College Algebra Real Mathematics Real People 7th Edition Larson Solutions Manual by clarakv54

CHAPTER 2 Solving Equations and Inequalities

Section 2.1 Linear Equations and Problem Solving

1. equation

2. 0 axb+=

3. extraneous

4. formulas

5. The equation 13 x += is a conditional equation.

6. To clear the equation 1 1 24 x += of fractions, multiply both sides of the equation by the least common denominator of all the fractions, which is 4.

7. 54 3 2 xx −=

−= −≠

(b) () ? 54 3 244 3 3 8

4 x = is not a solution.

(d) () ? 54 3 21414 63

−= −≠ 1 4 x = is not a solution.

(a) () () ? 54 3 21212 −= 33 = 1 2 x =− is a solution.

()

2714 142419 1414 3819 1414 += = = 2

=−

solution.

1 x = ()

2714 1919 1414 += = 1 x = is a solution. (c) 1 2 x = () += + = ≠ ? ? 612 1219 2714 72619

1 2 x = is not a solution.

7 x = () ? 67

2714 719 6 214 1919 214 += += = 7 x = is not a solution. NOT FOR SALE INSTRUCTOR USE ONLY © Cengage Learning. All Rights Reserved. College Algebra Real Mathematics Full Download: http://testbanktip.com/download/college-algebra-real-mathematics-real-people-7th-edition-larson-solutions-manual/ Download all pages and all chapters

8. 619 2714 xx += (a)

x =−

? ? ?

219

is not a

(b)

119

1414 1919 2814

(d)

719

NOT FOR SALE

11. () 2122 xx−=− is an identity by the Distributive Property. It is true for all real values of x

12. ()() 5151 xx −−=−+ is a contradiction. There are no real values of x for which it is true. 5555 55 xx −+=−−

13. ()()() 2 3527xxxx +−=−+ is a contradiction. There are no real values of x for which it is true. 22215214 1514 xxxx −−=−−

14. () 2 2 85411xxx−+=−− is an identity since () 2 22 4118161185. xxxxx −−=−+−=−+ It is true for all real values of x.

15. ()()() 2 682xxx+=++ is a conditional. There are real values of x for which the equation is not true (for example, 0 x = ).

16. ()()()() 1531xxxx +−=+− is a conditional. There are real values of x for which the equation is not true (for example, 0 x = ).

17. 14 3 11 x xx += ++ is conditional. There are real values of x for which the equation is not true (for example, 0) x =

18. 53 24 xx += is conditional. There are real values of x for which the equation is not true (for example, 1) x =

Section2.1LinearEquationsandProblemSolving 139 © 2016 Cengage Learning. All Rights Reserved. May not be scanned, copied or duplicated, or posted to a publicly accessible website, in whole or in part. 9. 4 34 6 x + += (a) ? 34 34 6 19 4 6 −+ += ≠ 3 x =− is not a solution. (b) ? 04 34 6 10 4 3 + += ≠ 0 x = is not a solution. (c) ? 214 34 6 23 4 6 + += ≠ 21 x = is not a solution. (d) ? 324 34 6 44 + += = 32 x = is a solution. 10. 3 8 1 3 x =− (a) 3 ? 3 4 48 1 3 4 1 3 x = =− ≠− 4 x = is not a solution. (b) 3 ? 3 ? 0 08 1 3 8 1 3 2 1 3 x = =− =− ≠− 0 x = is not a solution. (c) 3 ? 3 ? 19 198 1 3 27 1 3 3 1 3 x =− =− =− =− 19 x =− is a solution. (d) 3 ? 3 ? 16 168 1 3 8 1 3 2 1 3 x = =− =− ≠− 16 x = is not a solution.

≠−

−≠−

Method

NOT FOR SALE

Method

in the same viewing

The lines intersect at 240

Method 2: Graph 12 416 2 and 35 x yxy=−= in the same viewing window. These lines intersect at

83 932 4 24 2396 96 23 xx xx x x −= = −= =−

19. Method 1: 34 4

=−=

96 4.1739. 23 x ≈−≈−

Method 1: () 3 6 810 31 6 810 22 6 80 680 240 21.8182 2211 zz z z z −=  −=   =   ==≈

Method 2: Graph 12 34 and 4 83 xx yy

in the same viewing window. These lines intersect at

20.

=−=

window.

21.8182. 11 x ≈≈

Method 1: () 24 5 53 2254 53 20 27 3 2020 32781 x x xx x x += + = = ==

2: Graph 12 3 and 6 810 xx yy

21.

yxy

0.2469. 81 x ≈≈

Method 1: 416 2 35 4616 35 48 2 5 24 5 y y yy y y −= = −= =−

2: Graph 12 24 5 and 53 x

=+= in the same viewing window. These lines intersect at 20

22.

