Solutions for Elementary Algebra for College Students 10Th Us Edition by Angel

Page 1


Chapter 2

Exercise Set 2.1

1. Intheexpression5x –3y +17–2x,17iscalleda constantterm.

2. Whenweapplythedistributivepropertyto–2(2x –3y –9),weobtain–4x +6y +18.

3. Intheexpression5x –3y +17–2x,5x and–2x arecalledliketerms.

4. Intheexpression5x –3y +17–2x,–3iscalled thecoefficientofthesecondterm.

5. Intheexpression5x –3y +17–2x,5x and–3y arecalledunliketerms.

6. Intheexpression12x +17,12and x arefactorsof thefirstterm.

7. Intheexpression4x2+17x –90,theparts–4x2 , 17x,and–90arecalledterms.

8. Intheexpression17x, x iscalledavariable.

9. 3912  xxx

10. 5813  yyy

11. Therearenoliketerms. 36 x 

12. Therearenoliketerms. 43xy 

13. 3443 53 yyyy y  

14. 47434 xxx

15. –2w –3w +5=–5w +5

16. 847127 yyy 

17. 35745  abaab

18. 54234 cdccd

19. 2222 2 xxxx x  

20. 3431333413 9 aaaa 

21. 3+6x –3–6x =6x –6x +3–3=0

22. 57755577 0 yyyy

23. 5241624516 221 tttt t  

24. 71355713 46 dddd d  

25. 4p –6–16p –2=4p –16p –6–2 =–12p –8

26. –6t +5+2t –9=–6t +2t +5–9 =–4t –4

27. –2x +4x –8=2x –8

28. 448448 34 xxxx x  

29. 27522572 35 xyxyxxyy xy 



30. 37943479 716 xxxx x  

31. 222 2 xxyxy 

32. 2222 2 163316 27 xxxx x  

33. Therearenoliketerms.

32 523 www

34. Therearenoliketerms.

32 3772 mmm

  

35. 2222 2222 22 39752 37952 10107 xyxy xxyy xy

36. 22 22 2 46361 43661 775 xyxy xxyy xy   

37. 22 2 43923249 3213 nnnn nn 



38. 2222 2 513531 81 xxxxxx xx  

Chapter 2: Solving Linear Equations and Inequalities

39. 363324 4114444 9 44

aaaa a

40. 32218 472828 13 28

bb

41. 3203 4 555 23 5

42. 33 22 44 34 2 44 7 2 4

43. Therearenoliketerms. 251 399 xy

44. Therearenoliketerms. 311 474 pq

45. 3737 3232 5454 1235 5 2020 23 5 20 xxxx

46. 131311 44 245425 352 4 41010 33 4 410 yxyxyy xyy

47. 5.16.424.35.14.36.42 0.86.42 nnnn n

48. 2.538.19.12.539.18.1 11.638.1 cccc c

49. 13.41.28.314.68.3 xxx

50. 22 43.15.248.3 xx

51. 19.3640.0212.2518.3 40.0218.319.3612.25 21.727.11 xx xx x  



 

52. 3.413.011.0917.3 3.41.0913.0117.3 4.494.29 kk kk k



53. 332332 32 252522 72 zzzzzzzz zzz 



54. 32233322 32 74255427 427 cccccccc cc  

55. 22 2255235 xxxxx 

56. x2–3xy –2xy +6= x2–5xy +6

57. 322 322 32 26231 26231 251 aaaaa aaaaa aaa    58 322 322 32 336224 332624 344 bbbbb bbbbb bbb   

59.     52552 510 xx x  

60.     54554 520 xx x  

61.  3413431 123 

62.

 xx x

  bb b b b



   725725 7275 1435 1435  

63.  2342324 68   cc c

64.  5925952 4510 

 dd d

 

 vv v v v



65.    434434 4344 1216 1216



66. 

825825 8285 1640 1640

67.  3273237 621

68.  9589598 4572

69.

532532 5352 1510

70. 

867867 8687 4856

71.

2392329 618 618

5435453 2015 2015

467467 4647 2428

77.

1 11 pqpq pq pq

24–8124(–8) 12141–8 2–48 2–48 xyxy xy xy xy

32373[23(7)] 3(2)(3)3(3)7 6921 abab ab ab

2952595 1045 xyxy xy 

8178717 567 bb b

 3243(3)2(3)4 3(6)(12) 3612 xyxy xy xy

86. 4(238)4(2)(4)(3)(4)(8) 81232 mnmn mn

335335 38 tt t

93.   62496818 1418 xxxx x 

32332

2232223

96.

653653

42334812312 831212 5 cccc cc

98.  53264515102430 15241030 3940 dddd dd d 



99.   8383 73 xxxx x  

100.   534534 354 49 xxxx xx x   

101.   (34)2342 342 46 ssss ss s   

102.   622356465 4566 wwww ww w   

103.

412344412324 44624 42464 22 xxxx xx xx x 

104.

432546 43425546 1285206 1265208 6512 bcb bcb bcb bbc bc

105.     43412443412 412412 441212 0 mmmm mm mm  

   

106.       323233(2)332 3636 3366 0 abababab abab aabb

107.   43454345 3445 35 xx x x   

108.   2623266(2)3 26123 415 yyyy yy y   

109.

110.

111.

112.

3224636286 32686 146 20 tttt tt t t

113. 3131 22 4343 31 2 43 381 443 51 43 xxxx xx xx x

114. 7171 33 8282 71 3 82 7241 882 311 82 xxxx xx x

 41431 31 57577 28151 35357 431 357 xxxx xx x

117.   0.450.620.450.62 0.450.62 6 yy y y

 1.0380.80.031.2 1.0380.80.031.2 0.81.21.0380.03 27 aa aa aa a

119.

0.2640.4 0.260.2440.4 1.20.241.6 0.241.21.6 0.242.8 xy xy xy xy xy 

120.  1.054641.53.05 4.26.3612.2 4.266.312.2 4.265.9 xy xy xy xy

121.   111336 336 232233 13 2 22 13 2 22 1234 2222 37 22 xxxx xx xx xx x 

122.   212414 24 323322 214 2 323 4346 6633 110 63 rrrr rr rr r

123. 23 

Chapter 2: Solving Linear Equations and Inequalities

124. 24

125. 232 xyxyy xxyyy xy

126. 222 222 26 xy xy xy

127. 118,29,36 positivefactors:1,2,3,6,9,18

128. 124,212,38,46

positivefactors:1,2,3,4,6,8,12,24

129. a. Toremovetheparentheses,multiplytheterms insidetheparenthesesby–1.Thesignsofall termsinsidetheparentheseschangewhenthe parenthesesareremoved.

b. –(x –8)=–1(x)–1(–8)=–x +8

130. a. Liketermsaretermsthathavethesame variablewiththesameexponent.

b. 3x and4y arenotliketermsbecausethe variablesaredifferent.

c. 7and–2areliketermsbecausetheyareboth constants.

d. 5x2and5x arenotliketermsbecausethe variableshavedifferentexponents.

e. 4x and–5xy arenotliketermsbecausethe variablesaredifferent.

131.

234523 2342023

11 32126 43 11 32226 43 11 128 43 1281 4433 2118 4343 yy yy yy yy yy

32332 661212 129 612 y y

77

161616

137. 413(6)4136 176 11

138. Answerswillvary.Theanswershouldinclude thattheorderisparentheses,exponents, multiplicationanddivisionfromlefttoright,and additionandsubtractionfromlefttoright.

139. Substitute–1foreach x intheexpression.

22561516 156 66 12

Exercise Set 2.2

1. Tosolvetheequation6= x –2,firstadd2toboth sidesoftheequation.

2. Numbers,whichwhensubstitutedforthevariable, maketheequationatruestatement,arecalled solutions.

3. Thestrategyforsolvingalinearequationinone variableincludesaprocesstoisolatethevariable.

4. Thesymbol ?  isusedwhencheckingthesolution inthegivenequation.

5. Theequation x –3=–19isanexampleofalinear equationinthevariable x.

6. Twoequationswiththesamesolutionsarecalled equivalentequations.

7. Astatementofequalitybetweentwoexpressions, like x +7=19,iscalledanequation.

8. Tosolvetheequation x –6=–2,firstadd6to bothsidesoftheequation.

9. Whensolvingtheequation x + a = b,theopposite of a isaddedtobothsides.

10. (TrueorFalse)Theequations x +19=–14and x +60=27areequivalentequations.True

11. Substitute2for x =2.

435 4235 835 55True x 

Sinceweobtaintruestatement,2isasolution.

12. Substitute5for x,5  x .

x

61713 651713 301713 1313True

Sinceweobtainatruestatement,5isasolution.

13. Substitute–3for x, x =–3.

2552 235532 6551 115False

Sinceweobtainafalsestatement,–3isnota solution.

14. Substitute–7for q,7  q

31175 3711775 318495 54495 55False

Sinceweobtainafalsestatement,–7isnota solution.

15. Substitute–15for p, p =–15.

25710 215515710 305810 304010 1010True pp

Sinceweobtainatruestatement,–15isa solution.

16. Substitute–2for k, k =–2.

5618 526218 10638 10188 88True kk

Sinceweobtainatruestatement,–2isasolution.

17. Substitute3for m,3  m .

347535 33473535 377235 211435 3535False

 mm

Sinceweobtainafalsestatement,3isnota solution.

