Section 1.1
Check Point Exercises
Chapter 1
4. The meaning of a [100,100,50]by[100,100,10] viewing rectangle is as follows: distance between -axis
minimummaximumtick -value-valuemarks [100,100,50] x xx by distance between -axis
minimummaximumtick -value-valuemarks [100,100,10] y yy
3,7 2,6 1,5 0,4 1,3 2,2 3,1 xy xy xy xy xy xy xy =−=
3. 4,3 3,2 2,1 1,0 0,1 1,2 2,3 xy xy xy xy xy xy xy
5. a. Thegraphcrossesthe x-axisat(–3,0). Thus,the x-interceptis–3. Thegraphcrossesthe y-axisat(0,5). Thus,the y-interceptis5.
b. Thegraphdoesnotcrossthe x-axis. Thus,thereisno x-intercept. Thegraphcrossesthe y-axisat(0,4). Thus,the y-interceptis4.
c. Thegraphcrossesthe x-and y-axesat theorigin(0,0). Thus,the x-interceptis0andthe y-interceptis0.
6. Thenumberoffederalprisonerssentencedfor drugoffensesin2003isabout57%of159,275. Thiscanbeestimatedbyfinding60%of 160,000.
96,000 N ≈ =× =
60%of160,000
0.60160,000
3,7 2,2 1,1 0,2 1,1 2,2 3,7 xy xy xy xy xy xy xy
14.
16.
x = –3, y = 11
x = –2, y = 6
x = –1, y = 3
x = 0, y = 2
x = 1, y = 3
x = 2, y = 6
x = 3, y = 11 15. 3,5 2,4 1,3 0,2 1,1 2,0 3,1 xy xy xy xy xy xy xy
x = –3, y = –1
x = –2, y = 0
x = –1, y = 1
x = 0, y = 2
x = 1, y = 3
x = 2, y = 4
x = 3, y = 5
17. 3,5 2,3 1,1 0,1 1,3 2,5 3,7 xy xy xy xy xy xy xy
18.
19. 3 3, 2 2,1 1 1, 2 0,0 1 1, 2 2,1 3 3, 2 xy xy xy xy xy xy xy
20.
x = –3, y = –10
x = –2, y = –8
x = –1, y = –6
x = 0, y = –4
x = 1, y = –2
x = 2, y = 0
x = 3, y = 2
x = –3, y = 7 2
x = –2, y = 3
x = –1, y = 5 2
x = 0, y = 2
x = 1, y = 3 2
x = 2, y = 1
x = 3, y = 1 2
21.
x =−3, y = 6
x =−2, y = 4
x =−1, y = 2
x = 0, y = 0
x = 1, y = 2
x = 2, y = 4
x = 3, y = 6
22. 3,6 2,4 1,2 0,0 1,2 2,4 3,6 xy xy xy xy xy xy xy
23. 3,4 2,3 1,2 0,1 1,2 2,3 3,4 xy xy xy xy xy xy xy
3,2 2,1 1,0 0,1 1,0 2,1 3,2 xy xy xy xy xy xy xy
25.
x =−3, y = 0
x =−2, y = 5
x =−1, y = 8
x = 0, y = 9
x = 1, y = 8
x = 2, y = 5
x = 3, y = 0
26. 3,9 2,4 1,1 0,0 1,1 2,4 3,9 xy xy xy xy xy xy xy =−=−
27. 3,27 2,8 1,1 0,0 1,1 2,8 3,27 xy xy xy xy xy xy xy =−=−
28. 3,28 2,9 1,2 0,1 1,0 2,7 3,26 xy xy xy xy xy xy xy =
29. (c) x-axis tick marks –5, –4, –3, –2, –1, 0, 1, 2, 3, 4, 5; y-axis tick marks are the same.
30. (d) x-axis tick marks –10, –8, –6, –4, –2, 0, 2, 4, 6, 8, 10; y-axis tick marks –4, –2, 0, 2, 4
31. (b); x-axis tick marks –20, –10, 0, 10, 20, 30, 40, 50, 60, 70, 80; y-axis tick marks –30, –20, –10, 0, 10, 20, 30, 40, 50, 60, 70
32. (a) x-axis tick marks –40, –20, 0, 20, 40; y-axis tick marks –1000, –900, –800, –700, . . . , 700, 800, 900, 1000
33. Theequationthatcorrespondsto2 Y inthetable is(c),22yx = .Wecantellbecauseallof thepoints(3,5),(2,4),(1,3),(0,2), (1,1),(2,0),and(3,1)areontheline 2 yx = ,butallarenotonanyoftheothers.
34. Theequationthatcorrespondsto1 Y inthetable is(b),2 1 yx = .Wecantellbecauseallofthe points(3,9),(2,4),(1,1),(0,0),(1,1), (2,4),and(3,9)areonthegraph2 yx = ,but allarenotonanyoftheothers.
35. No.Itpassesthroughthepoint(0,2).
36. Yes.Itpassesthroughthepoint(0,0)
37. (2,0)
38. (0,2)
39. Thegraphsof1 Y and2 Y intersectatthepoints (2,4)and(1,1)
40. Thevaluesof1 Y and2 Y arethesamewhen 2 x =− and1 x =
41. a. 2; The graph intersects the x-axis at (2, 0).
b. –4; The graph intersects the y-axis at (0,–4).
42. a. 1; The graph intersects the x-axis at (1, 0).
b. 2; The graph intersects the y-axis at (0, 2).
43. a. 1, –2; The graph intersects the x-axis at (1, 0) and (–2, 0).
b. 2; The graph intersects the y-axis at (0, 2).
44. a. 1, –1; The graph intersects the x-axis at (1, 0) and (–1, 0).
b. 1; The graph intersect the y-axis at (0, 1).
45. a. –1; The graph intersects the x-axis at (–1, 0).
b. none; The graph does not intersect the yaxis.
46. a. none; The graph does not intersect the x-axis.
b. 2; The graph intersects the y-axis at (0, 2).
47.
3(3,5) 2(2,5) 1(1,5) 0(0,5) 1(1,5) 2(2,5) 3(3,5) xxy
52. (,)
3(3,1)
2(2,1)
1(1,1)
0(0,1)
1(1,1)
2(2,1)
3(3,1) xxy
53. () () (,) 1 22, 2 11,1 11 ,2 22 11 ,3 33 11 ,3 33 11 ,2 22 11,1 1 22, 2 xxy
⎛⎞ ⎜⎟ ⎝⎠ ⎛⎞ ⎜⎟ ⎝⎠ ⎛⎞ ⎜⎟ ⎝⎠ ⎛⎞ ⎜⎟ ⎝⎠ ⎛⎞ ⎜⎟ ⎝⎠ ⎛⎞ ⎜⎟ ⎝⎠
⎛⎞ ⎜⎟ ⎝⎠ ⎛⎞ ⎜⎟ ⎝⎠
⎛⎞ ⎜⎟ ⎝⎠ ⎛⎞ ⎜⎟ ⎝⎠ ⎛⎞ ⎜⎟ ⎝⎠
54. () () (,) 1 22, 2 11,1 11 ,2 22 11 ,3 33 11 ,3 33 11 ,2 22 11,1 1 22, 2 xxy ⎛⎞ ⎜⎟ ⎝⎠
55. Therewereapproximately65democracies in1989.
56. Therewere12040=80more democraciesin2002thanin1973.
57. Thenumberofdemocraciesincreasedat thegreatestratebetween1989and1993.
58. Thenumberofdemocraciesincreasedat theslowestratebetween1981and1985.
59. Therewere49democraciesin1977.
60. Therewere110democraciesin1997.
61. 1650.75;40RAA=−= ()1650.751650.7540
16530135 RA−−=− =−=
Thedesirableheartrateduringexercisefor a40-yearoldmanis135beatsperminute. Thiscorrespondstothepoint(40,135)on thebluegraph.
62. 1430.65;40RAA=−= ()1430.651430.6540
14326117 RA−−=− =−=
Thedesirableheartrateduringexercisefor a40-yearoldwomanis117beatsper minute.Thiscorrespondstothepoint(40, 117)ontheredgraph.
63. a. Atbirthwehave0 x = () 2.936 2.9036 2.9036 36 yx=+ =+ =+ = Accordingtothemodel,thehead circumferenceatbirthis36cm.
b. At9monthswehave9 x = . () 2.936 2.9936 2.9336 44.7 yx=+ =+ =+ = Accordingtothemodel,thehead circumferenceat9monthsis44.7cm.
c. At14monthswehave14 x = 2.936 2.91436 46.9 yx=+ =+ ≈ Accordingtothemodel,thehead circumferenceat14monthsisroughly 46.9cm.
d. Themodeldescribeshealthychildren.
64. a. Atbirthwehave0 x = () 435 4035 4035 35 yx=+ =+ =+ = Accordingtothemodel,thehead circumferenceatbirthis35cm.
b. At9monthswehave9 x = () 435 4935 4335 47 yx=+ =+ =+ = Accordingtothemodel,thehead circumferenceat9monthsis47cm.
c. At14monthswehave14 x = 435 41435 50 yx=+ =+ ≈ Accordingtothemodel,thehead circumferenceat14monthsisroughly 50cm.
d. Themodeldescribessevereautistic children.
71. 2 45.48334.351237.9yxx =−+
The discharges decreased from 1990 to 1994, but started to increase after 1994. The policy was not a success.
72. a. False; (x, y) can be in quadrant III.
b. False; when x = 2 and y = 5, 3y – 2x = 3(5) – 2(2) = 11.
c. False; if a point is on the x-axis, y = 0.
d. True; all of the above are false. (d) is true.
73. (a)
74. (d)
75. (b)
76. (c)
77. (b)
78. (a)
Section 1.2
Check Point Exercises
1. 4529 x += 455295 424 424 44 6 x x x x +−=− = = =
Check: 4529
4(6)529 24529 2929true x += += += = Thesolutionsetis{6}.
2. 4(21)293(25) xx +−=− 8429615 825615 82566156 22515 225251525 210 210 22 5 xx xx xxxx x x x x x +−=− −=− −−=−− −=− −+=−+ = = =
Check: 4(21)293(25) 4[2(5)1]293[2(5)5] 4[101]293[105] 4[11]293[5] 442915 1515true xx +−=− +−=− +−=− −= −= = Thesolutionsetis{5}. 3. 355 4147 xx + =− ()() 355 2828 4147 732(5)45 72110420 721410 741021 1111 1111 1111 1 xx xx xx xx xx x x x
−=−− −=−− +=−+ = = = Check: 355 4147 13515 4147 256 4147 11 22 xx + =− + =− =− −=− Thesolutionsetis{1}.
4. 5171 ,0 2183 x xx = −≠ 5171 1818 2183 5171 181818 2183 45176 4561766 5117 5117 1717 3 xx xx xx xx x x x x x ⎛⎞ ⋅=−⎜⎟ ⎝⎠ =⋅−⋅ =− +=−+ = = = Thesolutionsetis{3}.
5. 22 ,2 223 x x xx =−≠ 22 3(2)3(2)223 22 3(2)(32)3(2) 223 36(2)2 362(2) 3624 3102 321022 510 510 55 2 x xx xx x xxx xx xx xx xx xx xxxx x x x ⎡⎤
Thesolutionsetistheemptyset,. ∅
6. Set12 yy = . 2 1122 4416 1122 44(4)(4) (4)(4)(4)(4)22(4)(4) 44(4)(4) (4)(4)22 4422 222 11 xxx xxxx xxxxxx xxxx xx xx x
Check: 2 2 1122 4416 1122 1141141116 1122 157105 2222 true 105105 xxx += +−− += +− += =
7. 474(1)3 xx−=−+ 474(1)3 47443 4741 71 xx xx xx −=−+ −=−+
Theoriginalequationisequivalenttothe statement–7=–1,whichisfalseforeveryvalue of x.Thesolutionsetistheemptyset, ∅ Theequationisaninconsistentequation.
Exercise Set 1.2
1. 7x – 5 = 72 7x = 77 x = 11
Check: 7572 7(11)572 77572 7272 x = −= −= =
The solution set is {11}.
2. 6x – 3 = 63 6x = 66 x = 11
The solution set is {11}.
Check: 6363 6(11)363 66363 6363 x = −= = = 3. 11(65)40 116540 5540 535 7 xx xx x x x −= −+= += = =
The solution set is {7}.
Check: 11(65)40
11(7)[6(7)5]40 77(425)40 77(37)40 4040 xx −= −= −= = =
4. 5x – (2x – 10) = 35 5x – 2x + 10 = 35 3x + 10 = 35 3x = 25 25 3 x =
The solution set is 25 3 ⎧⎫ ⎨⎬ ⎩⎭
Check: 5(210)35 2525 521035 33 12550 1035 33 12520 35 33 105 35 3 3535 xx−−=
5. 2x – 7 = 6 + x x – 7 = 6 x = 13
The solution set is {13}.
Check: 2(13)7613 26719 1919 −=+ −= =
6. 3x + 5 = 2x + 13 x + 5 = 13 x = 8
The solution set is {8}.
Check: 35213 3(8)52(8)13 2451613 2929 xx+=+ +=+ +=+ = 7. 7x + 4 = x + 16 6x + 4 = 16 6x = 12 x = 2
The solution set is {2}.
Check: 7(2)4216 14418 1818 + =+ += = 8. 13x + 14 = 12x – 5 x + 14 = –5 x = –19
The solution set is {–19}.
Check: 1314125
13(19)1412(19)5 247142285 233233 xx + =− +=−− −+=−−
9. 3(x – 2) + 7 = 2(x + 5) 3x – 6 + 7 = 2x + 10 3x + 1 = 2x + 10 x + 1 = 10 x = 9
The solution set is {9}.
Check: 3(92)72(95) 3(7)72(14) 21728 2828 +=+ += += =
10. 2(x – 1) + 3 = x – 3(x +1) 2x – 2 + 3 = x – 3x – 3 2x +1 = –2x – 3 4x + 1 = –3 4x = –4 x = –1
The solution set is {–1}.
Check: 2(1)33(1) 2(11)313(11) 2(2)313(0) 4310 11 xxx+=−+ −+=−−−+ −+=−− −+=−+ =−
11. 3(x – 4) – 4(x – 3) = x + 3 – (x – 2) 3x – 12 – 4x + 12 = x + 3 – x + 2 –x = 5 x = –5
The solution set is {–5}.
Check: 3(54)4(53)53(52)
3(9)4(8)2(7) 273227 55 −−−−−=−+−−−
12. 2 – (7x + 5) = 13 – 3x 2 – 7x – 5 = 13 – 3x –7x – 3 = 13 – 3x –4x = 16 x = –4
The solution set is {–4}.
Check: 2(75)133 2[7(4)5]133(4) 2[285]1312 2[23]15 22325 2525 xx−+=− −−+=−−
13. 16 = 3(x – 1) – (x – 7) 16 = 3x – 3 – x + 7 16 = 2x + 4 12 = 2x 6 = x
The solution set is {6}.
Check:
163(61)(67) 163(5)(1) 16151 1616 = =−− =+ =
14. 5x – (2x + 2) = x + (3x – 5) 5x – 2x – 2 = x + 3x – 5 3x – 2 = 4x – 5 –x = –3 x = 3
The solution set is {3}.
Check: ( ) ( ) 52235 5(3)[2(3)2]3[3(3)5] 15[62]3[95]
15834
77 xxxx+=+− +=+− −+=+− −=+ =
15. 25 – [2 + 5y – 3(y + 2)] = –3(2y – 5) – [5(y – 1) – 3y + 3]
25 – [2 + 5y – 3y – 6] = –6y + 15 – [5y – 5– 3y + 3]
25 – [2y – 4] = –6y + 15 – [2y – 2]
25 – 2y + 4 = –6y + 15 – 2y + 2 –2y + 29 = –8y + 17 6y = –12 y = –2
The solution set is {–2}.
Check:
25[253(2)3(25)[5(1)33]
25[25(2)3(22)3[2(2)5][5(21)3(2)3] 25[2103(0)]3[45][5(3)63] 25[8]3(9)[159]
yyyyy −+−+=−−−−−+ −+−−−+=−−−−−−−−+ −−−=−−−−−++ −−=−−−−+
25827(6) 33276 3333 +=−− =+ =
16. 45 – [4 – 2y – 4(y + 7)] = –4(1 + 3y) – [4 – 3(y + 2) – 2(2y – 5)]
45 – [4 – 2y – 4y – 28] = –4 – 12y – [4 – 3y – 6 – 4y + 10]
45 – [–6y – 24] = –4 – 12y – [–7y + 8] 45 + 6y + 24 = –4 – 12y + 7y – 8 6y + 69 = –5y – 12 11y = –81 81 11 y =− The solution set is 81 11 ⎧⎫ ⎨⎬
17. 2 32 62 32 xx xx =− ⎡⎤ =− ⎢⎥ ⎣⎦ 2312 1232 12 xx xx x =− =− = The solution set is {12}.
18. 1 56 301 56 6530 6530 30 xx xx xx xx x =+ ⎡⎤ =+ ⎢⎥ ⎣⎦ =+ −= = The solution set is {30}.
19. 20 32 620 32 12023 12032 1205 120 5 24 xx xx xx xx x x x −= ⎡⎤ −= ⎢⎥ ⎣⎦ −= =+ = = = The solution set is {24}. 20. 1 526 1 30 526 6155 6515 15 xx xx xx xx x −= ⎡ ⎤ −= ⎢ ⎥ ⎣ ⎦ −= −= = The solution set is {15}.
21. 32 1 53 32 151 53 91015 91015 15 15 xx xx xx xx x x =+ ⎡ ⎤ =+ ⎢ ⎥ ⎣ ⎦ = + −= −= =−
The solution set is {–15}.
22. 3 5 24 3 45 24 2320 2320 20 20 Thesolutionsetis{20}. xx xx xx xx x x =+ ⎡⎤ =+ ⎢⎥ ⎣⎦ =+ −= −= =−
23. 35 5102 35 10 5102 61025 425 525 5 xx x xx x xxx xx x x −=− ⎡⎤ −=− ⎢⎥ ⎣⎦ −=− −−=− −=− =
The solution set is {5}.
