1803. xx−=

Solve the equation.

1804

180 45 4 x x = ==

The measure of the angle is 45°. The supplement is 18045135, °−°=° which is three times 45°.

65. Let the x = measure of the angle. Then 90the x −= measure of its complement, and 180the x −= measure of its supplement. Itsmoretwiceits supplementmeasures39thancomplement.

180392(90) xx ° ↓↓↓↓↓ −=+−

Solve the equation.

180392(90) 180391802 1802192 180219 39 xx xx xx x x −=+− −=+− −=− += = The measure of the angle is 39°. The complement is 903951. °−°=° Now 39° more than twice its complement is 392(51)141, °+°=° which is the supplement of 39° since 18039141. °−°=°

66. Let the x = measure of the angle. Then 180the x −= measure of its supplement, and 90the x −= measure of its complement. The38lessthan3times supplementisitscomplement. 1803(90)38 xx

Solve the equation. 180270338

The measure of the angle is 26. °

67. Let the x = measure of the angle. Then 180the x −= measure of its supplement, and 90the x −= measure of its complement. Remember that difference means subtraction.

The3 times its supplementminuscomplementis10.

Solve the equation. ()() 18039010 180270310

The measure of the angle is 50° The supplement is 18050130 °−°=° and the complement is 905040 °−°=° The answer checks since . 1303(40)10 °−°=°

68. Let the x = measure of the angle. Then 90the x −= measure of its complement, and 180the x −= measure of its supplement. “The sum of the measures of its complement and supplement is 160” ° can be written as (90)(180)160. xx −+−=

Solve the equation. 2270160 2110 55 x x x −+= −=− =

The measure of the angle is 55° The sum of the measures of its complement (905535) °−°=° and its supplement (18055125) °−°=° is 160. °

2.5 Formulas and Additional Applications from Geometry

Classroom Examples, Now Try Exercises

1. 22

12622(25) Let 126 and 25. 126250

762 Subtract50.

38 Divideby2. PLW LPW L L L =+ =+== =+ = =

N1. 22

782(12)2 Let 78 and 12. 78242

542 Subtract24.

27 Divideby2. Pab bPa b b b =+ =+== =+ = =

2. The fence will enclose the perimeter of the rectangular field, so use the formula for the perimeter of a rectangle. Find the length of the field by substituting 800 P = and 50 WL=− into the formula and solving for L 22

80022(50) 80022100 9004

225 PLW LL LL L L =+ =+− =+− = = Since 225,22550175LW==−= The dimensions of the field are 225 m by 175 m. N2. The fence will enclose the perimeter of the rectangular garden, so use the formula for the perimeter of a rectangle. Find the width of the garden by substituting 160 P = and 210LW=− into the formula and solving for W. 22

1602(210)2 1604202 1806

30 PLW WW WW W W =+ =−+ =−+ = = Since 30,2(30)1050WL==−= The dimensions of the garden are 50 ft by 30 ft.

3. Let the s = length of the shortest side, in inches; 5the s += length of the medium side, and (5)16the ss++=+= length of the longest side.

The perimeter is 32 inches, so (5)(6)32.

31132

321 7 sss s s s ++++= += = =

Since 7,512ss=+= and 613. s += The lengths of the sides are 7, 12, and 13 inches. The perimeter is 7121332, ++= as required.

N3. Let the s = length of the medium side, in feet; 1the s += length of the longest side, and 7the s −= length of the shortest side. The perimeter is 30 feet, so (1)(7)30. 3630

336 12 sss s s s +++−= −= = = Since 12,113ss=+= and 75. s −= The lengths of the sides are 5, 12, and 13 feet. The perimeter is 5121330, ++= as required.

4. Use the formula for the area of a triangle.

1 2 1 120(24) Let120,24. 2 12012 10 Abh bAh b b = === = = The length of the base is 10 meters.

N4. Use the formula for the area of a triangle. 1 2 1 77(14) Let77,14 2 777 11 Abh hAh. h h = === = = The height is 11 centimeters.

5. The sum of the measures of the two angles is 180° because together they form a straight angle. (629)(11)180 740180 7140 20 xx x x x +++= += = = If 20,6296(20)29149xx=+=+= and 11201131. x +=+=

The measures of the angles are 149° and 31°

N5. Since the marked angles are vertical angles, they have equal measures. 6288 228 102 5 xx x x x +=− =− = = If 5,626(5)232xx=+=+= and 888(5)832. x −=−=

The measure of the angles is 32°

6. Solve Iprt = for t Divide by , or Iprt pr. prpr II tt prpr = ==

N6. Solve WFd = for F. Divide by , or WFdd. dd WW FF dd = ==

7. Solve 2 22 Srhr =+ππ for h 22 2 2 22 Subtract2. 2 , Divide by2. 2 2 or 2 Srrhr Srhr r hSr r πππ π π π π π −= = =

N7. Solve AxByC += for A Subtract .

Divide by . AxCByBy AxCBy x xx ACBy x =− = =

8. Solve 9 32 5 FC=+ for C. () () () 9

323232Subtract 32. 5 9 32 5 5595

32Multiply by . 9959 532 9 532 5160 ,or 99 FC FC FC F C FF CC −=+− −= ⎛⎞ −=

= ==

N8. Solve 1 () 2 Sabc =++ for a 2Multiply by2.

2 Subtract and . Sabc Sbcabc =++ −−=

9. (a) Solve 24 xy −+= for y 2242Add 2. 24 xyxxx yx −++=+ =+

(b) Solve 39xy+= for y.

39Subtract . 9 Divide by 3. 3 11 3,or 3 33 xyxxx x y yxyx +−=− = =−=−+

N9. (a) Solve 53 xy+= for y. 5535Subtract 5. 35 xyxxx yx +−=− =−

(b) Solve 28xy−= for y 28Subtract . 8 Divide by2. 2 11 4,or 4 22 xyxxx x y yxyx −−=− =− =−+=−

Exercises

1. Carpeting for a bedroom covers the surface of the bedroom floor, so area would be used.

2. Sod for a lawn covers the surface of the lawn, so area would be used.

3. To measure fencing for a yard, use perimeter since you would need to measure the lengths of the sides of the yard.

4. The baseboards for a living room go around the edges of the room. The amount of baseboard needed will be the sum of the lengths of the sides of the room, so perimeter would be used.

5. Tile for a bathroom covers the surface of the bathroom floor, so area would be used.

6. Fertilizer for a garden covers the surface of the garden, so area would be used.

7. To determine the cost of replacing a linoleum floor with a wood floor, use area since you need to know the measure of the surface covered by the wood.

8. Determining the cost of planting rye grass in a lawn for the winter requires finding the amount of surface to be covered, so area would be used.

9. 22;8,5PLWLW =+== 22 2(8)2(5) 1610 26 PLW P =+ =+ =+ =

10. 22; 6, 4 PLWLW =+== 22 2(6)2(4) 128 20 PLW P =+ =+ =+ = 11. 1 ; 8, 16 2 Abhbh=== 1 2 1 (8)(16) 2 64 Abh A = = = 12. 1 ; 10, 14 2 Abhbh=== 1 2 1 (10)(14) 2 (5)(14) 70 Abh A = = = = 13. ; 12, 3, 5 PabcPac =++=== 1235 128 4 Pabc b b b =++ =++ =+ =

14. ; 15, 3, 7 PabcPab =++=== 1537 1510 5 Pabc c c c =++ =++ =+ =

15. ; 252, 45 drtdr=== 25245 25245 4545 5.6 drt t t t = = = =

16. ; 100, 2.5 drtdt===

20. 2; 8.164, 3.14 CrC=== ππ 2 8.1642(3.14) 8.1646.28 1.3

Cr r r r = = = = π

21. 2; 20 CrC==ππ 2 202 10Divide by 2. Cr r r π ππ π = = =

22. 2; 100 CrC==ππ 2 1002

50Divide by 2. Cr r r π ππ π = = =

100(2.5)

1002.5 1002.5 2.52.5 40 drt r r r r = = = = =

17. 1 (); 91, 7, 12 2 AhbBAhb =+=== 1 () 2 1 91(7)(12) 2 182(7)(12) 1 12(182) 7 261214 AhbB B B B B =+ =+ =+ += =−=

18. 1 (); 75, 19, 31 2 AhbBAbB =+=== 1 () 2 1 75(1931) 2 150(50) 150 3 50 AhbB h h h =+ =+ = ==

19. 2; 16.328, 3.14 CrC=== ππ 2 16.3282(3.14) 16.3286.28 2.6 Cr r r r = = = = π

23. 2 ; 4, 3.14 Arr=== ππ 2 2 3.14(4) 3.14(16) 50.24 Ar A = = = = π

24. 2 ; 12, 3.14 Arr=== ππ 2 2 3.14(12) 3.14(144) 452.16 Ar A = = = = π

25. 2; 120, 10 SrhSh === ππ () 2 120210 12020 6Divide by 20. Srh r r r π ππ ππ π = = = =

26. 2; 720, 30 SrhSh === ππ () 2 720230 72060 12Divide by 60. Srh r r r π ππ ππ π = = = =

27. ; 10, 5, 3 VLWHLWH ==== (10)(5)(3) 150 VLWH V = = =

28. ; 12, 8, 4 VLWHLWH ====

(12)(8)(4) 384 VLWH V = = =

29. 1 ; 12, 13 3 VBhBh=== 1 3 1 (12)(13) 3 52 VBh V = = =

30. 1 ; 36, 4 3 VBhBh=== 1 3 1 (36)(4) 3 48 VBh V = = =

31. 3 4 ; 12, 3.14 3 Vrrππ === 3 3 4 3 4 (3.14)(12) 3 4 (3.14)(1728) 3 7234.56 Vr V π = = = =

32. 3 4 ; 6, 3.14 3 Vrrππ === 3 3 4 3 4 (3.14)(6) 3 4 (3.14)(216) 3 904.32 Vr V π = = = =

33. ; $7500, 4%, 2yr Iprtprt ====

()()() $75000.042 $600 Iprt = = =

34. ; $3600, 3%, 4yr Iprtprt ====

$432 Iprt = = =

()()() $36000.034

35. ; $33, 2%, 3yr IprtIrt ====

$550 Iprt p p p = = = =

$330.023

$330.06

()() ()()

36. ; $270, 5%, 6yr IprtIrt ====

$2700.056

$2700.3

$900 Iprt p p p = = = =

()()() ()()

37. ; $180, $4800,2.5%IprtIpr ====

$180$120 1.5yr Iprt t t t = = = =

$180$48000.025

()()() ()()

38. ; $162, $2400,1.5%IprtIpr ====

$162$36 4.5yr Iprt t t t = = = =

()()() ()() $162$24000.015

39. 22

542(9)2Let9. 542182 54418 364 9 PLW WWLW WW W W W =+ =++=+ =++ =+ = =

The width is 9 inches and the length is 9918 += inches.

40. 22

6222(3)Let3. 62226 6246 684 17 PLW LLWL LL L L L =+ =+−=− =+− =− = =

The length is 17 feet and the width is 17314 −= feet.

41. 22

362(32)2Let 32. 36642 3684 328 4 PLW WWLW WW W W W =+ =++=+ =++ =+ = =

The width is 4 meters and the length is 3(4)214 += meters.

42. 22

3622(218)Let 218. 362436 36636 726 12 PLW LLWL LL L L L =+ =+−=− =+− =− = =

The length is 12 yards and the width is 2(12)186 −= yards.

43. Let the s = length of the shortest side, in inches; 2the s += length of the medium side, and 3the s += length of the longest side. The perimeter is 20 inches, so (2)(3)20.

45. Let the s = length of the two sides that have equal length, in meters, and 24the s −= length of the third side. The perimeter is 24 meters, so (24)24.

4424

428 7 sss s s s ++−= −= = = Since 7, 2410.ss=−= The lengths of the sides are 7, 7, and 10 meters. The perimeter is 771024, ++= as required.

46. Let the s = length of the shortest side, in yards; 2the s = length of the medium side, and 37the s −= length of the longest side. The perimeter is 47 yards, so 2(37)47.

6747

654 9 sss s s s ++−= −= = =

Since 9, 218ss== and 3720. s −= The lengths of the sides are 9, 18, and 20 yards. The perimeter is 9182047, ++= as required.

3520

315 5 sss s s s ++++= += = =

Since 5,27ss=+= and 38. s += The lengths of the sides are 5, 7, and 8 inches. The perimeter is 57820, ++= as required.

44. Let the s = length of the shortest side, in feet. 4the s += length of the medium side, and 2the s = length of the longest side. The perimeter is 28 feet, so (4)(2)28.

47. The page is a rectangle with length 1.5 m and width 1.2 m, so use the formulas for the perimeter and area of a rectangle. 22

PLW P ALW A =+

2(1.5)2(1.2) 32.4 5.4 meters (1.5)(1.2) 1.8 square meters

4428

424 6 sss s s s +++= += = =

Since 6,410ss=+= and 212. s = The lengths of the sides are 6, 10, and 12 feet. The perimeter is 6101228, ++= as required.

48. Since the sand painting is a square with side 12.24 m, use the formulas for the perimeter and area of a square. 4 4(12.24) 48.96 meters Ps P = = = 2 2 (12.24) 149.8176 square meters

As A = = =

To the nearest hundredth of a square meter, the area is 149.82 square meters.

49. Use the formula for the area of a triangle with 70 A = and 14. b = 1 2 1 70(14) 2 707 10 Abh h h h = = = = The height of the sign is 10 feet.

50. Use the formula for the area of a triangle with 96 A = and 12. h = 1 2 1 96(12) 2 966 16 Abh b b b =

The base of the banner is 16 feet.

51. The diameter of the circle is 443 feet, so its radius is 443 221.5ft. 2 = Use the area of a circle formula to find the enclosed area. 2 2 2 (221.5) 154,133.6 ft, Ar π π = = ≈ or about 2 154,000ft (If 3.14 is used for ,π the value is 154,055.465.)

52. The diameter of the circular dome is 630 feet, so its radius is 630 315 ft. 2 = Use the circumference of a circle formula. 2 2(315) 1979.2 ft Cr = = ≈ π π If we use 3.14 for ,π the answer is 1978.2 ft, or about 1978 ft.

53. To find the area of the drum face, use the formula for the area of a circle, 2 Ar = π 2 2 (3.14)(7.87) (3.14)(61.9369) 194.48 Ar A =

The area of the drum face is about 194.48 square feet.

Use the circumference of a circle formula. 2(3.14)(7.87)

49.4236

The circumference is about 49.42 feet.

54. To find the area of the drum face, use the formula for the area of a circle, 2 Ar = π

132.665 Ar A

2 (3.14)(6.5) (3.14)(42.25)

The area of the drum face is about 132.665 square feet.

Use the circumference of a circle formula.

2(6.5)

If we use 3.14 for ,π the answer is about 40.82 feet.

55. Use the formula for the area of a trapezoid with 115.80,171.00,Bb== and 165.97. h = 1 () 2 1 (115.80171.00)(165.97) 2 1 (286.80)(165.97) 2

23,800.098

ABbh =+ =+ = =

To the nearest hundredth of a square foot, the combined area of the two lots is 23,800.10 square feet.

56. Let the A = area of Lot A. Use the formula for the area of a trapezoid with 82.05,26.84,Bb== and 165.97. h = 1 () 2 1 (82.0526.84)(165.97) 2 9036.23665

ABbh =+ =+ =

To the nearest hundredth of a square foot, the area is 9036.24 square feet.

57. The girth is 41872 ⋅= inches. Since the length plus the girth is 108, we have 108 72108 36in. LG L L += += =

VLWH = = =

The volume of the box is 3 (36)(18)(18) 11,664in.

58. To find the volume of the sandwich, use the formula for the volume of a rectangular solid. 11 (12)(12)1 24 12 VLWH = ⎛⎞ = ⎜⎟ ⎝⎠ = 6 12 1 ⋅ 1 35 1 24 ⋅ 2 1

210 / =

The volume of the sandwich was 210 cubic feet.