18. Substitute4for n,4  n .

8163424 841634424 83612424 246824 244824 2424False



 nn

Sinceweobtainafalsestatement,4isnota solution.

19. Substitute1 2 for x, 1 2 x  . 4422 11 4422 22 2412 21False xx

Sinceweobtainafalsestatement,1 2 isnota solution.

Chapter 2: Solving Linear Equations and Inequalities

20. Substitute 2 3 for x, 2 3  x . 61125 22 61125 33 1224 15 33 4185

Sinceweobtainatruestatement, 2 3 isasolution.

21. Substitute3.4for x, x =3.4.

Sinceweobtainatruestatement,3.4isasolution.

22. Substitute5.5for x,5.5 x 

35.5355.5418 38.551.518 25.57.518 1818True

Sinceweobtainatruestatement,5.5isasolution.

23. x +2=7 2272 05 5 x x x

Check:27 527 77True x

24. x +6=14 66146 08 8 x x x

Check:614 8614 1414True x

25. –6= x +1 x +1=–6 1161 07 7 x x x

Check:16 716 66True x 

26. –5= x +4 x +4=–5 4454 09 9 x x x

Check:45 945 55True x

27. 48 x  4484 04 4 x x x

Check:48 448 88True x 

28. p –5=–12 55125 07 7 p p p

Check:512 7512 1212True p

29. t +9=52 99529 043 43 t t t

Check:952 43952 5252True

ISM: Elementary Algebra

30. x +8=17 817 88178 09 9 x x x x  

 Check:817 9817 1717True x 

31. –6+ w =9 6696 015 15 w w w

Check:69 6159 99True w

32. –10+ n =5 1010510 015 15 n n n 

Check:1015 10155 55True n

33. 27= x +16 27161616 110 11 x x x

Check:2716 271116 2727True x 

34. 50= x –35 50353535 850 85 x x x

Check:5035 508535 5050True x

35. 1814 x  18141414 40 4 x x x

Check:1814 18414 1818True x 

2: Solving Linear Equations and Inequalities

36. –4= w +5 4555 90 9 w w w 

Check:45 495 44True w 

37. 94 x  9949 05 5 x x x 

Check:94 9(5)4 44True x 

38. 12+ w =9 1212912 03 3 w w w 

Check:129 1239 99True w   

39. 4+ x =–8 4484 012 12 x x x   

Check:48 4128 88True x 

40. 7+ z =–3 7737 010 10 z z z 



Check:73 7103 33True z   

41. 7+ r =–23 723 77237 030 30 r r r r 

Check:723 7(30)23 2323True r

Chapter 2: Solving Linear Equations and Inequalities

42. 7+ v =–7 7777 014 14 v v v

Check:77 7147 77True v

43. 8=8+ v 8888 00 0 v v v

Check:88 880 88True v

44. –5=–5+ m 5555 00 0 m m m

Check:55 550 55True m

45. 750 x  77507 057 57 x x x

Check:750 75750 5050True x

46. –5+ c =–23 55235 018 18 c c c

Check:523 51823 2323True

47. 1216 x  12161616 40 4 x x x

Check:1216 12164 1212True x 

48. 62= z –15 62151515 770 77 z z z

Check:6215 627715 6262True z

49. 15+ y =–50 15155015 065 65 y y y

Check:1550 156550 5050True y 

50. 204 x  20444 240 24 x x x 

Check:204 20424 2020True x  

51. 1515 x  15151515 00 0 x x x 

  Check:1515 15015 1515True x  



52. 77 x  7777 00 0 x x x    Check:77 707 77True x   

53. 512 x  5121212 170 17 x x x    Check:512 51712 55True x   

54. 8= x –9 8999 170 17 x x x

55. 5024 x  50242424 260 26 x x x

Check:5024 502624 5050True x

56. 3712 x  37121212 250 25 x x x

57. 43=15+ p 43151515 280 28 p p p

Check:4315 431528 4343True p

58. 2574 x  25747474 990 99 x x x

Check:2574 257499 2525True x 

a a

59. 71 88  a 7717 8888 6 0 8 3 4

Check:71 88 371 488 671 888 11 True 88

60. 51 66  b 5515 6666 4 0 6 2 3

 b b b

Check:51 66 251 366 451 666 11 True 66

Chapter 2: Solving Linear Equations and Inequalities

61. 15 96  c 1151 9969 152 0 1818 17 18

64. 3.4111 t  3.413.41113.41 07.59 7.59 t t t

65. –5.9= x +4.01 –5.94.014.014. 9.910 01 9.91 x x x 

 Check:5.94.01 5.99.914.01 5.95.9True x 

66. –13.7= x +7.28 13.77.287.287.28 20.980 20.98 x x x 

 Check:13.77.28 13.720.987.28 13.713.7True x 

67. No;therearenorealnumbersthatcan make12 xx

68. Yes;infinitenumber;theequation44 xx is truenomatterwhatrealnumberissubstitutedfor x

69. Toisolatethevariable,usepropertiesthatallow ustogetthevariablebyitselfononesideofthe equation.

Forexample,intheequation x +8=–12,you wouldsubtract8frombothsidessothatthe variable x remainsontheleftbyitself.

70. Allthreeequationshavethesamesolution,where x =1.

71. x x

72.

73.

74.

75. a. yes b. yes c. yes d. yes e. Sincetheleftsideoftheequationisequalto therightsideoftheequation,2x +6=2x +6, thesolutionisallrealnumbers.

76. 751425142511 15630303030

77.

229 11322931 12824242424

4325743657 43567

Exercise Set 2.3

1. Theexpression1 3 x canberewrittenas 3 x

2. Tosolvetheequation3x =5,multiplybothsides oftheequationby1 3

3. Tosolvetheequation7 3 x  ,wemultiplyboth sidesoftheequationby3.

4. Thestrategyforsolvingalinearequationinone variableincludesaprocesstoisolatethevariable.

5. Thesymbol ?  isusedwhencheckingthesolution inthegivenequation.

6. Anothernameforthereciprocalofanumberis themultiplicativeinverse.

7. Thereciprocalof7 x is7 x .

8. Twonumbersequaltotheirownreciprocalare1 and–1.

9. 520 520 55 4 x x x   

Check:520 5420 2020True x   

10. 550 550 55 10 x x x   

Check:550 51050 5050True x    11. 412 412 44 3 x x x   

Check:412 4312 1212True x 

65. a. In5+ x =10,5isaddedtothevariable, whereasin5x =10,5ismultipliedbythe variable.

66. a. In3+ x =6,3isaddedtothevariable,whereas in3x =6,3ismultipliedbythevariable.

Elementary Algebra

67. Multiplyingby32iseasierbecausetheequation involvesfractions.

68. Multiplyingby1 4 is easierbecausetheequationinvolvesafraction.

69. Multiplyingby73iseasierbecausetheequation involvesfractions.

70.

b. Dividebothsidesoftheequationby 

c.

71.

b. Dividebothsidesoftheequationby

c.

b. Multiplybothsidesoftheequationby

75. 23 42636168636 164836 16166 06 6

76. Associativepropertyofaddition

Exercise Set 2.4

1. Themultiplicationpropertyofequalityallowsus tomultiplyordividebothsidesofanequationby thesamenonzeronumber.

2. Theadditionpropertyofequalityallowsustoadd orsubtractthesamenumberfrombothsidesofan equation.

3. Intheequation0.51.21.8 x  wecaneliminate thedecimalnumbersbymultiplyingbothsidesof theequationby10.

4. Intheequation0.02175335.25 x  wecan eliminatethedecimalnumbersbymultiplying bothsidesoftheequationby100.

5. Wecaneliminatefractionsfromanequationby multiplying both sidesoftheequationbytheleast commondenomination(LCD)ofthefractions withintheequation.

6. Intheequation255 386 x  wecaneliminatethe decimalnumbersbymultiplyingbothsidesofthe equationby24.

7. 3612

2921

5619

578

5210

y y y y y

14. 359 35595 314 314 33 14 3

15. 5919 599199 510 510 55 2 k k k k k

16. 7620 766206 714 714 77 2

 z z z z z

17. 125 1212512 7 (1)()(1)(7) 7 x x x x x

18.

138 1313813 21 1121 21 x x x x x

19. 8319 883198 311 311 33 11 3 x x x x x

ISM: Elementary Algebra

20. 5818 558185 813 813 88 13 8 x x x x x

21. 16514 1655145 1619 1619 1616 19 16 x x x x x

 x x x x x

22. 13827 1388278 1319 1319 1313 19 13

23. 959 99599 50 50 55 0 x x x x

24. 241624 2424162424 160 160 1616 0 x x x x x

25. 7162 71616216 714 714 77 2 r r r r r

26. 248 24484 212 212 22 6 w w w w w

27. 6059 609599 515 515 55 51 5 s s s s s

28. 4568 458688 536 536 66 53 6 v v v v v

29. 14583 14538 1428 148288 62 62 22 3 xx xx x x x x x

      

30. 15633 15633 1593 153933 189 189 99 2 xx xx x x x x x       

31. 2.39.346.3 2.39.349.346.39.34 2.315.64 2.315.64 2.32.3 6.8 x x x x x     

Chapter 2: Solving Linear Equations and Inequalities

32. 1.51.053.9 1.51.051.053.91.05 1.54.95 1.54.95 1.51.5 3.3 q q q q q

33. 0.0716.05

1.0716.05 1.0716.05 1.071.07 15

34. 0.0817.28 1.0817.28 1.0817.28 1.081.08 16     xx x x x

38. 