24. 217 2 722 217 142 722 2847119 247119 17119 7
Thesolutionsetis{7}. xx x xx x xxx xx x x −=+ ⎡⎤ −=+ ⎢⎥ ⎣⎦ −=+ −= = =
25. 335 684 335 24 684 4129630 462112 233 33 2 xx xx xx xx x x +− =+ +− ⎡⎤ =+ ⎢⎥ ⎣⎦ +=+− −=−− −=− =
The solution set is 33 2 ⎧⎫ ⎨⎬ ⎩⎭
26. 112 463 112 12 463 33284 34103 77 1
Thesolutionsetis{1}. xx xx xx xx x x +− =+ +− ⎡⎤ =+ ⎢⎥ ⎣⎦ +=+− +=− =
–12 The solution set is {–12}.
28. {} 23 5 38 23 245 38 12081639 839104 595 19
Thesolutionsetis19. xx xx xx xx x x −+ += −+ ⎡⎤ += ⎢⎥ ⎣⎦ +−=+ −=− =− =−
29. 12 5 37 xx + + =− 12 215 37 xx++ ⎡ ⎤ =− ⎢ ⎥ ⎣ ⎦ 7x + 7 = 105 – 3x – 6 7x + 3x = 99 – 7 10x = 92 92 10 46 5 x x = =
The solution set is 46 . 5 ⎧ ⎫ ⎨ ⎬ ⎩⎭
30. 332 523 332 30 523 1815451020 3102045 725 25 7
The25solutionsetis. 7 xxx xxx xxx xx x x −+ −= −+ ⎡ ⎤ −= ⎢ ⎥ ⎣ ⎦ −+=+ −=− −=− = ⎧ ⎫ ⎨ ⎬ ⎩⎭
31. a. 45 23(0) x xx =+≠
b. 45 3 2 856 36 1 2 xx x x x =+ =+ = =
The solution set is 1 2 ⎧⎫ ⎨⎬ ⎩⎭
32. a. 5104(0) 3 x xx =+≠
b. 510 4 3 151012 512 5 12 xx x x x =+ =+ = =
The solution set is 5 12 ⎧⎫ ⎨⎬ ⎩⎭
33. a. 25133(0) 24 x xx +=+≠
b. 2513 3 24 8121013 2 2 xx xx x x +=+ +=+ −= =−
The solution set is {–2}.
34. a. 7522 233(0) x xx −=≠
b. 7522 233 211044 1144 1 4 xx x x x −= −= = =
The solution set is 1 4 ⎧⎫ ⎨⎬ ⎩⎭
35. a. 21111 3463(0) x xx +=−≠
b. 21111 3463 83224 714 2 xx xx x x + =− + =− = =
The solution set is {2}.
36. a. 5811 29183(0) x xx =−≠
b. 5811 29183 45166 1751 3 xx xx x x −=− −=− −=− = The solution set is {3}.
37. a. 21 1 2 xx xx + += (x ≠ 0)
b. 21 1 2 xx xx + += x – 2 + 2x = 2x + 2 x – 2 = 2 x = 4
The solution set is {4}.
38. a. 4974 55 x xx =− (x ≠ 0)
b. 4974 55 20974 162 8 x xx xx x x =− = −+ = =
The solution set is {8}.
39. a. 111 5(1)11 x xx + =≠
b. 111 5 11 15(1)11 15511 5411 515 3 xx x x x x x += +−= +−= −= = =
The solution set is {3}.
40. a. 34 7(4)44 x xx −=≠− ++
b. 34 7 44 37(4)4 37284 721 3 xx x x x x −= ++ −+=− −−=− −= =−
The solution set is {–3}.
41. a. 884(1)11 x x xx =−≠− ++
b. 88 4 11 84(1)8 8448 44 1nosolution x xx xx xx x x =− ++ =+− =+− =− =−⇒
The solution set is the empty set, ∅
42. a. 2 222(2) x x xx =−≠
b. 2 2 22 22(2) 224 2nosolution x xx xx xx x =− =−− =−+ =⇒
The solution set is the empty set, ∅
43. a. 312(1)2221 x xx +=≠
b. 312 2221 312 2(1)21 31(1)4 314 2 xx xx x x x += += +−= +−= =
The solution set is {2}.
44. a. 351 3262(3,2) xx xxx = +≠−≠ ++−
b. 351 32(3)2 6(2)5(2)2(3) 61251026 8 8 xxx xxx xxx x x =+ ++− =−++ −=−++ −= =−
The solution set is {–8}.
45. a. 328;(2,2) 22(2)(2) x xxxx +=≠− +−+−
b. 328 22(2)(2) (2,2) 3(2)2(2)8 36248 510 2nosolution xxxx xx xx xx x x += +−+− ≠≠− −++= −++= = =⇒
The solution set is the empty set, ∅
46. a. 5312 22(2)(2) (2,2) xxxx xx += +−+− ≠≠−
b. 5312 22(2)(2) 5(2)3(2)12 5103612 816 2nosolution xxxx xx xx x x += +−+− −++= −++= = =⇒
The solution set is the empty set, .∅
47. a. 2 212 111(1,1) x xx xxx =≠≠− +−−
b. 2 212 111 212 11(1)(1) 2(1)1(1)2 2212 3 3 x xxx x xxxx xxx xxx x x −= +−− −= + −+− −−+= −−−= −= =−
The solution set is {–3}.
48. a. 2 4232 5525xxx += +−− ; 5,5 x ≠−
b. 4232 55(5)(5) (5,5)
4(5)2(5)32 42021032 642 7 xxxx xx xx xx x x += +−+− ≠≠− −++= −++= = =
The solution set is {7}.
49. a. 156 ;(2,4) 42(4)(2) x xxxx −=≠− −+−+
b. 156 42228 156 42(4)(2) (4,2)
1(2)5(4)6 25206 416 4nosolution xxxx xxxx xx xx xx x x −= −+ −= −+−+ ≠≠− +−−= +−+= −=− =⇒
The solution set is the empty set, ∅
50. a. 2 6520 326xxxx −= +−+− ; 3,2 x ≠
b. 6520 32(2)(3) (3,2)
6(2)5(3)20 61251520 7 xxxx xx xx xx x −= +−−+ ≠−≠ −−+=− −−−=− =
The solution set is {7}.
51. Set 12yy = 5(28)25(3)3 104025153 1042512 1051242 530 6 xx xx xx xx x x −−=−+ −−=−+ −=− −=−+ = = The solution set is {6}.
52. Set 12yy = 7(32)56(21)24 2114512624 2191218 2112189 927 3 xx xx xx xx x x +=−+ −+=−+ −=+ −=+ = =
The solution set is {3}.
53. Set 121.yy = ()() 35 1 54 35 2020201 54 435520 41252520 1320 7 7 xx xx xx xx x x x −= −⋅=⋅ −−−= −−+= −+= −= =−
The solution set is {–7}.
54. Set 124.yy =− ()() 12 4 43 12 121212(4) 43 314248 334848 1148 59 59 xx xx xx xx x x x +− −=− +− −⋅=− +−−=− +−+=− −+=− −=− =
The solution set is {59}.
55. Set 123 yyy += 2 531219 43712 531219 43(4)(3) 531219(4)(3)(4)(3)43(4)(3) 5(3)3(4)1219 5153121219 8271219 48 2 x xxxx x
The solution set is {2}.
56. Set 123. yyy += 2 2121 2842 2121 (4)(2)42 2121 (4)(2)(4)(2)(4)(2)42 212(2)4 21244 454 39 3 x xxxx x
The solution set is {3}.
57. 04[(3)]7(1) xxx =−−−+ 04[3]77 04[23]77 081277 019 19 19 xxx xx xx x x x =−+−− =−−− =−−− =− −=− =
The solution set is {19}.
58. 02[3(46)]5(6) xxx =−−−− 02[346]530 02[6]530 0212530 0742 742 6 xxx xx xx x x x =−+−+ =−+−+ =−+−+ =−+ = =
The solution set is {6}.
59. 652 0 31243 x xx + =−− 652 0 3(4)43
3(4)03(4)652 3(4)43
0(6)152(4) 061528 01 1 x xx x xx xx xxxx xx xx xx x x + =−− ⎛⎞ + −⋅=−−−⎜⎟ ⎝⎠
03(4)(6)53(4)23(4) 3(4)43
The solution set is {–1}.
60. 137 0 5515 xx =−+ ++ 137 0 5(1)15 137 5(1)05(1)5(1)15
015(1)35(1)75(1) 5(1)15 01157(1) 011577 077 77 1 xx xx xx xxx xx x x x x x =−+ ++
The solution set is {1}.
61. 4(x – 7) = 4x – 28 4x – 28 = 4x – 28
The given equation is an identity.
62. 4(x – 7) = 4x + 28 4x – 28 = 4x + 28
The given equation is an inconsistent equation.
63. 2x + 3 = 2x – 3 3 = –3
The given equation is an inconsistent equation.
64. 7 7 77 x x xx = =
The given equation is an identity.
65. 4x + 5x = 8x 9x = 8x x = 0
The given equation is a conditional equation.
66. 8x + 2x = 9x 10x = 9x x = 0
The given equation is a conditional equation.
67. 26 4 33 264(3) 26412 26 3nosolution x xx xx xx x x =+ =+− =+− −=− =⇒
The given equation is an inconsistent equation.
68. 3 3 33 33(3) 339 412 3nosolution x xx xx xx x x =+ =+− =+− −=− =⇒
The given equation is an inconsistent equation.
69. 521 4 23
3(5)242(21) 3152442 7 7 xx xx xx x x +− −= +−=− +−=− −= =−
The solution set is {–7}. The given equation is a conditional equation.
70. 21 5 73
3(2)1057(1) 3610577 1092 92 10 46 5 xx xx xx x x x ++ =− +=−+ +=−− = = =
The solution set is 46 5 ⎧⎫ ⎨⎬ ⎩⎭
The given equation is a conditional equation.
71. 2 3 22 23(2) 236 48 2nosolution x xx xx xx x x =+ =−+ =−+ −=− =⇒
The solution set is the empty set, ∅ The given equation is an inconsistent equation.
72. 62 2 33
62(3)2 6262 412 3nosolution x xx xx xx x x += ++ ++=− ++=− =− =−⇒
This equation is not true for any real numbers. The given equation is an inconsistent equation.
73. 8x – (3x + 2) + 10 = 3x 8x – 3x – 2 + 10 = 3x 2x = –8 x = –4
The solution set is {–4}. The given equation is a conditional equation.
74. 2(x + 2) + 2x = 4(x + 1) 2x + 4 + 2x = 4x + 4 0 = 0
This equation is true for all real numbers. The given equation is an identity.
75. 213 24 823 8 8 x xx x x += += =− = The solution set is {8}. The given equation is a conditional equation.
76. 311 63 182 318 6 x xx x x −= −= =− = The solution set is {6}. The given equation is a conditional equation.
77. 437 25(5)(2)
4(5)3(2)7 420367 77 1 xxxx xx xx x x += ++− ++−= ++−= =− =−
The solution set is {–1}. The given equation is a conditional equation.
78. 114 1(23)(1)23
1(23)14(1) 23144 26 3 xxxx xx xx x x =+ +−+ +=+− +=+− −=− = The solution set is {3}. The given equation is a conditional equation.
79. ()() 2 2 2 22 22 412436 ;3,3 339 43123436 4121236436 42436436 243636 2472 3Nosolution xx x xxx xxxx xxxx xxx x x x
The solution set is { }. The given equation is an inconsistent equation.
80. 222 413 310612xxxxxx −= +−+−−−
= = The solution set is 1 7 ⎧⎫ ⎨⎬ ⎩⎭
The given equation is a conditional equation.
81. Theequationis3(4)3(22) xx −=− ,andthe solutionis2 x =
82. Theequationis3(25)52 xx−=+ ,andthe solutionis17 x = .
83. Theequationis3(3)5(2) xx−−=− ,andthe solutionis0.5 x = .
84. Theequationis254(31)2 xx−=+− ,andthe solutionis0.7 x =− .
85. Solve:4(2)242(2) 482442 4664 264 22 1 xxx xxx xx x x x −+=−− −+=−+ −=− −−=− −= =− Now,evaluate2xx for1 x =− : 22(1)(1) 1(1)112 xx−=−−− =−−=+=
86. Solve:2(6)32(21) 212342 21272 5122 510 2 xxx xxx xx x x x =+− −=+− −=− −−=− −= =− Now,evaluate2xx for2 x =− : 22(2)(2) 4(2)426 xx−=−−− = −−=+=
87. Solvefor x: 3(3)26 5 3(3)5(26) 391030 7930 721 3 x x xx xx x x x + =+ +=+ +=+ −+= −= =−
88.
Solvefor y:210518 71018 728 4 yy y y y −−=+ −−= −=
Now,evaluate2() xxyy for3 x =− and 4 y =− : [] [] 2 2 2 () (3)3(4)(4) (3)12(4) 9(124)9167 xxyy
Solvefor x: 136 52 4 1364(52) 136208 768 714 2 x x xx xx x x x =+ −=+
Solvefor y:57(4)1 57281 5729 5829 824 3 yy yy yy y y y −=++ −=++ −=+ −= −= =−
Now,evaluate2() xxyy for2 x =− and 3 y =− : [] [] 2 2 2 () (2)2(3)(3) (2)6(3) 4(63)495 xxyy
89. () () () 2 2 363454 93454 813454 27454 10854 2 x x x x x x ⎡⎤+÷⋅=− ⎣⎦
÷⋅=− ÷⋅=− ⋅=−
Thesolutionsetis {} 2
90. () () 33 3 24538 8428 8488 8328 248 3 x x x x x x ⎡⎤−=− ⎣⎦ ⎡⎤ =− ⎣⎦
⋅=− =− =− = Thesolutionsetis { } 3.
91. ( ) () [] [] [] 3 5128763255 512876321255 51287631275 5128721275 512872545 512872545 51212246 5246 xxx xxx xxx xxx xxx xxx xx ⎡ ⎤ −=−−÷++ ⎣ ⎦ −=−−÷++ ⎡ ⎤ ⎣ ⎦ −=−−÷⋅+
Thefinalstatementisacontradiction,sotheequation hasnosolution.Thesolutionsetis ∅
92. ( ) ( ) () () 2558104213.511 10116104611 101161045 101161020 11620 xx xx xx xx +=+÷− +=+− +=+− +=− =−
Thefinalstatementisacontradiction,sotheequation hasnosolution.Thesolutionsetis ∅
93. 0.70.4(20)0.5(20) 0.780.510 0.2810 0.22 10 xx xx x x x + =+ +=+ += = = Thesolutionsetis{10}.
94. 0.5(2)0.13(0.10.3) 0.510.10.30.9 0.510.31 0.211 0.20 0 xx xx xx x x x + =++ +=++ +=+ += = = Thesolutionsetis{0}.
95.
Thesolutionsetis {} 2
96.
Thesolutionsetis
97. LetT=4421.Then 44211652771 1650165 10 x x x =+ = =
Tuitionwillbe$4421tenyearsafter1996,which istheschoolyearending2006.
98. LetT=4751.Then 47511652771 1980165 12 x x x =+ = =
Tuitionwillbe$4751twelveyearsafter1996,which istheschoolyearending2008.
99. 1267 ; 992 DND = += 7126 299 71261818 299 63252 112 112 22 5.5 N N N N N N =+
=+ ⎜⎟⎜⎟ ⎝⎠⎝⎠ =+ = = = Ifthehigh-humorgroupaveragesalevelof depressionof3.5inresponsetoanegativelife event,theintensityofthateventwouldbe5.5.The solutionisthepointalongthehorizontalaxis wherethegraphforthehigh-humorgrouphasa valueof3.5ontheverticalaxis.Thiscorresponds tothepoint ( ) 5.5,3.5onthehigh-humorgraph.
100. Substitute10for D inthelowhumor formula.TheLCDis9.
() 101053 99 105391099 99 901053 9053105353 3710 3710 1010 3.7 N N N N N N N =+ =+⎛⎞⎛⎞
=+ −=+− = = = Theintensityoftheeventwas3.7.This isshownasthepoint(3.7,10)onthe low-humorgraph.
101. 12;500,1000 DA CCD A === + 1000 500 12 A A = + ()() 1000 1250012 12 50060001000 6000500 12 A AA A AA A A ⎛⎞ +⋅=+ ⎜⎟ + ⎝⎠ += = = Thechild’sageis12yearsold.
102. ;300,1000 12 DA CCD A === + 1000 300 12 A A = + LCD=12 A + ()() 1000 1230012 12
30036001000
3600700 3600 700 36 5.14 7 A AA A AA A A A ⎛⎞ +⋅=+ ⎜⎟
Tothenearestyear,thechildis5yearsold.
103. The solution is the point (12, 500) on the blue graph.
104. The solution is the point (5, 300) on the blue graph..
105. No,becausethegraphscross,neitherformulagives aconsistentlysmallerdosage.
106. Yes,thedosagegivenbyCowling’sRulebecomes greateratabout10years.
107. 11learningtrials;representedbythe point () 11,0.95 on the graph.
108. 1learningtrial;representedbythe point () 1,0.5 on the graph.
109. 0.1(500)
500 Cx x + = + 0.280.1(500)
500 0.28(500)0.1(500) 0.2814050
0.7290 0.7290 0.720.72 125 x x xx xx x x x + = + +=+ +=+ −=− = = 125litersofpureperoxidemustbeadded.
110. a. 0.35(200) 200 Cx x + = +
b. 0.740.35(200)
200 x x + = +
0.74(200)0.35(200)
300 xx xx x x x + =+ +=+ −=− = = 300litersofpureacidmustbeadded. 120. { }3
0.7414870 0.2678 0.2678 0.260.26
121. { }5 122. { }7 123. { }5
124. a. False; –7x = x –8x = 0 x = 0
The equation –7x = x has the solution x = 0.
b. False; 4 44 x xx = (x ≠ 4) x = 4 ⇒ no solution
The equations 4 44 x xx = and x = 4 are not equivalent.
c. True; 3y – 1 = 11 3y – 7 = 5 3y = 12 3y = 12 y = 4 y = 4
The equations 3y – 1 = 11 and 3y – 7 = 5 are equivalent since they are both equivalent to the equation y = 4.
d. False; if a = 0, then ax + b = 0 is equivalent to b = 0, which either has no solution (b ≠ 0) or infinitely many solutions (b = 0).
(c) is true.