59. The two angles are supplementary, so the sum of their measures is 180. ° (1)(456)180 555180 5235 47 xx x x x ++−= −= = = Since 47,147148xx=+=+= and 4564(47)56132. x −=−=

The measures of the angles are 48° and 132. °

60. In the figure, the two angles are supplementary, so their sum is 180. ° (107)(73)180 1710180 17170 10 xx x x x +++= += = = Since 10,10710(10)7107xx=+=+= and 737(10)373. x +=+=

The two angle measures are 107° and 73. °

61. In the figure, the two angles are complementary, so their sum is 90. ° (81)590 13190 1391 7 xx x x x −+= −= = = Since 7,818(7)155xx=−=−= and 55(7)35. x ==

The two angle measures are 55° and 35. °

62. In the figure, the two angles are complementary, so their sum is 90. ° 4(313)90 71390 777 11 xx x x x ++= += = =

Since 11,44(11)44xx=== and 3133(11)1346. x +=+=

The two angle measures are 44° and 46. °

63. The two angles are vertical angles, which have equal measures. Set their measures equal to each other and solve for x 5129221 312921 3108 36 xx x x x −=− −=− = = Since 36, 51295(36)12951xx=−=−= and 2212(36)2151. x −=−=

The measure of each angle is 51. °

64. The two angles are vertical angles, which have equal measures. Set their measures equal to each other and solve for x. 75345 4545 440 10 xx x x x +=+ += = =

The measure of the first angle is 7(10)575; +=° the measure of the second angle is 3(10)45, + which is also 75. °

65. The angles are vertical angles, so their measures are equal. 1231015 2315 218 9 xx x x x −=+ −= = =

Since 9,12312(9)3105xx=−=−= and 101510(9)15105. x +=+=

The measure of each angle is 105. °

66. The angles are vertical angles, which have equal measures. Set 1137 x equal to 727 x + and solve.

74. Solve VLWH = for W

1137727

43727

464 16 xx x x x −=+ −= = =

Since 16,113711(16)37139xx=−=−= and 7277(16)27139. x +=+=

The angles both measure 139. °

67. Solve drt = for t.

Divide by . ,or drt r rr dd tt rr = ==

68. Solve drt = for r.

Divide by . ,or drt t tt dd rr tt = ==

69. Solve Abh = for b

Divide by . ,or Abh h hh AA bb hh = ==

70. Solve ALW = for L. Divide by . ,or ALW W WW AA LL WW = ==

71. Solve Cd = π for d

Divide by . ,or Cd CC dd π π ππ ππ = ==

72. Solve 4 Ps = for s. 4

Divide by 4. 44 ,or 44 Ps PP ss = ==

73. Solve VLWH = for H.

Divide by . ,or VLWH LW LWLW VV HH LWLW = ==

Divide by . ,or VLWH LH LHLH VV WW LHLH = ==

75. Solve Iprt = for r

Divide by . ,or Iprtpt ptpt II rr ptpt = ==

76. Iprt = for p.

Divide by . ,or Iprtrt rtrt II pp rtrt = ==

77. Solve 1 2 Abh = for h 1 22Multiply by2. 2 2 2

Divide by . 22 , or Abh Abh Abh b bb AA hh bb ⎛⎞ = ⎜⎟ ⎝⎠ = = ==

78. Solve 1 2 Abh = for b 1 22Multiplyby2. 2 2 2

Divide by . 22 , or Abh Abh Abh h hh AA bb hh ⎛⎞ = ⎜⎟ ⎝⎠ = = ==

79. Solve 2 1 3 Vrh π = for h. 2 2 2 2 22 22 1 33Multiplyby3. 3 3 3

Divide by . 33 , or Vrh Vrh Vrh r rr VV hh rr π π π π ππ ππ ⎛⎞ = ⎜⎟ ⎝⎠ = = ==

80. Solve 2 Vrh = π for h. 2 2 22 22 Divide by . ,or Vrh r rr VV hh rr π π ππ ππ = ==

81. Solve Pabc =++ for b Subtractand. ,or Pacabcacac PacbbPac −−=++−− −−==−−

82. Solve Pabc =++ for a. Subtractand. ,or Pbcabcbcbc PbcaaPbc −−=++−− −−==−−

83. Solve 22 PLW =+ for W 2222Subtract 2. 22 22 Divide by2. 22 22 ,or 22 PLLWLL PLW PLW PLPL WW −=+− −= = ==

84. Solve Apprt =+ for r Subtract. Divideby. ,or Appprtpp Apprt Apprtpt ptpt ApAp rr ptpt −=+− −= = ==

85. Solve ymxb =+ for m Subtract . Divide by . , or ybmxbbb ybmx ybmx x xx ybyb mm xx −=+− −= = ==

86. Solve ymxb =+ for x

Subtract . Divide by . , or ybmxbbb ybmx ybmx m mm ybyb xx mm −=+− −= = ==

87. Solve AxByC += for y

Subtract . Divide by . ByCAxAx ByCAxB BB CAx yB =− = =

88. Solve AxByC += for x Subtract . Divide by . AxCByBy AxCByA AA CBy x A =− = =

89. Solve () 1 MCr =+ for r Distributive property Subtract . Divideby. , or MCCr MCCrC MCCr C CC MCMC rr CC =+ −= = == Alternativesolution: () 1 1Divide by . 1Subtract1. MCr M rC C M r C =+ =+ −=

90. Solve () 1 Aprt =+ for t Distributive property Subtract. Divideby. ,or Apprt Appprtpp Apprt Apprt pr prpr ApAp tt prpr =+ −=+− −= = ==

91. Solve 2()Pab =+ for a. 22Distributive property 22Subtract 2. 22 Divideby2. 22 2 2 Pab Pbab Pba Pb a =+ −= = =

92. Solve 2()Pab =+ for b () 2 Divideby2. 22 2

Subtract. 2 2 , or 22 Pab P ab P aba PPa abb + = =+ −= −==

The second form of the answer results from first using the distributive property, then subtracting 2, a and then dividing by 2— similar to the method used in Exercise 91.

93. Solve 1 () 2 Sabc =++ for b.

2Multiply by2. 2 Subtract and . Sabc Sacbac =++ −−=

94. Solve 1 () 2 Sabc =++ for c

2Multiply by2. 2 Subtract and . Sabc Sabcab =++ −−=

95. Solve () 5 32 9 CF=− for F () () 5 32 9 995 32 559 9 32 5 9 323232 5 9 32 5 CF CF CF CF CF =− =⋅− =− +=−+ +=

96. Solve () 1 2 AhbB =+ for b. () () 1 2 1 22 2

=+ −=+− = =

2Distributive Prop. 2Subtract . 2 Divide by . 2 AhbB AhbB AhbhB AhBhbhBhBhB AhBhb h hh AhB b h =+ ⎛⎞=+⎜⎟

97. Solve 64 xy+= for y 6646Subtract 6. 46 xyxxx yx +−=− =−

98. Solve 36 xy+= for y 3363Subtract 3. 63 xyxxx yx +−=− =−

99. Solve 52 xy−= for y. 5525Subtract 5. 25

Divide by1. 11 25,or 52 xyxxx yx yxyx −−=− =− =−+=−

100. Solve 41 xy−= for y 4414Subtract 4. 14

Divide by1. 11 14,or 41 xyxxx yx yxyx −−=− =− =−+=−

101. Solve 3515 xy −+=− for y 353153Add 3. 5153

Divide by 5. 55 33 3,or 3 55 xyxxx yx yxyx −++=−+ −+ = =−+=−

102. Solve 239 xy −+=− for y 23292Add 2. 392 Divide by 3. 33 22 3,or 3 33 xyxxx yx yxyx −++=−+ −+ = =−+=−

103. Solve 312xy−= for y 312Subtract . 312 Divide by 3. 33 11 4,or 4 33 xyxxx yx yxyx −−=− =− =−+=−

104. Solve 510xy−= for y 510Subtract . 510 Divide by 5. 55 11 2,or 2 55 xyxxx yx yxyx −−=− =− =−+=−

2.6 Ratio, Proportion,

and Percent

Classroom Examples, Now Try Exercises

1. (a) To find the ratio of 3 days to 2 weeks, first convert 2 weeks to days.

2weeks2714days =⋅=

The ratio of 3 days to 2 weeks is then 3days3days3 2weeks14days14 ==

(b) 4days42496hours =⋅=

The ratio of 12 hr to 4 days is then 12hr12hr121 4days96hr968 ===

N1. (a) The ratio of 7 inches to 4 inches is 7inches7 . 4inches4 =

(b) 45 453 seconds 604 == minute

The ratio of 45 seconds to 2 minutes is then 45seconds3313 2. 2minutes4428 =÷=⋅=

2. The results in the following table are rounded to the nearest thousandth.

N2. The results in the following table are rounded to the nearest thousandth.

Size Unit Cost (dollars per oz)

75 oz $8.94 $0.119() 75 = ∗

100 oz

$13.97 $0.140 100 = 150 oz $19.97 $0.133 150 =

Because the 75-oz size produces the lowest unit cost, it is the best buy. The unit cost, to the nearest thousandth, is $0.119 per oz.

3. (a) 2162 1545 =

Compare the cross products. 2145945 1562930 ⋅= ⋅=

The cross products are different, so the proportion is false.

(b) 1391 17119 =

Check to see whether the cross products are equal. 131191547 17911547 ⋅= ⋅=

The cross products are equal, so the proportion is true.

N3. (a) 133 3100 = Compare the cross products. 1100100 33399 ⋅= ⋅=

Because the 36-oz size produces the lowest unit cost, it is the best buy. The unit cost, to the nearest thousandth, is $0.108 per oz.

The cross products are different, so the proportion is false.

(b) 416 1352 =

Check to see whether the cross products are equal. 452208 1316208 ⋅= ⋅=

The cross products are equal, so the proportion is true.

4. 35 642 x =

N5. 332 64

42635Cross products

635 Divide by 42. 42 657 Factor. 67 5Cancel. x x =⋅ ⋅ = ⋅⋅ = ⋅ =

The solution set is {5}.

Note: We could have multiplied 635 to get 210 and then divided 210 by 42 to get 5. This may be the best approach if you are doing these calculations on a calculator. The factor and cancel method is preferable if you’re not using a calculator.

N4. 9 756 x = 7956Cross products

956 Divide by7. 7 978 Factor. 7 72Cancel. x x =⋅ = = =

The solution set is {72}.

Note: We could have multiplied 956 ⋅ to get 504 and then divided 504 by 7 to get 72. This may be the best approach if you are doing these calculations on a calculator. The factor and cancel method is preferable if you’re not using a calculator.

5. 62 25

5(6)2(2)Crossproducts

5304Distributive property 526Subtract30. 26 Divideby5. 5 x x x x x + = += += =− =− The solution set is 26 . 5 ⎧⎫

4(3)6(32)Crossproducts 4121812Distributive property 141212Subtract18. 1424Add12. 24 Divide by14. 14 12 7 kk

The solution set is 12 7 ⎧ ⎫ ⎨ ⎬ ⎩⎭

6. Let the x = cost for 16.5 gallons. $49.68

1216.5

68.31Divide by12. x x x x = = = = It would cost $68.31.

1216.5(49.68)Cross products 12819.72Multiply.

N6. Let the x = cost for 27 gallons. $69.80 2027

2027(69.80)Cross products 201884.60Multiply. 94.23Divide by20. x x x x = = = = It would cost $94.23.

7. (a) What is 3% of 80? 3%800.03802.4 ⋅=⋅=

(b) 16% of what number is 12? As in Example 7(b), let n denote the number. 0.1612 12 Divideby0.16. 0.16

75Simplify. n n ⋅= = = 16% of 75 is 12.

As in Example 7(c), let p denote the percent. 9075 90 1.2120% 75 p p =⋅ === 90 is 120% of 75.

N7. (a) What is 20% of 70?

20%700.207014 ⋅=⋅=

(b) 40% of what number is 130?

As in Example 7(b), let n denote the number.

0.40130

n n ⋅= = = 40% of 325 is 130.

130 Divide by0.40. 0.40 325Simplify.

As in Example 7(c), let p denote the percent.

121484

121 0.2525% 484 p p =⋅ === 121 is 25% of 484.

8. We can think of this problem as “What percent of $3800 is $912?”

2. Since there are 14 days in 2 weeks, the ratio of 4 days to 2 weeks can be represented by 4 , 14

choice C, or 2 , 7 choice E.

3. The ratio of 40 miles to 30 miles is 40 miles404 . 30 miles303 ==

4. The ratio of 60 feet to 70 feet is 60 feet606 . 70 feet707 ==

5. The ratio of 120 people to 90 people is 120 people4304 . 90 people3303 == ⋅

6. The ratio of 72 dollars to 220 dollars is 72 dollars7218418 220 dollars22055455 ===

Let p denote the number. 3800912 912 Divide by3800. 3800 0.2424%Simplify.

p p ⋅= = ==

Mark’s rent is 24% of his monthly income.

N8. We can think of this problem as “48 is what percent of 120?”

Let p denote the percent. 48120

48 0.4040% 120 p p =⋅ ===

The coat costs 40% of the regular price, so the savings is 100%40%60% −= of the regular price.

Exercises

1. (a) 75 to 100 is , 753 1004 = or 3 to 4.

The answer is C.

(b) 5 to 4, or , 55315 44312 == ⋅ or 15 to 12.

The answer is D.

The answer is B.

(d) 4 to 5, or , 442080 5520100 == ⋅ or 80 to 100.

The answer is A.

7. To find the ratio of 20 yards to 8 feet, first convert 20 yards to feet. 3feet 20 yards20 yards60 feet 1yard =⋅=

The ratio of 20 yards to 8 feet is then 60 feet6015415 . 8 feet8242 === ⋅

8. First convert 8 feet to inches. 8 feet81296 inches =⋅= The ratio of 30 inches to 8 feet is then 30 inches30565 . 96 inches9616616 ⋅ === ⋅

9. Convert 2 hours to minutes. 60 minutes 2 hours2 hours 1 hour 120 minutes =⋅ = The ratio of 24 minutes to 2 hours is then 24 minutes241241 120 minutes1205245 ⋅ ===

10. To find the ratio of 16 minutes to 1 hour, first convert 1 hour to minutes. 1 hour = 60 minutes

The ratio of 16 minutes to 1 hour is then 16 minutes16444 60 minutes6015415 ⋅ ===

11. 2 yards236 feet 6 feet61272 inches =⋅= =⋅=

The ratio of 60 inches to 2 yards is then 60 inches5125 . 72 inches6126 ⋅ == ⋅

12. 1hour3600 seconds =

The ratio of 720 seconds to 1 hour is then 720 seconds17201 3600 seconds57205 ⋅ ==

13. Find the unit price for each size.

The 10-lb size is the best buy.

14. Find the unit price for each size.

Size

per lb) 23

17. Find the unit price for each

The 48-oz size is the best buy.

15. Find the unit price for each size.

The

The 33.8-oz size is the best buy.

19. Find the unit price for each size.

The 32-oz size is the best buy.

20. Find the unit price for each size.

The

The 32-oz size is the best buy.

The

21. Find the unit price for each size.

The 263-oz size is the best buy.

22. Find the unit price for each size.

The 24-oz size is the best buy.

23. Check to see whether the cross products are equal.

556280 358280 ⋅= ⋅=

The cross products are equal, so the proportion is true.

24. Check to see whether the cross products are equal. 42184 12784 ⋅= ⋅=

The cross products are equal, so the proportion is true

25. Check to see whether the cross products are equal. 120101200 827574 ⋅= ⋅=

The cross products are different, so the proportion is false

26. Check to see whether the cross products are equal.

271102970 160182880 ⋅= ⋅= The cross products are different, so the proportion is false

27. Check to see whether the cross products are equal. 1 105 2 515 ⋅= ⋅=

The cross products are equal, so the proportion is true.