24 4.8. 5 x =−=−

3275 12

xx x −=+ −=+ =

5263 73 3 7 xx xx x x +=− +=− = =

() 3535 31535 182 9 yy yy y y −=+ −=+ −= =−

() 43318 412318 71218 13 13 zzz zzz zz z z −+=+ −+=+ −=+ −= =−

3 52 25 3 10 330 10 xx xx x x −= = −= =−

() 3 5 42 3 445 42 3220 520 4 xx xx xx x x +=− +=−  +=− =− =− 29. ()() 542 543 354254 1512108 520 4 x x xx xx x x = + −=+ −=+ = = 30. 1031 562 20656 150 0 x x xx x x + = + +=+ = =

23. 3527

24. 5362

25.

26.

27.

28.

INSTRUCTOR

USE ONLY

NOT FOR SALE

In the original equation, 4 x = yields a denominator of 0. So, 4 x = is an extraneous solution, and the original equation has no solution.

In the original equation, 2 x = yields a denominator of 0. So, 2 x = is an extraneous solution, and the original equation has no solution.

Section2.1LinearEquationsandProblemSolving 141 © 2016 Cengage Learning. All Rights Reserved. May not be scanned, copied or duplicated, or posted to a publicly accessible website, in whole or in part. 31. () 23 44 510 283 4 5510 78 4 105 833 510 16 33 z zz z zz z z z z −+= −+= −= −= −= 32. () () () ()() () 31 210 24 31 442410 24 6240 7240 742 6 x x x x xx x x x +−=  +−=   +−= −= = = 33. () () () 1732 100 1732 100 1732100 492100 4998 1 2 yy yy yy yyy yy yyy yy y y ++ += ++ += +++= += = = 34. 119 2 1192 1139 1139 22 1 xx xx xxx xx xx xx xx x x =+ −−+ = = −=− −= −= 35. ()() 2 22 1110 339 33 10 99 210 5 xxx xx xx x x += −+− ++− = = = 36. () () () ()() += −++− +−++−=+− −++− ++−= ++−= −= = = 2 222 2 134 236 134 666 236 3324 3364 434 47 7 4 xxxx xxxxxx xxxx xx xx x x x 37. () 12 0 5 1520 350 35 5 3 xx xx x x x += −+= −= = = 38. 2 32 2 2 1 2 22 0 z z z z =+ + = + += = 39. ()() ()() ()() ()() ()() 212 4242 212 42 42 4242 21224 2228 123 4 xxxx xxxx xxxx xx xx x x =+   =+−−      =−+− =−+− = =

() () () () () 251 22 251 22 22 252 2510 48 2 xxxx xxxx xxxx xx xx x x +=   −+=−      +−= +−= = =

40.

INSTRUCTOR

USE ONLY

In the original equation, 3 x = yields a denominator of 0. So, 3 x = is an extraneous solution, and the original equation has no solution.

49. Female: 0.38619.20yx=−

For 43, y =

In the original equation, 3 x =− yields a denominator of 0. Thus, 3 x =− is an extraneous solution, and the original equation has no solution.

430.38619.20

62.20.386

=− =

161.14.

x x x

The height of the female is about 161.14 centimeters.

50. Male: () ?

yx=− =−

0.44229.37 480.44217529.37

4847.98

Yes, the estimated height of a male with a 48-centimeter thigh bone is about 175 centimeters.

51. (a)

(b) () 1.5 2221.525 lw Plwwww = =+=+=

52. (a)

Dimensions: 7.5 m5 m ×

l h w

(b) () 2 3 10 24 22222333

hw Phwwwwww

= =+=+=+=

==  == ==

30.9 m and 0.90.6 m Pww h

Dimensions: 0.6 m0.9 m ×

142 Chapter2SolvingEquationsandInequalities ©

+−= +−= = =

2016

341 33 343 3412 39 3 xxxx xx xx

x +=

42.

()

()()

6182315

xxx xxx

++ +−=+ +−=+ +=+ =−

() ()

() () () ()

35 62 33 35 62 333 33 63235

418315

x xxxx x xxxxxx xxxx

+ −= ++ + +−+=+

43. 1 2 2 2 Abh Abh A h b = = = 44. () 1 2 2 2 2 Aabh Aahbh Aahbh Aah b h =+ =+ −= = 45. 1 1 1 nt nt nt r AP n A P r n r PA n  =+  =  +    =+  46. APPrt APPrt AP r Pt =+ −= = 47. 2 2 Vrh hV r π π = = 48. 2 2 2 1 3 3 3 Vrh Vrh hV r π π π = = =

≈

NOT FOR SALE INSTRUCTOR USE ONLY

53. (a) test 1test 2test 3test 4 Test Average 4 +++ =

1test 2test 3test 4

You must earn at least 92 points on the fourth test to earn an

in the course.

54. Sales:Monday, $150

55. Model: ()() distanceratetime =

The salesperson drove 50 km in a half hour, therefore the rate is 50 kmkm 100 . 12 hrhr =

Since the salesperson continues at the same rate to travel a total distance of 250 km, the time required is distance250 km 2.5 hours. rate100 kmhr ==

56. Model: ()() distanceratetime =

Total distancerate #1time #1rate #2time #2

57. Model: ()()() 12 Distanceratetimetime =+

Labels:

miles, rate,

Equation: 200200 400

58.