1 23 5 1 525(3) 5 215 22152 17 x x x x x

39. 3 9 7 3 77(9) 7 363 33633 60 d d d d d

35. 0.912.250.015.85

0.910.012.255.85

0.902.255.85

0.902.252.255.852.25

0.903.60 0.903.60 0.900.90 4 yy yy y y y y y

36. 0.150.051.350.20

40. 6 2 5 6 55(2) 5 610 66106 16 m m m m m

0.150.050.201.35

0.150.151.35

0.151.350.151.351.35

1.500.15

1.500.15

0.150.15 10 xx xx x x x x x

1 64 7 1 7674 7 628 66286 22 x x x x x

1 77 3 1 373(7) 3 721 77217 14 t t t t t

1 97 5 5195(7) 15 935 99359 26

3 512 4 434 512 343 516 55165 11 x x x x x

44.

47. 42 77 42 77 77 42 4424 2 x x x x x

51.

554 62 t  554 66 62 53(54) 51512 512151212 1715 1715 1515 17 15 t t t t t t t

7(1)14 7114 77 1121 3 

x x x x 52. 3(2)12 3612 366126 318 318 33 6 x x x x x x

53. 2(3)26 2626 266266 220 220 22 10 x x x x x x

60. 2(8)51 21651 2211 22121121 222 222 22 11 x x x x x x x

qq qq q q q q

61. 25(7)13 253513 735351335 722 722 77 22 7 

62.  34211 34811 7811 788118 719 719 77 19 7 rr rr r r r r r

63. 53479 479 47797 416 416 44 4 xxx x x x x x

64. 7421126 51126 511112611 515 515 55 3 xxx x x x x x

65. 0.7(3)1.4

0.72.11.4

0.72.12.11.42.1

0.73.5 0.73.5 0.70.7 5 x x x x x x

66. 

0.596.5 0.596.5 0.50.5 913 99139 4 x x x x x

68. 0.15(61)4.8

0.90.154.8

0.90.150.154.80.15

69. 32(3)21 32621 23621 211 21111 22 22 22 1 x x x x x x x x

70.   734536 7312536 3712536 32436 324243624 312 312 33 4 x x x x x x x x

 cc cc cc c c c

71. 3(5)713 35713 35713 886138 65 66 5 6

72.  64513 64513 56413 41013 xx xx xx x 

410101310 43 43 44 3 4 x x x x

73. 4.856.41.1125.8 6.44.851.1125.8 6.45.9625.8 6.45.965.9625.85.96 6.419.84 6.419.84 6.46.4 3.1 x x x x x x x  

Chapter 2: Solving Linear Equations and Inequalities

Chapter 2: Solving Linear Equations and Inequalities

90.

73 5 168 7315 1688 7315 1616 1688 7630 73063030 h h h h h

95. 313 564 313 6060 564 361045 3610101045 2645 q q q q q

2645 4545 26 45 q q



96. 315 564 315 6060 564 361075 3610101075 2675 2675 7575 m m m m m m

97. a. Bysubtractingfirst,youwillnothavetowork withfractions.

b. 3211 322112 39 39 33 3 x x x x x

98. a. Byaddingfirst,youwillnothavetoworkwith fractions. b. 5312

102.  -  = @

100.

103. a. Let x =costofoneboxofstationery 3x +6=42

b. 3642 366426 336 12 x x x x 

Aboxofstationerycosts$12.00.

104. a. Let x =costofonerollofcandy. 3x +0.5=2.75 b. 30.52.75 30.50.52.750.5 32.25 0.75 x x x x     Arollofcandycosts$0.75.

105. False.2isarealnumberbutisirrational.

106.

107. Tosolveanequation,weneedtoisolatethe variableononesideoftheequation.

108. Tosolvetheequation,wedividebothsidesofthe equationby–4.

Mid-Chapter Test: Sections 2.1 – 2.4

1. 5912476 5794126 256 xyyx

2. 231231 88 542542 815161 202022 715 202

6. Substitute2forinthefollowingequation

Sinceweobtainedatruestatement,2isasolution.

7. 2 Substituteforinthefollowingequation 5

Sinceweobtainedafalsestatement,2 5 isnota solution.

20.   89462 893662 983662 9222 920 20 9 y y y y y y

Exercise Set 2.5

1. Ourgoalinsolvingalinearequationistoisolate thevariable.

2. Toeliminatedecimalnumbersfromanequation wecanmultiplybothsidesoftheequationbya poweroften.

3. Toeliminatefractionsfromanequationwecan multiplybothsidesoftheequationbytheleast commondenominatorofthefractionsinthe equation.

4. Aconditionalequationistrueforaspecificvalue orvaluesofthevariable.

5. Anidentityistrueforallvaluesofthevariable.

6. Acontradictionisnottrueforanyvalueofthe variable.

7. Whenwehaveanidentity,wewillstatethe solutionisallrealnumbers.

8. Whenwehaveacontraction,wewillstatethat thereisnosolution.

9. Whensolvinganequationandweobtainan equationthatisalwaystrue,wewritetheanswer as allrealnumbers

10. Whensolvinganequationandweobtainan equationthatisnevertrue,wewritetheansweras nosolution

11.

13. 3210 322210 510 510 55 2 xx xxxx x x x

14. 356 35556 26 26 22 3 xx xxxx x x x

15. 74225 722442254 31 31 33 1 3

16. 93272 927337723 41 41 44 1 4

17. 21632 216 21666 217 217 77 3 ppp pp pppp p p p

Chapter 2: Solving Linear Equations and Inequalities

18. 30859 3088598 306 306 66 5

  qqq qqqqq q q q

19. 2836 228326 86 8666 2 xx xxxx x x x

20. 83450 8334350 8750 85075050 427 427 77 6 xx xxxx x x x x x 

21. 62986 6292 622922 69False yyy yy yyyy

Sinceafalsestatementisobtained, thereisnosolution.

22. 78325 78352 7882 788882 72False yyy yyy yy yyyy

Sinceafalsestatementisobtained, thereisnosolution.

23.   5326 53212 5232212 3312 333123 39 39 33 3 xx xx xxxx x x x x x

24.   1432 1436 1436 1426 146266 202 202 22 10 xx xx xxxx x x x x x  

 

   

25. 4281953 4825193 42516 4425416 2916 21691616 189 yyy yyy yy yyyy y y y

189 99 2 y y  







26. 359249 395429 125411 12454411 8511 855115 816 xxx xxx xx xxxx x x x



   816 88 2 x x  

   

27.   22462 24426 2426 224226 46False xxx xxx xx xxxx

 Sinceafalsestatementisobtained, thereisnosolution.

28.

 952716 925776 75713 7757713 513False xxx xxx xx xxxx

Sinceafalsestatementisobtained, thereisnosolution.

29.

30.







3174 3174 33174344 132 132 22 13 2

31.

3255313 6155313 620313 66203613 20913 201391313

32.  7353106 mm 21353018 212135301821 35303 353030303 53 53 33 5 3 mm mmmm m m m m m  

33. 124.89.44.832.5 124.89.49.44.89.432.5 124.814.232.5 124.832.514.232.532.5 92.314.2 92.314.2 14.214.2 6.5 xx xxxx x x x x x  

34. 90.54.58.5

90.50.54.50.58.5 958.5 98.558.58.5 xx xxxx x x     0.55 0.55 55 0.5 5 0.1 x x x x   

35. 0.620.659.752.63 0.622.630.659.752.632.63 3.250.659.75

3.250.650.659.750.65

3.2510.4 3.2510.4 3.253.25 3.2 xx xxxx x x x x x    

36.

7 2 23 7 662 23 21212 2122122 2114 2114 1414 x x x x xx xxxx x x

5 2 73 5 21212 73 33542 342354242 x x x x xx xxxx

45350 453535035 4535 4535 4545 7 9 x x x x x

43. 511 842 511 88 842 524 52242 52 52 22 aa aa aa aaaa a a

44. 573 632 bb  573 66 632 5149 551495 144 144 44 7 2 bb bb bbbb b b b

45.   0.1100.34 0.110.34 0.10.110.30.14 140.244 50.2 50.2 0.20.2 25 xx xx xxxx x x x x

46.   0.3514.32 1.50.34.32 1.520.34.322 0.50.34.3

0.50.30.34.30.3 0.54 0.54 0.50.5 8 xx xx xxxx x x x x x

47.   244325 284325 2828 xxx xxx xx  



Sincetheleftsideoftheequationisidenticalto therightside,theequationistrueforallvaluesof x.Thusthesolutionisallrealnumbers.

 

48.   319865 33936 3636 yyy yy yy



Sincetheleftsideoftheequationisidenticalto therightside,theequationistrueforallvaluesof x.Thusthesolutionisallrealnumbers.

49.     5332546 15151086 1515168 15151516158 158 15888 23 nnn nnn nn nnnn n n n  

 6217324 12621144 1212614214121414 205 205 55 4

mmm mmm mmmmm m m m 51.

 323 323 32223 333 33333 30 30 33 0 pp pp pppp p p p p p   

Chapter 2: Solving Linear Equations and Inequalities

52.  3773 3773 33777733 06 06 66 0

53. 1695373 1695219 169169

Sincetheleftsideoftheequationisidenticalto therightside,theequationistrueforallvaluesof z.Thusthesolutionisallrealnumbers.