28. Check to see whether the cross products are equal. 1 186 3 616 ⋅= ⋅=

The cross products are equal, so the proportion is true.

29. 204(175)Cross products 20700 20700 Divide by20. 2020 35 k k k k = = = = The solution set is { }35

30. 18 64 4618Cross products 4108 4108 Divide by 4. 44 27 x x x x x = ⋅=⋅ = = =

The solution set is { }27

31. 49 568

5649(8)Cross products 56392 56392 Divide by56. 5656 7 z z z z z = = = = =

The solution set is { }7

32. 20 10080

1002080Cross products are equal. 1001600 1001600 Divide by 100. 100100 16 z z z z z = ⋅=⋅ = = = The solution set is { }16

33. 15 2416

1624(15)Cross products

16360 16360

Divide by 16. 1616 45845 282 x x x x x = = = = ==

The solution set is 45 2 ⎧⎫ ⎨⎬ ⎩⎭

34. 12 430

38. 2839 43

39.

304(12)Cross products

3048

3048 Divide by 30.

3030 868 565 x x x x x = = = = ==

The solution set is 8 5 ⎧⎫ ⎨⎬ ⎩⎭

35. 1 23

40.

3(28)4(39)Cross products 6241236Distributive property 62436Subtract 12.

660Subtract 24. 10Divide by6. rr rr rr rr r r +− =

The solution set is { }10

()() 5132 63 351632Cross products 1531812Distributive property 3312Subtract 18. 315Subtract 3. 5Divide by 3. kk kk kk kk k k +− = +=− +=− −+=− −=− =−

The solution set is { }5

32(1)Cross products

322Distributive property

2Subtract 2. zz zz zz zz + = =+ =+ =

The solution set is { }2

36. 2 52

()() 410 68 84610Cross products 832660Distributive property 23260Subtract 6. 228Subtract 32. 14Divide by 2. xx xx xx xx x x ++ = +=+ +=+

The solution set is { }14

41.

25(2)Cross products

2510Distributive property

310Subtract 5. 10

Divide by3. 3 mm mm mm mm m =

The solution set is 10 . 3 ⎧⎫ ⎨⎬

37. 3265 511

()() 271 34 42731Cross products 82833Distributive property 5283Subtract 3. 531Subtract 28. 31 Divide by 5. 5 pp pp pp pp p p +− = +=− +=− +=− =− =−

The solution set is 31 5 ⎧ ⎫ ⎨ ⎬ ⎩⎭

11(32)5(65)Cross products

33223025Distributive property

32225Subtract30. 33Add22. 1Divideby3. yy yy yy yy y y = −=−

The solution set is { }1

42. ()() 324 53 33254Cross products 96205Distributive property 14620Add 5. 1426Add 6. 2613 Divide by 14. 147 mm mm mm mm m m

The solution set is 13 7 ⎧⎫ ⎨⎬ ⎩⎭

43. () () ()() 2443 35 28412

Distributive property 35 5283412Cross products 10401236Dist. prop. 40236Subtract 10. 42Add 36. 4 2Divide by 2. 2 xx xx xx xx xx x x

The solution set is { } 2.

44. () () ()() 9362 62 927612 Distributive property 62 29276612Cross products 18543672Dist. prop. 541872Subtract 18. 1818Add 72. 1Divide by 18. xx xx xx xx xx x x

The solution set is { } 1.

45. Let the x = cost of 24 candy bars. Set up a proportion. $20.00 2416 1624(20) 16480 30 x x x x = = = = The cost of 24 candy bars is $30.00.

46. Let the x = cost of 8 ring tones. Set up a proportion. $30.00 812 128(30) 12240 20 x x x x = = = = The cost of 8 ring tones is $20.00.

47. Let the x = cost of 5 quarts of oil. Set up a proportion.

$14.00 58 85(14) 870 8.75 x x x x = = = = The cost of 5 quarts of oil is $8.75.

48. Let the x = cost of 7 tires. Set up a proportion.

$398.00 74 47(398) 42786 696.50 x x x x = = = = The cost of 7 tires is $696.50.

49. Let the x = cost of 5 pairs of jeans. 9pairs5pairs

$121.50

95(121.50) 9607.5

9607.5 99 67.5 x x x x x = = = = = The cost of 5 pairs is $67.50.

50. Let the x = cost of 11 shirts. 7shirts11shirts

$87.50

711(87.50)

7962.5

7962.5 77 137.5 x x x x x = = = = = The cost of 11 shirts is $137.50.

51. Let the x = cost for filling a 15-gallon tank.

Set up a proportion.

dollars15 gallons

6338.4 56.4 x x x x = = = = It would cost $56.40 to completely fill a 15-gallon tank.

$22.566 gallons

615(22.56)

55. Let the x = distance between Memphis and Philadelphia on the map (in feet). Set up a proportion with one ratio involving map distances and the other involving actual distances.

feet1000 miles

2.4 feet600 miles

1000 2.4600

600(2.4)(1000)

52. Let the x = sales tax on a $120.00 DVD player.

$1.32$16 $$120 1.3216

16120(1.32)

16158.4 9.90 x x x x x = = = = =

The sales tax on a $120 DVD player would be $9.90.

53. Let the x = number of fish in North Bay. Set up a proportion with one ratio involving the sample and the other involving the total number of fish.

7 fish500 fish

700 fish fish

6002400 4 x x x x x = = = = = The distance on the map between Memphis and Philadelphia would be 4 feet.

56. Let the x = number of inches between Mexico City and Cairo on the map.

Set up a proportion. 11inches 3300 miles

7700 miles inches 113300 7700 330011(7700)

7(700)(500)

7350,000 50,000 x x x x =

We estimate that there are 50,000 fish in North Bay.

54. Let the x = number of fish in West Okoboji Lake.

Set up a proportion with one ratio involving the sample and the other involving the total number of fish.

18 fish840 fish

1000 fish fish

330084,700 84,700 3300 847772 ,or25 3333 x x x x x = = = = = ==

Mexico City and Cairo are 2 25 3 inches apart on the map.

57. Let the x = number of inches between St. Louis and Des Moines on the map. Set up a proportion. 8.5 inches1040 miles

inches333 miles

10408.5(333)

18(1000)(840)

18840,000

46,666.6 x x x x = = = =

The approximate fish population of West Okoboji Lake, to the nearest hundred, is 46,700 fish.

10402830.5 2.72 x x x x = = = ≈

St. Louis and Des Moines are about 2.7 inches apart on the map.

58. Let the x = number of inches between Milwaukee and Seattle on the map. Set up a proportion.

8.0 inches912 miles

91215,520 17.0 x x x x = = = ≈

inches1940 miles

9128.0(1940)

Milwaukee and Seattle are about 17.0 inches apart on the map.

59. Let the x = number of inches between Moscow and Berlin on the globe. Set up a proportion.

12.4 inches10,080 km

10,08019,964 1.98 x x x x = = = ≈

inches1610 km

10,08012.4(1610)

Moscow and Berlin are about 2.0 inches apart on the globe.

60. Let the x = number of inches between Paris and Stockholm on the globe. Set up a proportion.

21.5 inches17,615 km

inches1605 km

17,61534,507.5 1.96 x x x

17,61521.5(1605)

Paris and Stockholm are about 2.0 inches apart on the globe.

61. Let the x = number of cups of cleaner. Set up a proportion with one ratio involving the number of cups of cleaner and the other involving the number of gallons of water. 1 10 gallons cups

62. Let the x = number of cups of cleaner. Set up a proportion with one ratio involving the number of cups of cleaner and the other involving the number of gallons of water. 1 15 gallons cups 2 1 1 gallons cup 2 11

The amount of cleaner needed is 3 7 4 cups.

63. Let the x = number of U.S. dollars Ashley exchanged. Set up a proportion. $1.34921 euro dollars300 euros 11.3492(300) 404.76 x x x

⋅= = She exchanged $404.76.

64. Let the x = number of pesos that can be obtained for $65. Set up a proportion. 103.0 pesos8 dollars pesos65 dollars 8103.0(65) 86695 836.875 x x x x = ⋅= = =

To the nearest tenth, one can obtain 836.9 pesos for $65.

65. 3 129 912336 4 x x x = =⋅= = Other possibilities for the proportion are 12951215 , , and 312155 x xx

The amount of cleaner needed is 5 2 8 cups.

8612 872

Other possibilities for the proportion are 64612 , , and 66468 xx x

67. 12 23 321224 8 x x x = =⋅= =

68. 6 48 84624 3 x x x = =⋅= =

69. 1212 158178 815(12)180817(12)204

70. 77 6141114

71. (a)

(b) These two triangles are similar, so their sides are proportional. 18 124 418(12) 4216 54 x x x x = = = = The chair is 54 feet tall.

72. (a)

(b) These two triangles are similar, so their sides are proportional. 5 322 25(32) 2160 80 x x x x = = = = The candle was 80 feet tall.

73. Let the 2001 x = price of electricity. () () 1999 price2001 price 1999 index2001 index 225

166.6177.1 166.6225177.1

225177.1 $239 166.6 x x x = = = =≈

The 2001 price would be about $239.

74. Let the x = 2003 price of electricity. () () 1999 price2003 price 1999 index2003 index 225

166.6184.0 166.6225184.0

225184.0 $248 166.6 x x x = = = =≈

The 2003 price would be about $248.

75. Let the x = 2007 price of electricity. () () 1999 price2007 price 1999 index2007 index 225

166.6207.3 166.6225207.3 225207.3 $280 166.6 x x x = = = =≈

The 2007 price would be about $280.

76. Let the x = 2011 price of electricity. () () 1999 price2011 price 1999 index2011 index 225

166.6224.9 166.6225224.9

225224.9 $304 166.6 x x x = = = =≈

The 2011 price would be about $304.

77. (a) Find the total amount of medication by multiplying.

()() 375mg/day7days2625mg =

(b) Let the x = number of mL of suspension. mg of Amoxiltotal mg of Amoxil mL of suspensiontotal mL of suspension

81. 42% of what number is 294?

As in Example 7(b), let n denote the number.

0.42294

294

n n ⋅= = = 42% of 700 is 294.

Divide by 0.42. 0.42

700Simplify.

125mg2625mg

5mL x = =

5mL

() 125mg2625mg

82. 18% of what number is 108?

As in Example 7(b), let n denote the number. 0.18108

108

12552625 52625 105mL

125 x x x = = ==

Logan’s pharmacist will make 105 mL of Amoxil suspension for the total course of treatment.

78. (a) Find the total amount of medication by multiplying.

()() 900mg/day10days9000mg =

(b) Let the x = number of mL of suspension. mg of Amoxiltotal mg of Amoxil mL of suspensiontotal mL of suspension

n n ⋅= = = 18% of 600 is 108.

Divide by 0.18. 0.18 600Simplify.

83. 120% of what number is 510?

As in Example 7(b), let n denote the number. 1.20510

510

n n ⋅= = = 120% of 425 is 510.

Divide by 1.20. 1.20

425Simplify.

84. 140% of what number is 315?

As in Example 7(b), let n denote the number. 1.40315

315

250mg9000mg

5mL x = =

5mL

() 250mg9000mg

n n ⋅= = = 140% of 225 is 315.

Divide by 1.40. 1.40

225Simplify.

85. 4 is what percent of 50?

As in Example 7(c), let p denote the percent.

25059000 59000 180mL

250 x x x = = ==

Ava’s pharmacist will make 180 mL of Amoxil suspension for the total course of treatment.

79. What is 18% of 780? 18%7800.18780140.4 ⋅=⋅=

80. What is 23% of 480? 23%4800.23480110.4 ⋅=⋅=

450 4 0.088% 50 p p =⋅ === 4 is 8% of 50.

86. 8 is what percent of 64?

As in Example 7(c), let p denote the percent. 864

8 0.12512.5% 64 p p =⋅ === 8 is 12.5% of 64.

87. What percent of 30 is 36?

As in Example 7(c), let p denote the percent.

3630

36 1.2120% 30 p p =⋅ === 36 is 120% of 30.

88. What percent of 48 is 96?

As in Example 7(c), let p denote the percent. 9648

96 2200% 48 p p =⋅ === 96 is 200% of 48.

89. 48 is what percent of 60?

As in Example 7(c), let p denote the percent. 4860

48 0.0880% 60 p p =⋅ === He earned 80% of the total points.

90. 63 is what percent of 75?

As in Example 7(c), let p denote the percent. 6375

63 0.8484% 75 p p =⋅ ===

She scored 84% of the total points.

91. Find the amount of the savings.

$700$504$196 −=

What percent of $700 is $196? As in Example 7(b), let n denote the number. 700196

196 Divide by 700 700 0.28Simplify. n n. ⋅= = =

0.2828(0.01)281%28% ==⋅=

92. Find the amount of the savings.

$980$833$147 −=

What percent of $980 is $147? As in Example 7(b), let n denote the number. 980147

147

Divide by 980 980 0.15Simplify.

n n. ⋅= = = 0.1515(0.01)151%15% ==⋅=

93. What percent of $1500 is $480?

As in Example 7(b), let n denote the number. 1500480

480

94. What percent of $2200 is $154? As in Example 7(b), let n denote the number. 2200154

154

Divide by 2200 2200 0.07Simplify.

n n. ⋅= = = 0.077(0.01)71%7% ==⋅= Lily budgeted 7% of her income for entertainment.

95. 65% of what number is 1950? As in Example 7(b), let n denote the number. 0.651950

1950

n n ⋅= = = She needs $3000 for the car.

Divide by 0.65. 0.65

3000Simplify.

96. 70% of what number is 525? As in Example 7(b), let n denote the number. 0.70525

525

Divide by 0.70 0.70 750Simplify.

n n. ⋅= = = She needs $750 for the apartment deposit.

2.7 Further Applications of Linear Equations

Classroom Examples, Now Try Exercises

1. (a) The amount of pure acid in 40 L of a 16% acid solution is 40L0.166.4 L.

AmountRateAmount ofofof pure solutionconcentrationacid = × ↑↑↑

(b) If $5000 is invested for one year at 4% simple interest, the amount of interest earned is $50000.04$200.

Divide by 1500 1500 0.32Simplify.

n n. ⋅= = = 0.3232(0.01)321%32% ==⋅=

Tyler pays 32% of his income in rent.

InterestInterest Principal rateearned = × ↑↑↑

N1. (a) The amount of pure alcohol in 70 L of a 20% alcohol solution is 70L0.2014 L.

AmountRateAmount ofofof pure solutionconcentrationalcohol = × ↑↑↑

(b) If $3200 is invested for one year at 2% simple interest, the amount of interest earned is $32000.02$64.

InterestInterest Principal rateearned = ×

2. Let the x = number of kilograms of metal that is 40% copper.

Then 80the x += number of kilograms of metal that is 50% copper.

Copper incopper incopper in plus 40% metal70% metalis50% metal.

0.400.70(80)0.0580 xx

Solve the equation.

0.400.70(80)0.5040 xx +=+

Multiply by 10 to clear decimals.

47(80)5400

45605400 560400 160 xx xx x x +=+ +=+ =+ = 160 kilograms of metal that is 40% copper is needed.

Check 160: x =

LS and RS refer to the left side and right side of the original equation.

Solve the equation.

0.700.10(30)0.5015 xx +=+

Multiply by 10 to clear decimals.

71(30)5150 7305150

230150 2120 60 xx xx x x x +=+ +=+ += = = 60 ounces of seasoning that is 70% salt is needed.

Check 60: x =

LS and RS refer to the left side and right side of the original equation.

LS:0.70(60)0.10(30)42345

RS:0.50(6030)0.50(90)45 +=+= +==

3. Let the x = number of gallons of 12% indicator solution.

Then 10the x −= number of gallons of 20% indicator solution. 12%20%14% solutionplussolutionissolution.