The average speed for the round trip was approximately 46.3 miles per hour.

Distance50 kilometers Rate100 kilometers/hour

59. Let height of the pine tree.

The

60. (a)

h 30 ft 5 ft 6 ft

(b) Model:

30 feet tall.

The salesperson traveled for 4 hours at 58 mph and then 2 hours at 52 mph.

INSTRUCTOR

height of pole

()

height of pole's shadow

height of person

height of person's shadow =

h =

Labels: height of pole, height of pole's shadow

= =

h h

= =⋅=

USE ONLY

Section2.1LinearEquationsandProblemSolving 143 ©

not be scanned, copied or duplicated, or posted to a publicly accessible website, in whole or in part.

(b)

Test Average 4 939184 90 4 268 90 4 360268 92 x x x x +++ = +++ = + = =+ =

test

Tuesday,

Wednesday,

Thursday, $180 Friday, $x Average: 15012575180 150 5 530 150 5 530750 $220 x x x x ++++ = + = += =

$125

$75

()()()() ()

mimi 33658526 hrhr 3365831252 246 4 tt tt t t =+  =+−   =+− = =

rate40

== ==

1 1 2 2 Distance2200400

distance200 time hours, rate55 distance200 time hours

r =⋅==

160022003800 400 440440440 43.6 r rr r  =+   =+=   ≈

5540

1 2

Time hours Total distance300 kilometers Total time Rate100 kilometers/hour 3 hours === == =

36 2024 24720 30 feet x x x x = = = =

pine tree is approximately

()

30535 feet, height of person6 feet, height of person's shadow5 feet =+=

Equation: 6 355 6 3542 5

The pole is 42 feet tall.

NOT FOR SALE

61. ()20080002

20016,000 200 16,000 0.0125, or 1.25%

IPrt r r r r

= = = = =

62. Let 1 2 amount in 4% fund. x = Then 12,000amount in 5% fund. x −=

5600.0450.0512,000

5600.0456000.05

You must invest $8000 in the 1 2 4% fund and 12,0008000$4000 in the 5% fund. −=

63. Model: Total pounds at $5.25 = pounds at $2.50pounds + at $8.00

Labels: pounds of peanuts at $2.50

100pounds of walnuts at $8.00 x x

Equation: ()()()() 1005.252.501008.00

5252.58008

2755.5 50

The mixture contains 50 pounds at $2.50 and 1005050 −= pounds at $8.00.

64. Initially, the forester has 64 33 gallons of gas and 2 33 gallons of oil. 642 2 3333 6433 32 233

+= = Suppose she adds x gallons of gas.

643340 2331 x + = which gives 16 33 gallon. x =

65. Model: Total profit = profit on notebooks + profit on tablet

Labels: amount invested in notebook computers 40,000amount invested in tablet computers x x

= −=

Equation: ()() ()0.2440,0000.200.2540,000

So, $8000 is invested in notebook computers and $40,000$8000$32,000, −= invested in tablet computers.

xy xx x

67. ()

Abh A h b

= ===

1 2 2182.25 2 27 ft 13.5

69.

 lwhw Vlwhwww w w w

= Hence, square III has side of length 58 =13. + 2 Area13169 square inches == == === = = = ×× 70. 3 3 3 3

h w l

Vr r r r

(a) 4 3 4 6255 3 18,765 4 18,765 11.43 cm 4

π π π π

= = = =≈

144 Chapter2SolvingEquationsandInequalities

be scanned, copied or duplicated, or posted to a publicly accessible website,

whole or

part.

()

=+−

0.00540 8000 xx x x x =+−

= =

= −=

xx xx x x =+− =+− = =

−=− =

96000.210,0000.25 4000.05 8000 xx xx x x =+− =+−

66. Let amount invested x = in 810 × frames, and amount invested y = in 57 × frames. 45004500 xyyx +=  =− () () 0.250.220.2445001080 0.250.2245001080 0.0390 3000 +== +−= = =

So, $3000 is invested in 810 × frames and $4500$3000$1500 −= is invested in 57 × frames.

68. Let length of side of square I, x = and length of side of square II. y = 4205 4328 xx yy (b) () ()() () 1 2 3 2 3 9 2 3

=  = = 3,1 32304 2304 512 8 inches Dimensions: 24812 inches

NOT FOR SALE

73. Let the wind speed, x = then the rate to the city600, x =+ the rate from the city600, x =− the distance to the city1500 kilometers, = the distance traveled so far in the return trip15003001200 kilometers.