125. Answers may vary.
126. 74 13 7(6)4136
Section 1.3
Check Point Exercises
1. Let x = the number of football injuries
Let x + 0.6 = the number of basketball injuries
Let x + 0.3 = the number of bicycling injuries ( ) ( ) 0.60.33.9 0.60.33.9
In 2004 there were 1 million football injuries, 1.6 million basketball injuries, and 1.3 million bicycling injuries.
2. Let x = the number of years after 2004 that it will take until Americans will purchase 79.9 million gallons of organic milk. 40.75.679.9 5.679.940.7 5.639.2
Americans will purchase 79.9 million gallons of organic milk 7 years after 2004, or 2011.
3. Let x = the number of minutes at which the costs of the two plans are the same.
PlanAPlanB
150.0830.12 150.081530.1215 0.080.1212 0.080.120.12120.12 0.0412 0.0412 0.040.04 300 xx xx xx xxxx
The two plans are the same at 300 minutes.
4. Let x = the computer’s price before the reduction.
0.30840 0.70840 840 0.70 1200 xx x x x −= = = = Before the reduction the computer’s price was $1200.
5. Let x =theamountinvestedat9%. Let5000– x =theamountinvestedat11%.
0.090.11(5000)487
0.095500.11487
$3150wasinvestedat9%and$1850was investedat11%.
6. Let x =thewidthofthecourt. Let x +44=thelengthofthecourt. 22 2(44)2288 2882288 488288 4200 200 4 50 4494 lwP xx xx x x x x x += ++= ++= += = = = += Thedimensionsofthecourtare50by94.
7. 22lwP += 2222 22 22 22 2 2 lwlPl wPl wPl Pl w +−=− =− = =
8. PCMC = + (1) (1) 11 1 1 PCM PCM MM P C M P C M = + + = ++ = + = +
Exercise Set 1.3
1. Let x =thenumber 5426 530 6 x x x = = = Thenumberis6.
2. Let x =thenumber 2311 214 7 x x x = = = Thenumberis7.
3. Let x =thenumber 0.2020 0.8020 25 xx x x = = = Thenumberis25.
4. Let x =thenumber 0.3028 0.7028 40 xx x x = = = Thenumberis40.
5. Let x =thenumber 0.60192 1.6192 120 xx x x + = = = Thenumberis120.
6. Let x =thenumber 0.80252 1.8252 140 xx x x + = = = Thenumberis140.
7. Let x =thenumber 0.70224 320 x x = = Thenumberis320.
8. Let x =thenumber 0.70252 360 x x = = Thenumberis360.
9. Let x =thenumber 26 x + =theothernumber ()2664 2664 22664 238 19 xx xx x x x ++= ++= += = = If x =19,then2645 x += . Thenumbersare19and45.
10. Let x =thenumber, Let x +24=theothernumber ()2458 2458 22458 234 17 xx xx x x x ++= ++= += = = If x =17,then x +24=41. Thenumbersare17and41.
11. 122yy−= (134)(510)2 1345102 8142 816 816 88 2 xx xx x x x x −−+= −−−= −= = = =
12. 123yy−= (106)(127)3 1061273 2133 210 210 22 5 xx xx x x x x +−−= +−+= −+= −=− = =
13. 12814yy = + 10(21)8(21)14 201016814 20101622 432 432 44 8 xx xx xx x x x =++ −=++ −=+ = = = 14. 121251yy = 9(35)12(31)51 2745361251 27453663 918 918 99 2 xx xx xx x x x =−− −=−− −=− −=− = =
15. 123 3522 yyy=− 3(26)5(8)()22 61854022 2222 2222 00 xxx xxx xx xx + −+=− +−−=− −=− =−+ = Thesolutionsetisthesetofallrealnumbers.
16. 123 2348 yyy=− 2(2.5)3(21)4()8 56348 6248 1010 1010 1010 1 xx xx xx x x x +=− −−=− −+=− −=− = =
17. 123344yyy += 111 344 21 324 1 54 1
5(1)4(1) 1 5(1)4 554 5 xxx xxx xx xxxx xx xx xx x
18. 123637yyy −= 2 2 111 637 1 637 1 637 (1)1 637 (1)(1)(1)1 6(1)3(1)7(1) (1)1 6(1)37 6637 697 679 9 9 11 9 xx xx xx xx xxxx xxxx xxxx xxxxxx xxxx xx xx xx xx x x x
19. Let x =thenumberofbirths(inthousands) Let229 x =thenumberofdeaths(in thousands). ()229521 229521 2229521 2229229521229 2750 2750 22 375 xx xx x x x x x +−= +−= −= −+=+ = = = Thereare375thousandbirthsand 375229146 −= thousanddeathseachday.
20. Let x = the number responding yes. Let 82 – x = the number responding no. (82)36 8236 82236 23682 246 246 22 23
23% responded yes and 59% responded no.
21. Let x = the number of Internet users in China. 10 x + = the number of Internet users in Japan. 123 x + = the number of Internet users in the United States. ( ) ( ) 10123271 3133271 3138 46 xxx x x x ++++=
= = If 46 x = , then 1056 x + = and 123169 x += . Thus, there are 46 million Internet users in China, 56 million Internet users in Japan, and 169 Internet users in the United States.
22. Let x = energy percentage used by Russia. 6 x + = energy percentage used by China. 16.4 x + = energy percentage used by the United States. ( ) ( ) 616.440.4 616.440.4 322.440.4 318 318 33 6 612 16.422.4 xxx xxx x x x
Thus, Russia uses 6%, China uses 12%, and the United States uses 22.4% of global energy.
23. Let x = the percentage of Conservatives. Let 2x + 4.4 = the percentage of Liberals. () 24.457.2
The percentage of Conservatives is 17.6% and the percentage of Liberals is 39.6%
24. Let x = the number of hate crimes based on sexual orientation.
3127 x + = the number of hate crimes based on race.
(3127)13431026337485 312713431026337485 425297485 44956 44956 44 1239 31273844
Thus, there were 3844 hate crimes based on race and 1239 based on sexual orientation.
25. Let L =thelifeexpectancyofanAmericanman. y =thenumberofyearsafter1900.
550.2
85550.2
300.2 150 Ly y y y =+ =+ =
Thelifeexpectancywillbe85yearsintheyear 19001502050 +=
26. Let L =thelifeexpectancyofanAmericanman, Let y =thenumberofyearsafter1900
550.2
91550.2
Thelifeexpectancywillbe91yearsintheyear 1900+180=2080.
27. a. 1.739.8yx = +
b. 1.739.844.98.5 x + =+ 1.739.853.4 1.713.6 1.713.6 1.71.7 8 x x x x +
= = = The number of Americans without health insurance will exceed 44.9 million by 8.5 million 8 years after 2000, or 2008.
28. a. 1.739.8yx = + b. 1.739.844.910.2 x + =+ 1.739.855.1 1.715.3 1.715.3 1.71.7 9 x x x x + = = = =
The number of Americans without health insurance will exceed 44.9 million by 10.2 million 9 years after 2000, or 2009. c.
29. Let v =thecar’svalue. y =thenumberofyears(after2003).
80,5008705
19,56580,5008705 60,9358705 7 vy y y y =− =− −=− =
Thecar’svaluewillbe$19,565after7years.
30. Let v =thecar’svalue. y =thenumberofyears(after2003).
80,5008705
36,97580,5008705 43,5258705 5 vy y y y =−
Thecar’svaluewillbe$36,975after5years.
31. Let x =thenumberofmonths. ThecostforClubA:2540 x + ThecostforClubB:3015 x + 25403015 54015 525 5 xx x x x +=+ −+= −=− =
Thetotalcostfortheclubswillbethesameat5 months.Thecostwillbe 25(5)4030(5)15$165 +=+=
32. Let g =thenumberofvideogamesrented 9450 550 10 gg g g =+ = =
Thetotalamountspentateachstorewillbethe sameafter10rentals. 99(10)90 g == Thetotalamountspentwillbe$90.
33. Let x =thenumberofuses. Costwithoutcouponbook:1.25 x Costwithcouponbook:150.75 x + 1.25150.75 0.5015 30 xx x x =+ = = Thebusmustbeused30timesinamonthforthe coststobeequal.
34. Costpercrossing:$5x Costwithcouponbook:$30+$3.50x 5303.50 1.5030 20 xx x x = + = = Thebridgemustbeused20timesinamonthfor thecoststobeequal.
35. a. Let x =thenumberofyears(after2005). CollegeA’senrollment:13,3001000 x + CollegeB’senrollment:26,800500 x 13,300100026,800500 13,300150026,800 150013,500 9 xx x x x + =− += = = Thetwocollegeswillhavethesame enrollmentintheyear20059 + 2014 = Thatyeartheenrollmentswillbe 13,3001000(9) 26,800500(9) 22,300students + =− =
b. Checkpointstodeterminethat 113,3001000 yx = + and 226,800500 yx =
36. Let x =thenumberofyearsafter2000 10,600,00028,00010,200,00012,000 16,000400,000 25 xx x x =− −=− = Thecountrieswillhavethesamepopulation25 yearsaftertheyear2000,ortheyear2025. 10,200,00012,00010,200,00012,000(25) 10,200,000300,000 9,900,000 x =− =− = Thepopulationintheyear2025willbe9,900,000.
37. Let x =thecostofthetelevisionset. 0.20336 0.80336 420 xx x x = = = Thetelevisionset’spriceis$420.
38. Let x =thecostofthedictionary 0.3030.80 0.7030.80 44 xx x x = = = Thedictionary’spricebeforethereductionwas$44.
39. Let x =thenightlycost 0.08162 1.08162 150 xx x x += = = Thenightlycostis$150.
40. Let x =thenightlycost 0.05252 1.05252 240 xx x x += = = Thenightlycostis$240.
41. Let x= theannualsalaryformenwhosehighest educationalattainmentisahighschooldegree. 0.2244,000 1.2244,000 36,000 xx x x += = ≈
Theannualsalaryformenwhosehighest educationalattainmentisahighschooldegreeis about$36,000.
42. Let x= theannualsalarywithahighschool degree 34,0000.26 34,0001.26 26984.13 xx x x =+ = Theannualsalaryforwomenwithahighschool degreeisapproximately$27,000.
43. Letc=thedealer’scost 5840.25 5841.25 467.20 cc c c =+ = = Thedealer’scostis$467.20.
44. Letc=thedealer’scost 150.25 151.25 12 cc c c =+ = = Thedealer’scostis$12.
45. Let x =theamountinvestedat6%. Let7000– x =theamountinvestedat8%. 0.060.08(7000)520 0.065600.08520 0.02560520 0.0240 40 0.02 2000 70005000 xx xx x x x x x + −= +−= −+= −=− = = −= $2000wasinvestedat6%and$5000was investedat8%.
46. Let x =theamountinvestedinstocks. Let11,000– x =theamountinvestedinbonds. 0.050.08(11,000)730 0.058800.08730 0.03880730 0.03150 150 0.03 5000 11,0006000 xx xx x x x x x + −= +−= −+= −=− = = −= $5000wasinvestedinstocksand$6000was investedinbonds.
47. Let x = amount invested at 12% 8000 – x = amount invested at 5% loss .12.05(8000)620 .12400.05620 .171020 6000 8000-2000 xx xx x x x −= −+= = = = $6000 at 12%, $2000 at 5% loss
48. Let x = amount at 14% 12000 – x = amount at 6% () .140.612000680 .14720.06680 .21400 7000 1200070005000 xx xx x x −−= −+= = = −=
$7000 at 14%, $5000 at 6% loss
49. Let w =thewidthofthefield
Let2w =thelengthofthefield ()() 2length2width P =+ ()()300222ww=+
30042
3006
50 ww w w =+ = = If50 w = ,then2100 w = .Thus,the dimensionsare50yardsby100yards.
50. Let w =thewidthoftheswimmingpool, Let3w =thelengthoftheswimmingpool ()() ()() 2length2width
320232
32062 3208 40 P ww ww w w =+ =+ =+ = = If w =40,3w =3(40)=120. Thedimensionsare40feetby120feet.
51. Let w =thewidthofthefield
Let2w +6=thelengthofthefield 228612 2166 36 w w w =+ = = If36 w = ,then262(36)678 w +=+= .Thus, thedimensionsare36feetby78feet.
52. Let w =thewidthofthepool, Let2w– 6=thelengthofthepool ()() ()() 2length2width 1262262 1264122 126612 1386 23 P ww ww w w w =+ =−+ =−+ =− = = Findthelength. 262(23)646640 w −=−=−= Thedimensionsare23metersby40meters.
53. Let x =thewidthoftheframe. Totallength:162 x + Totalwidth:122 x + ()() 2(length)2(width) 7221622122
72324244
72856 168 2 P xx xx x x x = + =+++ =+++ =+ = = Thewidthoftheframeis2inches.
54. Let w =thewidthofthepath
Let40+2w =thewidthofthepoolandpath
Let60+2w =thelengthofthepoolandpath 2(402)2(602)248 8041204248 2008248 848 6 ww ww w w w + ++= +++= += = = Thewidthofthepathis6feet.
55. Let x = number of hours
35x = labor cost
35x + 63 = 448
35x = 385 x = 11 It took 11 hours.
56. Let x = number of hours
63x = labor cost
63x + 532 = 1603 63x = 1071 x = 17 17 hours were required to repair the yacht.
57. Let x = inches over 5 feet 100 + 5x = 135 5x = 35 x = 7 A height of 5 feet 7 inches corresponds to 135 pounds.
58. Let g =thegrossamountofthepaycheck ( ) YearlySalary212750 3315024750 3240024 1350 g g g g =+ =+ = = Thegrossamountofeachpaycheckis$1350.
59. Let x = the weight of unpeeled bananas. 7weightofpeeledbananas 8 x = 77 88 17 88 7 xx x x =+ = = The banana with peel weighs 7 ounces.
60. Let x =thelengthofthecall. () 0.430.3212.105.73
66. 2 ; 2 Cr C r = π = π circumferenceofacircle
67. 2 2; Emc E m c = = Einstein’sequation
0.430.320.322.105.73
0.322.215.73 0.323.52 11 x x x x x +−+= +−+= += = =
Thepersontalkedfor11minutes.
61. A = lw A w l = areaofrectangle
62. DRT D R T = = distance,rate,timeequation
63. 1 2 2 2 ; Abh Abh A b h = = = areaoftriangle
64. 1 3 3 3 VBh VBh V B h = = = volumeofacone
65. I = Prt ; I P rt = interest
68. 2 2; Vrh hV r =π = π volumeofacylinder
69. TDpm = + TDpm TDpm mm TD p m = = = totalofpayment
70. PCMC PCMC PC M C = + −= = markupbasedoncost
71. 1() 2 Ahab = + 2() 2 2 Ahab Aab h Aba h = + =+ −= areaoftrapezoid
72. 1() 2 Ahab = + 2() 2 2 Ahab Aab h Aab h = + =+ −= areaoftrapezoid
73. S = P + Prt S – P = Prt ; SP r Pt = interest
74. S = P + Prt S – P = Prt ; SP t = Pr interest
75. () F B SV BSVF F SV B F SV B = −= −= =+
76. 1 (1) 1 1 1 C S r SrC
markupbasedonsellingprice
77. () IRIrE IRrE E I Rr += += = + electriccurrent
78. 222 Alwlhwh =++ ()222 2 22 Alwhlw Alw h lw −=+ = + surfacearea
79. 111 () pqf qfpfpq fqppq fpq pq += += += = + thinlensequation
80. 111 12RRR =+ () 1221 1212 122 2 1 2 RRRRRR RRRRRR RRRRR RR R RR =+ −= −= = resistance
88. ConsiderExercise25.
Thelifeexpectancywillbe85yearsintheyear 1900+150=2050.
89. a. F = 30 + 5x F = 7.5x b.
c. Calculator shows the graphs to intersect at (12, 90); the two options both cost $90 when 12 hours court time is used per month.
d. 3057.5 302.5 12 xx x x + = = = Rent the court 12 hours per month.
90. .1.9(1000)420 .1900.9420 .8480 600 xx x x x +−= +−= −=− = 600 students at the north campus, 400 students at south campus.
91. Let x = original price
x – 0.4x = 0.6x = price after first reduction 0.6x – 0.4(0.6x) = price after second reduction 0.60.2472 0.3672 200 xx x x −= = =
The original price was $200.
92. Let x = woman’s age
3x = Coburn’s age
3x + 20 = 2(x + 20)
3x + 20 = 2x + 40 x + 20 = 40 x = 20
Coburn is 60 years old the woman is 20 years old.
93. Let x = correct answers 26 – x = incorrect answers 85(26)0 813050 131300 13130 10 xx xx x x x −−= −+= −= = = 10 problems were solved correctly.
94. Let x = mother’s amount 2x = boy’s amount 2 x = girl’s amount 214,000 2 7 14,000 2
$4,000 x xx x x ++= = =
The mother received $4000, the boy received $8000, and the girl received $2000.
95. Let x = the number of plants originally stolen
After passing the first security guard, the thief has: 111 222 222 xxxxx ⎛⎞ ⎜⎟+=−−=− ⎝⎠
After passing the second security guard, the thief has: 1 2 11 2 223 224 x xx ⎛⎞ ⎜⎟ −+=− ⎜⎟ ⎜⎟ ⎜⎟ ⎝⎠
After passing the third security guard, the thief has: 1 3 117 4 32 4282 x xx ⎛⎞ ⎜⎟ −+=− ⎜⎟ ⎜⎟ ⎜⎟ ⎝⎠
Thus, 17 1 82 x = 288 36 x x = =
The thief stole 36 plants.