0.120.2(10)0.14(10) xx

Solve the equation. ()() 0.120.2100.1410 xx+−= Multiply by 100 to clear decimals. ()() 1220101410 1220020140 8200140 860 60151 7 822 xx xx x x x +−= +−= −+= −=− ===

There needs to be 1 7 2 gallons of the 12% indicator solution.

RS:0.50(16080)0.50(240)120 +=+= +==

LS:0.40(160)0.70(80)6456120

N2. Let the x = number of ounces of seasoning that is 70% salt.

Then 30the x += number of ounces of seasoning that is 50% salt.

Salt insalt insalt in 70%10%50% seasoningplusseasoningisseasoning.

0.700.10(30)0.50(30) xx

Check 1 7: 2 x =

LS and RS refer to the left side and right side of the original equation. () 11 LS:127201079050140 22 RS:1410140

+−=+=

N3. Let the x = number of liters of 25% saline solution.

Then 15the x −= number of liters of 10% saline solution. 25%10%15% solutionplussolutionissolution.

N4. Let the x = amount invested at 3%.

Then 5000the x += amount invested at 4%.

Amount Invested (in dollars) Rate of Interest Interest for One Year

x 0.03 0.03 x

5000 x + 0.04 0.04(5000) x +

0.250.1(15)0.15(15) xx

↓↓↓↓↓ +−=

Solve the equation. ()() 0.250.1150.1515 xx+−=

Multiply by 100 to clear decimals. ()() 2510151515 2515010225 15150225 1575 5 xx xx x x x +−= +−= += = =

There needs to be 5 liters of the 25% saline solution.

Since the total annual interest is $410, the equation is 0.030.04(5000)410. xx++=

Multiply by 100 to eliminate the decimals.

34(5000)100(410)

3420,00041,000 720,00041,000 721,000 3000 xx xx x x x ++= ++= += = =

The financial advisor should invest $3000 at 3% and $8000 at 4%.

5. Let the x = number of quarters.

Then 9the x += number of nickels.

RS:1515225 +−=+= =

Check 5: x = LS and RS refer to the left side and right side of the original equation. ()()() () LS:25510155125100225

4. Let the x = amount invested at 5%.

Then 23000the x += amount invested at 8%.

Amount Invested (in dollars) Rate of Interest Interest for One Year

x 0.05 0.05 x

23000 x + 0.08 0.08(23000) x +

Since the total annual interest was $1710, the equation is 0.050.08(23000)1710. xx++=

Multiply by 100 to eliminate the decimals. 58(23000)100(1710)

Thevaluethevalue ofquartersofnickelsplusis$2.55.

0.250.05(9)2.55 xx

Multiply by 100 to eliminate the decimals. 255(9)255 25 545255 3045255 30210 7 xx xx x x x ++= ++= += = =

Since 7,916xx=+=

The man has 16 nickels and 7 quarters.

N5. Let the x = number of dimes.

Then 10the x += number of quarters.

51624,000171,000

2124,000171,000

21147,000 7000 xx xx x x x ++= ++= += = =

The engineer invested $7000 at 5%.

Thevaluethevalue ofdimesofquartersplusis$5.65.

0.100.25(10)$5.65 xx

++=

Multiply by 100 to eliminate the decimals.

1025(10)565 1025250565 35250565 35315 9 xx xx x x x ++= ++= += = = Since 9,1019.xx=+=

Clayton has 19 quarters and 9 dimes.

6. 100meters 10.3842 9.63seconds d r t ==≈

His rate was about 10.38 meters per second.

N6. 400miles 66.6667 6hours d r t ==≈

The rate was about 66.67 miles per hour.

7. Let the t = time it takes for the planes to be 3290 miles apart. Use the formula . drt = fasterslowertotal 5304103290 9403290 3290 3.5 940 ddd tt t t += += = ==

It will take 3.5 hours for the planes to be 3290 miles apart.

N7. Let the t = time it takes for the bicyclists to be 5 miles apart. Use the formula . drt = fasterslowertotal 20185 25 5 2.5 2 ddd tt

It will take 2.5 hours for the bicyclists to be 5 miles apart.

8. Let the x = rate of the slower bus. Then 10the x += rate of the faster bus. Use the formula drt = and the fact that each bus travels for 1 5 hour.

slowerfastertotal 11 (10)12 55 11 212 55 2 212 5 2 10 5 525 (10) 252 25 ddd xx xx x x x x += ⎛⎞⎛⎞ ++= ⎜⎟⎜⎟ ⎝⎠⎝⎠ ++= += = ⎛⎞ = ⎜⎟ ⎝⎠ =

Since 25,1035.xx=+= The slower bus had a rate of 25 mph and the faster bus had a rate of 35 mph.

Check: The slower bus traveled 1 255 5 ⎛⎞ = ⎜⎟ ⎝⎠ miles and the faster bus traveled 1 357 5 ⎛⎞ = ⎜⎟ ⎝⎠ miles. The total miles traveled is 5712, += as required.

N8. Let the x = rate of the slower car. Then 6the x += rate of the faster car. Use the formula drt = and the fact that each car travels for 1 4 hour.

slowerfastertotal 11 (6)35 44 13170 4242 2370 422 167 22 21267 1212 67 ddd xx xx x x x x += ⎛⎞⎛⎞ ++= ⎜⎟⎜⎟ ⎝⎠⎝⎠ ++= += = ⎛⎞⎛⎞ = ⎜⎟⎜⎟ ⎝⎠⎝⎠ =

Since 67,673.xx=+= The slower car had a rate of 67 mph and the faster car had a rate of 73 mph.

Check: The slower car traveled

167 67 44 ⎛⎞ = ⎜⎟ ⎝⎠ miles and the faster car traveled

173 73 44 ⎛⎞ = ⎜⎟ ⎝⎠ miles. The total miles traveled is 6773140 35, 444 +== as required.

Exercises

1. The amount of pure alcohol in x liters of a 75% alcohol solution is 0.75 times the volume of solution, or 0.75 x liters. So choice A is the correct answer.

2. Each quarter is worth $0.25. The value of x quarters is 0.25 x dollars. So choice C is the correct answer.

3. Use , drt = where 55 r = and t is the number of hours. ()5555 dtt == miles.

So choice C is the correct answer.

4. Use , drt = where 6 t = and r is the rate of the car. ()66 drr == miles.

So choice D is the correct answer.

5. The concentration of the new solution could not be more than the strength of the stronger of the original solutions, so the correct answer is D, since 32% is stronger than both 20% and 30%.

6. Because pure alcohol (100% concentration) is to be added, the new solution must be stronger than the original one. Therefore, the concentration of the new solution must be greater than 24%, so the correct answer is A, 22%.

7. To estimate the average rate of the trip, round 405 to 400 and 8.2 to 8.

Use d r t = with 405 d = and 8.2. t =

r t ==≈ The best estimate is choice A, 50 mph.

8. The distance traveled cannot be found by multiplying 45 and 30 because the rate is given in miles per hour, while the time is given in minutes. To find the correct distance, start by converting the time to hours. 1 30 minutes hour 2 1 45 2 22.5 drt = = ⎛⎞ = ⎜⎟ ⎝⎠ =

The car traveled 22.5 1 or 22 2 ⎛⎞ ⎜⎟ ⎝⎠ miles.

9. The amount of pure alcohol in 150 liters of a 30% alcohol solution is 1500.3045 liters.

AmountRateAmount ofofof pure solutionconcentrationalcohol ×= ↑↑↑

10. The amount of pure acid in 250 milliliters of a 14% acid solution is 2500.1435milliliters.

AmountRateAmount ofofof pure solutionconcentrationacid = × ↑↑↑

11. If $25,000 is invested at 3% simple interest for one year, the amount of interest earned is $25,0000.031$750.

InterestInterest PrincipalTimerateearned = ×× ↑↑↑↑

12. If $10,000 is invested for one year at 3.5% simple interest, the amount of interest earned is $10,0000.035$350.

InterestInterest Principal rateearned = × ↑↑↑

13. The monetary value of 35 half-dollars is 35$0.50$17.50.

NumberDenominationMonetary of coinsvalue = ×

14. The monetary value of 283 nickels is 283$0.05$14.15.

NumberDenominationMonetary of coinsvalue = × ↑↑↑

15. Step2

Let the x = number of liters of 25% acid solution to be used.

Step3

Use the box diagram in the textbook to write the equation.

Pure acidpure acidpure acid in 25%in 40%in 30% solutionplussolutionissolution.

Multiply by 10 to clear decimals.

52(80)4(80) 51604320 160320 160 xx xx x x +=+ +=+ += = Step5

160 gallons of 50% antifreeze is needed.

Step6

50% of 160 gallons plus 20% of 80 gallons is 80 gallons plus 16 gallons, or 96 gallons, of pure antifreeze, which is equal to 40% of (16080) + gallons. [] 0.40(240)96 =

17. Let the x = number of liters of 5% drug solution.

0.250.40(80)0.30(80) xx

Step4

+=+

Multiply by 100 to clear decimals.

2540(80)30(80)

253200302400

2580030

8005 160 xx xx xx x x +=+ +=+ += = =

Step5

160 liters of 25% acid solution must be added.

Step6

25% of 160 liters plus 40% of 80 liters is 40 liters plus 32 liters, or 72 liters, of pure acid; which is equal to 30% of (16080) + liters.

[0.30(240)72] =

16. Step2

Let the x = number of gallons of 50% solution needed.

Then 80the x += number of gallons of 40% solution.

Step3

Use the box diagram in the textbook to write the equation.

Purepurepure antifreezeantifreezeantifreeze in 50%in 20%in 40% plus is solutionsolutionsolution. + 0.500.20(80) 0.4080xx

Step4

Solve the equation. 0.500.20(80)0.40(80) xx +=+

Pure drugpure drugpure drug in 10%in 5%in 8% solutionplussolutionissolution. 0.050.08(20) 0.1020 xx

+=+

Solve the equation. 0.10(20)0.050.08(20) 10(20)58(20) 20058160 2003160 403 401 13 33 xx xx xx x x x +=+ +=+ +=+ =+ = ==

The pharmacist needs 1 13 3 liters of 5% drug solution.

Check 1 13: 3 x =

LS and RS refer to the left side and right side of the original equation. 12

LS: 0.10(20)0.05132 33 12 RS: 0.0813202 33 ⎛⎞ += ⎜⎟ ⎝⎠

18. Let the x = number of kilograms of metal that is 20% tin.

Then 80the x += number of kilograms of metal that is 50% tin.

Tin intin intin in 20% metalplus70% metalis50% metal.

20. Let the x = number of liters of the 10% solution.

0.200.70(80)0.50(80) xx

+=+

Solve the equation.

0.200.70(80)0.5040 xx +=+

Multiply by 10 to clear decimals.

27(80)5400 25605400 5603400 1603 1601 53 33 xx xx x x x +=+ +=+ =+ = == 1 53 3 kilograms of metal that is 20% tin is needed.

Check 1 53: 3 x =

LS and RS refer to the left side and right side of the original equation. 12

LS: 0.20530.70(80)66 33 12 RS: 0.50538066 33

19. Let the x = number of liters of the 20% alcohol solution.

Complete the table. Strength Liters of Solution Liters of Pure Alcohol 12% 12 0.12(12)1.44 = 20% x 0.20 x 14% 12 x + 0.14(12) x +

From the last column, we can formulate an equation that compares the number of liters of pure alcohol. The equation is 1.440.200.14(12). xx +=+

Solve the equation.

1.440.200.14(12) 1.440.200.141.68 0.060.24 4 xx xx x x +=+ +=+ = = 4 L of the 20% alcohol solution is needed.

Strength Liters of Solution Liters of Pure Alcohol 10% x 0.10x 50% 40 0.50(40)20 = 40% 40 x + 0.40(40) x + Alcoholalcoholalcohol in 10%in 50%in 40% solutionplussolutionissolution. 0.10200.40(40) xx

+=+

Multiply by 10 to clear decimals. ()1200440 2004160 340 40401 , or 13 333 xx xx x x +=+ +=+ −=− == 1 13 3 L of 10% alcohol solution should be added.

21. Let the x = amount of water to be added. Then 20the x += amount of 2% solution. There is no minoxidil in water.

Purepurepure minoxidilminoxidilminoxidil in mLin 4%in 2% solutionplussolutionissolution.

0()0.04(20)0.02(20+) x xx

Solve the equation. 00.04(20)0.02(20) 4(20)2(20) 80402 402 20 xx x x x x +=+ =+ =+ = = 20 milliliters of water should be used.

Check 20: x = LS: 0(20)0.04(20)0.8 RS: 0.02(2020)0.8 += += This answer should make common sense—that is, equal amounts of 0% and 4% solutions should produce a 2% solution.

22. Let the x = number of milliliters of 4% solution.

Purepurepure minoxidilminoxidilminoxidil in 1%in 4%in 2% solutionplussolutionissolution.

24. Let the x = number of gallons of 50% juice fruit drink.

Then 12the x −= number of gallons of 20% juice fruit drink.

Juice injuice injuice in 50% drinkplus20% drinkis40% drink.

0.50.20(12)0.40(12) xx ↓↓↓↓↓ +−=

+=+

0.01(50)0.040.0250xx

Multiply by 10 to clear decimals.

52(12)4(12)

1(50)42(50)

5041002

502100 250 25 xx xx x x x +=+ +=+ += = =

The pharmacist must add 25 milliliters of 4% solution.

Check 25: x =

524248 32448 324 8 xx xx x x x +−= +−= += = =

8 gallons of the 50% juice fruit drink must be added.

LS: 0.01(50)0.04(25)1.5

RS: 0.02(5025)1.5 += +=

23. Let the x = number of liters of 60% acid solution.

Then 20the x −= number of liters of 75% acid solution.

Pure acidpure acidpure acid in 60%in 75%in 72% solutionplussolutionissolution.

Check 8: x = ()() 0.580.201284.8 LS: RS: 0.4(12)4.8 +−= =

25. Let the x = amount invested at 5% (in dollars). Then 1200the x −= amount invested at 4% (in dollars).

Amount Invested (in dollars)

Rate of Interest Interest for One Year x 0.05 0.05 x 1200 x 0.04 0.04(1200) x

0.60 0.75200.7220xx

Solve the equation.

0.600.75(20)0.72(20)

6075(20)72(20)

601500751440 1500151440 1560 4 xx xx xx x x x +−= +−= +−= −= −=− =

4 liters of 60% acid solution must be used.

Check 4: x =

Since the total annual interest was $141, the equation is 0.050.04(1200)141.

RS: 0 .72(20)14.4 +−= =

LS: 0.60(4)0.75(204)14.4

54(1200)100(141) 54480014,100 9480014,100 918,900 2100 xx xx xx x x x +−= +−= +−= −= = = Since 2100,1200900.xx=−= Arlene invested $2100 at 5% and $900 at 4%.

26. Let the x = amount invested at 2%. Then 3000the x += amount invested at 3%.

Amount Invested (in dollars) Rate of Interest Interest for One Year

x 0.02 0.02 x

3000 x + 0.03 0.03(3000) x +

Since the total annual interest was $390, the equation is 0.020.03(3000)390. xx++=

Multiply by 100 to clear decimals.

23(3000)39,000

239,00039,000 59,00039,000 530,000 6000 xx xx x x x ++= ++= += = = Since 6000,30009000.xx=+=

Margaret deposited $6000 at 2% and $9000 at 3%.

27. Let the x = amount invested at 6%.

Then 36000the x += amount invested at 5%.

0.060.05(36000)825

65(36000)100(825)

61530,00082,500

29. Let the x = amount Jamal invested at 8%.

Then 2500the x −= amount invested at 2%. () () 0.080.022500152 82250015,200 85000215,200 6500015,200 610,200 1700 xx xx xx x x x +−= +−= +−= += = = Since 1700,2500800.xx =−= Jamal invested $1700 at 8% and $800 at 2%.