Section2.1LinearEquationsandProblemSolving 145 © 2016 Cengage Learning. All Rights Reserved. May not be scanned, copied or duplicated, or posted to a publicly accessible website, in whole or in part. 71. Solve the temperature for C () 9 32 5 9 32 5 5 32 9 FC FC FC =+ −= −= () () () () () () () 5 73:733222.8 9 5 74:743223.3 9 5 76:763224.4 9 5 78:783225.6 9 5 82:823227.8 9 5 79:793226.1 9 5 74:743223.3 9 FCC FCC FCC FCC FCC FCC FCC =°=−  ≈° =°=−  ≈° =°=−  ≈° =°=−  ≈° =°=−  ≈° =°=−  ≈° =°=−  ≈° 72. Solve the temperature for C () () () 9 32 5 9 32 5 5 32 9 5 74.332 9 5 42.3 9 23.5 FC FC FC C C C =+ −= −= =− = =°

=−= ()() Distance Time Rate 15001200 600600 15006001200600 900,0001500720,0001200 180,0002700 200 3 xx xx xx x x = = +− −=+ −=+ = = Wind speed: 200 kmh 3

Let height of the building in feet. h = feet4 feet 80 feet3.5 feet 4 803.5 3.5320 91.4 feet h h h h = = = ≈ 75. () () 12 507510 5075075 125750 WxWLx xx xx x =− =− =− = 6 feet from the 50-pound child x = 76. () () 12 1 2 11 3 200 pounds 550 pounds 5 feet 2005505 2002750550 7502750 feet WxWLx W W L xx xx x x =− = = = =− =− = = 30 20 22 24 26 28 Time Temperature ( ° C) 10:00 A.M. 11:00 A.M. 12:00 P.M. 1:00 P.M. 2:00 P.M. 3:00 P.M. 4:00 P.M. 5:00 P.M. 6:00 P.M. h 3 ft 1 2 4 ft 80 ft Not drawn to scale

74.

NOT FOR SALE

77. False. ()310 is a quadratic equation. xx−= 22 310 or 3100xxxx −+=−+=

78. False. () () 3 3 4 3

Volume of cube9.5857.375 cubic inches

Volume of sphere5.9860.29 cubic inches π

79. You need ()()()()() 33 or 33.abcbcaac −+=−=−−=−

One answer is 2, 1 ac== and () 3 23 bxx =+= and another is () 6, 1, and 15 615.acbxx ===+=

80. You need () 0 or .abcbc +== So, a can be any real number except 0. One answer is 1 a = and 4:44.bcx==+=

81. You need 1 or 44. 4 abcabc  +=+=  

One answer is 2 a = and 1 2 b = So, 1 24441. 2 cc  +==  =  

The equation is 1 21. 2 x +=

82. You need ()2.5. abc −+= One answer is 1 and 2. ab=−= So, () 12.524.5. c −−+==

The equation is 24.5. x −+=

83. In the original equation, 1 x = yields a denominator of zero. So, 1 x = is an extraneous solution and therefore cannot be a solution to the equation.

84. (a) Height of building Height of pole Length of building's shadowLength of pole's shadow =

85.

(b) 4 303 x =

()()

634 3131xxxx =+

To clear this equation of fractions, find the least common denominator (LCD) of all terms in the equation and multiply every term by this LCD. It is possible to introduce an extraneous solution because you are multiplying by a variable. To determine whether a solution is extraneous, substitute the answer for the variable in the original equation or graph the original equation.

== =≈

5 4 3 2 1 3 4 5

y 2 4246810 2 4 6 8

y x

4 5 123 1 2 5

+=+−= +=+− +=− 1 2 3 4 5 12354 1 2 3 5 4

5 3 2 1 3 4

2 5

y 1 1 2 3 5 6 7 8 9

521242, 2 52212422 102202 410 x

xcxcx cc cc c c y 1 2 3 4123456 2 3 4

3 2 1 2 4 5 6 7

146 Chapter2SolvingEquationsandInequalities ©

2016

5 2

86. = =

()() 87. 54 yx =− 88. 353531 22 22222 x yxx =+=−+=− 89. () 2 37 yx=−− 90. 2 1 4 3 yx =− 91. 1 2 41 yx=−+−

NOT FOR SALE

Section 2.2 Solving Equations Graphically

1. x-intercept, y-intercept

2. zero

3. The x-intercepts of ()yfx = are () 1, 0 and () 1, 0.

4. The y-intercept of ()ygx = is ()0,1.

5. The zeros of the function f are 1 x =− and 1. x =

6. The solutions of the equation ()() fxgx = are

2-interceptyy =++= 

10. 2 4 yx =−

Let 0: y =

()() 2 042, 22, 0, 2, 0-interceptsxxx =−  =− 

Let 0: x = () 2 4040, 4-interceptyy =−= 

11. 2 yxx=+

Let 0: y =

()() 020, 20, 0, 2, 0-interceptsxxxx=+  =− 

Let 0: x = ()00200, 0-interceptyy =+= 

12. 1 2 31 yxx=−++

Let 0: y =

11 22 0313132 xxxxxx=−++  ++  +=

 +=  +−=

23 2 2 2

 −++=  −+=

() ()

 =  =−

xxxx xxxxx xx

34340 1440120

11, 0 2 is impossible

Let ()010, 1-interceptxyy =  = 

13. 48 x y x =

Let 0: y =

48 0 048 84 22,0-intercept

= =− = = 

x x x x xx

()