96. CS VCN L =− () VLCLCNSN VLSNCLCN VLSNCLN VLSN C LN = −+ −=− −=− =
Section 1.4
Check Point Exercises
1. a. (52)(33) ii−++ 5233 (53)(23) 8 ii i i =−++ =++−+ =+
b. (26)(12) ii+−− 2612 (212)(61) 107 ii i i =+−+ =−++ =−+
2. a. 2 7(29)7(2)7(9) 1463 1463(1) 6314 iiiii ii i i −=− =− =−− =+
b. (54)(67)303524282 30352428(1) 30283524 5811 iiiii ii ii i +−=−+− =−+−− =+−+ =−
3. 54544 444 iii iii +++ =⋅ −−+ 2 2 205164 1644 20214 161 1621 17 1621 1717 iii iii i i i +++ = +−− +− = + + = =+
4. a. 27482748 93163 3343 73 ii ii ii i −+−=+ =⋅+⋅ =+ =
b. 22 22 2 (23)(23) (2)2(2)(3)(3) 4433 4433(1) 143 i ii ii i i −+−=−+ =−+−+ =−+ =−+− =−
c. 14121412 22 1423 2 1423 22 73 i i i i −+−−+ = −+ = =+ =−+
Exercise Set 1.4
1. (7 + 2i) + (1 – 4i) = 7 + 2i + 1 – 4i = 7 + 1 + 2i – 4i = 8 – 2i
2. (–2 + 6i) + (4 – i) = –2 + 6i + 4 – i = –2 + 4 + 6i – i = 2 + 5i
3. (3 + 2i) – (5 – 7i) = 3 – 5 + 2i + 7i = 3 + 2i – 5 + 7i = –2 + 9i
4. (–7 + 5i) – (–9 – 11i) = –7 + 5i + 9 + 11i = –7 + 9 + 5i + 11i = 2 + 16i
5. 6(54)(13)65413 iiii −+−−−=+−++ 243i =
6. 7(92)(17)79217 iiii −+−−−=+−++ 33 i =
7. 8i – (14 – 9i) = 8i – 14 + 9i = –14 + 8i + 9i = –14 + 17i
8. 15i – (12 – 11i) = 15i – 12 + 11i = –12 + 15i + 11i = –12 + 26i
9. –3i(7i – 5) 2 2115 ii=−+ = –21(–1) + 15i = 21 + 15i
10. –8i (2i – 7) 2 1656 ii=−+ = –16(–1) + 56i 2 92592534 i =−=+= = 16 + 56i
11. (54)(3)1551242 iiiii −++=−−++
1574 197 i i =−+− =−+
12. (48)(3)1242482 iiiii −−+=−−−− 12288 428 i i =−−+ =−−
13. (7 – 5i)(–2 – 3i) 142110152 iii=−−++ = –14 – 15 – 11i = –29 – 11i
14. (8 – 4i)(–3 + 9i) 247212362 iii=−++− = –24 + 36 + 84i = 12 + 84i
15. (3 + 5i)(3 – 5i)
16. ()() 2 272744944953 iii +−=−=+=
17. (5)(5)25552 iiiii −+−−=+−− 251 26 =+ =
18. (7)(7)49772 iiiii −+−−=+−− 491 50 =+ =
19. () 22234129 4129 512 iii i i +=++ =+− =−+
20. () 22 5225204 25204 2120 iii i i −=−+ =−− =−
21. () 223 333 2(3) 91 23 10 3 5 31 55 i iii i i i i + =⋅ −−+ + = + + = + = =+
22. () () 2 334 444 34 16 34 17 123 1717 i iii i i i i =⋅ + +− = = =−
23. 2 2212222 1 111112 iiiiii i iii −−+ = ⋅===+ ++−+
24. 552 222 iii iii + =⋅ −+ 1052 41 510 5 12 ii i i + = + −+ = =−+
25. 2 8843 434343 3224 169 2432 25 2432 2525 iii iii ii i i + =⋅ −+ + = + −+ = =−+
26. 663218122 32323294 12181218 131313 iiiii iii i i −−−−+ =⋅= ++−+ ==−−
27. 2 23232 222 443 41 74 5 74 55 iii iii ii i i + +− =⋅ + +− +− = + + = =+
28. 2 343443 434343 122512 169 25 25 iii iii ii i i =⋅ ++− −+ = + = =−
29. 64256425 853 ii iii −−−=− =−=
30. 8114481144 ii=− = 9i – 12i = –3i
31. 516381 −+− = 5(4i) + 3(9i) = 20i + 27i = 47i
32. 58318 −+− 58318542392 10292 192 iiii ii i =+=⋅+⋅ =+ =
33. () 2 24−+− ()222 i =−+ 4842 ii=−+ = 4 – 8i – 4 = –8i
34. () 2 592 (59) i =−− ()532 i =−− 253092 ii=++ = 25 + 30i – 9 = 16 + 30i
35. () () () 22 2 3737 9677 9767 267 i ii i i −−−=−− =++ =−+ =+
36. () () 22 211211 i −+−=−+ () 2 441111 411411 7411 ii i i =−+ =−− =−−
37. 832832 2424 8162 24 842 24 12 36 i i i i −+−−+ = +⋅ = −+ = =−+
38. 122812281247 323232 ii +−−+−+⋅ == 122737 32816 i i −+ ==−+
39. 612612 4848 643 48 623 48 13 824 i i i i = = = =−−
40. 151815181592 333333 ii −−−−−−⋅ == 153252 331111 i i ==−−
41. ( ) () 8358(35) 2235 26210 ii ii i −−−=− =− =−−
42. ( ) 124212(42) ii =− () 2 2322 4326 4326 ii ii i =− =− =−−
43. ( ) ( ) ( ) () 2 354123583 2415 2415 ii i −−−=− =− =
44. () () () () 2 3728 (37)(28)(37)(242) 374212141214 iiii iii ==⋅ ===−
45. ()()()() ()() () 222 22 2 23133 22333 2539 754 7541 115 iiii iiii iii ii i i −−−−+ =−−+−− =−+−+ =−−+ =−−+− =−−
46. ()()()() ()() () 222 22 2 89211 1681891 161091 15108 151081 2310 iiii iiii iii ii i i +−−−+ =−+−−− =+−−+ =+− =+−− =+
47. ()() ()() 22 22 22 23 4496 4496 510 ii iiii iiii i +−− =++−−+ =++−+− =−+
48. ()() ()() () 22 22 22 2 412 168144 168144 15123 151231 1812 ii iiii iiii ii i i −−+ =−+−++ =−+−−− =−− =−−− =−
49. 516381 51613811 5439 2027 47or047 ii ii ii −+− =−+− =⋅+⋅ =+ =+
50. () 58318 54213921 522332 10292 1092 192or0192 ii ii i ii −+− = −+− =⋅+⋅ =+ =+ =+
51. ( ) 222fxxx = −+ ()()() 2 2 2 11212 12222 1 11 0 fiii iii i + =+−++ = ++−−+ =+ =− =
52. ( ) 225fxxx = −+ ()()() 2 2 2 12122125 144245 44 44 0 fiii iii i =−−−+ = −+−++ =+ =− =
53. () () () 2 2 2 2 19 2 319 3 23 919 23 919 23 10 23 1023 2323 2030 49 2030 49 2030 13 2030 1313 fxx x i fii i i i i i ii i i i i i + = + = + = −+ = = + =⋅ + + = + = + + = =+
x ii
i i i ii i i i i i i + = + + == −+ = = −+ =⋅ −+ = = + = =− =−−
54. () () () 2 22 2 11 3 4111611 4 3434 1611 34 5 34 534 3434 1520 916 1520 916 1520 25 1520 2525 34 55
55. ()() 4537 EIRii ==−+ 122815352 Eiii =+−− ()1213351Ei=+−− 1235134713 Eii =++=+
Thevoltageofthecircuitis (47+13i)volts.
56. ()() () 2 2335 6109156151 61521 EIRii iiii ii ==−+ =+−−=+−− =++=+
Thevoltageofthecircuitis ( ) 21 i + volts.
57. Sum: () ()515515 515515 55 10 ii ii ++− =++− =+ = Product: () () 2 515515 2551551515 2515 40 ii iii +− =−+−
67. a. False; all irrational numbers are complex numbers.
b. False; (3 + 7i)(3 – 7i) = 9 + 49 = 58 is a real number.
c. False; 737353 535353 iii iii + +− =⋅ + +− 446223 341717 i i ==−
d. True; ()()() 2 222 xyixyixyixy + −=−=+ (d) is true.
68. ()() 2 2 44 23623 4 61 4 7 47 77 284 49 284 491 284 50 284 5050 142 2525 iiiii i i i ii i i i i i i = +− +− = ++ = + =⋅ +− = = + = =− =−
69. ()() ()() ()() ()() ()()()() ()() 22 2 11 1212 112112 12121212 112112 1212 122122 14 122122 14 6 5 60 5 ii ii iiii iiii iiii ii iiiiii i iiii i +− + +− +−−+ =+ +−+− +−+−+ = +− −+−++−− = −++++−+ = + = =+
70. 2 2 88 22 1 8 2 8 2 82 22 168 4 168 41 816 5 816 55 i iii i i i i ii ii ii i i i i = ++ = + = + =⋅ +− = + = + + = =+
Section 1.5
Check Point Exercises
1. a. 2 390 3(3)0 xx xx −= −= 30or30 03 xx xx =−= ==
The solution set is {0, 3}.
b. 2 2 21 210 (21)(1)0 xx xx xx += +−= −+= 210or10 211 1 2 xx xx x −=+= ==− =
The solution set is 1 1, 2 ⎧⎫ ⎨⎬ ⎩⎭
2. a. 2 2 2 321 321 33 7 7 x x x x = = = =± The solution set is {} 7,7
b. 2 5450 x + = 2 2 545 9 9 3 x x x xi = =− = ±− =±
c. 2 (5)11 511 511 x x x += +=± =−± The solution set is {} 511,511. −+−−
3. a. The coefficient of the x-term is 6. Half of 6 is 3, and 23 is 9. 9 should be added to the binomial. 22 69(3)xxx++=+
b. The coefficient of the x-term is –5. Half of –5 is 5 2 , and 52 2 ⎛⎞ ⎜⎟ ⎝⎠ is 25 4 25 4 should be added to the binomial. 2 2255 5 42 xxx ⎛⎞ −+=−⎜⎟ ⎝⎠
c. The coefficient of the x-term is 2 3 . Half of 2 3 is 1 3 , and 12 3 ⎛⎞ ⎜⎟ ⎝⎠ is 1 9 . 1 9 should be added to the binomial. 2 2211 393 xxx ⎛⎞ ++=+⎜⎟ ⎝⎠ 4. 2410xx + −= () 2 2 2 41 4414 25 25 25 xx xx x x x += ++=+ += +=± =−±
8. a. 1,6,9abc = == 224(6)4(1)(9) 3636 0 bac−=− =− = Since 240bac = , the equation has one real solution.
b. 2,7,4abc = =−=− 224(7)4(2)(4) 4932 81 bac =−−− =+ = Since 240bac > , the equation has two real solutions. Since 81 is a perfect square, the two solutions are rational.
c. 3,2,4abc = =−= 224(2)4(3)(4) 448 44 bac−=−− =− = Since 240bac < , the equation has two imaginary solutions that are complex conjugates.
9. 2 0.010.05107PAA =++
2 2 1150.010.05107
00.010.058 AA AA =++ =+−
0.01,0.05,8abc ===−
2 2 4 2
(0.05)(0.05)4(0.01)(8) 2(0.01)
0.050.3225 0.02
0.050.32250.050.3225 0.020.02
Age cannot be negative, reject the negative answer.
Thus, a woman whose normal systolic blood pressure is 115 mm Hg is 26 years old.
10. 222 2 2 915 81225 144 144 12 w w w w w += += = =± =±
The width of the television is 12 inches.
Exercise Set 1.5
1. 23100xx−−= (x + 2)(x – 5) = 0 20or50 2or5 xx xx +=−= =−=
The solution set is {–2, 5}.
2. 213360xx−+= (x – 4)(x – 9) = 0 40or90 4or9 xx xx −=−= ==
The solution set is {4, 9}.
3. 2815xx=− 28150xx−+= (x – 3)(x – 5) = 0 30or50 3or5 xx xx −=−= ==
The solution set is {3, 5}.
4. 21110xx = 211100xx + += (x + 10)(x + 1) = 0 100or10 10or1 xx xx + =+= = −=−
The solution set is {–10, –1}.
5. 2 611100 xx + −= (2x + 5)(3x – 2) = 0 250or320 2532 52 or 23 xx xx xx + =−= =−= =−=
The solution set is 52 , 23 ⎧⎫ ⎨⎬ ⎩⎭ .
6. 2 9920 xx + += (3x + 2)(3x + 1) = 0 3x + 2 = 0 or 3x + 1 = 0 2 3 x = or 1 3 x =
The solution set is 21 , 33 ⎧⎫ ⎨⎬ ⎩⎭
7. 2 328 xx = 2 3280 xx −= (3x + 4)(x – 2) = 0 340or20 34 4 or2 3 xx x xx + =−= =− =−=
The solution set is 4 ,2 3 ⎧⎫ ⎨⎬ ⎩⎭ .
8. 2 4133 xx =− 2 41330 xx += (4x – 1)(x – 3) = 0 410or30 41 1 or3 4 xx x xx =−= = ==
The solution set is 1 ,3 4 ⎧⎫ ⎨⎬ ⎩⎭ .
9. 2 3120 3(4)0 xx xx += += 30or40
0or4 xx xx =+= ==−
The solution set is {–4, 0}.
10. 2 5200 5(4)0 xx xx −= −= 50or40
0or4 xx xx =−= ==
The solution set is {0, 4}.
11. () 2 2357 xxxx −=−
22 26570 xxxx−−+=
2 30 xx −+=
x(–3x + 1) = 0 0or310 31 1 3 xx x x =−+= −=−
The solution set is 1 0, 3 ⎧⎫ ⎨⎬
12. 16x(x – 2) = 8x – 25
2 16328250 xxx−−+= 2 1640250 xx−+=
(4x – 5)(4x – 5) = 0 4x – 5 = 0 4x = 5 5 4 x =
The solution set is 5 4 ⎧⎫ ⎨⎬ ⎩⎭ .
13. 7 – 7x = (3x + 2)(x – 1) 2 7–732 xxx=−−
2 77320 xxx −−++=
2 3690 xx −−+= –3(x + 3)(x – 1) = 0 30or10 3or1 xx xx +=−= =−=
The solution set is {–3, 1}.
14. () 101212 xx=+
2 101441 xxx=++
2 1014410 xxx−−−= 2 4620 xx+−=
–2(2x – 1)(x – 1) = 0 210or10 21 1 or1 2 xx x xx =−= = ==
The solution set is 1 ,1 2 ⎧ ⎫ ⎨ ⎬ ⎩⎭
15. 2 327 x = 29 x = 93 x = ±=±
The solution set is {–3, 3}.
16. 2 545 x = 29 x = 93 x = ±=±
The solution set is {–3, 3}.
17. 2 5151 x + = 2 550 x = 210 x = 10 x =±
The solution set is { } 10,10 .
18. 2 2 2 3147 348 16 164 x x x x −= = = = ±=±
The solution set is {–4, 4}.
19. 2 2555 x =− 2 2 250 25 255 x x xi =− =− = ±−=±
The solution set is { } 5,5. ii 20. 2 2715 x =− 2 2 28 4 42 x x xi =− =− = ±−=±
The solution set is { } 2,2. ii
21. () 2 225 x +=
225 25 25 x x x +=± +=± =−± 25or25 37 xx xx =−+=−− ==−
The solution set is {}7,3.
22. () 2 336 x −=
336 36 36 x x x −=± −=± =± 36or36 93 xx xx =+=− ==−
The solution set is {}3,9.
23. () 2 3415 x −= () 2 45 45 45 x x x −= −=± =±
The solution set is {} 45,45. +−
24. () 2 3421 x += () 2 47 47 47 x x x += +=± =−±
The solution set is {} 47,47.−+−−
25. () 2 316 x +=−
316 34 34 x xi xi +=±− +=± =−±
The solution set is {} 34,34. ii−+−−
26. () 2 19 x −=− 19 13 13 x xi xi −=±− −=± =±
The solution set is { } 13,13. ii+−
27. () 2 35 x =− 35 35 35 x xi xi =±− −=± =±
The solution set is {} 35,35. ii+−
28. () 2 27 x + =− 27 27 27 x xi xi +=±− +=± =−±
The solution set is {} 27,27. ii−+−−
29. () 2 329 x + = 3293 x + =±=± 323or323 3531 51 or 33 xx xx xx + =−+= =−= =−=
The solution set is 51 , 33 ⎧⎫ ⎨⎬ ⎩⎭
30. 2 (41)16 x = 41164 x =±=± 414or414 4345 35 or 44 xx xx xx =−−= =−= ==
The solution set is 35 , 44 ⎧⎫ ⎨⎬ ⎩⎭ .
31. () 2 517 x = 517 x −=± 517 x =± 17 5 x ± =
The solution set is 1717 , 55 ⎧⎫ −+ ⎪⎪ ⎨⎬ ⎪⎪ ⎩⎭
32. () 2 835 835 835 35 8 x x x x −= −=± =± ± =
The solution set is 3535 ,.88 ⎧⎫ −+ ⎪⎪ ⎨⎬ ⎪⎪ ⎩⎭
33. () 2 348 x −=
34822 x −=±=±
3422 x =± 422 3 x ± =
The solution set is 422422 , 33 ⎧⎫ −+ ⎪⎪ ⎨⎬
34. 2 (28)27 282733 2833 833 2 x x x x += +=±=± =−+ −± =
The solution set is 833833 ,.22 ⎧⎫ −−−+ ⎪⎪ ⎨⎬ ⎪⎪ ⎩⎭
35. 2 2 2 22 12 12 636 2 1236(6) xx xxx + ⎛⎞ == ⎜⎟ ⎝⎠ ++=+
36. 2 2 2 22 16 16 864; 2 1664(8) xx xxx + ⎛⎞ == ⎜⎟ ⎝⎠ ++=+
37. 2 2 2 22 10 10 525 2 1025(5) xx xxx ⎛⎞ == ⎜⎟ ⎝⎠ −+=−
38. 2 2 2 22 14 14 2(7)49; 1449(7) xx xxx ⎛⎞ =−= ⎜⎟ ⎝⎠ −+=−
39. 2 2 2 2 3 39 24 93 3 42 xx xxx + ⎛⎞ = ⎜⎟ ⎝⎠ ⎛⎞ ++=+⎜⎟ ⎝⎠
40. 2 2 2 2 5 525 ; 24 255 5 42 xx xxx + ⎛⎞ = ⎜⎟ ⎝⎠ ⎛⎞ ++=+⎜⎟ ⎝⎠
41. 2 2 2 2 7 749 24 497 7 42 xx xxx ⎛⎞ = ⎜⎟ ⎝⎠
⎛⎞ −+=−⎜⎟ ⎝⎠ 42. 2 2 2 2 9 981 ; 24 819 9 42 xx xxx ⎛⎞ = ⎜⎟ ⎝⎠
⎛⎞ −+=−⎜⎟ ⎝⎠
43. 2 222 3 2 2 2 3 11 239 211 393 xx xxx ⎛⎞ ⎛⎞ == ⎜⎟ ⎜⎟ ⎝⎠ ⎝⎠
⎛⎞ −+=−⎜⎟ ⎝⎠ 44. 2 422 5 2 2 4 5 24 ; 2525 442 5255 xx xxx + ⎛⎞ ⎛⎞ == ⎜⎟ ⎜⎟ ⎝⎠ ⎝⎠
⎛⎞ ++=+⎜⎟ ⎝⎠
45. 2 122 3 2 2 1 3 11 2636 111 3366 xx xxx ⎛⎞ ⎛⎞ == ⎜⎟ ⎜⎟ ⎝⎠ ⎝⎠ ⎛⎞ −+=−⎜⎟ ⎝⎠
46. 2 122 4 2 2 1 4 11 ; 2864 111 4648 xx xxx
47. 2 2 2 67 6979 (3)16 34 34 xx xx x x x += ++=+ += +=± =−± The solution set is {–7, 1}.