54.

3821138 3168138 138138 yyy yyy yy

Sincetheleftsideoftheequationisidenticalto therightside,theequationistrueforallvaluesof y.Thusthesolutionisallrealnumbers.

55.     3521743 70357283 70351028 701035101028 603528 6035352835 6063 xxx xxx xx xxxx x x x

 6063 6060 21 20 x x  

56.  24217523 482435143 482435314 484824384814 241014 xxx xxx xxx xxxx x  

2414101414 3810 3810 1010 19 5 x x x x

0.40.280.62.52

0.40.40.280.60.42.52 0.280.22.52 0.282.520.22.522.52 2.80.2 2.80.2 0.20.2 14 xx xx xxxx x x x x x

58. 0.5(68)1.4(5)0.2 341.470.2 341.47.2 31.441.41.47.2 1.647.2 1.6447.24 1.63.2 1.63.2 1.61.6 xx xx xx xxxx x x x x 

2 x

59. 31 2 52 31 10210 52 620105 66201065 2045 205455 154 xx xx xx xxxx x x x

154 44 15 4 x x

1090 10109010 xx xxxx

354371 3541277 3547127 35319 3553519 xxx xxx xxx xx xxxx

3219 31921919 222 222 22 11 x x x x x

  

69.      752363 7526618 12412 121241212 424 44424 524 yyy yyy yy yy yy yyyy y

524 55 24 5 y y 

70.  3427521 34214105 7210145 7819 788819 7919 ccc ccc ccc cc cccc c

779197 912 912 99 4 3 c c c c

71.   32 635 53 31810 2 553 31810 15152 553 9543050 995430950 542150 5450215050 xx xx xx xx xxxx x x      

  421 421 2121 4 21 x x x  

72.

11 2444 23 44 2 33 44 323 33 3644 336434 64 dd d d d d dd dddd

73.

32536 53 61536 53 61536 1515 53 3615536 18451530 181545151530 34530

345453045 315 315 33 5

74.

34523 43 3121015 43 3121015 1212 43 331241015 9364060 993640960 363160 xx xx xx xx xx xxxx x

3660316060 2431 2431 3131 24 31 x x x x

75.

21 54341 72 1083 21 772 1083 1 772 1083 14141 772 20162114 202016212014 xx xx xx xx xx xxxx

1614 16141414 30 x x x 

521 221 1236 510421 1212336 510421 1212 1212336 5101682 5101610 551016510 xx xx xx xx xx xxxx

101110 1010111010 x x 

x

614141614

79. a. Oneexampleis12 xxx b. Ithasasinglesolution.

c. Answerswillvary.Forequationgiveninpart a): 12 212 212 12 1121 1 xxx xx xxxx x x x

80. a. Oneexampleis21 xxx .

b. Ithasasinglesolution.

c. Answerswillvary.Forequationgiveninpart a): 21 221 221 21 2111 1 xxx xx xxxx x x x      

81. a. Oneexampleis121 xxx

b. Bothsidessimplifytothesameexpression.

c. Thesolutionisallrealnumbers.

82. a. Oneexampleis2x +1= x + x +1.

b. Bothsidessimplifytothesameexpression.

c. Thesolutionisallrealnumbers.

83. a. Oneexampleis122 xxx

b. Itsimplifiestoafalsestatement.

c. Thesolutionisthatthereisnosolution.

84. a. Oneexampleis221 xxxx

b. Itsimplifiestoafalsestatement.

c. Thesolutionisthatthereisnosolution.

85. 5145 55195 14 14 44 1 4







 



 

86. 2435 2425 224225 45False



 Nosolution

87. 3525 3535 





 Theleftsideoftheequationisidenticaltothe rightside.Thesolutionisallrealnumbers.

88.

89.

28 28 22 4 x x x

90. a. Let x bethecostofonechocolatebar. 2x = x +6

b. 2x = x +6 2x – x = x +6 x =6 Achocolatebarcosts$6.00.

91. a. |4|=4

b. |–7|=7

c. |0|=0

92. a abab b x xxx x

93. Factorsareexpressionsthataremultiplied together;termsareexpressionsthatareadded together.

94.

23442644

23440 26440 7100 71010010

Exercise Set 2.6

1. Anequationusedtoexpressaspecific relationshipmathematicallyiscalledaformula.

2. Theprocessofsubstitutingvaluesandperforming indicatedoperationsonaformulaiscalled evaluatingit.

3. Afour-sidedfigureiscalledaquadrilateral.

4. Twicetheradiusofacircleisthediameterofthe circle.

5. Simpleinterestisfoundbymultiplyingthe principal,therate,andthetime.

6. Whencalculatingsimpleinterest,therateshould beexpressedasadecimalnumber.

7. Thecircumferenceofacircleisfoundby multiplying2πbytheradius.

8. Distanceisfoundbymultiplyingtheratebythe time.

9. Apossibleunitofmeasureforvolumeiscubic feet.

10. Apossibleunitofmeasureforareaissquare inches.

11. Substitute75for r and3for t

12. Substitute6for s

13.

17. Substitute10for

20. Substitute6for r and7for h.

2 2 1 3 1 67 3 263.89 Vrh V V

21. Substitute16for m and56for n

22. Substitute72for a,81for b,and93for c. 3 728193 3 246 3 82 abc A A A A

23. Substitute100for x,80for m,and10for s 10080 10 20 2

24. Substitute135for s and100for d. 135100 100 0.35 sd m d

25. Substitute56for P

26. Substitute28for P and6for w

27. Substitute678.24for V and6for r 2 2

678.24(6)

678.2436

678.2436 3636

28. Substitute188.5for V and6for r

29. a)Substitute3for s

30. a)Substitute7for s

 As A A

2 2 2 7 49mi

b)Substitute7fors.

4 47 28mi    Ps P P

31. a)Substitute11for l and7for w  2 117 77yd    Alw A A

b)Substitute11for l and7for w

 22 21127 2214 36yd     Plw P P P

32. a)Substitute7for l and15for w

 2 715 105mm    Alw A A

b)Substitute7for l and15for w

 Plw P P P

 22 27215 1430 44yd

33. a)Substitute17for l and12for h

 2 1712 204m    Alh A A

b)Substitute17for l and13for w

 Plw P P P

 22 217213 3426 60m

34. a)Substitute7for l and3for h.  2 73 21mm    Alh A A

b)Substitute7for l and5for w

 22 2725 1410 24mm Plw P P P

35. a)Substitute4for h,4for b and10for d

   Ahbd A A A A

2 1 2 1 4410 2 1 414 2 1 56 2 28cm 

Chapter 2: Solving Linear Equations and Inequalities

b)Substitute5for a,4for b,5for c,and10for d 54510 24cm

36. a)Substitute12for h,12for b and22for d.

Ahbd A A A A

b)Substitute13for a,12for b,13for c,and22 for d 13121322 60yd

37. a)Substitute10for b and12for h

39. a)Substitute3for r  2 2 2 3 928.27m Ar A A 

r  2 23 618.85m       Cr C C

40. a)Substitute 8 4 2  for r  2 2 2 4 1650.27ft Ar A A 

Substitute 8 4 2  for r

41. a)Substitute147 2  for r

Ar A A

2 2 2 7 49153.94m

b)Substitute147 2  for r

2 24 825.13ft

b)Substitute13for a,10for b,and13for c. 131013 36ft

Pabc P P

38. a)Substitute19for b and5for h

Abh A A A

2 27 1443.98m

42. a)Substitute9for r.

Ar A A

2 2 2 9 81254.47in.

b)Substitute9for r

Cr C C

2 29 1856.55in.

Pabc P P

b)Substitute8.6for a,19for b,and13for c 8.61913 40.6m

43. Substitute4for l,3for w,and5for h V = lwh V =(4)(3)(5) V =60ft3

ISM: Elementary Algebra

44. Substitute7for l,2for w,and5for h

2: Solving Linear Equations and Inequalities

49. Substitute6for r

45. Substitute4for r and9for h

V =πr2h

V =π(4)2(9)

V =π(16)(9)

V =144π≈452.39cm3

46. Substitute 12 6 2  for r and12for h

50. Substitute126 2  for r

47. Substitute4for

and8for h.

51. Substitute50for F

Theequivalenttemperatureis10°C.

52. Substitute86for F.

48. Substitute73.5

Theequivalenttemperatureis30°C.

53. Substitute25for C.

9 32 5 9 2532 5 4532

Theequivalenttemperatureis77°F.

Chapter 2: Solving Linear Equations and Inequalities

54. Substitute–40for C

62. 4525 445254 5254 5254 55 254425 55 mn mmnm nm nm mm n

63. 32 3323 23 23 22 33 22 rn rn rn rn nn r

64. 75 7757 57 pq pq pq

 pq q p q p q p

57 55 7 5 17 15 7 5

65.