Let 0: x = () 408 0 y = is impossible. No y-intercepts

14. 31 4 x y x =

Let 0: y =

()11 33 031 , 0-interceptxxx=−  = 

Let 0: 01 x ==− is impossible. No y-intercepts

Section2.2SolvingEquationsGraphically 147 © 2016 Cengage Learning. All Rights Reserved. May not be scanned, copied or duplicated, or posted to a publicly accessible website, in whole or in part. 92. 210 yx=−+

1 2 and 1. xx=−=

() Let

==−  () Let 0: 0550, 5-interceptxyyy ==−  =− 

3 4 3 yx=−− () 33 44 Let 0: 03344, 0 -intercept y xxxx = =−−  =−  =−  ()() 3 4 Let 0: 0330, 3 -intercept x yy = =−−=− 

5 yx=−

0: 055, 0-interceptyxx

Let 0: x = ()()() 2

9. 2 22yxx=++ Let 2 0: 220no -intercepts yxxx =++= 

020220,

() ()() ()()

18 16 12 10 8 6 4 2 108642 2 4 6 8 10 x y

INSTRUCTOR

Learning.

Rights Reserved.

USE ONLY © Cengage

All

15.

NOT

16.

17.

19. ()20310yx=−−

=+− =

=−− =−+

x-intercept: () ()

 

  = 

x-intercept: () ()

x x x x x 

y-intercept: () () 10202 60, 6 y y

−= −= =

=+− =



22. ()()359 31590

36 2

x x x =−+ −+= = = 16 12 18 4 14 2 10 14 35 10 40 40 12 4 12 12 3 15 12 12 4 4 8 4

148 Chapter2SolvingEquationsandInequalities ©

Cengage

2016

Learning.

 =  Let 0: x = ()

-intercept yyy−+=  = 

210xyyx−−+= Let 0: y = ()01011, 0-interceptxxx=−+=

11 22 210 0,

Let 0: y =

xx x xx −+= −= =−  Let 0: x = ()()

44 yy y yy −+= =  =   

43xyxy−+=

()() () 0403 3 33,0-intercept

0043

330,-intercept

x-intercept: () 3250 3650 311 1111 ,0 33 x x x x −−= −−= =  =    y-intercept: () () 3025 65 110,11 y y y =−− =−− =− 

() 325yx=−−

x x x x =+− =+ =− =− 

-intercept:

18. () 432yx=+− x-intercept: () () 

22 0432 0410 410 , 0

() () 4032 100, 10 y y

020310 020310 0303 330 1010, 0 =−

x x x x x = 

y-intercept: () () 203010 300, 30 y y

=−−

20. ()1022yx=+−

01022 01024 062 26 33, 0

=+− =+− =+ −= =−

21. ()() () 43 430 30

fxx

x x x =−

fxx

Section2.2SolvingEquationsGraphically 149 © 2016 Cengage Learning. All Rights Reserved. May not be scanned, copied or duplicated, or posted to a publicly accessible website, in whole or in part. 23. () () ()() 32 32 2 65 650 650 510 0, 5, 1 fxxxx xxx xxx xxx x =−+ −+= −+= −−= = 24. () ()() 32 32 918 9180 360 0, 3, 6 fxxxx xxx xxx x =−+ −+= −−= = 25. () ()() 12 1 27 12 10 27 7122140 7724140 5250 525 5 fxxx xx xx xx x x x +− =−+ +− −+= +−−+= +−++= += =− =− 26. () ()() 2 10 3 10 30 3100 520 5, 2 fxx x x x xx xx x =−− −−= −−= −+= =− 27. 2.70.41.2 2.31.2 1.212 0.522 2.323 xx x x −= = ==≈ () 2.70.41.20 2.31.20 0.522 fxxx x x =−−= −= ≈ 28. 3.68.20.5 3.18.2 8.282 3.131 xx x x −= = == () 3.68.20.50 3.18.20 82 31 fxxx x x =−−= −= = 29. ()() 1221543 122415603 387 29 xx xx x x +=−− +=−− −=− = ()()()12215430 3870 29 fxxx x x =+−−+= −+= = 30. ()12003002500 90021000 19002 950 x x x x =+− =− = = ()()300250012000 3002100012000 219000 950 fxx x x x =+−−= +−−= −= = 31. () 31 210 24 61 10 442 719 42 38 7 x x xx x x ++= +=− = = () () 31 2100 24 311 100 242 719 0 42 7380 38 7 x fxx xx x x x =++−= ++−= −= −= = 2 15 6 2 4 12 10 12 7 15 9 9 6 4 6 4