48. 2 2 2 68 6989 (3)1 31 31 xx xx x x x +=− ++=−+ += +=± =−±
The solution set is {–4, –2}.
49. 2 2 2 22 2121 (1)3 13 13 xx xx x x x −= −+=+ −= −=± =±
The solution set is {} 13,13. +−
50. 2 2 2 412 44124 (2)16 24 24 xx xx x x x += ++=+ += +=± =−±
The solution set is {–6, 2}.
51. 2 2 2 2 6110 611 69119 (3)20 320 325 xx xx xx x x x −−= −= −+=+ −= −=± =±
The solution set is {} 325,325. +−
52. 2 2 2 2 250 25 2151 (1)6 16 16 xx xx xx x x x −−= −= −+=+ −= −=± =±
The solution set is {} 16,16. +−
53. 2 2 2 2 410 41 4414 (2)3 23 23 xx xx xx x x x ++= +=− ++=−+ += +=± =−±
The solution set is {} 23,23. −+−−
54. 2 2 2 2 650 65 6959 (3)14 314 314 xx xx xx x x x +−= += ++=+ += +=± =−±
The solution set is {} 314,314. −+−−
55. 2560xx−+= 2 2 2 56 2525 56 44 51 24 51 24 51 22 51 22 xx xx x x x x −=− −+=−+ ⎛⎞ ⎜⎟−= ⎝⎠ −=± −=± =± 5151 or 2222 32 xx xx =+=− ==
The solution set is {2, 3}.
56. 2780xx+−= 2 2 2 78 4949 78 44 781 24 781 24 79 22 79 22 xx xx x x x x += ++=+ ⎛⎞ ⎜⎟+= ⎝⎠ +=± +=± =−± 7979 or 2222 18 xx xx =−+=−− ==−
The solution set is {–8, 1}.
57. 2 2 2 2 310 31 99 31 44 313 24 313 22 313 2 xx xx xx x x x +−= += ++=+ ⎛⎞ ⎜⎟+= ⎝⎠ +=± −± = The solution set is 313313 ,.22 ⎧ ⎫ −+−− ⎪ ⎪ ⎨ ⎬ ⎪ ⎪ ⎩⎭ 58. 2 2 2 2 350 35 99 35 44 329 24 329 22 329 2 xx xx xx x x x −−= −= −+=+ ⎛⎞ ⎜⎟−= ⎝⎠ ± −= ± = The solution set is 329329 ,.22 ⎧ ⎫ +− ⎪ ⎪ ⎨ ⎬ ⎪ ⎪ ⎩⎭
59. 2 2 2 2 2 2730 73 0 22 73 22 749349 216216 725 416 75 44 75 44 xx xx xx xx x x x −+= −+= −= −+=−+ ⎛⎞ ⎜⎟−= ⎝⎠ −=± =±
The solution set is 1 ,3. 2 ⎧ ⎫ ⎨ ⎬ ⎩⎭
64. 2 35100 xx−−=
2 2 2 2 510 0 33 510 33 5251025 336336 5145 636 5145 66 5145 6 xx xx xx x x x −−= −= −+=+ ⎛⎞ ⎜⎟−= ⎝⎠ ± −= ± = The solution set is 51455145 ,.66 ⎧⎫ ±− ⎪⎪
65. 28150xx++= 2 884(1)(15) 2(1) 86460 2 84 2 82 2 x x x x −±− = −±− = −± = −± =
The solution set is {5,3}.
66. 2 2 8120 884(1)(12) 2(1) 86448 2 816 2 84 2 xx x x x x ++= −±− = −±− = −± = −± = The solution set is {–6, –2}.
67. 2 2 530 554(1)(3) 2(1) 52512 2 513 2 xx x x x ++= −±− = −±− = −± = The solution set is 513513 ,.22 ⎧ ⎫ −+−− ⎪ ⎪ ⎨ ⎬ ⎪ ⎪ ⎩⎭
68. 2520xx + += 2 554(1)(2) 2(1) 5258 2 517 2 x x x −±− = −±− = −± = The solution set is 517517 ,.22 ⎧ ⎫ −+−− ⎪ ⎪ ⎨ ⎬ ⎪ ⎪ ⎩⎭
69. 2 2 3340 3(3)4(3)(4) 2(3) 3948 6 357 6 xx x x x −−= ± = ±+ = ± = The solution set is 357357 , 66 ⎧ ⎫ +− ⎪ ⎪ ⎨ ⎬ ⎪ ⎪ ⎩⎭
70. 2 520 xx + −= 2 114(5)(2) 2(5) 1140 10 141 10 x x x ±−− = −±+ = −± = The solution set is 141141 ,.1010 ⎧ ⎫ −+−− ⎪ ⎪ ⎨ ⎬ ⎪ ⎪ ⎩⎭
71. 2 2 427 4270 xx xx =+ −−= 2 2(2)4(4)(7) 2(4) 24112 8 2116 8 2229 8 129 4 x x x x x ±−−− = ±+ = ± = ± = ± = The solution set is 129129 ,.44 ⎧⎫ +−
72. 2 2 361 3610 xx xx =− −+= 2 6(6)4(3)(1) 2(3) 63612 6 624 6 626 6 36 3 x x x x x ±−− = ±− = ± = ± = ± =
The solution set is 3636 ,.33 ⎧⎫ +− ⎪⎪
i
xi −+= ±−−
73. 2 2 6100 6(6)4(1)(10) 2(1) 63640 2 64 2 62 2 3 xx x
The solution set is {} 3,3. ii+− 74. 22170xx += 2 2(2)4(1)(17) 2(1) 2468 2 264 2 28 2 14 x x x i x xi ±−− = ±− = ±− = ±
The solution set is {14,14}. ii+−
75. 2 2 450 (4)4(1)(5) xx−−= = 16 + 20 = 36; 2 unequal real solutions
76. 2 2 4230 (2)4(4)(3) xx += = 4 – 48 = –44; 2 complex imaginary solutions
77. 2 2 21130 (11)4(2)(3) xx += = 121 – 24 = 97; 2 unequal real solutions
78. 2 2 21160 114(2)(6) xx + −= = 121 + 48 = 169; 2 unequal real solutions
79. 2 2 210 (2)4(1)(1) xx += = 4 – 4 = 0; 1 real solution
80. 2 2 2 321 3210 (2)4(3)(1) xx xx =− += = 4 – 12 = –8; 2 complex imaginary solutions
81. 2 2 370 (3)4(1)(7) xx−−= = 9 + 28 = 37; 2 unequal real solutions
82. 2 2 3420 44(3)(2) xx+−= = 16 + 24 = 40; 2 unequal real solutions
83. 2 2 21 210 (21)(1)0 210or10 21 1 or1 2 xx xx xx xx x xx −= −−= +−= +=−= =− =−=
The solution set is 1 ,1. 2 ⎧⎫ ⎨⎬ ⎩⎭
84. 2 2 344 3440 (32)(2)0 xx xx xx −= −−= +−= 32or20 32 2 or3 3 xx x xx +−= =− =−=−
The solution set is 2 ,2. 3 ⎧⎫ ⎨⎬ ⎩⎭
85. 2 2 5211 51120 (51)(2)0 510or20 51 1 or2 5 xx xx xx xx x xx += −+= −−= −=−= = ==
The solution set is 1 ,2. 5 ⎧⎫ ⎨⎬ ⎩⎭
86. 2 2 5613 51360 (52)(3)0 xx xx xx =− +−= −+= 520or3 52 2 or3 5 xx x xx =+ = = =−
The solution set is 2 3,. 5 ⎧ ⎫ ⎨ ⎬ ⎩⎭
87. 2 2 360 20 20 25 x x x x = = =± =±
The solution set is { } 25,25. 88. 2 2 2250 125 125 55 x x x x = = =± =±
The solution set is { } 55,55.
89. 2 2 2 21 2111 (1)2 12 12 xx xx x x x −= −+=+ −= −=± =±
The solution set is { } 12,12. +−
90. 2 2 231 2310 xx xx + = + −= 2 334(2)(1) 2(2) 398 4 317 4 x x x ±−− = −±+ = −± =
The solution set is 317317 ,.44 ⎧ ⎫ −+−− ⎪ ⎪ ⎨ ⎬ ⎪ ⎪ ⎩⎭
91. 2 2 (23)(4)1 283121 211110 xx xxx xx ++= +++= ++= 2 11114(2)(11) 2(2) 1112188 4 1133 4 x x x −±− = −±− = −± =
The solution set is 11331133 ,.44 ⎧⎫ −+−−
92. 2 2 (25)(1)2 22552 2370 xx xxx xx −+= +−−= −−= 2 3(3)4(2)(7) 2(2) 3956 4 365 4 x x x ±−−− = ±+ = ± =
The solution set is 365365 ,.44 ⎧⎫ +−
93. 2 (34)16 3416 344 344 38or30 8 or0 3 x x x x xx xx −= −=± −=± =± == ==
The solution set is 8 0,. 3 ⎧⎫ ⎨⎬ ⎩⎭
94. 2 (27)25 275 275 x x x += +=± =−± 212or22 6or1 xx xx =−=− ==−
The solution set is {–6, –1}.
95. 2 2 312120 440 (2)(2)0 20 2 xx xx xx x x += += −= = =
The solution set is {2}.
96. 2 2 960 690 (3)(3)0 30 3 xx xx xx x x += += −= = =
The solution set is {3}.
97. 2 2 2 4160 416 4 2 x x x x −= = = = ±
The solution set is {}2,2.
98. 2 2 2 3270 327 9 3 x x x x −= = = = ±
The solution set is {–3, 3}.
99. 2 2 2 2 6130 613 69139 (3)4 32 32 xx xx xx x xi xi −+= −=− +=−+ −=− −=± =±
The solution set is {} 32,32. ii+−
100. 2 2 2 2 4290 429 44294 (2)25 25 25 xx xx xx x xi xi −+= −=− +=−+ −=− −=± =±
The solution set is {25,25}. ii+−
101. 2 2 2 2 47 47 4474 (2)3 23 23 xx xx xx x xi xi =− −=− −+=−+ −=− −=± =±
The solution set is {} 23,23. ii+−
102. 2 2 523 5230 xx xx =− −+= 2 2(2)4(5)(3) 2(5) 2460 10 256 10 2214 10 114 5 x x x i x i
The solution set is 114114 ,.55 ii ⎧⎫ +− ⎪⎪ ⎨⎬ ⎪⎪ ⎩⎭
103. 2 270 (27)0 0or270 xx xx xx −= −= =−= 27 7 0or 2 x xx = ==
The solution set is 7 0,. 2 ⎧⎫ ⎨⎬ ⎩⎭
104. 2 2 253 2530 xx xx + = + −= 2 554(2)(3) 2(2) 52524 4 549 4 57 4 1 3, 2 x x x x x ±−− = −±+ = −± = −± = =−
The solution set is 1 3,. 2 ⎧ ⎫ ⎨ ⎬ ⎩⎭
105. ()()()() () 111 ;0,2 23 2 3632 2 046 2 44416 21 41624 2 440 2 4210 2 210 x xx xxxx xx x x x x x +=≠− + ++=+ =−− −±−−− = ±+ = ± = ± = =±
The solution set is {210,210}. +−
106. 111 ;0,3 34 x xx + =≠− + ()()()() () 2 41243 2 0512 2 554112 21 52548 2 573 2 xxxx xx x x x ++=+ =−− −−±−−− = ±+ = ± =
The solution set is 573573 ,.22 ⎧ ⎫ +− ⎪ ⎪ ⎨ ⎬ ⎪ ⎪ ⎩⎭
107. ()() 2 2 2 2 2628 ;3,3 339 236328 2661828 212100 650 (1)(5)0 x x xxx xxx xxx xx xx xx +=≠− −+ ++−=− ++−=− ++= ++= ++=
The solution set is {5,1}.
108. 2 2 3520 ;3,4 34712 x x xxxx +=≠ −+ ()() 2 2 31251520 087 071 71 xxx xx xx xx −+−=− =−+ =−− ==
The solution set is {1, 7}.
109. 2450xx−−= (1)(5)0 xx+−= 1050 or 15 xx xx +=−= =−=
This equation matches graph (d).
110. 2670xx−+= 1,6,7abc==−= 2 2 4 2 (6)(6)4(1)(7) 2(1) 68 2 32 1.6,4.4 bbac x a x x x xx −±− = −−±−− = ± = =± ≈≈
This equation matches graph (a).
111. 2 0(1)4 x =−++ 2 (1)4 12 12 3,1 x x x xx += +=± =−± =−=
This equation matches graph (f).
112. 2 0(3)1 x = −++ 2 (3)1 31 31 4,2 x x x xx += +=± =−± = −=−
This equation matches graph (e).
113. 2220xx += 1,2,2abc = =−= 2 2 4 2 (2)(2)4(1)(2) 2(1) 24 2 22 2 1 bbac x a x x i x xi −±− = −−±−− = ±− = ± = =±
This equation has no real roots. Thus, its equation has no x-intercepts. This equation matches graph (b).
114. 2690xx + += (3)(3)0 xx + += 30 3 x x + = =
This equation matches graph (c).
115. 2 23 yxx = 2 2 223 0232 0(21)(2) xx xx xx =− =−− = +− 1 ,2 2 xx = −=
116. 2 53 yxx = + 2 2 253 0532 0(1)(52) xx xx xx =+ =+− = +− 2 1, 5 xx = −=
Chapter 1 ISM: College Algebra
117. 1214yy = 2 2 (1)(4)14 3414 3180 (6)(3)0 xx xx xx xx −+= +−= +−= +−= 6,3xx=−=
119. 121yy+= ()() ()() ()() ()() ()() ()()()() 22 2 23 1 24 23 24124 24 224324 24 24 243224 283668 520 x xx x xxxx xx xxxxx xx xx xxxxx xxxxx xx += ++ ⎛⎞ +++=++ ⎜⎟ ++ ⎝⎠ ++++ +=++ ++ +++=++ +++=++ +−= 2 2 4 2 (5)(5)4(1)(2) 2(1) 533 2 bbac x a x x −±− = −±−− = −± = The solution set is 533533 ,.22 ⎧⎫ −+−− ⎪⎪ ⎨⎬
120. 123yy+= () ()() () () () () 2 2 2 2 38 3 1 38 131 1 3181 31 1 38133 38833 11833 03148 0(32)(4) xx xxxx xx xxxx xx xx xxxx xxxx xxx xx xx += ⎛⎞ −+=− ⎜⎟ ⎝⎠ +=− +−=− +−=− −=− =−+ =−− 2 ,4 3 xx==
The solution set is 2 ,4. 3 ⎧⎫ ⎨⎬ ⎩⎭
118. 1230yy = 2 2 (3)(8)30 52430 560 (3)(2)0 xx xx xx xx +=− + −=− ++= ++= 3,2xx = −=−
121. 120yy−= ()() 22 22 2 25415100 25415100 31060 xxxx xxxx xx +−−−+−= +−+−+= −+= 2 2 4 2 (10)(10)4(3)(6) 2(3) 1028 6 1027 6 57 3
bbac x a x x x x −±− = −−±−− = ± = ± = ± =
The solution set is 5757 ,.33
122. 120yy−= ()() 22 22 2 42310 42310 2310 xxxx xxxx xx −+−−−+−= −+−+−+= +−= 2 2 4 2 (3)(3)4(2)(1) 2(2) 317 4 bbac x a x x −±− = −±−− = −± =
The solution set is 317317 ,.44 ⎧⎫ −+−−
123. Values that make the denominator zero must be excluded. 2 2490 xx + −= 2 2 4 2 (4)(4)4(2)(9) 2(2) 488 4 4222 4 222 2 bbac x a x x x x −±− = ±−− = −± = −± = −± =
124. Values that make the denominator zero must be excluded. 2 2850 xx += 2 2 4 2 (8)(8)4(2)(5) 2(2) 824 4 826 4 46 2 bbac x a x x x x −±− = −−±−− = ± = ± = ± =
125. ( ) 2620xx+= 2260xx −= Applythequadraticformula. 126abc = =−=− ()()()() () () 2 22416 21 2424 2 228 2 247227 17 22 x −−±−−− = ±−− = ± = ±⋅± ===± Wedisregard17becauseitisnegative,and wearelookingforapositivenumber. Thus,thenumberis17 +
1
126. Let x =thenumber. () 2 2 2120 2210 xx xx −+= −−=
Applythequadraticformula. 221abc==−=− ()()()() () () 2 2 4 2 22421 22 248 4 21224322313 4442 bbac x a x −±− = −−±−−− = ±−− = ±±⋅±± ====
Wedisregard13 2 + becauseitispositive, andwearelookingforanegativenumber.The numberis13 2
127. 22 115 2 324 115
(1)(2)2(2)(2) x xxx xxxxx =+ + −+− =+ −−++−
Multiplybothsidesoftheequationbytheleast commondenominator,(1)(2)(2) xxx−−+ .This resultsinthefollowing: 2 2 2 2(1)(2)5(1) 22255 223 05 xxxx xxxxx xxx xx +=−−+− +=−−++− +=+− =+−
Applythequadraticformula: 115abc===− ()() () () 2 114151120 212 121 2 x −±−−−±−− == −± = Thesolutionsare 121 2 −± ,andthesolutionsetis 121 2 ⎧⎫ −± ⎪⎪ ⎨⎬ ⎪⎪ ⎩⎭
128. 2 11 2356 11 23(2)(3) xx xxxx xx xxxx += −−−+ +=
Multiplybothsidesoftheequationbytheleast commondenominator,(2)(3) xx .Thisresults inthefollowing: 22 2 2 (3)(1)(2)1 3321 2631 2620 xxxx xxxxx xx xx −+−= −++−= += +=
Applythequadraticformula: 262abc = =−= . ()() () 2 (6)(6)422 22 63616620 44 645625 44 35 2 x −−±−− = ±−± == ±⋅± == ± = Thesolutionsare35 2 ± ,andthesolutionsetis 35 2 ⎧ ⎫ ± ⎪ ⎪ ⎨ ⎬ ⎪ ⎪ ⎩⎭
129. 2 23220 xx + −= Applythequadraticformula: 2322abc ===− ( ) ( ) () 2 334222 22 39(16) 22 32535 2222 x −±−− = −±−− = −±−± == Evaluatetheexpressiontoobtaintwosolutions.