Chapter 2: Solving Linear Equations and Inequalities

74. 396 39996 396 33963 xy xyyy xy xxyx

75. 5103 5103 55 103 55 3 2 5 3 2 5 yx yx x y yx yx

76. 7219 7219 77 219 77 9 3 7 yx yx x y yx

77. 1536 1536 66 153 66 51 22 15 22 xy xy x y xy yx

78. 2318 2318 22 318 22 3 9 2 yx yx x y yx

79.  1 34 3 14 3 33 14 333 33 149 333 15 33 yx yx yx yx yx

80.  2 34 3 28 3 33 28 333 33 289 333 217 33 yx yx yx yx yx

81. 11 2 53 12 2 53 1121 2 5535 103 2 1515 13 2 15 yx yx yx yx yx

82. 31 5 42 33 5 48 33 555 48 3340 488 337 48 yx yx yx yx yx

83. a)

15,0000.04542700

Hewillpay$2700interest. b)15,000270017,700 pi

Hewillpayatotalof$17,700.

84. a)

35,0000.03755 6562.50

Shewillpay$6562.50interest. b)35,0006562.5041,562.50 pi Shewillpayatotalof$41,562.50.

85. iprt 

4500.033 4500.09 4500.09 0.090.09 5000 p p p p     Sheplaced$5000inthesavingsaccount.

86. iprt 

84060000.035 840210 840210 210210 4 t t t

Hismoneyhadbeenintheaccountfor4years.

87. 7630.02 15.26 drt d d    Thecartravels15.26mi.

88.  3.42 6.8 drt d d   

Shewalked6.8mi.

89. 1803 60 drt r r    Heraveragespeedwas60mph.

90.

39000.3 13,000    drt r r Theaveragespeedwas13,000mph.

91. a)

20090 18,000    Alw A A

Theareaofthepastureis18,000squareyards. b)

 22 2200290 400180 580     Plw P P P

Shewillneed580yardsoffencing.

92. a) 3557.4 2009    Alw A A

Theareais2009squareinches. b)

 22 235257.4 70114.8 184.8     Plw P P P

Theperimeteris184.8inches.

93. a)  2 30 2 2 215 94.25 

  Cr C C C

Hewillneedapproximately94.25feetoffencing. b)   2 152 225706.86       Ar A A

Theareaofthecoverisapproximately706.86 squarefeet.

Theareaofthehelipadisapproximately4071.50 squarefeet.

Thecircumferenceofthehelipadisapproximately 226.19feet. 95. 2 62.832

Theradiusofthegardenisapproximately10.00 feet.

Theradiusofthetabletopisabout3.00feet.

Thediameteristwicetheradius,so d =2(3.00)=6.00

Thediameterofthetabletopisabout6.00feet.

97. Totalarea=Areaoftoptriangle+areaofbottom triangle

TotalArea=0.5b1h1+0.5b2h2

TotalArea=0.5(2)(1)+0.5(2)(2)

TotalArea=1+2=3

Theareaofthekiteis3squarefeet.

98. 1 2 1 2(36)(31)558

Abh A  

Theareais558in.2

100.  17146 1428 Vlwh V V   

Thevolumeofthecarryingcaseis1428in.3 .

101.V =πr2h =π(4)2(3) =π(16)(3) =48π≈150.80

Thevolumeofwaterinthehottubisabout 150.80ft3 .

102. 2 Vrh 

Theradiusis1 2 thediameteror

 1 24in.12in.1foot. 2  2 (1)(4)412.57 V 

Thevolumeofthedrumisabout12.57ft3

103.  1 2 Ahbd     1 243 2 1 27 2 17 7 A    

Theareaofthesignis7ft2 .

104.  1 2 Ahbd 

   1 10080200 2 1 100280 2 50280 14,000 A    

Theseatingareais14,000ft2 .

99.  60301.5 2700

Vlwh V V   

Thevolumeofsandis2700cubicfeet.

105. Theradiusishalfthediameter. 3 3 4 3 49 32 381.7 Vr V V   

Thevolumeofthebasketballisabout381.7in.3

106. Theradiusishalfthediameter.

32 32 183 18737 6174147 6027 Vxx V

Volumeis6027cm3 . c.

Thevolumeoftheballisabout904.78ft

107. Theradiusishalfthediameter.

Thevolumeoftheconeisabout18.85cubicfeet. 108.

d.

2 2 548 54787 264656 2590 Sxx S

Surfaceareais2590cm2 .

112. Whenusingthedistanceformula,iftherateis doubledandthetimeishalved,thedistancestays thesame.Todoubletherate,multiply r by2.To halvethetime,divide t by2.

2 2 22 drt trt drrt

113. Whenusingthesimpleinterestformula,ifthe principalandratearedoubledandthetimeis halved,simpleinterestdoubles.Todoublethe principal,multiply p by2.Todoubletherate, multiply r by2.Tohalvethetime,divide t by2.

4 222 22 Iprt Itprtprprt

114 Whenusingtheformulaforareaofasquare,ifthe sideofthesquareisdoubled,thenthearea becomes4timesaslarge.Todoubletheside length,multiply s by2.

2 2222224 As Asss

115. Whenusingtheformulaforvolumeofacube,if thesideofthecubeisdoubled,thenthevolume becomes8timesaslarge.Todoubletheside length,multiply s by2.

3 3333228 Vs Vsss

Chapter 2: Solving Linear Equations and Inequalities

Therefore,asquarewithsidelength s incheshasa greaterareathanacirclewithdiameter s inches. 117.

Firstfindthesidelength x ofthesquareinterms of s.UsethePythagoreantheoremtosolve.Then comparetheareasofthesquareandcircle.

Therefore,acirclewithdiameter s incheshasa greaterareathanasquarewithdiagonallength s inches.

a. 2 B b. 2 A c. 22AB d. 2 AC e.

Exercise Set 2.7

1. Aproportionisastatementofequalityoftwo ratios.

2. Intheproportion ap bq  , a and q arecalled extremes.

3. Intheproportion ap bq  , b and p arecalled means.

4. Amethodforsolvingproportionsiscross multiplying.

5. Twofiguresthathaveequalcorrespondingangles andhavetheircorrespondingsidesinproportion arecalledsimilarfigures.

6. Anotherwayofexpressingtheratio c d is c : d

7. Whensolvinganappliedprobleminvolving proportions,itiscrucialthatthetworatioshave thesameunits.

8. Thequotientoftwoquantitiesiscalledaratio.

9. 6:9=2:3

Elementary Algebra

10. 6:4=3:2

11. 6:24=1:4

12. 9:24=3:8

13. GradesbetterthanC=6+4=10

RatioofgradesbetterthanCtototal grades=10:24=5:12

14. GradesbetterthanC6410 

GradeslessthanC=3+2=5

RatioofgradesbetterthanCtogradeslessthan C=10:5=2:1

15. 7:4

16. 3:47

17. 5:15=1:3

18. 18:24=3:4

19. 3hours360180  minutes

Ratiois1806 301  or6:1.

20. 7minutes760420seconds  Ratiois42028 151  or28:1.

21. 3feet31236inches  Ratiois369 82  or9:2.

22. 4yardsis4312  feet

Ratiois61 122  or1:2.

23. 12quartersis 5 1230 2

dimes

Ratiois7:30.

24. 5quarters5525nickels 

Ratiois27 25 or27:25.

25. Gearnumberofteethondrivinggear ratio numberofteethondrivengear 408 51  

Gearratiois8:1.

26. Gearnumberofteethondrivinggear ratio numberofteethondrivengear 3015 84  

Thusthegearratiois15:4.

Dividingbothpartsoftheratioby4,ratiocanalso beexpressedas3.75:1.

27. Gearnumberofteethondrivinggear ratio numberofteethondrivengear  455 182 

Thusthegearratiois5:2.

Dividingbothpartsoftheratioby2,ratiocanalso beexpressedas2.5:1.

28. Gearnumberofteethondrivinggear ratio numberofteethondrivengear  567 162 

Thusthegearratiois7:2.

Dividingbothpartsoftheratioby2,ratiocan alsobeexpressedas3.5:1.

29. a) 663,000:1200=1105:2

b)Dividingbothpartsoftheratioby2,ratiocan alsobeexpressedas552.5:1.

30. a. 270:75=18:5

b. Dividingbothpartsoftheratioby5,ratiocan alsobeexpressedas3.6:1.

31. a. 2320:1450=8:5

b. Dividingbothpartsoftheratioby5,ratiocan alsobeexpressedas1.6:1.

32. a. 19,320:7850=1932:785

b. Since1932÷785≈2.46,1932:785≈2.46:1.

33. a. 35:17

b. 25:19

34. a. 2500:2200=25:22

b. 3500:2500=7:5

35. a. 40:32or5:4

b. 15:11

36. a. 80:11

b. 1:5

Thusthesideis11.2feetinlength.

1626

53.

54.

Thusthesideis19.5inchesinlength.

Thusthesideis5.6inchesinlength.

55. Let x =numberofloadsonebottlecando. 4flounces100flounces 1loadloads 4100 1 4100 25 x x x x

Onebottlecando25loads.

56. Let x =amountoftimeittakestolaycablein hours. 42feet252feet 1hourhours 42252

Itwilltakethem6hours.

57. Let x =numberofmilesthatcanbedrivenwitha fulltank.

25miles 1gallon21.1gallons

Itcantravel527.5milesonafulltank.

58. Let x =numberofsharesthatcanbepurchased.

2sharesshares $38.25$344.25 2 38.25344.25 2344.2538.25 688.538.25 18 x x x x x

18sharescanbepurchased.

59. Let x =lengthofmodelinfeet. 1footmodelfootmodel 20foottrain30foottrain 1 2030 2030 1.5 x x x x

Themodelshouldbe1.5feetlong.