INSTRUCTOR

NOT FOR SALE

USE ONLY

150 Chapter2SolvingEquationsandInequalities © 2016 Cengage Learning. All Rights Reserved. May not be scanned, copied or duplicated, or posted to a publicly accessible website, in whole or in part. 32. () 21 56 32 215 6 322 717 62 51 7.286 7 x x x x x +−=  +=+  = =≈ () () () 21 560 32 435360 4315360 7510 7.286 fxxx xx xx x x =+−−= +−−= +−−= −= ≈ 33. () 0.600.401001.2 0.60400.401.2 0.2038.8 194 xx xx x x +−= +−= =− =− () () 0.600.401001.20 0.60400.401.20 0.2038.80 194 fxxx xx x x =+−−= +−−= += =− 34. () 0.750.2803.9 0.75160.23.90 0.5512.10 551210 22 xx xx x x x +−= +−−= += =− =− () () 0.750.2803.90 0.5512.10 5512100 22 fxxx x x x =+−−= += += =− 35. ()() 335 32 23335 26915 97 9 1.286 7 xx xx xx x x = −=− −=− = =≈ () ()() 335 0 32 233350 269150 790 790 1.286 fxxx xx xx x x x =−= −−−= −−+= −+= −= ≈ 36.

35 2512 123255 123625125

= −=− −=− = =≈ ()

1232550 1236251250

fxxx

−−−= −−+= −+= −= ≈

= = () 5 100 42 52400 3450 15 fxxx xx x x =+−= −+−= −= =

()()

x −= −−−= −−+= −= =− () () 53 10

526100 90 90 9 fxxx xx xx x x x =−−= −−−−= −−+−= −−= += =−

0.3

=− =−=−

22 2610 44610 1030 0.3 fxxxx xxxx x x =+−+−= ++−+−= += =−

()()

1389 89 6.846 13 xx xx xx x x

()() 35

2512

13890 13890 6.846

xx xx x x x =−=

37. () 5 10 42 5240 345 15 xx xx x x

−+=

38.

53 1 105 52310 52610 9 9 xx xx xx x

105 523100

39. () 2 2 22 3 10 261 4461 103

xxx xxxx x x +=−+ ++=−+

()()

NOT FOR SALE

Section2.2SolvingEquationsGraphically 151 © 2016 Cengage Learning. All Rights Reserved. May not be scanned, copied or duplicated, or posted to a publicly accessible website, in whole or in part. 40. ()()()() 2 22 1 5 12212 21242 51 0.2 xxxx xxxxx x x ++−=+− +++−=−− = == ()()()()() () 2 22 22 122120 212420 4320 510 0.2 fxxxxx xxxxx xxxx x x =++−−+−= +++−−−−= +−−++= −= = 41. 3 40 1.379 xx x ++= ≈− 42. 3 240 1.128 xx x ++= ≈− 43. () 2 1 4 10170 2.172, 7.828 xx x −+= ≈ 44. () 2 1 2 660 1.268, 4.732 xx x −−+= ≈ 45. 32 21890 3.0, 0.5. 3.0 xxx x −−+= =− 46. 32 41226240 4.206, 0.735, 1.941 xxx x +−−= ≈−− 47. 53 53 33 330 1.861 xx xx x =− −+= ≈− 48. 53 53 32 320 xx xx =+ −−= 1.638 x ≈ 49. 2 3 2 2 30 2 x x = + −= + 1.333 x ≈− 50. 1 2 3 1 20 3 x x =− += 2.5 x = −6 1 6 7 8 8 12 12 3 3 12 7 5 7 10 3 5 20 5 30 6 40 4 60 2 33 7 6 6 6 2 8 6 4 2 2 6 48 NOT

SALE INSTRUCTOR USE

FOR

ONLY

152 Chapter2SolvingEquationsandInequalities © 2016 Cengage Learning. All Rights Reserved. May not be scanned, copied or duplicated, or posted to a publicly accessible website, in whole or in part. 51. 53 1 2 53 130 2 xx xx =+ + −−= + 3.162, 3.162 x ≈− 52. 33 1 2 33 10 1 xx xx += + +−= 1.303, 2.303 x ≈− 53. 13 130 4,2 x x x −+=− −++= =− 54. 6 2 260 4,8 x x x −−=− −+= =− 55. 3214 3250 x x −−= −−= 1.0, 2.333 x ≈− 56. 4128 4160 x x ++= +−= 1.75, 1.25 x =− 57. 23 230 x x −= −−= 11 x = 58. 48 480 x x −= −−= 68 x = 59. 251 150 4 x x x −+= −+= =− 60. 896 290 5 x x x −+= −+= =− 6 16 6 6 45 4 2 6 10 8 6 4 711 8 66 6 2 87 6 4 1 5 14 5 0100 10 6 2 63 2 3 10 5 3

NOT FOR SALE

Because of the sign change, 1.81.9. x << To improve accuracy, evaluate the expression for values in this interval and determine where the sign changes. Let 1 3.25.8.yx=− The graph of 1y crosses the x-axis at 1.8125. x =

Because of the sign change, 56. x <<

Because of the sign change, 5.15.2. x << To improve accuracy, evaluate the expression in this interval, and determine where the sign changes. Let () 1 0.31.81.yx=−− The graph of 1y crosses the x

Section2.2SolvingEquationsGraphically 153 © 2016 Cengage Learning. All Rights Reserved. May not be scanned, copied or duplicated, or posted to a publicly accessible website, in whole or in part. 61. (a) x 1 0 1 2 3 4 3.25.8 x 9 5.8 2.6 0.6 3.8 7.0 Because of the sign change, 12.