30. Let the x = amount Carter invested at 1%.

Then 9000the x −= amount invested at 4%. () () 0.010.049000285 14900028,500 36,000428,500 336,00028,500 37500 2500 xx xx xx x x x +−= +−= +−= −+= −=− = Since 2500,90006500.xx =−= Carter invested $2500 at 1% and $6500 at 4%.

31. Let the x = number of nickels. Then 2the x += number of dimes. The valuethe value of nickelsplusof dimesis$1.70.

Since 2500,3600013,500.xx=+=

The artist invested $2500 at 6% and $13,500 at 5%.

28. Let the x = amount invested at 3%.

Then 230,000the x += amount invested at 4%.

0.030.04(230,000)5600

34(230,000)100(5600)

38120,000560,000

11120,000560,000

11440,000 40,000 xx xx xx x x x ++= ++= ++= += = =

Since 40,000,230,000110,000.xx=+=

The actor invested $40,000 at 3% and $110,000 at 4%.

0.05+0.10(2)=1.70 xx ↓↓↓↓↓ + 510(2)100(1.70) 51020170 1520170 15150 10 xx xx x x x ++= ++= += = =

The collector has 10 nickels.

32. Let the x = number of $5 bills. Then 5the x += number of $20 bills. The valuethe value of fivesplusof twentiesis$725.

5+20(5)=725 xx ↓↓↓↓↓ + 520100725 25100725 25625 25 xx x x x ++= += = =

The teller has twenty-five $5 bills.

33. Let the x = number of 46-cent stamps. Then 45the x −= number of 20-cent stamps. The value of the 46-cent stamps is 0.46x and the value of the 20-cent stamps is 0.20(45). x

The total value is $15.50, so 0.460.20(45)15.50

The total value is $87.50, so 88.5(2)87.50.

3.5 xx xx x x += += = =

81787.50

2587.50

Since 3.5, 27.xx==

4620(45)1550

46900201550

26650 25 xx xx xx x x x +−= +−= +−= += = =

269001550

Since25, 4520.xx=−=

She bought twenty-five 46-cent stamps valued at $11.50 and twenty 20-cent stamps valued at $4.00, for a total value of $15.50.

34. Let the x = number of adult tickets sold. Then 600the x −= number of children’s tickets sold.

The value of the adult tickets is 8x and the value of the children's tickets is 5(600). x The total value was $4116, so 85(600)4116.

She can buy 3.5 pounds of Colombian Decaf and 7 pounds of Arabian Mocha.

36. Let the x = number of pounds of Italian Espresso beans.

Then 4the x = number of pounds of Kona Deluxe beans.

8300054116

330004116

31116 372 xx xx x x x +−= +−= += = =

Since 372,600228.xx=−=

There were 372 adult tickets valued at $2976 and 228 children’s tickets valued at $1140, for a total value of $4116.

35. Let the x = number of pounds of Colombian Decaf beans. Then 2the x = number of pounds of Arabian Mocha beans.

The sum of the values of the Kona Deluxe beans and the Italian Espresso beans must equal the value of the mixture, so () () 9.0011.504247.50

4.5 xx xx xx x x += += += = =

9011542475

904602475

5502475

Since 4.5, 418.xx==

The customer can buy 4.5 pounds of Italian Espresso beans and 18 pounds of Kona Deluxe beans.

37. Use the formula drt = with 53 r = and 10. t = (53)(10) 530 drt = = =

The distance between Memphis and Chicago is 530 miles.

38. Use drt = with 164 r = and 2. t = 164(2) 328 drt = = =

The distance from Warsaw to Rome is 328 miles.

39. Use drt = with 500 d = and 187.433. r =

500187.433

500 2.668 187.433 drt t t = = =≈

His time was about 2.668 hours.

40. Use drt = with 400 d = and 153.485. r =

400153.485

400 2.606 153.485 drt t t = = =≈

His time was about 2.606 hours.

41. 200meters 9.14 21.88seconds d r t ==≈

Her rate was about 9.14 meters per second.

42. 400meters 8.07 49.55seconds d r t ==≈

Her rate was about 8.07 meters per second.

43. 110meters 8.51 12.92seconds d r t ==≈

His rate was about 8.51 meters per second.

44. 200meters 10.35 19.32seconds d r t ==≈

His rate was about 10.35 meters per second.

45. Let the t = number of hours Marco and Celeste traveled. Make a chart using the formula . drt = rtd

Marco 10 t 10t

Celeste 12 t 12t

Marco’sCeleste’s distanceminusdistanceis15. 121015 tt

Solve the equation. 121015 215 151 or7 22 tt t t −= = = They will be 15 miles apart in 1 7 2 hours.

46. Let the t = number of hours until the steamboats will be 9 miles apart. Make a chart using the formula . drt = rtd

Slower Boat 18 t 18t

Faster Boat 24 t 24t

Distancedistance traveled bytraveled by faster boatminusslower boatis9.

24189 tt

Solve the equation. 24189 69 31 or1 22 tt t t −= = = In 1 1 2 hours, the steamboats will be 9 miles apart.

47. Let the t = number of hours until John and Pat meet. The distance John travels and the distance Pat travels total 440 miles. John’sPat’stotal distanceanddistanceisdistance.

6028440 tt

Solve the equation. 6028440 88440 5 tt t t += = = It will take 5 hours for them to meet.

48. Let the t = time each plane travels. Use the chart in the text to help write the equation. Distancedistancedistance of planeof planebetween leavingleavingPortland PortlandplusSt. Louisisand St. Louis.

901162060 tt

Solve the equation.

901162060 2062060 2060 206 10 tt t t t += = = =

It will take the planes 10 hours to meet.

49. Let the t = number of hours until the trains are 315 kilometers apart. Distance ofdistance of northboundsouthboundtotal trainplustrainisdistance.

Solve the equation.

8595315 180315 3157 1804 tt t t += = ==

It will take 3 1 4 hours for the trains to be 315 kilometers apart.

50. Let the t = number of hours until the steamers are 110 miles apart. Each steamer will travel 22t miles and the total distance traveled will be 110 miles.

2222110 44110 1105 442 tt t t += = ==

It will take 1 2 2 hours for the steamers to be 110 miles apart.

51. Let the x = rate of the westbound plane. Then 150the x −= rate of the eastbound plane. Using the formula drt = and the chart in the text, we see that westeasttotal (3)(150)(3)2250 334502250 62700 450 ddd xx xx x x += +−= +−= = =

Since 450, 150300. xx=−=

The rate of the westbound plane is 450 mph and the rate of the eastbound plane is 300 mph.

52. Let the x = rate of the northbound train.

Then 20the x += rate of the southbound train. Using the formula drt = and the chart in the text, we see that northsouthtotal (2)(20)(2)280

2240280

4240 60 ddd xx xx x x += ++= ++= = =

Since 60, 2080. xx=+=

The rate of the northbound train is 60 mph and the rate of the southbound train is 80 mph.

53. Let the x = rate of the slower car.

Then 20the x += rate of the faster car.

Use the formula drt = and the fact that each car travels for 4 hours.

fasterslowertotal (20)(4)()(4)400 4804400 8320 40 ddd xx xx x x += ++= ++= = =

The rate of the slower car is 40 mph and the rate of the faster car is 60 mph.

54. Let the x = rate of the faster car.

Then 15the x −= rate of the slower car.

Use the formula drt = and the fact that each car travels for 2 hours.

fasterslowertotal (2)(15)(2)230 2230230 4260 65 ddd xx xx x x += +−= +−= = =

Since 65, 1550. xx=−=

The rate of the faster car is 65 km per hour and the rate of the slower car is 50 km per hour.

55. Let Bob’s x = current age.

Then 3Kevin’s x = current age.

Three years ago, Bob’s age was 3 x and Kevin’s age was 33, x and this sum was 22.

(3)(33)22 4622 428 7 xx x x x −+−= −= = =

Bob is 7 years old and Kevin is 3(7)21 = years old.

56. Let the x = number of pint cartons. Then 6the x = number of quart cartons. Since 1 1 quart2 pints, 1 pint quart, 2 == and 1 2 x is the number of quarts contained in pint cartons. The total number of quarts is 39, so 1 639. 2

60. Let the x = amount of the sales. salestaxtotal 0.052394 1.052394 2394 2280 1.05 xx x x += += = == The amount of sales was $2280.

2.8 Solving Linear Inequalities

Classroom Examples, Now Try Exercises

There are 6 pint cartons and 6(6)36 = quart cartons.

57. Let the w = width of the table. Then 3the w = length of the table. If we subtract 3 feet from the length (33) w and add 3 feet to the width (3), w + then the length and the width would be equal. 333 36 26 3 ww ww w w −=+ =+ = = The width is 3 feet and the length is 3(3)9 = feet.

58. Let the x = number of hours worked. Her gross pay (pay before deductions) is 8x gross paydeductiontake-home pay

80.25(8)450 82450 6450 450 75 6 xx xx x x −=

She must work 75 hours to take home $450.

59. Let her x = gross pay (pay before deductions). gross paydeductiontake-home pay 0.10()585 0.90585 585 650 0.90 xx x x −= −= = == She was paid $650 before deductions.

1. (a) The statement 3 x ≥ says that x can represent any number greater than or equal to 3. The interval is written as [3,) ∞ (the parenthesis at ∞ shows that ∞ is not part of the graph). To graph the inequality, place a bracket at 3 and draw an arrow extending to the right.

(b) The statement 4 x −> is the same as 4. x <− The interval is written as (,4). −∞−

Graph this inequality by placing a parenthesis at 4 on a number line and drawing an arrow to the left.

N1. (a) The statement 1 x <− says that x can represent any number less than 1. The interval is written as (,1). −∞− Graph this inequality by placing a parenthesis at 1 on a number line and drawing an arrow to the left.

(b) The statement 2 x −≤ is the same as 2. x ≥− The interval is written as [2,). −∞ Graph this inequality by placing a bracket at 2 on a number line and drawing an arrow to the right.

2. 1872

181721Add1. 873

87737Subtract7. 3 xx xx xx xxxxx x −+<+ −++<++ <+ −<+− <

Graph the solution set (,3). −∞

To graph this inequality, place a parenthesis at 3 on a number line and draw an arrow to the left.

N2. 5543 555435Subtract 5. 542

54424Subtract 4. 2 xx xx xx xxxxx x +≥+ +−≥+− ≥− −≥−− ≥−

Graph the solution set [2,). −∞

To graph this inequality, place a bracket at 2 on a number line and draw an arrow to the right.

3. 212 212 Divide by 2. 22 6 r r r −>− <− <

Graph the solution set (,6). −∞

N3. 515 515 Divide by 5. 55 3 k k k −≥ ≤− ≤−

Graph the solution set (,3]. −∞−

4. 5275 4275

427757Subtract 7. 325 32252Subtract 2. 37 37 Divide by3. 33 7 3 xxx xx xxxxx x x x x x −+<− +<− +−<−− −+<− −+−<−− −<− >− >

Graph the solution set 7 ,. 3 ⎛⎞ ∞ ⎜⎟ ⎝⎠

N4. 62584 3684 368848Subtract8 564 56646Subtract6. 510 510 Divide by5. 55 2 ttt tt ttttt. t t t t t −+≤− +≤− +−≤−− −+≤− −+−≤−− −≤− ≥− ≥

Graph the solution set [2,). ∞

5. 4(1)315(21) xxx −−>−−+ 4431521Dist. Prop. 4162 421622Add 2. 3416 344164Add 4. 312 312 Divide by 3. 33 4 xxx xx xxxxx x x x x x −−>−−− −>−− −+>−−+ −>− −+>−+ >− > >−

Graph the solution set (4,). −∞

N5. 23(6)4(7) xxx −−<+ 2318428Dist. prop. 18428 18428Add . 18528

182852828Subtract 28. 105 105

Divide by5. 55 2 2 xxx

Graph the solution set (2,). −∞

6. () () 11 314 25 xx−>+ ()() 53124Multiply by 10. 15528Dist. prop. 1552282Subtract 2. 1358 135585Add 5. 1313 1313

Divide by 13. 1313 1 xx xx xxxxx

Graph the solution set (1,). ∞

N6. () () ()() 11 428 86 34428Multiply by 24. 312832Dist. prop. 31238323Subtract 3. 12532 123253232Subtract 32. 205 205

Divide by 5. 55 4 4

Graph the solution set ( ] ,4.−∞−

7. Let Maggie’s x = score on the fourth test. The averageis at least90. 988688 90 4 x ↓↓↓ +++ ≥

Solve the inequality. Combine like terms in the numerator, and multiply by 4 to eliminate the fraction.

() 272 4490 4 272360

272272360272Subtract 272. 88Combine terms. x x x x + ⎛⎞ ≥ ⎜⎟ ⎝⎠ +≥ +−≥− ≥

She must score 88 or more on the fourth test to have an average of atleast 90.

N7. Let Will’s x = score on the third test. The averageis at least90.

9885 90 3 x ↓↓↓ ++ ≥

Solve the inequality. Combine like terms in the numerator, and multiply by 3 to eliminate the fraction.

3390 3 183270

183183270183Subtract 183. 87Combine terms. x x x x + ⎛⎞ ≥

+−≥− ≥ He must score 87 or more on the third test to have an average of atleast 90.

8. The statement 24 x <≤ says that x can represent any number between 2 and 4, excluding 2 and including 4. To graph the inequality, place a parenthesis at 2 and a bracket at 4 and draw a line segment between them. The interval is written as (2,4].

N8. The statement 02 x ≤< says that x can represent any number between 0 and 2, including 0 and excluding 2. To graph the inequality, place a bracket at 0 and a parenthesis at 2 and draw a line segment between them. The interval is written as [0,2).

Graph the solution set [1,3].

Graph the solution set 2 2, . 3

Exercises

1. When graphing an inequality, use a parenthesis if the inequality symbol is > or < Use a square bracket if the inequality symbol is ≥ or ≤

Examples: A parenthesis would be used for the inequalities 2 x < and 3. x > A square bracket would be used for the inequalities 2 x ≤ and 3. x ≥

2. False. A parenthesis is always used with the symbols −∞ and .∞

3. In interval notation, the set { }|0xx > is written ( ) 0, ∞

4. In interval notation, the set of all real numbers is ( ) , −∞∞

5. The set of numbers graphed corresponds to the inequality 4. x >−

6. The set of numbers graphed corresponds to the inequality 4. x ≥−

7. The set of numbers graphed corresponds to the inequality 4. x ≤

8. The set of numbers graphed corresponds to the inequality 4. x <

9. The statement 4 z ≤ says that z can represent any number less than or equal to 4. The interval is written as (,4]. −∞ To graph the inequality, place a square bracket at 4 (to show that 4 is part of the graph) and draw an arrow extending to the left.

11. The statement 3 x <− says that x can represent any number less than 3. The interval is written as (,3). −∞− To graph the inequality, place a parenthesis at 3 (to show that3 is not part of the graph) and draw an arrow extending to the left.

12. The statement 11 r <− says that r can represent any number less than 11. The interval is written (,11). −∞− To graph the inequality, place a parenthesis at11 (to show that 11 is not part of the graph) and draw an arrow extending to the left.

13. The statement 4 t > says that t can represent any number greater than 4. The interval is written (4,). ∞ To graph the inequality, place a parenthesis at 4 (to show that 4 is not part of the graph) and draw an arrow extending to the right.

14. The statement 5 m > can be written as (5,). ∞

10. The statement 3 x ≤ can be written as (,3]. −∞

15. The statement 0 x ≥ (or 0 x ≤ ) says that x can represent any number less than or equal to 0. The interval is written as (,0]. −∞ To graph the inequality, place a bracket at 0 (to show that 0 is part of the graph) and draw an arrow extending to the left.