3535 or 2222 8222 222222 8222 44 222 2 xx −+ = = =⋅ =⋅ = = =− =
Thesolutionsare22and 2 2 ,andthesolution
setis 2 22, 2 ⎧⎫ ⎪⎪ ⎨⎬ ⎪⎪ ⎩⎭ .
130. 2 36730 xx++=
Applythequadraticformula: 3673abc === () () () 2 664373 23 63684 23 648 23 x −±− = −±− = −±− = 6163(1) 23 643 23 643 32 2323 i i i −±⋅⋅− = −± = =±=−±
Thesolutionsare32i −± ,andthesolution setis {} 32i −± .
131. () 2 2 2 0.0131.1928.24 30.0131.1928.24 00.0131.1925.24 fxxx xx xx =−+ =−+ =−+
Applythequadraticformula: 0.0131.1925.24abc ==−= x = ()()()() () 2 1.191.1940.01325.24 20.013 −−±−−
58.15or33.39 ±− = ± = ± ≈ ≈
1.191.41611.31248 0.026
1.190.10362 0.026 1.190.32190 0.026
Thesolutionsareapproximately33.39and58.15. Thus,33yearoldsand58yearoldsareexpectedto bein3fatalcrashesper100millionmilesdriven. Thefunctionmodelstheactualdatawell.
132. ( ) 2 2 2 0.0131.1928.24 100.0131.1928.24 00.0131.1918.24
0.0131.1918.24
1.191.41610.94848 0.026 1.190.467621.190.68383 0.0260.026 x −−±−− = ±− = ±± =≈
()()()() () 2 1.191.1940.01318.24 20.013
Evaluatetheexpressiontoobtaintwosolutions. 1.190.683831.190.68383 or 0.0260.026 1.873830.50617 0.0260.026 72.119 xx xx xx +− == == ≈≈
Driversofapproximatelyage19andage72are expectedtobeinvolvedin10fatalcrashesper100 millionmilesdriven.Theformuladoesnotmodel thedataverywell.Theformulaoverestimatesthe numberoffatalaccidents.
133. Let210.010.76.1 yxx =−++
UsingtheTRACEfeature,wefindthattheheight oftheshotputisapproximately0feetwhenthe distanceis77.8feet.Graph(b)showstheshot’ path.
134. Let210.042.16.1 yxx =−++
UsingtheZEROfeature,wefindthattheheightof theshotputisapproximately0feetwhenthedistance is55.3feet.Graph(a)showstheshot’spath.
135. Ignoringthethicknessofthepanel,weessentially needtofindthediameteroftherectangular opening.
139. Let w =thewidth Let w +3=thelength () ()() 2 2 Area 543 543 0354 096 lw ww ww ww ww = =+ =+ =+− = +−
Applythezeroproductprinciple. 9060 96 ww ww + =−= = −=
Thesolutionsetis { } 9,6.Disregard–9 becausewecan’thaveanegativelength measurement.Thewidthis6feetandthelength is639 + = feet.
Sincewearelookingforalength,wediscardthe negativesolution.Thesolutionis458.9 ≈ and weconcludethatapanelthatisabout8.9feet longisthelongestthatcanbetakenthroughthe doordiagonally.
136. 222 2 2 9090 81008100 16200 127.28 x x x x += += = ≈±
The distance is 127.28 feet.
137. 222 2 2 1520 225400 175 13.23 x x x
13.23
The ladder reaches 13.23 feet up.
138. 222 2 2 1030 100900 800 x x x += += = Applythesquarerootproperty. 8004002202 x =±=±⋅=± Wedisregard202becausewecan’thavea negativelengthmeasurement.Thesolutionis 202.Weconcludethattheladderreaches 202feet,orapproximately28.3feet,upthe house.
140. Let w =thewidth Let w +3=thewidth () ()() 2 2 Area 1803 1803 03180 01512 lw ww ww ww ww = =+ =+ =+− =+−
150 15 w w + = =− 120 12 w w = = Thewidthis12yardsandthelengthis12yards+ 3yards=15yards.
141. Let x =thelengthofthesideoftheoriginalsquare Let x +3=thelengthofthesideofthenew,larger square ( ) ()() 2 2 2 364 6964 6550 1150 x xx xx xx + = + += + −= + −= Applythezeroproductprinciple. 11050 115 xx xx + =−= = −=
Thesolutionsetis { } 11,5.Disregard–11 becausewecan’thaveanegativelength measurement.Thismeansthat x,thelengthofthe sideoftheoriginalsquare,is5inches.
142. Let x =thesideoftheoriginalsquare, Let x +2=thesideofthenew,largersquare () ()() 2 2 2 236 4436 4320 840 x xx xx xx += ++= +−= +−= 80 8 x x += =− 40 4 x x −= = Thelengthofthesideoftheoriginalsquare,is4 inches.
143. Let x =thewidthofthepath ()() 2 2 202102600 20040204600 200604600 xx xxx xx ++= +++= ++=
() ()() 2 2 2 460200600 4604000 4151000 42050 xx xx xx xx ++= +−= +−= +−=
Applythezeroproductprinciple. () 420050 2005 20 xx xx x += −= +== =−
Thesolutionsetis {}20,5.Disregard–20 becausewecan’thaveanegativewidth measurement.Thewidthofthepathis5meters.
144. Let x =thewidthofthepath ()() () ()() 2 2 2 2 122152378 18024304378 454180378 4541980 2227990 223330 xx xxx xx xx xx xx ++= +++= ++= +−= +−= +−= () 22330 2330 233 33 2 x x x x += += =− =− 30 3 x x −= =
Thewidthofthepathis3meters.
145. 2 2 ()(2)200 2200 100 10 xx x x x = = = =± The length and width are 10 inches.
146. 2 2 ()(3)75 375 25 5 xx x x x = = = = ± The length and width is 5 inches.
147. ( ) ()()()() () 2 2 2 20213 20213 022013 20204213 22 20296 4 1017.2 4 9.3,0.7 xx xx xx x x x x −= −= =−+ −−±−− = ± = ± = = 9.3 in and 0.7 in
148. 22 8 2 44 xx ⎛⎞⎛⎞ + = ⎜⎟⎜⎟ ⎝⎠⎝⎠ ()() 22 22 2 2 6416 2 1616 641632 216320 8160 440 4in xxx xxx xx xx xx x −+ += +−+= −+= −+= −−= = Both are 4 inches.
160. a. False; () 2 2325 x −= 235 x =±
b. False; Consider x 2 = 0, then x = 0 is the only distinct solution.
c. True
d. False; 20axc + = 0042 22 aciaciac x aaa ±− === (c) is true.
161. 2 2 (3)(5)0 53150 2150 xx xxx xx +−= −+−= −−=
162. 2 0 2 0 0 16 016 16,, stvt tvts abvcs =−+ =−+− =−==− ()()() () () 2 00 2 00 2 00 416 216 64 32 64 32 vvs t vvs t vvs t −±−−− = −±− = ±− =
163. Thedimensionsofthepoolare12metersby8 meters.Withthetile,thedimensionswillbe 12+2x metersby8+2x meters.Ifwetake theareaofthepoolwiththetileandsubtract theareaofthepoolwithoutthetile,weareleft withtheareaofthetileonly.
()()() 12282128120 96 xx ++−= 2 2416496 xxx +++− 2 2 120 4401200 10300 11030 xx xx abc = +−= +−= ===− ()() () 2 10104130 21 10100120 2 102201014.8 22 x −±−− = −±+ = −±−± =≈
Evaluatetheexpressiontoobtaintwo solutions. 1014.81014.8 or 22 4.824.8 22
2.412.4 xx xx xx −+−− == == ==− Wedisregard–12.4becausewecan’thavea negativewidthmeasurement.Thesolutionis 2.4andweconcludethatthewidthofthe uniformtileborderis2.4meters.Thisismore thanthe2-meterrequirement,sothetilemeets thezoninglaws.
Mid-Chapter 1 Check Point
1. 53(5)2(34) xx ++=− 531568 31068 318 318 33 6 xx xx x x x ++=− + =− −=− = =
Thesolutionsetis { } 6.
2. 2 527 xx = 2 5270 (57)(1)0 xx xx −= += 570or10 571 7 5 xx xx x =+= = =− =
Thesolutionsetis71,.
3. 35 1 54 xx −= () () () 35 20120 54 203205 201 54 4(3)205(5) 41220525 432525 7 7 xx xx xx xx xx x x ⎛⎞⎛⎞ −= ⎜⎟⎜⎟ ⎝⎠⎝⎠ −= −−=− −−=− −=− −= =−
Thesolutionsetis { } 7.
4. 2 3620 xx−−= 2 2 4 2 (6)(6)4(3)(2) 2(3) 660 6 6215 6 315 3 bbac x a x x x x −±− = −−±−−− = ± = ± = ± =
Thesolutionsetis 315315 ,.33 ⎧⎫ +− ⎪⎪
5. 42(1)3(21)5 xxx −−=+− 42(1)3(21)5 422635 6262 00 xxx xxx xx −−=+− −+=+− −=− =
Thesolutionsetisallrealnumbers.
6. 2 5137 x += 2 2 2 536 536 55 36 5 36 5 6 5 65 55 65 5 x x x x x x x = = = =± =± =±⋅ =±
Thesolutionsetis 6565 ,.55 ⎧⎫
7. (23)4xx =− 2 2 234 2340 xx xx =− −+= 2 2 4 2 (3)(3)4(2)(4) 2(2) 323 4 323 4 bbac x a x x i x −±− = −−±−− = ±− = ± = Thesolutionsetis 323323 ,.44 ii ⎧⎫ +−
8. 343 1 43520 xxx −+=− 343 1 43520 343 60160 43520 60(3)6060(4)60(3) 60(1) 43520 452060489 2560489 2369 2369 2323 3 xxx xxx xxx xxx xx x x x −+=− ⎛⎞⎛⎞ −+=− ⎜⎟⎜⎟ ⎝⎠⎝⎠ −+=− −+=− +=− −=− = =
Thesolutionsetis {}3. 9. 2 (3)24 x + = 324 326 x x +=± =−±
Thesolutionsetis {} 326,326. −+−−
10. 2 14 10 xx −+= 22 2 22 2 2 2 2 14 1(0) 4 0 140 410 xx xx xx x xx xx xx ⎛⎞ ⎜⎟−+= ⎝⎠ −+= −+= −+= 2 2 4 2 (4)(4)4(1)(1) 2(1) 412 2 423 2 23 bbac x a x x x x −±− = −−±−− = ± = ± = =±
Thesolutionsetis {} 23,23. +−
11. 31(5)24 xxx +−−=− 2624 64 xx+=− =−
Thesolutionsetis ∅
12. 2 22 ,2,4 6842 xx xx xxxx =−≠−≠− ++ ++ 2 2 22 (4)(2)42
2(4)(2)2 (4)(2) (4)(2)42
2(2)2(4) 2228 028 0(2)(4) xx xxxx xxxx xx xxxx xxxxx x xx xxxx xxxx xx xx =− ++++ ++ ⎛⎞ ++++=++−⎜⎟ ⎝⎠ ++++ =− ++ =+−+ =+−− =−− =+−
2(4)(2)2(4)(2) 42
20or40 24 xx xx +=−= =−= –2mustberejected.
Thesolutionsetis { } 4.
13. Let0. y = 2 062 xx =++ 2 2 4 2 (6)(6)4(1)(2) 2(1) 628 2 627 2 37 bbac x a x x x x −±− = −±− = −± = −± = =−± x-intercepts:37and37. −+−−
14. Let0. y = 04(1)3(6) 04436 022 22 1 xxx xxx x x x =+−−− =+−−+ =− −=− = x-intercept:1.
15. Let0. y = 2 2 2 0226 226 13 13 13 x x x x xi =+ −= =− =±− =± Thereareno x-intercepts.
16. Let0. y = 2 2 2 2 2 0 323 2 6(0)6323 6662 0 323 0234 xx xx xx xx =+− ⎛⎞ =+−⎜⎟ ⎝⎠ ⋅ ⋅⋅ =+− =+− 2 2 4 2 (3)(3)4(2)(4) 2(2) 341 4 bbac x a x x −±− = ±−− = −± = x-intercepts:341341 and. 44 −+−−
17. Let0. y = 2 058 xx = −+ 2 2 4 2 (5)(5)4(1)(8) 2(1) 57 2 57 2 bbac x a x x i x −±− = −−±−− = ±− = ± = Thereareno x-intercepts.
18. 12yy = 3(25)2(41)5(3)2 615825152 217517 30 0 xxx xxx xx x x −+=−+− −−=−−− −−=−− = = Thesolutionsetis { } 0.
19. 1210yy = 2 2 (23)(2)10 27610 2740 (21)(4)0 xx xx xx xx ++= ++= +−= −+= 210or40 14 2 xx x x += −= =− =
Thesolutionsetis 1 4,. 2 ⎧⎫ ⎨⎬ ⎩⎭
20. 2 2 1030 103 xx xx +−= +=
Since10 b = ,weadd 2 102525 2 ⎛⎞
. 2 2 1025325 (5)28 xx x ++=+ += Applythesquarerootprinciple: 528 54727 527 x x x +=± +=±⋅=± =−±
Thesolutionsare527 −± ,andthesolutionsetis {} 527−±
21. 2 2540 254 xx abc ++= === ()() 2245424 25327 bac−=− =−=−
Sincethediscriminantisnegative,therearenoreal solutions.Therearetwoimaginarysolutionsthat arecomplexconjugates.
22. () 2 2 1041515 10401515 1025150 xxx xxx xx +=− +=− −+= 102515abc ==−= ()()()2242541015 62560025 bac−=−− =−=
Sincethediscriminantispositiveandaperfect square,therearetworationalsolutions.
25. (,) 26 13 02 13 26 xxy
26. (1) Land =+− 1 1 1 Ladnd dnadL dnadL dddd aL n dd La n dd La n d
27.
Alwlhwh
29. Let x =thedefensebudgetofJapaninbillions Let x+ 4 = thedefensebudgetofRussiain billions
Let x+ 251 = thedefensebudgetofU.S.in billions (4)(251)375
ThedefensebudgetofJapanis$40billion,of Russia$44billion,andoftheU.S.$291billion.
30. Let x =thenumberofmonthsittakesforthe averagefemaleinfanttoweigh16pounds 71.516 1.59 1.59 1.51.5 6 x x x x + = = = = Ittakes6monthsfortheaveragefemaleinfant toweigh16pounds.
31. Let x =theamountinvestedat8%. Let25,000– x =theamountinvestedat9%. 0.080.09(25,000)2135 0.0822500.092135 0.0122502135 0.01115 115 0.01 11,500 25,00013,500 xx xx x x x x x + −= +−= −+= −=− = = −= $11,500wasinvestedat8%and$13,500was investedat9%.
32. Let x =thenumberofprints. PhotoShopA:0.111.60 x + PhotoShopB:0.131.20 x + 0.131.200.111.60 0.021.201.60
0.020.40 20 xx x x x +=+ += = = Thecostwillbethesamefor20prints. Thatcommonpriceis
0.11(20)1.600.13(20)1.20 $3.80 +=+ =
33. Let x =theaverageweightforanAmerican womanaged20through29in1960. 0.22157 1.22157 1.22157 1.221.22 129 xx x x
TheaverageweightforanAmericanwoman aged20through29in1960was129pounds.
34. Let x =theamountinvestedat4%. Let4000– x =theamountinvestedthatlost3%.
0.040.03(4000)55 0.041200.0355 0.0712055
$2500wasinvestedat4%and$1500lost3%.
35. Let x =thewidthoftherectangle Let2x +5=thelengthoftherectangle 22 2(25)246 410246 61046 636 636 66 6 2517 lwP xx xx x x x x x += ++= ++= += = = = += Thedimensionsoftherectangleare6by17.
36. Let x =thewidthoftherectangle Let2x –1=thelengthoftherectangle
(21)28
2mustberejected. If4,then217 xx = −= Thedimensionsoftherectangleare4by7.
37. Let x =theheightupthepoleatwhichthewires areattached.
–12mustberejected. Thewiresareattached12feetupthepole.
38. 2 62.27000Nx=+
–25mustberejected. Theequationpredictsthattherewere46,000 multinationalcorporations25yearsafter1970, or1995.Themodeldescribestheactualdata showninthegraphquitewell.
39. 2 0.00490.35911.78Pxx =−+ 2 2 150.00490.35911.78 00.00490.3593.22 xx xx =−+ =−− 2 2 2 00.00490.3593.22 4 2
0.3590.191993 0.0098 81,8(rejected) xx bbac x a x x xx =−−
(0.359)(0.359)4(0.0049)(3.22) 2(0.0049)
ThepercentageofforeignbornAmericanswillbe15%about81yearsafter1930,or2011.