<< (b) x 1.5 1.6 1.7 3.25.8 x 1 0.68 0.36 x 1.8 1.9 2.0 3.25.8 x 0.04 0.28 0.6

62. () 0.31.810 x −−=

63. 6 32 yx yx =− =− ()() 632 84 2624 ,2,4 xx x xy xy −=− = =  =−= = 64. 23 9 yx yx =− =− ()() 239 312 4945 ,4,5 xx x xy xy −=− = =  =−= = 65. 2662 0 xyyx xyyx +=  =− −+=  = ()() 62 63 22 , 2, 2 xx x xyx xy −= = =  == = 66. 44 2525 xyxy xyxy −=−  =− +=  =−+ ()() 425 39 3341 , 1, 3 yy y yx xy −=−+ = =  =−=− =− x 2 3 4 5 6 7 8 () 0.31.81 x 0.94 0.64 0.34 0.04 0.26 0.56 0.86 x 5.0 5.1 5.2 5.3 5.4 () 0.31.81 x 0.04 0.01 0.02 0.05 0.08 8 2 510 5 6 10 4

-axis at 5.13. x ≈

INSTRUCTOR

Cengage

USE ONLY ©

Rights Reserved. May not be scanned, copied or duplicated, or posted to a publicly accessible website, in whole or in part. 67. 1 2 1010 242 xyyx xyyx −=  =− +=  =−+ ()() 1 2 102 2204 324 88102 , 8, 2 xx xx x xy xy −=−+ −=−+ = =  =−=− =− 68. 11 44 4444 41 xyyx xyyx −=  =− −=  =− ()() 11 44 44 16161 1515 1440 , 1, 0 xx xx x xyx xy −=− −=− = =  =−= = 69. 2 2 1 24 yxx yxx =−+ =++ ()() ()() 22 2 124 33 1 1113 , 1, 3 xxxx x x y xy −+=++ −= =− =−−−+= =− 70. 2 2 31 24 yxx yxx =−++ =−−− ()() ()() 22 2 3124 3124 55 1 13113 , 1, 3 xxxx xx x x y xy −++=−−− +=−− =− =− =−−+−+=− =−− 71. 92 3 yx yx =− =− ()() , 4,1 xy =

2016 Cengage Learning. All

1 331 3 511 5211 22 ,3,2 xyyx xyyx xy −=−  =+ −=  =− = 73. 2 2 yx yxx = =− ()()() , 0, 0, 1, 1 xy = 74. 2 4 21 yx yx =− =− ()() () ,1.449, 1.898, 3.449, 7.899 xy = 75. 3333 2552 xyyx xyyx −=  =− +=  =− ()() ,1.670,1.660xy = 4 4 8 4 (4, 1) y = 9 2x y = x 3 8 99 4 y = x + 1 1 3 y = x 5 2 11 2 (3, 2) 3 3 45 (0, 0) (1, 1) y = x y = 2 x − x 2 11 9 10 5 y = 4 x 2 y = 2x 1 ( 3.449, 7.899) (1.449, 1.898) 6 4 6 4 y = x 3 3 y = 5 2x

INSTRUCTOR

72. ()()

NOT FOR SALE

USE ONLY

NOT FOR SALE

80. (a) ()0.3355 Axx =+−

(b) Domain: 055 x ≤≤

13.9287153813.93

79. (a) () 280 6354 xx t =+

(b) Domain: 0280 x ≤≤

and 164.5 x = miles.

81. (a) Divide into two regions. Then find the area of each region and add.

Total areaArea 1Area 2 =+

(b)

Axxx xx x

() 442 48 12

=⋅+⋅ =+ =

82. (a) 1 2 Abh =

() () () 2 1211 2332 1 Axxxxx =+=+

(b)

4 4 4 12 y = 2x 2 y = x 4 2x ( 2, 8)(2, 8) (0, 0) 0 0 280 10 0 0 55 55 x x 2x 4 4 Area 1 Area 2 8 0 0 20 240 0 0 25 240

83. (a) 11 10,00010,00022 TISxxx =+=+−=+

==−  =

(d) If 1 2 12,50010,000 Tx ==+ then 5000. x = Thus, 1 2 10,000$7500.Sx=−=

yxx = =− ()()()() ,0, 0, 2,8, 2, 8 xy =−

(a) 10.732051.73205 10.732050.26795 6.4640796.46 + = ≈≈ (b) 10.732051.73205 10.732050.26795 1.73 0.27 6.4074076.41 + = ≈ ≈≈

76. 2 42 2 2 yx

77.

Yes, the more rounding performed, the less accurate the result.