60. Let x =Jane’spropertytax.

$1000assessedvalue$333,716assessedvalue $15.9933taxdollarstax 1000333,716 15.9933 100015.9933333,716 10005,337,220.103 5337.22 x x x x x      ThusJane’spropertytaxis$5337.22.

61. Let x =numberofteaspoonsneededforsprayer. 3teaspoonsteaspoons 1gallonwater8gallonswater 3 18 381 24 x x x x 

 Thus24teaspoonsareneededforthesprayer.

Chapter 2: Solving Linear Equations and Inequalities

62. Let x =numberofpoundsoffertilizerneeded.

5000squarefeet26,000squarefeet 40poundsfertilizerpoundsfertilizer

500026,000 40

50004026,000

50001,040,000 208 x x x x x

Thus208poundsoffertilizerareneededfor 26,000squarefeet.

63. Let x =actualdistanceinmiles. 1inch3.75inches 4milesmiles 13.75 4 143.75 15 x x x x

Theactualdistanceis15miles.

64. Let x =lengthonamapininches. 0.5inchesinches 22miles55miles 0.5

66. Let x =timeinhoursitwilltakeKarentoreadthe entirenovel. 72pages656pages 1.3hourshours

72656 1.3 720.3656 72852.8 11.84 x x x x x

Itwilltakeherabout11.84hours.

67. Let x =lengthofthemodelbullinfeet. 2.95feetmetalbull28feetmetalbull 1feetrealbullxfeetrealbull 2.9528 1 2.95128 2.9528 9.49 x x x x

Themodelbullisabout9.49feetlong.

68. Let x =time,inminutes,itwilltaketo removethewater. 3inches12inches 30minutesminutes

Thelengthonthemapwillbe1.25inches.

65. Let x =numberofcupsneeded. 1 2 3 2 1cupscups 6servings15servings 615 3 615 2 622.5 3.75 x x x x x

3.75cupsofonionsareneeded.

Itwilltake120minutesor2hours.

69. Let x =numberofmilliliterstobegiven. 1millilitermilliliter 400micrograms220micrograms 1 400220 1220400 220400

0.55 x x x x x x

Thus0.55millilitershouldbegiven.

70. Let x =numberofgramsofmeprobanate tobegiven.

1sqmbodysurface0.6sqmbodysurface 0.7gramsgrams

Thus0.42gmeprobamateshouldbegiven.

71. Let x =numberofminutestoreadentirebook.

40pages760pages 30minutesminutes 40760 30 4030760 4022,800

Thusitwilltakeher570minutesor 9hours30minutestoreadtheentirebook.

72. Let x =time,inminutes,ittakesJasontoswim30 laps.

3laps30laps 2.3minutesminutes

Itwilltakehim23minutes.

73. Let x =numberofchildrenbornwithPrader-Willi Syndrome.

12,000births4,315,000births 1babywithsyndromebabieswithsyndrome

74. Let x =timeittakeshertocompletethe scrapbook. 4pages36pages 20.5minutesminutes

Thus,about360childrenwerebornwithPraderWilliSyndrome.

Itwilltakeher184.5minutesor3hr.4min.30 sec.tocompletethescrapbook.

75. 12inches78inches 1footfeet 1278

Thus78inchesequals6.5feet.

76. 1milemiles 5280feet22,704feet 1 528022,704 22,7045280 22,704 5280 4.3

Thus22,704feetequals4.3miles.

77. 9squarefeet26.1squarefeet 1squareyardsquareyards 926.1 1 926.1

Thus26.1squarefeetequals2.9squareyards.

Chapter 2: Solving Linear Equations and Inequalities

78. 1poundpounds 16ounces146.4ounces 1 16146.4 146.416 146.4 16 9.15 x x x x x   

Thus146.4ouncesequals9.15pounds.

79. 2.54cm50.8cm 1inchinches 2.5450.8 1 2.5450.8 50.8 20 2.54 x x x x

Thusthelengthofthenewbornis20inches.

80.

1mile520miles 1.6kilometerskilometers 1520 1.6 1.6520 832 x x x x 

ThusthedistancefromSanDiegotoSan Franciscois832kilometers.

81. $41.08$513.50 1gramgrams 41.08513.5 1 41.081513.5 41.08513.5 513.5 12.5 41.08

Briecanpurchase12.5gramsofgold.

82. $31.08$1359.75 1gramgrams

31.081359.75 1

31.0811359.75 31.081359.75 1359.75

Jacquecanpurchase43.75gramsofplatinum.

Timneeds7 1 8cupsofsugar. 84. 1 2teaspoonofcayenneteaspoonsofcayenne 3avocados8avocados

Anthonyneeds1 1 3teaspoonsofcayennepepper.

85. $1.00U.S.$U.S. $1.33Canadian$1995Canadian 1 1.331995 19951.33 19951.33 1.331.33 1500 x x x x x 

Willywillneed$1500U.S.dollars.

86. $1.00U.S.$400U.S. 0.92euroseuros 1400 0.92 10.92400 368  

x x x x Andrewwillreceive368euros.

87. TheratioofJavier’stotalcholesteroltohigh densitycholesterolis17556.Ifwedivide175by56

weobtain3.125.ThusJavier’sratioisequivalent to3.125:1.Thereforehisratioislessthanthe desired3.5:1ratio.

88. a. TheratioofCameron’stotalcholesteroltohigh densitycholesterolis21648.Ifwedivide216by

48weobtain4.5.ThusCameron’sratiois equivalentto4.5:1.Thereforehisratioisnot lessthanorequaltothedesired3.5:1ratio.

89. In ac bd  ,if b and d remainthesamewhile a increases,then c increasesbecause ad=bc.If a increases,then ad increasesso bc mustincrease byincreasing c.

90. In ac bd  ,if a and c remainthesamewhile d decreases,then b decreasesbecause ad = bc.If d decreases,then ad decreasesso bc mustdecrease bydecreasing b

91. Let x =numberofmilesremainingonthelifeof eachtire.

Inchesremainingonthelifeofeachtire:

0.310.060.25

0.03inches0.25miles

5000milesmiles

0.030.25

5000

0.0350000.25

Thetireswilllastabout41,667moremiles.

92. Let x representtheamountoftheingredientthat shouldbeused.

3cupofflourshouldbeused.

nutmeg:

2 3teaspoonofcinnamonshouldbeused. salt:1

Chapter 2: Solving Linear Equations and Inequalities

1cupofsugarshouldbeused.

93. Let x =numberofcubiccentimetersoffluid needed.

0.625cubiccentimetersoffluidshouldbedrawn upintoasyringe.

94. a. – d. Answerswillvary.

95. Answerswillvary.

96. commutativepropertyofaddition

97. associativepropertyofmultiplication

98. distributiveproperty

99.

Sinceatruestatementisobtained,thesolution isallrealnumbers.

100. ymxb ybmxbb ybmx ybmx xx yb m x

Exercise Set 2.8

1. Amathematicalstatementthatincludesthe symbols<,>, ,or≥iscalledaninequality.

2. Wheneverwemultiplyordividebothsidesofan inequalitybyanegativenumberwemustchange thedirectionoftheinequalitysymbol.

3. Whenrepresentingasolutiononanumberline,an opencircle,  ,indicatesthattheendpointisnot includedinthesolution.

4. Whenrepresentingasolutiononanumberlinea closedcircle,  ,indicatesthattheendpointis includedinthesolution.

5. Whenwearesolvinganinequalityandweobtain aninequalitythatis always truethenwestatethat thesolutionisallrealnumbers.

6. Whenwearesolvinganinequalityandweobtain aninequalitythatis never truethenwestatethat thereisnosolution.

7.

12. 62 6626 4

13. 39 3393 6 x

14. 73 7737 4

4, 

15.

16.

38 3383 5 115 5 x x x x x 

5, 

42 4424 6 116 6 x x x x x

,6

17. 82 8222 6 (1)(6)(1)() 6 6 r r r r r r

 ,6

18.  1512 1512 3 131 3 3

s s s s s s

,3

31. 162169 16221692 161611 1616161611 011False ss ss ss ssss

Sinceafalsestatementisobtained,thereis nosolution.

32. 5357 537577 5105 551055 100False tt tt tt tttt

Sinceafalsestatementisobtained,thereis nosolution.

33.

Chapter 2: Solving Linear Equations and Inequalities

8361 2494 249949 2413 24242413 013 xx xx xx xx xxxx

Since0isalwayslessthan13,thesolutionisall realnumbers

Since3isalwayslessthan4,thesolutionisall realnumbers.

Since4isalwaysgreaterthanorequalto–3,the solutionisallrealnumbers.

 ,

41.   63215 186215 1815621515 362 36626 38 38 88 xx xx xx xx xxxx x x

Sinceatruestatementisobtained,thesolutionis allrealnumbers.

Elementary Algebra Chapter 2: Solving Linear Equations and Inequalities

42.   2346 6246 626 66266 212 212 22 6 xx xx x x x x x

43. 4445 44420 4444420 420 xx xx xxxx

Since–4isneverlessthan–20,thereisno solution.

44.

 25324 102364 102210 10222210 1010 xxx xxx xx xxxx

Since10isnevergreaterthan10,thereisno solution.