16. The statement 1 x ≥ (or 1 x ≤ ) says that x can represent any number less than or equal to l. The interval is written as (,1]. −∞ To graph the inequality, place a bracket at l (to show that l is part of the graph) and draw an arrow extending to the left.

17. The statement 1 2 x −≤ 1 or 2 x ⎛⎞ ≥− ⎜⎟ ⎝⎠ says that x can represent any number greater than or equal to 1 2 The interval is written 1 ,. 2 ⎡⎞ −∞ ⎟ ⎢ ⎣⎠ To graph the inequality, place a bracket at 1 2 (to show that 1 2 is part of the graph) and draw an arrow extending to the right.

18. The statement 3 4 x −≤ 3 or 4 x ⎛⎞ ≥− ⎜⎟ ⎝⎠ says that x can represent any number greater than or equal to 3 . 4 The interval is written 3 ,. 4 ⎡⎞ −∞ ⎟ ⎢ ⎣⎠ To graph the inequality, place a bracket at 3 4 (to show that 3 4 is part of the graph) and draw an arrow extending to the right.

19. 87 8878Add8. 1 z z z −≥− −+≥−+ ≥

Graph the solution set [ ) 1,. ∞

20. 311 33113Add3. 8 p p p −≥− −+≥−+ ≥−

Graph the solution set [ ) 8,.−∞

21. 238 238Subtract. 38 3383Subtract3. 5 kk kkkkk k k k +≥+ +−≥+− +≥ +−≥− ≥

Graph the solution set [ ) 5,. ∞

22. 37211 3722112Subtract2. 711 77117Subtract7. 4 xx xxxxx x x x +≥+ +−≥+− +≥ +−≥− ≥

Graph the solution set [ ) 4,. ∞

23. 3526 325226Subtract2. 56 5565Subtract5. 11 nn nnnnn n n n +<− −+<−− +<− +−<−− <−

Graph the solution set (,11). −∞−

24. 5245 524454Subtract4. 25 2252Add2. 3 xx xxxxx x x x −<− −−<−− −<− −+<−+ <−

Graph the solution set (,3). −∞−

25. The inequality symbol must be reversed when one is multiplying or dividing by a negative number.

26. For choice A, the second part of the multiplication property of inequality states that when both sides of an inequality are multiplied by a negative number, the inequality symbol is reversed. Since r is negative, this statement is false.

Look at choice B. Both sides of the inequality pq < have been multiplied by the negative number r and the inequality symbol has been reversed, so this statement is true.

Look at choice C. By the addition property of inequality, this statement is true When the same number is added to both sides of an inequality, the inequality symbol is not reversed.

Look at choice D. When the same number is subtracted from both sides of an inequality, the inequality symbol is not reversed. The statement is true. Therefore, only statement A is false.

27. 318 318

Divideby3. 33 6 x x x < < <

Graph the solution set (,6). −∞

28. 535 535

Divideby5. 55 7 x x x < < <

Graph the solution set (,7). −∞

29. 220 220

Divideby2. 22 10 y y y ≥− ≥ ≥−

Graph the solution set [10,). −∞

30. 624 624

Divideby6. 66 4 m m m ≥− ≥ ≥−

Graph the solution set [4,). −∞

31. 824 824

Divideby8. 88 3 t t t −> <− <−

Graph the solution set (,3). −∞−

32. 749 749

Divideby7. 77 7 x x x −> <− <−

Graph the solution set (,7). −∞−

33. 0 10 10

Divideby1. 11 0 x x x x −≥ −≥ ≤− ≤

Graph the solution set (,0]. −∞

34. 0 10 10

Divideby1. 11 0 k k k k −< −< >− >

Graph the solution set (0,). ∞

35. Multiply by 4 , 3 the reciprocal of 3 ; 4

reverse the inequality sign. 3 15 4 434 (15) 343 20 r r r −<− ⎛⎞⎛⎞⎛⎞ −−>−−

Graph the solution set (20,). ∞

Chapter 2 Linear Equations and Inequalities in One Variable

36. Multiply by 8 , 7 the reciprocal of 7 , 8 and reverse the inequality symbol. 7 14 8 878 (14) 787 16 t t t −<− ⎛⎞⎛⎞⎛⎞ −−>−− ⎜⎟⎜⎟⎜⎟ ⎝⎠⎝⎠⎝⎠ >

Graph the solution set (16,). ∞

37. 0.020.06

0.020.06 Divideby0.02. 0.020.02 3 x x x −≤ ≥− ≥−

Graph the solution set [3,). −∞

38. 0.030.12

0.030.12 Divide by 0.03. 0.030.03 4 v v v −≥− ≤− ≤

Graph the solution set (,4]. −∞

39. 8915

824Subtract 9.

824

Divide by 8. 88 3 x x x x +≤− ≤− ≤ ≤−

Graph the solution set (,3]. −∞−

40. 6717

624Subtract 7. 624 Divide by 6. 66 4 x x x x +≤− ≤− ≤ ≤−

Graph the solution set (,4]. −∞−

41. 431

44Add 3. 44

Divide by4. 44 1 x x x x −−< −< >− >−

Graph the solution set (1,). −∞

42. 546

510Add 4. 510

Divide by 5. 55 2 x x x x −−< −< >− >−

Graph the solution set (2,). −∞

43. 5139

219Subtract 3. 210Subtract 1. 5Divide by 2. rr rr r r +≥− +≥− ≥− ≥−

Graph the solution set [5,). −∞

44. 63312

3312Subtract3. 39Subtract3. 3 Divide by3. tt tt t r +<+ +< < <

Graph the solution set (,3). −∞

45. 5210

6210Add . 612Add2. 2 Divide by6. xx xx x x −≤−+ −≤ ≤ ≤

Graph the solution set ( ] ,2.−∞

46. 3926 596Add 2. 515Add9. 3 Divide by5. xx xx x x −≥−+ −≥ ≥ ≥

Graph the solution set [ ) 3,. ∞

47. 7432 442Add 7. 64Add2. 6 Divide by4. 4 3 2 3 2 xx xx x x x x −+>−− >− > > > <

Graph the solution set 3 ,. 2 ⎛⎞ −∞ ⎜⎟ ⎝⎠

48. 81411 1411Add 8. 104Subtract 11. 10 Divide by4. 4 5 2 xx xx x x x −+<−+ <+ −< −< >−

Graph the solution set 5 ,. 2 ⎛⎞ −∞ ⎜⎟ ⎝⎠

49. 63244

33Subtract 3. 1Divide by 3. xxx xx xx x x ++<++ +<+ +< < <

7346Combine like terms.

336Subtract 4.

Graph the solution set (,1). −∞

50. 41299 51229Combine terms. 3129Subtract 2.

33Subtract 12. 1Divide by 3. wwww ww ww w w −++≥++ +≥+ +≥ ≥− ≥−

Graph the solution set [1,). −∞

51. 47236 6443 344 30 0 xxx xx x x x −++≤−++ +≤+ +≤ ≤ ≤

Graph the solution set (,0]. −∞

52. 146741010 216106 1166 110 0 xxx xx x x x −+>+− −>− −>− > >

Graph the solution set (0,). ∞

53. 5(1)3(2) 5536 256 21 1 2 tt tt t t t −>− −>− −>− >− >−

Graph the solution set 1 ,. 2 ⎛⎞ −∞ ⎜⎟ ⎝⎠

Chapter 2 Linear Equations and Inequalities in One Variable

54. 7(2)4(4) 714416 31416 32 2 3 mm mm m m m −<− −<− −<− <− <−

Graph the solution set 2 ,. 3 ⎛⎞ −∞− ⎜⎟ ⎝⎠

55 5(3)63(21)4 5156634 1523 3153 312 312

Divide by 3. 33 4 xxxx xxxx xx x x x x +−≤+− +−≤+−

Graph the solution set [4,). ∞

56. 2(5)34(6)1 21034241 510423 1023 13 xxx xxx xx x x −+<−+ −+<−+ −<− −<− <−

Graph the solution set (,13). −∞−

57. () () () () ()() 12 543 35 12 1554153 35 55463 2520618 1938 2 xx xx xx xx x x −≥+ ⎛⎞⎛⎞−≥+⎜⎟⎜⎟ ⎝⎠⎝⎠ −≥+ −≥+ ≥ ≥

Graph the solution set [2,). ∞

58. () () () () ()() 55 575 126 55 1257125 126 557105 25351050 1515 1 xx xx xx xx x x −<− ⎛⎞⎛⎞ −<−⎜⎟⎜⎟ ⎝⎠⎝⎠ −<−

−<− <− <−

Graph the solution set (,1). −∞−

59. 25 36(3)(4)

25 6(3)6(4) 36 4(3)5(4) 412520 1220 32 32

Divide by 1. 11 32 pp pp pp pp p p p p +>− ⎛⎞⎛⎞+>−⎜⎟⎜⎟ ⎝⎠⎝⎠ +>− +>− −+>− −>− <− <

Graph the solution set (,32). −∞

60. 74 93(4)(5) 74 9(4)9(5) 93

7(4)12(5) 7281260 52860 588 588

Divide by 5. 55 88 5 xx xx xx xx x x x x −≤+ ⎛⎞⎛⎞−≤+⎜⎟⎜⎟ ⎝⎠⎝⎠ −≤+ −≤+ −−≤ −≤ ≥− ≥−

Graph the solution set 88 ,. 5 ⎡⎞ −∞ ⎟ ⎢ ⎣⎠

61. () () () 413 3 5210 413 10310 5210 8533 85153 3153 318 318 33 6 xx xx xx xx x x x x −+≤ ⎡⎤⎛⎞ −+≤ ⎜⎟ ⎢⎥ ⎣⎦⎝⎠ −+≤ −−≤ −≤ ≤ ≤ ≤

Graph the solution set ( ] ,6.−∞

62. () () () 111 1 632 111 616 632 213 223 323 35 35 33 5 3 xx xx xx xx x x x x +−> ⎡⎤⎛⎞ +−> ⎜⎟ ⎢⎥ ⎣⎦⎝⎠ +−> +−> −> > > >

Graph the solution set 5 ,. 3 ⎛⎞ ∞ ⎜⎟ ⎝⎠

63. 4(61)82(3) 461826 21106 1216 125 125 1212 5 12 Divide by 12. xxxx xxxx xx x x x x −+≤+− −−≤+− −−≤− −−≤− −≤− ≥ ≥

Graph the solution set 5 12 ,. ⎡⎞ ∞ ⎟ ⎢ ⎣⎠

64. 2(43)63(4) 2436312 23912 11312 1115 1115 Divide by 11. 1111 15 11 zzzz zzzz zz z z z z −+>++ −−>++ −−>+ −−> −> <− <−

Graph the solution set 15 ,. 11 ⎛⎞ −∞− ⎜⎟ ⎝⎠

65. 5(23)2(8)3(24)2 10152166122 831710 3110 21 kkkk kkkk kk k k +−−>++− +−+>++− +>+ +> >−

Graph the solution set (21,). −∞

66. 2(35)4(6)2(32)315 61042464315 1014911 1411 25 zzzz zzzz zz z z −++≥++− −++≥++− +≥− +≥− ≥−

Graph the solution set [25,). −∞

67. The statement “You must be at least 18 yr old to vote” translates as 18. x ≥

68. The statement “Less than 1 in. of rain fell” translates as 1. x <

69. The statement “Chicago received more than 5 in. of snow” translates as 5. x >

70. The statement “A full-time student must take at least 12 credits” translates as 12. x ≥

71. The statement “Tracy could spend at most $20 on a gift” translates as 20. x ≤

72. The statement “The car’s rate exceeded 60 mph” translates as 60. x >

73. Let the x = score on the third test. The average is at of the80. least three tests 7681 80 3 x

Then solve the inequality.

In order to average at least 80, Christy’s score on her third test must be 83 or greater.

74. Let the x = score on the third test. The average is at of the90. least three tests

9686 90 3 x ↓↓↓ ++ ≥

Then solve the inequality. 182 90 3

182 33(90) 3 182270 88 x x x x + ≥ + ⎛⎞ ≥ ⎜⎟ ⎝⎠

In order to average at least 90, Joseph’s score on his third test must be 88 or greater.

75. Let the x = amount of precipitation in December. The average exceeds4.6in. precipitation 5.74.3 4.6 3 x ↓↓↓ ++ >

Then solve the inequality. 10 4.6 3

33(4.6) 3 1013.8 3.8 x x x x + > + ⎛⎞ > ⎜⎟ ⎝⎠ +> > In order for the average monthly precipitation to exceed 4.6 in., more than 3.8 in. must fall in December.

76. Let the x = amount of precipitation in August. The average exceeds6.7in. precipitation 8.15.7 6.7 3 x ↓↓↓ ++ > Then solve the inequality. 13.8 6.7 3 13.8 33(6.7) 3 13.820.1 6.3 x x x x + > + ⎛⎞ > ⎜⎟ ⎝⎠ +> >

In order for the average monthly precipitation to exceed 6.7 in., more than 6.3 in. must fall in August.

77. Let the n = number.

“When 2 is added to the difference between six times a number and 5, the result is greater than 13 added to five times the number” translates to (65)2513. nn−+>+ Solve the inequality. 652513 63513 313Subtract5. 16Add3. nn nn nn n −+>+ −>+ −> > All numbers greater than 16 satisfy the given condition.

78. Let the n = number.

“When 8 is subtracted from the sum of three times a number and 6, the result is less than 4 more than the number” translates to (36)84. nn+−<+

Solve the inequality. 3684 324 224Subtract.

26Add 2.

3Divide by 2. nn nn nn n n +−<+ −<+ −< < <

All numbers less than 3 satisfy the given condition.

79. The Fahrenheit temperature must correspond to a Celsius temperature that is greater than or equal to 25.−° () () 5 3225 9 959(32)25 595 3245 13 CF F F F =−≥− ⎡⎤−≥−⎢⎥ ⎣⎦ −≥− ≥−

The temperature in Minneapolis on a certain winter day is never less than 13 −° Fahrenheit.

80. The Celsius temperature must give a Fahrenheit temperature that is less than or equal to 122°. 9 32122 5 9 90 5 595 (90) 959 50 FC C C C =+≤ ≤ ⎛⎞ ≤ ⎜⎟ ⎝⎠ ≤

The temperature of Phoenix has never exceeded 50° Celsius.

81. 22; 400 PLWP=+≥

From the figure, we have 43Lx=+ and 37. Wx=+ Thus, we have the inequality 2(43)2(37)400. xx+++≥

Solve this inequality. 86274400 1080400 10320 32 xx x x x +++≥ +≥ ≥ ≥

The rectangle will have a perimeter of at least 400 if the value of x is 32 or greater.

82. ; 72 PabcP=++≥

From the figure, we have , 11, axbx==+ and 25.cx=+ Thus, we have the inequality (11)(25)72.xxx ++++≥

Solve this inequality. 41672 456 56 14 4 x x x +≥ ≥ ≥=

The triangle will have a perimeter of at least 72 if the value of x is 14 or greater.

83. 20.305.60 10(20.30)10(5.60) 20356 336 12 x x x x x +≤ +≤ +≤ ≤ ≤

Alan can use the phone for a maximum of 12 minutes.

84. Let the number of gallons. x =

The amount she spends can be represented by $3$3.60. x + This must be less than or equal to $48.00.

33.648

3.63.6 12.5 x x x x +≤ ≤ ≤ ≤

3.645Subtract 3.

3.645

Divide by 3.6.

She can purchase 12.5 gallons of gasoline.

85. “Revenue from the sales of the DVDs is $5 per DVD less sales costs of $100” translates to 5100,Rx=− where x represents the number of DVDs to be produced.

86. “Production costs are $125 plus $4 per DVD” translates to 1254. Cx =+

87.