40. (62)(7)6271 iiiii −−−=−−+=−−
41. 2 3(2)6336 iiiii +=+=−+
42. (1)(43)43432 iiiii +−=−+− 43 7 i i =++ =+
43. 111 111 iii iii +++ =⋅ −−+ 2 2 1 1 121 11 2 2 iii i i i i +++ = +− = + = =
44. 7512532333 iii−−−=−=
45. () () 22 2323 i −−=− 44332 4433 143 ii i i =−+ =−− =−
Section 1.6
Check Point Exercises 1. () 42 42 22 412 4120 430 xx xx xx = −= −= 22 22 40or30 x03 03 03 xx x xx xx =−= == =±=± ==±
Thesolutionsetis {} 3,0,3 2. 32 2 2 23812 (23)4(23)10 (23)(4)0 xxx xxx xx +=+ + −+= +−= 2 2 230or40 2=34 3 2 2 xx xx xx + =−= −= = −=±
Thesolutionsetis32,,2 2 ⎧⎫ ⎨⎬ ⎩⎭
3. 33 xx ++= () () 22 2 2 33 33 369 076 0(6)(1) xx xx xxx xx xx +=− +=− +=−+ =−+ =−− 60or10 61 xx xx −=−= == 1doesnotcheckandmustberejected.
Thesolutionsetis {}6.
4. 532xx+−−= () () ()() () () () () 22 22 22 523 523 522233 54433 443 443 44 13 13 13 4 xx xx xxx xxx x x x x x x +=+− +=+− +=+−+− +=+−+− =− = =− =− =− =
Thecheckindicatesthat4isasolution.
Thesolutionsetis { } 4.
5. a. () () 3/2 3/2 3/2 3/22323 2/33 5250 525 5 5 5or25 x x x x x −= = = = = Check: () () 2/33/2 55250 55250 25250 00 −= −= −= = Thesolutionsetis {} 2/3 5or {}325.
b. ( ) ( ) 2 384 2/34 3/2 2/33/2 4 23/22 32 8 x x x x x x −=− = = = = = or 3(2) 8 x x =− =−
Thesolutionsetis{–8,8}. 6. () 42 2 22 560 560 xx xx += += Let t = x2 2560 (3)(2)0 tt tt += −= 22 30or20 3or2 3or2 3or=2 tt tt xx xx −=−= == == =±±
Thesolutionsetis { } 3,3,2,2 7. 2/31/3 31140 xx −= Let1/3 tx = 2 31140 (31)(4)0 tt tt −= + −= 310or40 3=1 1 =4 3 tt t tt + =−= −= 1/31/3 3 3 1 4 3 1 4 3 1 64 27 xx xx xx = −= ⎛⎞ = ⎜⎟−= ⎝⎠ = −=
Thesolutionsetis 1 ,64 27 ⎧ ⎫ ⎨ ⎬ ⎩⎭ 8. 215 x = 215or215 2624 32 xx xx xx =−=− = =− = =−
Thesolutionsetis{–2,3}.
9. 412200 x −−= 41220 125 x x −= −= 125or125 2426 23 xx xx xx −=−=− −=−=− =−=
Thesolutionsetis{–2,3}.
Exercise Set 1.6
1. 42 22 2 3480 3(16)0 3(4)(4)0 xx xx xxx −= −= +−= 2 2 30 0 0 x x x = = = 40 4 x x += =− 40 4 x x −= =
Thesolutionsetis{–4,0,4}.
2. 42 22 2 5200 5(4)0 5(2)(2)0 xx xx xxx −= −= +−= 2 2 502020 0 022 xxx x xxx =+=−= = ==−=
Thesolutionsetis{–2,0,2}.
3. 32 32 2 2 32128 321280 (32)4(32)0 (32)(4)0 xxx xxx xxx xx +=+ +−−= +−+= +−= 320 32 2 3 x x x += =− =− 2 2 40 4 2 x x x −= = =±
Thesolutionsetis22,,2. 3 ⎧⎫ ⎨⎬ ⎩⎭ 4. 32 32 2 2 412927 4129270 4(3)9(3)0 (3)(49)0 xxx xxx xxx xx =− −−+= −−−= −−= 2 2 2 30490 349 9 4 3 2 xx xx x x −=−= == = =±
Thesolutionsetis33,,3. 22 ⎧⎫ ⎨⎬ ⎩⎭ 5. 32 32 2 2 23812 812230 4(23)(23)0 (23)(41)0 xxx xxx xxx xx −=− −−+= −−−= −−= 230 23 3 2 x x x = = = 2 2 2 410 41 1 4 1 2 x x x x −= = = =±
Thesolutionsetis311,,.222 ⎧⎫ ⎨⎬ ⎩⎭ 6. 32 32 2 2 199 9910 9(1)(1)0 (1)(91)0 xxx xxx xxx xx +=+ +−−= +−+= +−= 2 2 2 10911 191 1 9 1 3 xx xx x x +=−= =−= = =±
Thesolutionsetis 11 1,,. 33 ⎧⎫ ⎨⎬ ⎩⎭
7. 32 32 2 2 428 4820 4(2)(2)0 (2)(41)0 yyy yyy yyy yy −=− +−−= +−+= +−= 20 2 y y += =− 2 2 2 410 41 1 4 1 2 y y y y −= = = =±
Thesolutionsetis 11 2,,. 22 ⎧⎫ ⎨⎬ ⎩⎭
8. 32 32 2 2 98418 918480 9(2)4(2)0 (2)(94)0 yyy yyy yyy yy +=+ −−+=
−−= 2 2 2 20940 294 4 9 2 3 yy yy y y −=−= == = =±
Thesolutionsetis 22 ,,2. 33 ⎧⎫ ⎨⎬ ⎩⎭
9. () 4 4 3 216 2160 280 xx xx xx = −= −= ()() () 3 2 2 2 2080 0(2)(22)0 20240 22414 2 21 212 2 xx xxxx xxx xx x =−= =−++= −=++= −±− == −±− = 223 2 13 i x xi −± = =−±
Thesolutionsetis {} 0,2,13. i −±
10. 4 4 3 381 3810 3(27)0 xx xx xx = −= −= 3x =03270 x = x =0; ()() () 2 2 2 (3)(39)0 30390 33419 3 21 3936 2 327 2 333 2 xxx xxx xx x x i x −++= −=++= −±− == −±− = −±− = −± = Thesolutionsetis 333 0,3, 2 i ⎧ ⎫ −± ⎪ ⎪ ⎨ ⎬
⎪
11. 2 2 318 318 3180 (3)(6)0 xx xx xx xx + = + = −= + −= x +3=0 x –6=0 x =–3 x =6 3(3)183 9183 93 +=− +=− = 3(6)186 18186 False366 += += = Thesolutionsetis{6}. 12. 2 2 208 208 8200 (10)(2)0 xx xx xx xx = = + −= + −= 10020 102 xx xx + =−= =−= 208(10)10 208010 10010 −=− + =− = False 208(2)2 20162 42 −= −= = Thesolutionsetis{2}.
13. 2 2 33 369 760 (1)(6)0 xx xxx xx xx +=− +=−+ −+= −−= x –1=0 x –6=0 x =1 x =6 1313 42 +=− =− 6363 False93 +=− = Thesolutionsetis{6}.
14. 2 2 2 102 10(2) 1044 560 (1)(6)0 xx xx xxx xx xx +=− +=− +=−+ −−= +−= 1060 16 xx xx +=−= =−= 11012 93False −+=−− =− 61062 164 +=− = Thesolutionsetis{6}.
15. 2 2 2 2 2137 213(7) 2131449 12360 (6)0 60 6 xx xx xxx xx x x x +=+ +=+ +=++ ++= += += =− 2(6)1367 12131 11 −+=−+ −+= = Thesolutionsetis{–6}.
16. 2 2 2 611 61(1) 6121 80 (8)0 800 8 xx xx xxx xx xx xx x +=− +=− + =−+ −= −= == = 6(0)101 011 11False +=− +=− =− 6(8)181 4817 497 +=− += = Thesolutionsetis{8}.
17. 2 2 2 255 525 (5)25 102525 12200 (2)(10)0 xx xx xx xxx xx xx −+= =+ −=+ −+=+ −+= −−= 20100 210 xx xx =−= = = 22(2)55 295 235 += = = 102(10)55 10255 False1055 −+= −= = Thesolutionsetis{10}.
18. 2 2 2 111 111 (1)11 2111 3100 (2)(5)0 xx xx xx xxx xx xx −+= =+ −=+ −+=+ −−= +−= 2050 25 xx xx + =−= =−= 22111 291 231False −−−+= −−= −−= 55111 5161 541 −+= −= = Thesolutionsetis{5}.
19. 2198xx +−= () () 22 2 2 2198 2198 2191664 01445 0(9)(5) xx xx xxx xx xx +=+ +=+ +=++ =++ =++ 90or50 95 xx xx +=+= =−=− –9doesnotcheckandmustberejected. Thesolutionsetis{–5}.
20. 2156xx +−= () () 22 2 2 2156 2156 2151236 01021 0(3)(7) xx xx xxx xx xx +=+ +=+ +=++ =++ =++ 30or70 37 xx xx +=+= =−=− –7doesnotcheckandmustberejected. Thesolutionsetis{–3}.
21. 2 2 2 3104 36 3(6) 31236 15360 (12)(3)0 xx xx xx xxx xx xx +=+ =− =− =−+ −+= −−= x –12=0 x –3=0 x =12 x =3 3(12)10124 361016 61016 +=+ += += 3(3)1034 9107 3107False +=+ += += Thesolutionsetis{12}.
22. 2 2 2 39 6 (6) 1236 13360 (9)(4)0 xx xx xx xxx xx xx −=− =− =− = −+ −+= −−= 9040 94 xx xx =−= = = 9399 3399 =− =− 4349 2349False −=− −=−
Thesolutionsetis{9}.
23. 2 842 842 8(42) 84444 844 844 24 44 8 xx xx xx xxx xxx x x x x +−−= +=−+ +=−+ + =−+−+ +=+− =− =− =− = 88842 1642 422 + −−= = = Thesolutionsetis{8}.
24. 532xx + −−= 2 532 5(32) 53434 5143 5143 443 13 13 4 xx xx xxx xxx x x x x x +=−+ +=−+ + =−+−+ +=++− =+− =− =− =− = 45432 912 312 + −−= = = Thesolutionsetis{4}.
25. 2 583 583 5(83) 58689 5168 668 18 18 9 xx xx xx xxx xxx x x x x
26.
Thesolutionsetis{2,6}.
27. 2 2 2 2 2322 2322 23(22) 234422 142 (1)16(2) 211632 14330 (11)(3)0 xx xx xx xxx xx xx xxx xx xx ++−= +=−− +=−− + =−−+− +=−− +=− ++=− −+= −−= x –11=0 x –3=0 x =11 x =3 2(11)31122 22392 532False ++−= ++= += 2(3)3322 6312 312False ++−= ++= += Thesolutionsetistheemptyset, ∅ 28. 2 2 2 2 2371 2137 2(137) 2123737 26237 337 (3)37 6937 320 (1)(2)0 xx xx xx xxx xx xx xx xxx xx xx +++= +=−+ +=−+ + =−+++ −−=−+ +=+ +=+ ++=+ ++= ++= 10 1 x x + = = 20 2 x x + = =− 123(1)71 141 121False −++−+= += += 223(2)71 011 011 ++−+= += += Thesolutionsetis{–2}.
29.
Checkproposedsolutions. Thesolutionsetis
x =0or x =4 Thesolutionsetis{0,4}.
31. 3/2
3/22/32/3 32 2 8 ()8 8 2 4 x x x x x = = = = = 3/2 3 3 48 48 28 = = = Thesolutionsetis{4}.
32. 3/2
3/22/32/3 32 2 27 ()27 27 3 9 x x x x x =
3/2 3 3 927 927 327 =
= Thesolutionsetis{9}. 33. 3/2 3/22/32/3 32 2 (4)27 ((4))27 427 43 49 13 x x x x x x −= −=
−= = 3/2 3/2 3 3 (134)27 927 927 327 = = = = Thesolutionsetis{13}.
(5)8 ((5))8
3/2 3/22/32/3
Thesolutionsetis{–1}.
35. 5/2 5/2 5/2 5/22/52/5 52 5 6120 612 2 ()2 2 4 x x x x x x −= = = = = = 55/2 1/55/2 1/2 6(4)120 6(4)120 6(4)120 6(2)120 = = = = Thesolutionsetis { } 54
36. 5/3 5/3 5/3 5/33/53/5 53 5 8240 824 3 ()3 3 27 x x x x x x −= = = = = = 55/3 1/55/3 1/3 8(27)240 8(27)240 8(27)240 8(3)240 −=
Thesolutionsetis { } 527.
37. () ()() (
Thesolutionsetis{–60,68}.
38. () () () 2 3 3 3 2 2 54 2 54 3 x x += ⎡⎤ ⎢⎥+= ⎢⎥ ⎣⎦ () 3 22 33 52 52or5(2) 5858 313 x xx xx xx += +=+=− +=+=− ==−
Thesolutionsetis{–13,3}.
39. 23/4 23/4 23/44/34/3 24 3 24 2 2 (4)26 (4)8 ((4))8 48 42 416 200 (5)(4)0 xx xx xx xx xx xx xx xx −−−= −−= −−= −−= −−= −−= −−= −+= x –5=0 x +4=0 x =5 x =–4 23/4 3/4 3/4 43 3 23/4 3/4 3/4 43 3 (554)26 (259)26 1626 1626 226 826 ((4)(4)4)26 (1644)26 1626 1626 226 826 −−= −= −= −= −= −= −−−−= + −−= −= −= −= −=
Thesolutionsetis{5,–4}.
40. 23/2 23/2 22/3 2 2 (33)10 (33)1 331 331 320 (1)(2)0 xx xx xx xx xx xx −+−= −+= −+= −+= −+= −−= 10 1 x x = = 20 2 x x −= = 23/2 3/2 3/2 (13(1)3)10 (133)10 110 110 +−= +−= = = 23/2 3/2 3/2 (23(2)3)10 (463)10 110 110 +−= +−= −= −=
Thesolutionsetis{1,2}.
41. 422 2 540let
(1)(4)0 xxtx
44. 42 422 4139 41390let xx xxtx =− +== 2 41390 (49)(1)0 tt tt += −=
Thesolutionsetis{1,–1,2,–2}
42. 422 13360let xxtx −+== 213360 (4)(9)0 tt tt
Thesolutionsetis{–3,–2,2,3}.
43. 42 92516 xx=− 422 2 925160let 925160 (916)(1)0 xxtx tt tt −+==
Thesolutionsetis33,1,1,. 22 ⎧ ⎫
45. () 13400Let. 213400 8(5)0 xxtx
Thesolutionsetis{25,64}.
46. 27300Let. xxtx −−== 2 27300 (25)(6)0 tt tt −= + −=
Thesolutionsetis 44 1,1,,. 33 ⎧⎫ ⎨⎬ ⎩⎭
Thesolutionsetis{36}since25/4doesnot checkintheoriginalequation.
47. 211 200Let xxtx −−== 2200 (5)(4)0 tt tt −−= −+= 50 5 15 1 5 15 1 5 t t x x x x −= =
Thesolutionsetis 11 ,.45 ⎧⎫ ⎨⎬ ⎩⎭
48. 211 60Lett.xxx −−== 2 11 60 (3)(2)0 3020 32 32 11 32 1312 11 32 tt tt tt tt xx xx xx xx −−= −+= −=+= ==−
Thesolutionsetis 11 ,.23 ⎧⎫ ⎨⎬ ⎩⎭
50. 2/31/31/327150let xxtx +−== 2 27150 (23)(5)0 tt tt + −= += 1/3 1/32 3 23050 235 3 5 2 3(5) 2 3 125 2 27 8 tt tt tx xx xx x −=+=
Thesolutionsetis27125,. 8 ⎧⎫ ⎨⎬
51. 3/23/43/4 210let xxtx −+== 2 3/4 4/3 210 (1)(1)0 10 1 1 1 1 tt tt t t x x x −+= −−= −= = = = =
Thesolutionsetis{1}.
60let
xxtx
49. 2/31/31/3
52. 2/51/51/5 60let xxtx +−== 260 (3)(2)0 tt tt + −= + −= 1/5 5 30 3 3 (3) 243 t t x x x + = =− =− = =− 1/5 5 20 2 2 2 32 t t x x x = = = = =
Thesolutionsetis{–243,32}.
Thesolutionsetis{27,–8}.
53. 1/21/2 2 2310let 2310 (21)(1)0 xxtx tt tt −+== −+= −−= 1/21/2 2 2 21010 21 1 1 2 1 1 2 1 1 2 1 1 4 tt t tt xx xx xx −=−= = == ==
Thesolutionsetis 1 ,1. 4
54. 1/21/2340let xxtx +−== 2340 (1)(4)0 tt tt +−= −+= 1/2 2 10 1 1 1 1 t t x x x −= = = = = 1/2 2 40 4 4 (4) 16 t t x x x += =− =− =− =
Thesolutionsetis{1}.
55. 2 2 (5)4(5)210let5 4210 (3)(7)0 xxtx tt tt −−−−==− −−= +−= 30 3 53 2 t t x x += =− −=− = 70 7 57 12 t t x x −= = −= = Thesolutionsetis{2,12}.
56. 2 (3)7(3)180let3 xxtx +++−==+ 27180 (9)(2)0 tt tt +−= +−= 90 9 39 12 t t x x += =− +=− =− 20 2 32 1 t t x x −= = += =− Thesolutionsetis{–12,–1}.
57. ( ) ( ) 2 2214240xxxx−−+=
Let2txx = 214240tt += (t –2)(t –12)=0 t =2or t =12 ()()()() 22 22 2or12 20120 210430 xxxx xxxx xxxx −=−= −−=−−= +=−+= Thesolutionsetis{–3,–1,2,4}.
58. ( ) ( ) 2 222112240xxxx−−+=
Let22txx = ()() ()()()() 2 22 22 11240 380 3or8 23or28 230280 310420 tt tt tt xxxx xxxx xxxx −+=
Thesolutionsetis{–2,–1,3,4}.
59. 2 88 5140yy yy
⎛⎞⎛⎞ ⎜⎟⎜⎟+−−= ⎝⎠⎝⎠
Let 8 ty y = 25140tt + −= (t +7)(t –2)=0 t =–7or t =2 ()()()() 22 88 7or2 780280 810420 yy yy yyyy yyyy −=−−= +−=−−= + −=−+=
Thesolutionsetis{–8,–2,1,4}.