45.     523622 1015622 10158 101515815 107 107 117 xxx xxx xx xx xx xxxx x

117 1111 7 11 x x

7 11,

46.     3212426 636486 636414 64364414 103614 1036361436 1022 xx xx xx xxxx x x x

   1022 1010 11 5 x x   11 ,5

47. 1.23.13.53.8 1.21.23.13.51.23.8 3.12.33.8 3.13.82.33.83.8 6.92.3 6.92.3 2.32.3 3 3 xx xxxx x x x x x x 

       3,  48. 5.36.72.36.5 5.36.56.72.36.56.5 1.26.72.3 1.26.76.72.36.7 1.29.0 1.29.0 1.21.2 7.5 rr rrrr r r r r r          7.5, 

Chapter 2: Solving Linear Equations and Inequalities

49.  

1.234.621.7

1.23.69.24.61.7

1.23.610.94.6

1.24.63.610.94.64.6

5.83.610.9

5.83.63.610.93.6

5.814.5 mm mm mm mmmm m m m

50.

4.642.430.2

18.44.62.47.20.2

18.44.62.47.4

18.44.62.42.42.47.4

18.42.27.4

18.418.42.27.418.4 2.211.0 xx xx xx xxxx x x x

2.211.0 2.22.2 5 x x

55.

59. a. Theaveragehightemperaturewasgreaterthan 65°FinMay,June,July,August,and September.

b. Theaveragehightemperaturewaslessthanor equalto59°FinJanuary,February,March, April,November,andDecember.

c. Theaveragelowtemperaturewaslessthan 29°FinJanuary,February,andDecember.

d. Theaveragelowtemperaturewasgreaterthan orequalto58°FinJune,July,andAugust.

60. a. KoreaandMexico

b. MexicoandtheUnitedStates

c. UnitedStatesandFrance

d. UnitedStates

61. ≠

62. Thedirectionoftheinequalitysymbolwasnot changedwhenbothsidesoftheinequalitywere dividedby–4.

63. Wecannotdividebothsidesofaninequalityby y becausewedonotknowthat y ispositive.If y is negative,wemustreversethesignofthe inequality.

Chapter 2: Solving Linear Equations and Inequalities

64.

69. Let x =numberofkilowatt-hoursofelectricity used. $0.174$87 1kilowatt-hourkilowatt-hours 0.17487 1 0.17487 87 500 0.174 x

ThustheVega'sused500kilowatt-hoursof electricityinJuly.

Review Exercises

67. Substitute–5for x

5.

5252 552 510 510 xx x x x

24224 28 28 xx x x

212 112 2 2 xx x x x

818 118 8 mm m m

6.

4444 444 164 xx x x

5555 555 255 255 pp p p p

8.

9.

11. 111 (24)(2)(4) 222

13.

32573257 323537 61521 61521 abab ab ab ab 

15. 10m –6m =4m

16. 533353 38 yy y  

17. 13215 51 xxx x 

18. 2333 xxyxy

19. 42464426 88 mnmnmmnn mn 

20. Therearenoliketerms. 932 xy cannotbefurthersimplified.

21. 6236436 xxyxy 

22. 893993 3 xxxxx   23. 222 483123 xxx 

24.   2222 22 234686868 6688 0 aaaa aa   

25.   234523125 57 xxxx x  

26.   4232464 23 32 bbbb b b   

27.   67676767 7766 0 xxxx xx   

28.   225104410104 44 xx x  

29.   6431842418184 1842418 2242 xxxx xx x   

30.   2 2 2 2 4.0321.661.09 4.0323.261.09 64.0323.21.09 62.032.11 xxx xxx xxx xx 

  

31. 1313 2525 4545 512 7 2020 7 7 20 dddd dd d   

32.     33 3 3 abababab aabb   

33.  5152 262 6363 54 2 66 1 2 6 xxxx xx x   

 326 366 36666 312 312 33 4 x x x x x x

45. 12328 12624 122462424 126 126 66 2 x x x x x x

46. 

4620 2480 24248024 824 824 88 3 x x x x x x

47. 6260 460 46606 46 46 44 63 42 nn n n n n n

48. 3346 3346 316 36166 31 31 11 3 ww ww w w w w w

49.  62346 62346 63246 366 33663 63 63 66 nn nn nn n n n n

50.  62343213 121812813 121218813 1013False xx xx xx

Sinceafalsestatementisobtained,thereisno solution.

51.  531311 533331 3232 332332 22True xx xx xx xxxx 

Thesolutionisallrealnumbers.

52. 8.46.36.32.1

8.42.16.36.32.12.1

6.36.36.3

6.36.36.36.36.3

6.312.6 6.312.6 6.36.3 2 rr rrrr r r r r r

53. 1.12.340.091.7 1.10.092.340.090.091.7 1.012.341.7 1.012.342.341.72.34 1.014.04 1.014.04 1.011.01 4 xx xxxx x x x x x 

0.3550.454 0.351.750.451.8 0.350.351.750.450.351.8 1.750.101.8 1.751.80.101.81.8 3.550.10 3.550.10 0.100.10 35.5 cc cc cccc c c c c c

0.260.321 0.21.20.60.3 0.20.61.20.60.60.3 0.81.20.3 0.81.21.20.31.2 0.80.9 0.80.9 0.80.8 1.125 xx xx xxxx x x x x x

 2.383.74 2.318.43.714.8 2.32.318.43.72.314.8 18.46.014.8 18.414.86.014.814.8 3.66.0 3.66.0 6.06.0 0.6 xx xx xxxx x x x x x

61. 85419 8454419 12519 1255195 1224 1224 1212 2 xx xxxx x x x x x

  2236 2612 21261212 106 106 107 ww ww ww ww wwww w

63. 26393 2636 226326 66 6666 0 xx xx xxxx x x x

64. 53210 531021010 572 55725 77 77 77 1 aa aa aa aaaa a a a

65.

52236 52612 5526512 212 2121212 10 pp pp pppp p p p

66. 312249 9249 992499 024False xxx xx xxxx

Sinceafalsestatementisobtained,thereisno solution.

67.

 423488 812488 8888 xx xx xx

Sincetheequationistrueforallvaluesof x,the solutionisallrealnumbers.

68.

424334 486312 54312 5343312 2412 244124 28 ccc ccc cc cccc c c c

69.

27694 214694 21429 2214229 149False xxx xxx xx xxxx

Sinceafalsestatementisobtained,thereisno solution.

70. 5(34)6209 15206209 15201520 xx xx xx

Thestatementistrueforallvaluesof x,thusthe solutionisallrealnumbers.

71.

  4350 41250 3170 31717017 317 317 33 17 3 xx xx x x x x x

72.

  24623 826123 82912 812291212 2029 202292 207 xxx xxx xx xx xx xxxx x

207 77 20 7 x x 

73.  3 22 232 262 22622 60False xx xx xx xxxx       Sinceafalsestatementisobtained,thereisno solution.

74.  5 72 275 2735 277735 535 535 55 7 yy yy yy yyyy y y y

Chapter 2: Solving Linear Equations and Inequalities

78.  1 3 462 3 4622 3 1212 4622 32618 5618 566618 118 xx x xxx xxx xxx xx xxxx x



21 222 56 4222 5566 4211 5533 4211 1515 5533 12655 1266565 xx xx xx xx xx xxxx

125 12555 7 x x x

79. Substitute7for y,2for x,and1for b.Thensolve for m.

721 721 71211 62 62 22 3 ymxb m m m m m

80. Substitute12for h,3for b,and5for d.Then solvefor A

1 2 1 1235 2 1 128 2 48 Ahbd A A A

2 1 2 1 83 2 12cm Abh A A

82.

83.

85.

87.

Hetraveled308.5miles.

88.

Theareaoftheflowergardenis240ft2 .

2 2 3 22 825.13in. Vrh V V

89. 15:20=3:4

90. 80ounces805 16  pounds

Theratioof80ouncesto12poundsisthus5:12. 91. 4minutes=240seconds

Theratioof4minutesto40secondsis2406 401 

Theratiois6:1.

92. 8 416 1684 1632 32 2 16 x x x x 

93. 5 2080 20805 20400 400 20 20 x x x x

94. 315 45 34515 13515 135 15 9 t t t t

95. 2015 45 201545 20675 675135 204 q q q q

96. 612 5 6125 660 60 10 6 x x x x

103. 410414 44104414 10814 101481414 248 248 88 3 3 aa aaaa a a a a a a

,3

104. 53215 515321515 1032 103323 105 105 55 2 2 rr rr rr rrrr r r r r

105.   2425 2825 228225 85 xx xx xxxx     Since8isneverlessthanorequalto–5,thereis nosolution.

106.  23644 26644 2624 226224 64 xxx xxx xx xxxx      Since6isalwaysgreaterthan4,theanswerisall realnumbers.

Chapter 2: Solving Linear Equations and Inequalities

112. Let t =thetimeitwilltakeinhourstomakethe boattrip. 40miles140miles 1.8hourshours 40140 1.8 401.8140 40252 40252 4040 6.3

Itwilltaketheboat6.3hourstotravel140miles.

113. Let x =numberofdisheshecanwashin21 minutes. 12dishesdishes 3.5minutes21minutes 12 3.521 12213.5 2523.5 2523.5 3.53.5 72 x x x x x x

Hecanwash72dishesin21minutes.

114. Let x =numberofpagesthatcanbecopiedin22 minutes. 1minutes22minutes 20pagespages 122 20 12220 440 x x x x

440pagescanbecopiedin22minutes.