(5100)(1254) 51001254 225 PRC xx xx x =− =−−+ =−−− =−

We can use this expression for P to solve the inequality. 0 2250

225 P x x > −> >

88. To make a profit, more than 225 DVDs must be produced and sold.

89. The graph corresponds to the inequality 12, x −<< excluding both 1 and 2.

90. The graph corresponds to the inequality 12. x −≤<

91. The graph corresponds to the inequality 12, x −<≤ excluding 1 but including 2.

92. The graph corresponds to the inequality 12. x −≤≤

93. The statement 810 x ≤≤ says that x can represent any number between 8 and 10, including 8 and 10. To graph the inequality, place brackets at 8 and 10 (to show that 8 and 10 are part of the graph) and draw a line segment between the brackets. The interval is written as [8, 10].

94. The statement 35 x ≤≤ can be written as [3, 5].

95. The statement 010 y <≤ says that y can represent any number between 0 and 10, excluding 0 and including 10. To graph the inequality, place a parenthesis at 0 and a bracket at 10 and draw a line segment between them. The interval is written as (0, 10].

96. The statement 30 x −≤< can be written as [3,0).

97. The statement 43 x >>− can be written as 34. x −<< Graph the solution set (3,4).

98. The statement 64 x ≥≥− can be written as 46. x −≤≤ Graph the solution set []4,6.

99. 844 844

Divide by 4. 444 21 x x x −<≤ <≤ −<≤

Graph the solution set ( ] 2, 1.

100. 3312 3312 Divide by 3. 333 14 x x x −≤< ≤< −≤<

Graph the solution set [ ) 1,4.

101. 5239 5323393Add 3. 2212 212 2Divide by 2. 22 16 x x x x x −≤−≤ −+≤−+≤+ −≤≤ ≤≤ −≤≤

Graph the solution set []1,6.

102. 7348 7434484Add 4. 3312 14Divide by 3. x x x x −≤−≤ −+≤−+≤+ −≤≤ −≤≤

Graph the solution set []1,4.

103. 107324 7721Subtract 3. 13Divide by 7. p p p <+< << <<

Graph the solution set (1, 3).

104. 8311

730Add 1. 7 0Divide by 3. 3 r r r −≤−≤− −≤≤ −≤≤

Graph the solution set 7 ,0. 3 ⎡ ⎤ ⎢ ⎥ ⎣ ⎦

105. 4212 4212

Divide by2. 222 26 62 x x x x −<−< >>− >>− −<<

Graph the solution set ()6,2.

106. 9315 9315

Divide by3. 333 35 53 x x x x <−< >>− −>>− −<<−

Graph the solution set ()5,3.

107. 51612

51161121Subtract 1. 4611 4611

Divide by6. 666 211 36 112 63 m m m m m m <−< −<−−<− <−< >>− −>>− −<<−

Graph the solution set 112 63,. ⎛⎞ ⎜⎟ ⎝⎠

Solving Linear Inequalities

108. 11516

11151161Subtract 1. 2515 2515

Divide by5. 555 2 3 5 2 3 5 q q q q q q −<−≤ −−≤−−≤− −≤−≤ ≥≥− ≥≥− −≤≤

Graph the solution set 2 3,. 5 ⎡⎤ ⎢⎥ ⎣⎦

109. () 63118

63318Distributive prop. 63333183Add 3. 9321 9321

Divide by3. 333 37 x x x x x x ≤−< ≤−< +≤−+<+ ≤< ≤< ≤<

Graph the solution set [ ) 3,7.

110. () 4216

Divide by2. 222 32 x x x x x x −<+≤ −<+≤ −−<+−≤− −<≤ <≤ −<≤

4226Distributive prop. 4222262Subtract 2. 624 624

Graph the solution set ( ] 3,2.

111.

Subtract 2. 1 1214Multiply by 2. 2 1 2(12)212(4) 2 2428Dist. prop. 2422282 266 z z z z z −≤+≤ ⎛⎞ −≤+≤ ⎜⎟ ⎝⎠ −≤+≤ −−≤+−≤− −≤≤

Note:Wecouldhavestartedthissolutionby subtracting1fromeachpart.

Graph the solution set [26, 6].

112. () () Multiply by 3. 1 635 3 1 633353Subtract 3. 3 1 92 3 1 39332 3 276 x x x x x −≤+≤ −−≤+−≤−

Note:Wecouldhavestartedthissolutionby multiplyingeachpartby3.

Graph the solution set [27, 6].

113. 2 137 3 2 3(1)333(7)Multiply by 3. 3 39221Dist. prop. 39929219Subtract 9. 6212Divide by 2. 36 p p p p p p ≤+≤

Graph the solution set [3, 6].

114. ( ) 3 2612 4 3 4(2)464(12)Multiply by 4. 4 824348Distributive prop. 16324Subtract 24. 16 8Divide by 3. 3 x x x x x <+< <+< <+< −<< −<<

Graph the solution set 16 , 8. 3 ⎛⎞ ⎜⎟ ⎝⎠

115. 5 711 4 5 711111Add 1. 4 5 60 4 44544 (6)(0)Multiply 55455by. 24 0 5 r r r r r −≤−≤−

−+≤−+≤−+

−≤≤ ⎛⎞ −≤≤ ⎜⎟ ⎝⎠

−≤≤

Graph the solution set 24 , 0. 5 ⎡⎤ ⎢⎥ ⎣⎦ 116. 3 1224 7 3 7(12)727(4)Multiply by 7. 7 8431428Dist. prop. 98342Subtract 14. 98 14Divide by 3. 3 x x x x x −≤+≤− ⎛⎞ −≤+≤− ⎜⎟ ⎝⎠

−≤+≤− −≤≤−

−≤≤−

To graph the solution set 98 ,14, 3 ⎡⎤ ⎢⎥ ⎣⎦ note that 982 32. 33 −=−

117. 3214 312 4 x x x += = =

Solution set: {4} 118. 3214 312 4 x x x +> > >

Solution set: (4,) ∞

119. 3214

312 4 x x x +< < <

Solution set: (,4) −∞

120. If you were to graph all the solutions from Exercises 117–120 on the same number line, the graph would be the complete number line, that is, all real numbers.

Chapter 2 Review Exercises

1. 51

6Add5. x x −= =

The solution set is {6}.

2. 84

12Subtract8. x x +=− =−

The solution set is {12}.

3. 3128

8. 1248

4Divideby12. r r =− =−

The solution set is {4}.

9. 27815

315 5Divideby3. ppp p p −+= = =

The solution set is {5}.

10. 1 12 12Multiplyby12. x x =− =−

The solution set is {12}.

11. 5 8 8 8588 (8)Multiplyby. 5855 64 5 q q q = ⎛⎞ = ⎜⎟ ⎝⎠ =

The solution set is 64 5 ⎧ ⎫ ⎨ ⎬ ⎩⎭

18Subtract2.

7Subtract1. tt tt t +=+ += =

The solution set is {7}.

4. 2 54 3 2 Subtract4. 3 zz zz =+ =

The solution set is 2 3 ⎧⎫ ⎨⎬ ⎩⎭

5. (42)(31)8 (42)1(31)8

38 11Add3. rr rr rr r r −−+= −−+= −−−= −= = The solution set is {11}.

42318Distributive property

6. 3(25)25

12. 121159 1248Subtract11. 4Divideby12. m m m += = =

The solution set is {4}.

13. 3(26)5(8)22 61854022 2222 xxx xxx xx +−+=− +−−=− −=−

This is a true statement, so the solution set is {all real numbers}.

14. 59(23)27 592327 31227 127 19 xxx xxx xx x x +−−=− +−+=− +=− +=− =−

The solution set is {19}.

61525Distributive property

152Subtract5. 17Add15. xx xx xx x −=+ −=+ −= =

The solution set is {17}.

7. 735

5Divideby7. x x = =

The solution set is {5}.

15. 1 236 1 666Multiplyby6. 236 32 rr r rr r rrr rr −= ⎛⎞⎛⎞⎛⎞ −= ⎜⎟⎜⎟⎜⎟ ⎝⎠⎝⎠⎝⎠ −= =

This is a true statement, so the solution set is {all real numbers}.

16. Multiply by 10 to clear decimals. 10[0.1(80)0.2]10(14) (80)2140Dist.prop. 380140 360 20 xx xx x x x ++= ++= += = =

The solution set is {20}.

17. 3(26)4(4) 326416 56516 616 xxxx xxxx xx −−+=−+ +−=−+ −=− −=−

This statement is false, so there is no solution set, symbolized by .∅

18. Multiply both sides by 6 to eliminate fractions. 12 6(3)(2)6(3) 23 12 6(3)6(2)18 23 3(3)4(2)18 394818 1718 1 1 xx xx xx xx x x x

+−−=

+−−= +−+= −+= −= =−

The solution set is { } 1.

19. Let x represent the number. 573

72Subtract 5. 7 Divide by 2. 2 xx xx x += =− −=−

The number is 7 2

20. Step2

Letthe x = number of Republicans. Then 24the x += number of Democrats.

Step3 (24)118xx++=

Step4 224118 294 47 x x x += = =

Step5

Since 47,2471.xx=+=

There were 71 Democrats and 47 Republicans.

Step6

There are 24 more Democrats than Republicans and the total is 118.

21. Step2

Let the x = land area of Rhode Island.

Then 5213the x += land area of Hawaii.

Step3

The areas total 7637 square miles, so (5213)7637.xx++=

Step4

252137637 22424 1212 x x x += = =

Step5

Since 1212,52136425.xx=+=

The land area of Rhode Island is 1212 square miles and that of Hawaii is 6425 square miles.

Step6

The land area of Hawaii is 5213 square miles greater than the land area of Rhode Island and the total is 7637 square miles.

22. Step2

Let the x = height of Twin Falls.

Then 5 the 2 x = height of Seven Falls.

Step3

The sum of the heights is 420 feet, so 5 420. 2 xx+=

Step4 5 22(420) 2 25840 7840 120 xx xx x x ⎛⎞+=⎜⎟ ⎝⎠ += = =

Step5

Since 55 120,(120)300. 22 xx===

The height of Twin Falls is 120 feet and that of Seven Falls is 300 feet.

Step6

The height of Seven Falls is 5 2 the height of Twin Falls and the sum is 420.

23. Step2

Let the x = measure of the angle. Then 90the x −= measure of its complement and 180the x −= measure of its supplement.

Step3

18010(90) xx−=−

Step4 18090010 9180900 9720 80 xx x x x −=− += = =

Step5

The measure of the angle is 80° Its complement measures 908010, °−°=° and its supplement measures 18080100. °−°=°

Step6

The measure of the supplement is 10 times the measure of the complement.

24. Step2

Let x = lesser odd integer. Then 2 x += greater odd integer.

Step3 2(2)(2)24xxx ++=++

Step4 2426 3426 2426 222 11 xxx xx x x x ++=+ +=+ += = =

Step5 Since 11,213.xx=+=

The consecutive odd numbers are 11 and 13. Step6

The lesser plus twice the greater is 11+2(13)=37, which is 24 more than the greater.

25. Solve for h

26. Solve for A 1 () 2 1 (8)(34) 2 1 (8)(7) 2 (4)(7) 28 AhbB A A =+ =+ = = =

27. Solve for r 2

29.832(3.14) 29.836.28 29.836.28 6.286.28 4.75 Cr r r r r π = = = = =

28. Solve for V. 3 3 4 3 4 (3.14)(6) 3 4 (3.14)(216) 3 4 (678.24) 3 904.32 Vr V π = = = = =

29. Solve for h Divide by . ,or Abh b bb AA hh bb = ==

30. Solve for h. 1 22()Multiplyby2. 2 2() 2() Divideby . ()() 22 ,or AhbB AhbB AhbB bB bBbB AA hh bBbB ⎡⎤=+⎢⎥ ⎣⎦ =+ + =+ ++ == ++

31. Solve for y 11 11 xy yx += =−+

32. Solve for y 3212 2312 3 6 2 xy yx yx −= −=−+ =−

33. Because the two angles are supplementary, (81)(36)180. 117180 11187 17 xx x x x −+−= −= = = Since 17,81135,xx=−= and 3645. x −=

The measures of the two angles are 135° and 45. °

34. The angles are vertical angles, so their measures are equal.

310420 1020 30 xx x x +=− =− = Since 30,310100xx=+= and 420100. x −=

Each angle has a measure of 100. °

35. Let the W = width of the rectangle. Then 12the W += length of the rectangle. “The perimeter of the rectangle is 16 times the width” can be written as 2216LWW += since the perimeter is 22. LW + Because 12, LW=+ we have 2(12)216. 224216 42416 12240 1224 2 WWW WWW WW W W W ++= ++= += −+= −=− = The width is 2 cm and the length is 21214cm +=

36. The sum of the three marked angles in the triangle is 180° 45(12.2)(32.8)180 460180 4120 30 xx x x x °++°++°=° += = =

Since 30,(12.2)42.2 and xx=+°=° (32.8)92.8. x +°=°

37. The ratio of 60 centimeters to 40 centimeters is 60cm3203 40cm2202 ⋅ == ⋅

38. To find the ratio of 90 inches to 10 feet, first convert 10 feet to inches. 10feet=1012=120inches

Thus, the ratio of 90 inches to 10 feet is 903303 1204304 ⋅ ==

39. 5 2130 30105Crossproducts 30105

Divide by 30. 3030 1057157 302152 p p p p = = = ===

The solution set is 7 2 ⎧ ⎫ ⎨ ⎬ ⎩⎭

40. ()() 52 36 6532Cross products 30663Distributive property 3096Add3. 924Subtract30. 248 93 xx xx xx xx x x +− = +=− +=− += =− ==−

The solution set is 8 . 3 ⎧ ⎫ ⎨ ⎬ ⎩⎭

41. Let the x = tax on a $36.00 item. Set up a proportion with one ratio involving sales tax and the other involving the costs of the items. dollars$36 $2.04$24 24(2.04)(36)73.44 73.44 3.06 24 x x x = == ==

The sales tax on a $36.00 item is $3.06.

42. Let the x = actual distance between the second pair of cities (in kilometers).

Set up a proportion with one ratio involving map distances and the other involving actual distances. kilometers80centimeters 150kilomters32centimeters 32(150)(80)12,000 12,000 375 32 x x x = == ==

The cities are 375 kilometers apart.

43. Let the x = number of gold medals earned by Italy. goldmedals2goldmedals 28medals7medals

72(28)56 8 x x x = == =

At the 2012 Olympics, 8 gold medals were earned by Italy.

44. To find the best buy, divide the price by the number of units to get the unit cost. Each result was found by using a calculator and rounding the answer to three decimal places. The best buy (based on price per unit) is the smallest unit cost. The results in the following table are rounded to the nearest thousandth.

Size Price Unit Cost (dollar per oz)

9 oz $3.49 $3.49 $0.388 9 ≈

14 oz $3.99 $3.99 $0.285 14 =

18 oz $4.49 $4.49 $0.249 18 ≈ () *

Because the 18-oz size produces the lowest unit cost, it is the best buy. The unit cost, to the nearest thousandth, is $0.249 per oz.

45. What percent of 12 is 21?

Let p denote the percent. 2112 21 1.75175% 12 p p =⋅ === 21 is 175% of 12.

46. 36% of what number is 900? Let n denote the number. 0.36900 900 Divideby0.36. 0.36 2500Simplify. n n ⋅= = = Thus, 36% of 2500 is 900.

Multiply by 10 to clear decimals. 1(15)62(15) 156230 15430 415 15 3.75 4 xx xx x x x +=+ +=+ += = ==

3.75 liters of 60% solution are needed.

48. Let the x = amount invested at 5%. Then 10,000the x −= amount invested at 3%. Interestinterest at 5%plusat 3%equals$400.

0.05+0.03(10,000)=400 xx

Solve the equation. 53(10,000)100(400) 530,000340,000 210,000 5000 xx xx x x +−= +−= = = Robert invested $5000 at 5% and 10,0005000$5000 −= at 3%.

49. Use the formula drt = or . d r t = 3150 8.203 384 d r t ==≈

Rounded to the nearest tenth, the Yorkshire’s average rate was 8.2 mph.