78. (a) 10.866031.86603 10.866030.13397

+ = ≈≈ (b) 10.866031.86603 10.866030.13397 1.87 0.13

+ = ≈ ≈≈

14.3846153814.38

Yes, the more rounding performed, the less accurate the result.

3 4 4 t =

(b) 11 22 If 660010,000 3400 $6800 Sxx x 

==+  =

INSTRUCTOR

USE ONLY

84. 16.9574,012ytt =+≤≤

(a) Let 0 t = and find y () 16.90574574 y =+=

The median weekly earnings of full-time workers was $574 in 2000.

(b) The slope m is 16.9. The median weekly earnings of full-time workers increases by $16.90 every year.

(d) Answers will vary. Sample answer: Algebraically: Let 800 y = and solve for t Graphically: Using the zoom and trace features, find t when 800. y =

86. (a)

() ()()() () ()()()

85. 43.45355,113

28.45398,113 Mtt Wtt =+≤≤ =+≤≤

(a)

501 4008 1 00 8 1 8 m yx yx == −=−

05.

The point of intersection is approximately () 2.9,5479.4. So, in 2002, both states had the same population.

tt t t

+=+ = ≈

(b) 43.4535528.45398 1543 2.9

The point of intersection is approximately () 2.9,5479.4. So, in 2002, both states had the same population.

(c) The slopes of the linear models represent the change in population per year. Since the slope of the model for Maryland is greater than that of Wisconsin, the population of Maryland is growing faster.

(d) Find 16. t =

43.41653556049.4

28.41653985852.4 M W =+= =+=

156 Chapter2SolvingEquationsandInequalities

2016 Cengage

Learning.

() ()

The population of Maryland will be 6,049,400 and the population of Wisconsin will be 5,852,400. Answers will vary. ()

TOTALRECTANGULAR SIDEWALLTRIANGULAR SIDEWALL 1 2 1 2 pool widthpool width 4402040520

=⋅⋅+⋅⋅⋅ =+ =

 =   = 3 3

5200 cubic feet VVV lwbh =+

(b) ()

gallons Number of gallons5200 ft7.48 ft 38,896 gallons

()

= So, () ()() 2 11 2082080,

22 Vbddddd ===≤≤ For 59, d <≤ ()()()()()() 1 5402054020 2 20008004000 8002000, 59. Vd d dd =+− =+− =−<≤ (e) (f) d 3 5 7 9 V 720 2000 3600 5200 (g) 4800: V = 80020004800 8006800 8.5 feet d d d −= = = 5000 113 6000 (0, 0) d b (0, 5) (40, 5) (0, 0) (0, 5) (0, 9) (40, 5) 231456789 1000 2000 3000 4000 5000 Depth (in feet) Volume (in cubic feet) d V

INSTRUCTOR

(d) For 05, d ≤≤ by similar triangle, 8. 540 db bd = 

NOT FOR SALE

USE ONLY

87. True.

NOT FOR SALE

(b) () 2 1 fxx=−

88. True. A line must intersect at least one axis.

89. 99 1100 1009999 99

x x xx x

= =− =−

The approximate answer 99.1 is not a good answer, even though the substitution yields a small error.

90. (a) From the table, () 0 fx = for 3. x =

(b) From the table, () 0 gx = for 2. x =−

(d) From the table, ()()6.fxgx =− In this case, ()() 12 and 2, fxgx=−= for 1. x =−

91. (a) 22yx=+

(c)

22, 1 yxyx=+=−

CollegeAlgebraRealMathematicsRealPeople7thEditionLarsonSolutionsManual FullDownload:http://testbanktip.com/download/college-algebra-real-mathematics-real-people-7th-edition-larson-solutions-manual/ Downloadallpagesandallchaptersat:TestBankTip.com

Section2.2SolvingEquationsGraphically 157

scanned, copied or duplicated, or posted to a publicly accessible website, in

or in part.

whole

21 1

x x += =− =− () 1, 0-intercept x Let 0: x = () 202 2 y y =+ = () 0, 2-intercept y Numerically: x 2 1 0 1 2 22yx=+ 2 0 2 4 6 The x-intercept is () 1, 0, The y-intercept is () 0, 2. Graphically:

Algebraically: Let 0: y = 220

Algebraically: 1

zeros. ()() ()() 2 2 1110 1110 f f −=−−= =−= Numerically: x 2 1 0 1 2 2 1 yx=− 3 0 1 0 3 So, the zeros are 1 and 1. Graphically:

and 1 xx=−= are

Algebraically: ()() 2 2 221 230 310 xx xx xx +=− −−= −+= 3010 31 xx xx −=+= ==− () () 32328 3, 8 xy =  =+= () () 12120 1, 0 xy =−  =−+= Numerically: x 1 0 1 2 3 22yx=+ 0 2 4 6 8 2 1 yx=− 0 1 0 3 8 The points of intersection are ()() 1, 0 and 3, 8. Graphically: 2 4 45 2 6 66 2 10 99 ( 1, 0) y1 y2 (3, 8)

INSTRUCTOR

USE ONLY