115. Let x =numberofinchesrepresenting380miles. 60miles380miles 1inchinches 60380 1 603801 60380 3801 6 603 x x x x x

1 6 3inchesonthemaprepresent380miles.

116. Let x =sizeofactualcarinfeet 1inch10.5inches 1.5feetfeet 110.5 1.5 11.510.5 15.75 x x x x

Thesizeoftheactualcaris15.75ft.

117. Let x =thevalueof1pesointermsofU.S. dollars.

$1U.S.dollars 20.52pesos1peso

x x x x x 1pesoequalsabout$0.049.

118. Let x =numberofbottlesthemachinecanfilland capin2minutes. 2minutes=120seconds 50seconds120seconds 80bottlesbottles

50120 80 5080120 509600 9600 192 50 x x x x x 

Themachinecanfillandcap192bottlesin 2minutes.

Practice Test

4. 42362346

5.

6.

12

Sinceafalsestatementisobtained,thereisno solution.

  3232317 69627 6969 xx xx xx

Sincetheequationistrueforallvaluesof x,the solutionisallrealnumbers.

13.0 0 axbyc axbybycby axcby axccbyc axbyc axbyc aa byc x a

14 652 66526 562 562 55 62 55 62 55 xy xxyx yx yx x y yx

Chapter 2: Solving Linear Equations and Inequalities

15

17. a. Anequationthathasexactlyonesolutionisa conditionalequation.

b. Anequationthathasnosolutionisa contradiction.

c. Anequationthathasallrealnumbersasits solutionisanidentity.

18. 24410 241041010

19.   34512 312512 3121251212 3245 332453 242 242 22 xx xx xx xx xxxx x x

20.   43263 412263 61263 6612663 123 xxx xxx xx xxxx  

Since12isneverlessthan–3,theanswerisno solution.

21.     23412 23442 4242True xxx xxx xx    Thisequationistrueforallrealnumbers.

 ,  22. 38 4 348 332 32 3 x x x x     Thelengthofside x is322 feetor10 33 feet.

23. Let r =thesimpleinterestrate.

Theinterestratewas4%.

24. Let C=thecircumferenceofthepie.

Thecircumferenceofthepieis28.27in.

25. Let x =numberofminutesitwilltake. 25miles125miles 35minutesminutes

Itwouldtake175minutesor2hours55minutes.

8. Substitute–2foreach x

9. distributiveproperty 10. 824842 12 xyxyxxyy xy 

11. 2332 916916 3443 98 25 1212 1 25 12 xxxx xx x 

12. 734 73343 77 77 77 1 t t t t t

19. LetA=theareaofthetrampoline.

Theareaofthetrampolineis380.13ft2

20. Let x =amountheearnsafter8hours.

2hours8hours

Heearns$42after8hours.

Chapter 1

Real Numbers

1.2 Problem Solving

Objectives

1. Learn the five-step problem-solving procedure.

2. Solve problems involving bar, line, and circle graphs.

3. Solve problems involving statistics.

Guidelines for Problem Solving (1 of 3)

1. Understand the problem.

• Read the problem carefully at least twice. In the first reading, get a general overview of the problem. In the second reading, determine (a) exactly what you are being asked to find and (b) what information the problem provides.

• Make a list of the given facts. Determine which are pertinent to solving the problem.

• Determine whether you can substitute smaller or simpler numbers to make the problem more understandable.

• If it will help you organize the information, list the information in a table.

• If possible, make a sketch to illustrate the problem. Label the information given.

Guidelines for Problem Solving (2 of 3)

2. Translate the problem to mathematical language.

• This will generally involve expressing the problem in terms of an algebraic expression or equation. (We will explain how to express application problems as equations in Chapter 3.)

• Determine whether there is a formula that can be used to solve the problem.

3. Carry out all necessary calculations.

4. Check the answer obtained in step 3.

• Ask yourself, “Does the answer make sense?” “Is the answer reasonable?” If the answer is not reasonable, recheck your method for solving the problem and your calculations.

Guidelines for Problem Solving (3 of 3)

• Check the solution using the original wording of the problem if possible.

5. Make sure you have answered the question.

• State the answer clearly.

Example 2 (1 of 3)

Larry Gilligan’s iMac computer can perform 200 billion operations per second (200 billion OPS).

How many operations can Larry’s computer perform in 0.8 second?

Understand We are given the computer owner’s name, the model of the computer, a speed of 200 billion (200,000,000,000) operations per second, and 0.8 second. To determine the answer to this problem, the owner’s name, Larry Gilligan, and the model of the computer, iMac, are not needed.

Example 2 (2 of 3)

Translate

Number of operations in 0.8 second = 0.8 ×the number of operations per second

Carry Out = 0.8 ×200,000,000,000 = 160,000,000,000

Check

The answer 160,000,000,000 operations is less than the 200,000,000,000 operations per second, which makes sense because the computer is operating for less than a second.

Example 2 (3 of 3)

Answer

In 0.8 second, the computer can perform 160,000,000,000 operations.

Example 4 (1 of 4)

Walking It Off Experts suggest that people walk 10,000 steps daily. Depending on stride length, each mile ranges between 2000 and 2500 steps.

The figure on the next slide is a bar graph that shows the number of steps it takes to burn off calories from a garden salad with fat-free dressing, a 12-ounce can of soda, a doughnut, and a cheeseburger.

Example 4 (2 of 4)

a) Using the bar graph, estimate the number of steps it takes to burn off calories from a cheeseburger.

We estimate that the number of steps it takes to burn off calories from a cheeseburger is about 7600 steps.

Example 4 (3 of 4)

b) If Cliff can walk a mile in 2000 steps, how many miles will he have to walk in order to burn off the calories from the cheeseburger he ate for lunch?

Understand Since it takes Cliff 2000 steps to walk a mile, it follows that he would need to take 4000 steps to walk 2 miles, 6000 steps to walk 3 miles, and so on. To determine how many miles Cliff will have to walk in order to burn off the calories from the cheeseburger, we need to divide as follows.

Example 4 (4 of 4)

Translate

Carry Out

Check and Answer The answer appears reasonable. Cliff will need to walk 3.8 miles in order to burn off the calories from the cheeseburger that he ate for lunch.

Example 5 (1 of 5)

Super Bowl Ads The cost of advertising during the annual Super Bowl has generally risen every year since Super Bowl 1 was held in 1967. The line graph shows the cost of a 30-second advertisement during Super Bowls 10, 20, 30, 40, and 50.

Example 5 (2 of 5)

a. Determine the cost of a 30second advertisement during Super Bowl 10 and during Super Bowl 50. A 30-second ad during Super Bowl 10 cost $0.11 million or $0.11(1,000,000) = $110,000.

A 30-second ad during Super Bowl 50 cost $5 million or $5(1,000,000) = $5,000,000.

Example 5 (3 of 5)

b. How much more was the cost of a 30-second advertisement during Super Bowl 50 than during Super Bowl 10? Understand To determine how much more a 30-second advertisement cost during Super Bowl 50 than during Super Bowl 10, we need to subtract, as shown below.

Translate Difference in the cost = SB 50 cost − SB 10 cost Carry Out = $5,000,000 − $110,000 = $4,890,000

Check and Answer The answer appears reasonable. The cost of a 30-second advertisement was $4,890,000 more during Super Bowl 50 than during Super Bowl 10.

Example 5 (4 of 5)

c. How many times greater was the cost of a 30-second advertisement during Super Bowl 50 than during Super Bowl 10? Round your answer to the nearest hundredth.

Understand To determine the answer, we will divide the cost during Super Bowl 50 by the cost during Super Bowl 10. Translate Carry Out cost during SB 50 number of times greater = cost during SB 10

Example 5 (5 of 5)

Check and Answer

By observing the graph, we see that the answer is reasonable. The cost of a 30-second advertisement during Super Bowl 50 was about 45.45 times greater than during Super Bowl 10.

Solve Problems Involving Statistics

The mean and median are two measures of central tendency, which are also referred to as averages. An average is a value that is representative of a set of data (or numbers). The mean of a set of data is determined by adding all the values and dividing the sum by the number of values.

The median is the value in the middle of a set of ranked data. The data may be ranked from smallest to largest or largest to smallest.

Example 7 (1 of 4)

The Mean Grade Jose’s first six exam grades are 90, 87, 76, 84, 78, and 62.

a. Determine the mean for Jose’s six grades.

Example 7 (2 of 4)

b. If one more exam is to be given, what is the minimum grade that Jose can receive to obtain at least a B average (a mean average of 80 or better)?

Understand For the mean average of seven exams to be 80, the total points for the seven exams must be 7(80) or 560. The minimum grade needed can be determined by subtracting the sum of the first six grades from 560.

Translate minimum grade = 560 –sum of first six exam grades

Example 7 (3 of 4)

Carry Out = 560 –(90 + 87 + 76 + 84 + 78 + 62) = 560 –477 = 83 Check

Example 7 (4 of 4)

c. If there is only one more exam, is it possible for Jose to obtain an A average (90 or better)? Explain. We can use the same reasoning as in part b). For a 90 average, the total points that Jose will need to attain is 90(7) = 630. Since his total points are 477, he will need 630 –477 or 153 points to obtain an A average. Since the maximum number of points available on most exams is 100, Jose would not be able to obtain an A in the course.

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