50. Let the t = number of hours until the planes are 1925 miles apart. Thethe distance the distancethe other distance one plane plane between flies northplusflies southequalsthe planes.

3504201925 tt

Solve the equation. 3504201925 7701925 192551 2 77022 tt t t += = === The planes will be 1925 miles apart in 1 2 2 hours.

0.1015+0.60()=0.2015 xx

47. Let the x = number of liters of the 60% solution to be used. Then 15the x += number of liters of the 20% solution. ()() 10%60%20% solutionplussolutionissolution.

51. The statement 4 x ≥− can be written as [ ) 4,.−∞

52. The statement 7 x < can be written as (,7). −∞

53. The statement 56 x −≤< can be written as [5,6).

54. By examining the choices, we see that 436 x −≤ is the only inequality that has a negative coefficient of x. Thus, B is the only inequality that requires a reversal of the inequality symbol when it is solved.

55. 63

3Subtract6. x x +≥ ≥−

Graph the solution set [3,). −∞

56. 542

2Subtract4. xx xx <+ <

Graph the solution set (,2). −∞

57. 618 618 Divide by6. 66 3 x x x −≤− ≥− ≥

Graph the solution set [3,). ∞

58. 8(5)(27)4 840274 424 46 xx xx x x −−+≥ −−−≥ −≥ ≥

Graph the solution set [46,). ∞

59. 431047 103 210 210 Divideby2. 22 5 xxxx xx x x x −>−+ >+ −> <− <−

Graph the solution set (,5). −∞−

60. 3(25)4(83)5(37) 61532121535 18471535 34735 312 4 xxx xxx xx x x x +++<+ +++<+ +<+ +< <− <−

Graph the solution set (,4). −∞−

61. 3214

423Subtract 1. 3 2Divideby2. 2 x x x −≤+≤ −≤≤ −≤≤

Graph the solution set 3 2,. 2 ⎡⎤ ⎢⎥ ⎣⎦

62. 93520

4315Subtract 5. 4 5Divideby3. 3 x x x <+≤ <≤ <≤

Graph the solution set 4 ,5. 3 ⎛⎤ ⎜ ⎥ ⎝⎦

63. Let the x = score on the third test. The average isat of the90. least threetests

9488 90 3 x ↓↓↓ ++ ≥ 182 90 3 182 33(90) 3 182270 88 x x x x + ≥ + ⎛⎞ ≥ ⎜⎟ ⎝⎠ +≥ ≥

In order to average at least 90, Awilda’s score on her third test must be 88 or more.

64. Let the n = number. “If nine times a number is added to 6, the result is at most 3” can be written as 963. n +≤

Solve the inequality. 963

93Subtract6. 3 Divideby9. 9 1 3 n n n n +≤ ≤− ≤ ≤−

All numbers less than or equal to 1 3 satisfy the given condition.

Chapter 2 Mixed Review Exercises

1. 5 72 27(5)Crossproducts 2735 535 7 xx xx xx x x = =− =− −=− = The solution set is {7}.

The solution set is (,2). −∞

4. 25413

9Divide by 2. kk kk. k k −=+ −−= −= =−−

The solution set is {9}.

5. 0.050.024.9 xx+=

2. Solve for . Iprtr = Divideby. ,or Iprtpt ptpt II rr ptpt = == 3. 24 24 Divideby2. 22 2 x x x −>− <− <

2513Subtract4 218Add 5.

To clear decimals, multiply both sides by 100. 100(0.050.02)100(4.9) 52490 7490 70 xx xx x x += += = =

The solution set is {70}.

6. 23(5)4 23154 1734 1744 413 1313 44 xx xx xx x x x −−=+ −+=+ −=+ −= −=− ==

The solution set is 13 4 ⎧ ⎫ ⎨ ⎬ ⎩⎭

7. 9(72)3(2) 97232 2222 22 xxxx xxxx xx −+=+− −−=+− −=+ −=

Because 22−= is a false statement, the given equation has no solution, symbolized by ∅

8. 115 752 326 115

Subtract 7. 326 sss sss ++=++ +=

The least common denominator is 6. 115 66 326 235 55 sss sss ss ⎛⎞⎛⎞ += ⎜⎟⎜⎟ ⎝⎠⎝⎠ += =

Because 55ss = is a true statement, the solution set is {all real numbers}.

9. Let the x = number of calories a 175-pound athlete can consume. Set up a proportion with one ratio involving calories and the other involving pounds. calories175pounds

50calories2.2pounds

13. Use the formula , drt = or , d t r = where 819 d = and 63. r = 819 13 63 d t r ===

It took Janet 13 hours to drive from Louisville to Dallas.

14. Let the x = rate of the slower train. Then 30the x += rate of the faster train.

To the nearest hundred calories, a 175-pound athlete in a vigorous training program can consume 4000 calories per day.

10. Let the x = sales for DiGiorno, in millions of dollars.

Then 399.9 x the sales for Red Baron, in millions of dollars.

(399.9)937.5 2399.9937.5 21337.4 668.7

DiGiorno sold $668.7 million worth of frozen pizza and Red Baron sold $268.8 million worth of frozen pizza.

11. The results in the following table are rounded to the nearest thousandth.

Because the 160-oz size produces the lowest unit cost, it is the best buy. The unit cost, to the nearest thousandth, is $0.062 per oz.

12. The angles make up a right angle, so the sum of their measures is 90. ° ()() 38290 11290 1188 8 xx x x x °++°=°

Slower Train x 3 3x Faster Train 30 x + 3 3(30) x +

The sum of the distances traveled by the two trains is 390 miles, so 33(30)390. xx++=

Solve the equation. 33(30)390 3390390 690390 6300 50 xx xx x x x ++= ++= += = =

Since 50,3080.xx=+=

The rate of the slower train is 50 miles per hour and the rate of the faster train is 80 miles per hour.

15. Let the x = length of the first side. Then 2the x = length of the second side. Use the formula for the perimeter of a triangle, , Pabc =++ with perimeter 96 and third side 30.

23096 33096 366 22 xx x x x ++= += = =

The sides have lengths 22 meters, 44 meters, and 30 meters. The length of the longest side is 44 meters.

16. Let the s = length of a side of the square. The formula for the perimeter of a square is4. Ps = cannotbe

The perimetergreaterthan200. 4200 4200 50 s s s ↓↓↓ ≤ ≤ ≤

The length of a side is 50 meters or less.

Chapter 2 Test

1. 59721

2921Subtract 7.

212Subtract 9.

6Divide by 2. xx xx x x +=+ −+= −= =−−

The solution set is { }6

2. () 4 12 7 747 12 474 21 x x x −=− ⎛⎞⎛⎞⎛⎞ −−=−−

The solution set is { } . 21

3. 7(4)32(1) 74322 112 112 xxx xxx xx −−=−++ −+=−++ −+=−+ =

Because the last statement is false, the equation has no solution set, symbolized by .∅

4. To clear decimals, multiply both sides by 100. ()()() ()() 1000.06200.08101004.6 620810460 6120880460 1440460 14420 30 xx xx xx x x x ++−= ⎡⎤ ⎣⎦ ++−= ++−= += = =

The solution set is { } 30.

5. ()() 824448 16321632 xx xx −+=−+ −−=−−

Because the last statement is true, the solution set is {all real numbers}.

6. ()() 23531 231531Distributive property 3174 1744Add 3. 134Subtract 4. 13 4 xx xx xx xx x x −−=++ −+=++ −+=+ =+ = =

The solution set is 13 4 ⎧ ⎫ ⎨ ⎬ ⎩⎭

7. Let the x = number of games the Cardinals lost. Then 233the x −= number of games the Cardinals won.

The total number of games played was 162. ()233162 333162 3195 65 xx x x x +−= −= = =

Since 65,23397xx=−=

The Cardinals won 97 games and lost 65 games.

8. Let the x = area of Kauai (in square miles). Then 177the x += area of Maui (in square miles), and ()17732933470thexx ++=+= area of Hawaii.

()()17734705300 336475300 31653 551 xxx x x x ++++= += = =

Since 551,177728xx=+= and 34704021 x +=

The area of Hawaii is 4021 square miles, the area of Maui is 728 square miles, and the area of Kauai is 551 square miles.

9. Let the x = measure of the angle. Then 90the x −= measure of its complement, and 180the x −= measure of its supplement. () 18039010 180270310 1802803 1802280 2100 50 xx xx xx x x x −=−+ −=−+ −=− += = =

The measure of the angle is 50° The measure of its supplement, 130, ° is 10° more than three times its complement, 40°

10. Step2

Let the x = lesser even integer. Then 2the x += greater even integer.

Step3 ()32022 xx=++

Step4 ()32022 32024 3242 24 xx xx xx x =++ =++ =+ = Step5 Since 24,226.xx=+=

The consecutive even numbers are 24 and 26.

Step6

Three times the lesser is () 32472, = and 20 more than twice the greater is () 20+22672. =

11. (a) Solve 22 PLW =+ for W. 22 22 ,or 22 PLW PLPL WW −= ==

(b) Substitute 116 for P and 40 for L in the formula obtained in part (a).

2 2 116240 2 1168036 18 22 PL W = = === 12. 548 54585Subtract 5. 485 485 Divide by4. 44 855 ,or2 44 xy xyxxx

(There are other correct forms.)

13. The angles are vertical angles, so their measures are equal. 31545 155 20 xx x x +=− =− = Since 20,31575and45=75.xxx =+=− Both angles have measure . 75°

14. 12 816 168(12)Cross products 1696 1696

Divide by 16. 1616 6 z z z z z = = = = = The solution set is {6}. 15. ()() 53 34 4533 42039 209 29 xx xx xx x x +− = +=− +=− +=− =−

The solution set is {29}.

16. What percent of 65 is 26? Let p denote the percent. 2665 26 0.440% 65 p p =⋅ === 26 is 40% of 65.

17. The results in the following table are rounded to the nearest thousandth.

Because the 16-oz size produces the lowest unit cost, it is the best buy. The unit cost, to the nearest thousandth, is $0.249 per oz.

18. Let the x = actual distance between Seattle and Cincinnati. () miles92inches 1050miles42inches

4292105096,600 96,600 2300 42 x x x = == ==

The actual distance between Seattle and Cincinnati is 2300 miles.

19. Let the x = amount invested at 3%. Then 6000 x += the amount invested at 4.5%.

Amount Invested (in dollars) Rate of Interest Interest for One Year x 0.03 0.03 x 6000 x + 0.045 0.045(6000) x + () 0.030.0456000870 xx++=

To clear decimals, multiply both sides by 1000. () 30456000870,000

23. 103414 6318Add 4. 26Divide by 3. x x x −<−≤ −<≤ −<≤

Graph the solution set ( ] 2,6.

3045270,000870,000 75270,000870,000

75600,000 8000 xxx xx x x x ++= ++= += = =

Since 8000,600014,000.xx=+= Carlos invested $8000 at 3% and $14,000 at 4.5%.

20. Use the formula drt = and let t be the number of hours they traveled. rtd

First Car 50 t 50t

Second Car 65 t 65t

First car’ssecond car’stotal distanceanddistanceisdistance.

24. 42(3)4(35)7 4264357 2610 610 4 14 11 4 xxxx xxxx xx x x x x −+−≥−+− −+−≥−−− −−≥−− −−≥− −≥− ≤ ≤

Graph the solution set ( ] ,4.−∞

25. Let the x = score on the third test. The average isat of the80. least three tests

5065460 tt

Solve the equation. 5065460 115460 4 tt t t += = =

The two cars will be 460 miles apart in 4 hours.

21. (a) The set of numbers graphed corresponds to the inequality 0. x < (b) The set of numbers graphed corresponds to the inequality 23. x −<≤

22. 333

11Divide by 3. x x −>− <−

Graph the solution set (),11.−∞

In order to average at least 80, Susan’s score on her third test must be 83 or more.

Chapters R–2 Cumulative Review Exercises 1. 517501528 6415606060 6528 60 37 60 +−=+− = =

Chapter 2 Linear Equations and Inequalities in One Variable

2. 91659168 838835 33168 835 48 5 ⋅÷=⋅⋅ = =

3. 4.812.516.7334.03 ++=

4. “The difference of half a number and 18” is written 1 18. 2 x

5. “The quotient of 6 and 12 more than a number is 2” is written 6 2. 12 x = +

6. () ()() 8(7)562 1 351 8758 1 351 5640 1 151 16 1 16 11 −+ ≥ ⋅+ ≥ ⋅+ ≥ + ≥ ≥ The statement is true.

7. 22 4(9)(2)36(2) 313 72 9 8 = −−⋅ = =−

8. ()()()()()() 7144844 32(4) 28 −−−+−=−−+− =+− =

9. ()()() ()()() () 22 6421264212 919 37 2424 919 0 0 28 = + + = + ==

10. Let 2, x =− 4, y =− and 3. z = 2323 33(2)(4) 44(3) 3(4)(64) 12 1264 12 76 12 19 3 xy z = = + = = =−

11. 7()77 pqpq +=+

The multiplication of 7 is distributed over the sum, which illustrates the distributive property.

12. () 770 +−=

A number added to its opposite is equal to 0. This illustrates the inverse property (of addition).

13. () 3.513.5 =

A number multiplied by 1 is equal to itself. This illustrates the identity property (of multiplication).

14. 268 66 1 rr r r −= −= −=

Check 1: 88True r =−−=−

The solution set is { } 1.

15. ()() 452311 4510331 5632 862 88 1 ss ss ss s s s −+=+− −−=+− −−=+ −−= −= =−

Check 1: 11True s =−−=−

The solution set is { } 1.

16. () 23 17 34 23 121217LCD12 34 89204 17204 12 xx xx xx x x +=− ⎛⎞+=−=⎜⎟ ⎝⎠ +=− =− =−

Check 12: 1717True x =−−=−

The solution set is { } 12.

17. 234 52 (23)(2)(5)(4) 46520 620 26 xx xx xx x x +− = +=− +=− =− =

Check 26: 1111True x ==

The solution set is { } 26.

18. Solve 3424for. 4243 243 4 xyy yx x y += =− =

19. Solve PabcB =+++ for c

Subtract a, b, and B PabcB PabBc =+++ −−−=

20. 6(1)2(35)4 666104 12164 1212 1 rr rr r r r −+−≤− −+−≤− −≤− ≤ ≤

Graph the solution set ( ],1−∞

21. 1899 21Divide;reverse the symbols. or 1<2 z z z −≤−< ≥>− −≤ Graph the solution set (1,2].

22. Let the x = length of the middle-sized piece. Then 3the x = length of the longest piece, and 5the x −= length of the shortest piece.

3(5)40 5540 545 9 xxx x x x ++−= −= = =

The length of the middle-sized piece is 9 centimeters, that of the longest piece is 27 centimeters, and that of the shortest piece is 4 centimeters.

23. Let the r = radius and use 3.14 for .π Using the formula for circumference, 2,Cr π = and 78, C = we have 278 78 12.4204 2 r r π π = =≈

To the nearest hundredth, the radius is 12.42 cm.

24. Let the x = rate of the slower car. Then 20the x += rate of the faster car. Use the formula drt = slowerfastertotal ()(4)(20)(4)400 4480400 880400 8320 40 ddd xx xx x x x += ++= ++= += = = The rates are 40 mph and 60 mph.

25. (a) The segment of the circle representing white cars is 19% of the circle. What is 19% of 2.8 million?

19%2.80.192.80.532 ⋅=⋅= 0.532 million cars, or 532,000 cars, are white.

(b) The segment of the circle representing silver cars is 18% of the circle. What is 18% of 2.8 million?

18%2.80.182.80.504 ⋅=⋅= 0.504 million cars, or 504,000 cars, are silver.

12%2.80.122.80.336 ⋅=⋅= 0.336 million cars, or 336,000 cars, are red.