60. 2 1010 6270yy yy ⎛⎞⎛⎞
Thesolutionsetis{–10,–2,1,5}
61. 8 x =
x =8, x =–8
Thesolutionsetis{8,–8}.
62. 6 x =
x =6, x =–6
Thesolutionsetis{–6,6}.
63. 27 x −=
x –2=7 x –2=–7 x =9 x =–5
Thesolutionsetis{9,–5}.
64. 15 x +=
x +1=5 x +1=–5 x =4 x =–6
Thesolutionsetis{–6,4}.
65. 215 x −=
2x –1=52x –1=–5 2x =62x =–4 x =3 x =–2
Thesolutionsetis{3,–2}.
66. 2311 x = 2311 214 7 x x x = = = 2311 28 4 x x x =− =− =−
Thesolutionsetis{–4,7}.
67. 2|3x –2|=14 |3x –2|=7 3x –2=73x– 2=7 3x =93x =5 x =3 x =5/3
Thesolutionsetis{3,5/3}
68. 3|2x –1|=21 |2x –1|=7 2x –1=7or2x –1=7 2x =82x =6 x =4 x =3
Thesolutionsetis{4,3}
69. 7|5x|+2=16 7|5x|=14 |5x|=2 5x =25x =2 x =2/5 x =2/5
Thesolutionsetis 22 , 55 ⎧⎫ ⎨⎬ ⎩⎭ .
70. 7|3x|+2=16 7|3x|=14 |3x|=2
3x =2or3x =2 x =2/3 x =2/3
Thesolutionsetis{2/3,2/3}
71.
Thesolutionsetis4,4
Thesolutionsetis17
73. |x +1|+5=3 |x +1|=2 Nosolution Thesolutionsetis{}.
74. |x +1|+6=2 |x +1|=4Thesolutionsetis{}.
75. |2x –1|+3=3 |2x –1|=0 2x –1=0 2x =1 x =1/2
Thesolutionsetis{1/2}.
Thesolutionsetis{2/3}.
77. |3x –1|=|x +5| 3x –1= x +53x –1= x –5
2x –1=54x– 1=5
2x =64x =4 x =3 x =1
Thesolutionsetis{3,1}.
78. () 273 273or273 10273 34 4 3 xx xxxx xxx x x −=+
The solution set is 4 10, 3 ⎧⎫
79. Set y = 0 to find the x–intercept(s). () () () () ()() () () 22 22 22 0213 213 213 212133 21619 21619 2861 661 661 66 11 11 11 2 xx xx xx xxx xxx xxx x x x x x x x
The x-interceptis2. The corresponding graph is graph (c).
80. Set y = 0 to find the x–intercept(s). () () () () ()() () () 22 22 22 0444 444 444 442444 448416 42084 2484 2484 88 34 34 94 5 xx xx xx xxx xxx x x x x x x x
=+ =+ = The x-interceptis5. The corresponding graph is graph (a).
81. Set y = 0 to find the x–intercept(s). 11 36 023 xx = +− Let t = 1 6 x ()() 11 36 2 11 66 2 230 230 230 310 xx xx tt tt + −= ⎛⎞ + −= ⎜⎟ ⎝⎠ + −= + −= 30or10 31 tt tt + =−= = −= Substitute 1 6for. xt () () 11 66 6611 6666 31 or 31 7291 xx xx xx =−= ⎛⎞⎛⎞ ⎜⎟⎜⎟=−= ⎝⎠⎝⎠ ==
729 does not check and must be rejected. The x-interceptis1. The corresponding graph is graph (e).
82. Set y = 0 to find the x–intercept(s). 21 06 xx =−−
Let t = 1 x ()
()() 21 2 11 2 60 60 60 230 xx xx tt tt −−= −−= −−= +−= 20or30 23 tt tt +=−= =−=
Substitute 1for.xt 113 2 or 11 23 xx xx = =− = =−
The x-interceptsare 1 2 and 1 3
The corresponding graph is graph (b).
83. Set y = 0 to find the x–intercept(s).
()() 2 292200xx+−++=
Let t = 2. x + ()() ()() 2 2 292200 9200 540 xx tt tt +−++= −+= −−= 50or40 54 tt tt −=−= ==
Substitute 2for. xt + 25or24 32 xx xx +=+= ==
The x-interceptsare2and3.
The corresponding graph is graph (f).
84. Set y = 0 to find the x–intercept(s).
()() 2 022523 xx =+++−
Let t = 2. x + ()() ()() 2 2 225230 2530 2130 xx tt tt +++−= +−= −+= 210or30 213 1 2 tt tt t −=+= ==− =
Substitute 2for. xt + 21or23 25 1 2 2 3 2 xx x x x + =− += =− =− =−
The x-intercepts are 3 52 and . The corresponding graph is graph (d).
85. 5411 x = 54115411 46or416 34 2 xx xx x x =−=− =−=− = =−
Thesolutionsetis3,4. 2 ⎧ ⎫ ⎨ ⎬ ⎩⎭
86. 2313 x = 23132313 311or315 115 3 xx xx x x =−=− =−=− = =−
Thesolutionsetis 11 ,5. 3 ⎧ ⎫ ⎨ ⎬ ⎩⎭
87. 57 xx + += () () 22 2 2 57 57 54914 01544 0(4)(11) xx xx xxx xx xx +=− +=− +=−+ =−+ =−− 40or110 411 xx xx =−= = = 11 does not check and must be rejected. Thesolutionsetis { } 4.
88. 24 xx−−= () () 22 2 2 24 24 2168 0918 0(6)(3) xx xx xxx xx xx −−=− −−=− −=−+ =−+ =−− 60or30 63 xx xx −=−= ==
3 does not check and must be rejected. Thesolutionsetis {}6.
89. 32 2826 xxx+−+= ()() () () ()()() 32 2 2 2840 214210 2140 21220 xxx xxx xx xxx +−−= +−+= +−= ++−= 2102020 oror 122 2 xxx xx x += +=−= =−= =−
Thesolutionsetis 1 ,2,2. 2 ⎧⎫ ⎨⎬ ⎩⎭
90. 324610xxx+−+= ()() () () ()()() 32 2 2 440 4140 410 4110 xxx xxx xx xxx +−−= +−+= +−= ++−= 40or10or10 411 xxx xxx +=+=−= =−=−=
Thesolutionsetis {} 4,1,1.
91. () 3 2 48 x += () () () () 2 32 3 23 2 3 2 48 48 42 44 0 x x x x x ⎛⎞ ⎜⎟+= ⎝⎠ += += += =
Thesolutionsetis {}0.
92. () 3 2 5125 x −= () () () () 2 32 3 23 2 3 2 5125 5125 55 525 30 x x x x x ⎛⎞ ⎜⎟−= ⎝⎠ −= −= −= =
Thesolutionsetis {}30.
93. 123yy=+ ( )() ()() 2 22 2 22 1213 12130 xx xx −=−+ −−−−=
Let21 tx = andsubstitute. 2230 (1)(3)0 tt tt −−= + −= 10or30 13 tt tt + =−= = −=
Substitute 21for. xt 22 22 1113 or 04 02 xx xx xx −=−−= == = =±
Thesolutionsetis {}2,0,2.
1
94. 126yy=+ 2 2 22 656 33 22 6560 33 xx xx xx xx ⎛⎞⎛⎞ ⎜⎟⎜⎟=+ ⎝⎠⎝⎠ ⎛⎞⎛⎞ ⎜⎟⎜⎟−−= ⎝⎠⎝⎠
Let 2 3 x t x = andsubstitute. 2 6560 (32)(23)0 tt tt −−= +−= 320230 or 23 32 tt tt += −= =− =
Substitute 2 for. 3 x t x or 2223 3332 xx xx =−=
Firstsolve 22 33 x x =− 2(3)(3)2(3)(3) 33
2(3)2(3) 626 86 3 4 xxx x xx xx x x =− =−− =−+ = =
Nextsolve 23 32 x x = 2(2)(3)3(2)(3) 32 2(2)3(3) 439 9 xxx x xx xx x = =− =− =−
Thesolutionsetis39,. 4 ⎧⎫ ⎨⎬ ⎩⎭
95. 223612xx+−= 22 22 2361223612 2480or2240 (8)(6)0(6)(4)0 xxxx xxxx xxxx +−=+−=− +−=+−= +−=+−=
Settingeachofthefactorsaboveequaltozero gives8,6,6,and4. xxxx =−==−=
Thesolutionsetis {} 8,6,4,6.
96. 2618xx + += 22 22 618618 or 670690 (7)(1)0(3)(3)0 xxxx xxxx xxxx + +=++=− +−=++= +−=++=
Settingeachofthefactorsaboveequaltozero gives7,3,and1. xxx = −=−=
Thesolutionsetis { } 7,3,1.
97. ()() 32 14210xxx + −+= ()() ( ) () () ()()() 2 22 2 11420 1420 1760 xxx xxx xxx + +−= + +−= + +−=
Settingeachofthefactorsaboveequaltozero gives7,1,and6. xxx = −=−=
Thesolutionsetis { } 7,1,6.
98. ()() 32 23520xxx−−= ()() ()()() () () ()()() 32 2 22 2 23520 22350 22350 2570 xxx xxx xxx xxx −−= −−= −−= +−=
Settingeachofthefactorsaboveequaltozero gives5,2,and7. xxx = −==
Thesolutionsetis { } 5,2,7.
99. Let x =thenumber.
542 xx−=− () () ()() 22 2 2 542 5444 098 081 xx xxx xx xx −=− −=−+ =−+ =−−
80 8 x x −= = or10 1 x x −= =
Check8 x = : () 58482 4046 366 66 −=− −= = =
Check1 x = : () 51412 541 11 −=− −=− −≠−
Discard1 x = .Thenumberis8.
100. Let x =thenumber.
35xx−=− () () ()() 22 2 2 35 31025 01128 074 xx xxx xx xx −=− −=−+ =−+ =−−
70 7 x x −= = or40 4 x x −= =
Check7 x = :7375 42 22 −=− = =
Check4 x = :4345 11 11 −=− =− ≠−
Discard4.Thenumberis7. 101.
102. 2 2 2 22 4 4 4 4or4 A r A r A r rAAr π π π π π = ⎛⎞ = ⎜⎟ ⎜⎟ ⎝⎠ = ==
103. Excludeanyvaluethatcausesthedenominator toequalzero. 2140 214 x x + −= + = 214214 or 1216 xx xx + =+=− = =− –16and12mustbeexcludedfromthedomain.
104. Exclude any value that causes the denominator to equal zero. () 32 2 2 330 (3)1(3)0 (3)10 (3)(1)(1)0 xxx xxx xx xxx + −−= + −+= + −= + +−= Setting each of the factors above equal to zero gives 3,1,and1.xxx = −=−= –3, –1, and 1 must be excluded from the domain.
105. Let 192. P
192 million computers will be sold 16 years after 1996, or 2012.
106. Let 143. P = () () 22 2880 1432880 6328 6328 2828 2.25 2.25 5 Pt t t t t
143 million computers were sold 5 years after 1996, or 2001. This matches the data shown in the figure quite well.
107. Fortheyear2100,weuse98 x = . () 0.0839857.9 66.034 H =+ = 0.369857.9 61.464 L =+ ≈
Intheyear2100,theprojectedhighend temperatureisabout66° andtheprojectedlowend temperatureisabout61.5° .
108. Fortheyear2080,weuse78 x = . () 0.0837857.964.4 H =+≈ 0.367857.961.1 L =+≈
Intheyear2080,theprojectedhighend temperatureisabout64.4° andtheprojectedlow endtemperatureisabout61.1°
109. Using H: 0.08357.957.91 0.0831 1 0.083 12 x x x x + =+ =
Theprojectedglobaltemperaturewillexceedthe 2002averageby1degreein2014(12yearsafter 2002).
Using L: () 2 2 0.3657.9157.9
Theprojectedglobaltemperaturewillexceedthe 2002averageby1degreein2010(8yearsafter 2002).
110. Using H: 0.08357.957.92 0.0832 2 0.083
Theprojectedglobaltemperaturewillexceedthe 2002averageby2degreesin2026(24yearsafter 2002).
Using
Theprojectedglobaltemperaturewillexceedthe 2002averageby2degreesin2033(31yearsafter 2002).
111. () 2 2 5000100 400005000100 400005000100 50005000 8100 8100 64100 36 36 yx
40,000peopleinthegroupwillsurvivetoage36. Thisisshownonthegraphasthepoint () 36,40000
112.
ivetoage51.This correspondstothepoint(51,35000)onthegraph.
TheaveragedistanceoftheEarthfromthesunis approximately149millionkilometers.
TheaveragedistanceofMercuryfromthesunis approximately58millionkilometers.
116. a. Distance from point 2222 63(12) Axx =+++− or 2236(12)9Axx = ++−+
b. Let the distance = 15.
The distance is 8 miles.
125. 32330xxx+−−=
The solution set is {–3, –1, 1}.
32 (3)3(3)(3)30 2727330 −+−−−−= −++−=
32 (1)3(1)(1)30 13130 −+−−−−= −++−=
32 13(1)(1)30 13130 +−−= +−−=
126. 432440xxx −+−=
The solution set is {0, 2}.
432 (0)4(0)4(0)0 00 −+−= =
432 (2)4(2)4(2)0 1632160 00 −+−= −+−= =
127. 21350 xx+−−=
The solution set is {–2}. 2(2)13(2)50 413250 930 330 −+−−−= −++−= −=
128. Tracing along the curve shows the point (36,40000)
129. a. False; ( ) 2 4141yyyy + +−≠++−
b. False; if 23(2) txx =− , the original equation can be written as 3560tt−+= , not a quadratic form.
c. False; the other value may be a solution.
d. True (d) is true
130.
The
020 (1)(2)0 xtt tt =+−= +=
64 does not check and must be rejected. The solution set is {0, 1}.
Section 1.7
Check Point Exercises 1. a. [ 2, 5) = x 2 ≤ x < 5 { }
b. [1, 3.5] = x 1 ≤ x ≤ 3.5 { }
c. [ −∞ , 1) = xx <− 1 {}
2. a. Graph [ ]1,3 :
Graph () 2,6 :
To find the intersection, take the portion of the number line that the two graphs have in common.
Numbers in both
[ ]1,3 and () 2,6 :
Thus, [ ]1,3 ∩ () 2,6 ( ]2,3 = .
b. Graph [ ]1,3 :
Graph () 2,6 :
To find the union, take the portion of the number line representing the total collection of numbers in the two graphs.
Numbers in either [ ]1,3 or () 2,6 or both:
Thus, [ ]1,3 ∪ () 2,6 [ ) 1,6 =
3. 235 33 1 x x x −≤ −≤ ≥−
The solution set is {}1or[1,) xx ≥−−∞
4. 31715 xx+>− 416 416 44 4 x x x −>− < <
The solution set is {} ( ] 4or-,4 xx <∞
5. a. 3(1)32 xx+>+
3332
32 xx+>+ >
32 > is true for all values of x
The solution set is {} isarealnumber. xx
b. 11xx+≤− 11 ≤− 11 ≤− is false for all values of x
The solution set is ∅
6. 12311 228 14 x x x ≤+< −≤< −≤<
The solution set is {}14or[1,4)xx−≤<−
7. 25 525 37 x x x −< −<−< −<<
The solution set is {}37or(3,7)xx−<<−
8. 3522019 x −−+≥− 35239 35239 33 5213 135213 11515 11515 555 11 3 5 x x x x x x x −−≥− ≤ −≤ −≤−≤ −≤≤ ≤≤ −≤≤ The solution set is 1111 3or,3 55 xx
9. 1863 x <− 63186318 or 324312 324312 3333 84 xx xx xx xx −<−−> −<−−> >< ><−
The solution set is {} 4or8xxx<−> or ()() ,48,.−∞−∞ ∪
10. Let x = the number of miles driven in a week. 260800.25 1800.25 720 x x x <+ < < Driving more than 720 miles in a week makes Basic the better deal.
Exercise Set 1.7
1. 1 < x ≤ 6
2. –2 < x ≤ 4
3. –5 ≤ x < 2
4. –4 ≤ x < 3
5. –3 ≤ x ≤ 1 6. –2 ≤ x ≤ 5
7. x > 2
x > 3
15. Graph () 3,0 :
Graph [ ] 1,2 :
To find the intersection, take the portion of the number line that the two graphs have in common.
Numbers in both
() 3,0 and [ ] 1,2 :
Thus, () 3,0 ∩ [ ] 1,2 [ ) 1,0=−
16. Graph () 4,0 :
Graph [ ]2,1 :
To find the intersection, take the portion of the number line that the two graphs have in common.
Numbers in both () 4,0 and [ ]2,1 :
Thus, () 4,0 ∩ [ ]2,1 [ ) 2,0=−
17. Graph () 3,0 :
Graph [ ] 1,2 :
To find the union, take the portion of the number line representing the total collection of numbers in the two graphs.
Numbers in either () 3,0 or [ ] 1,2 or both:
Thus, () 3,0 ∪ [ ] 1,2 ( ] 3,2=−
18. Graph () 4,0 :
Graph [ ]2,1 :
To find the union, take the portion of the number line representing the total collection of numbers in the two graphs.
Numbers in either () 4,0 or [ ]2,1 or both:
Thus, () 4,0 ∪ [ ]2,1 ( ]4,1=−
19. Graph () ,5−∞ :
Graph [ ]1,8 :
To find the intersection, take the portion of the number line that the two graphs have in common.
Numbers in both () ,5−∞ and [ ]1,8 :
Thus, () ,5 −∞ ∩ [ ]1,8 [ ) 1,5 =
20. Graph () ,6 −∞ :
Graph [ ] 2,9 :
To find the intersection, take the portion of the number line that the two graphs have in common. Numbers in both () ,6 −∞ and [ ] 2,9 : Thus, () ,6−∞ ∩ [ ] 2,9 [ ) 2,6 =
21. Graph () ,5−∞ :
Graph [ ]1,8 :
To find the union, take the portion of the number line representing the total collection of numbers in the two graphs.
Numbers in either () ,5−∞ or [ ]1,8 or both:
Thus, () ,5−∞ ∪ [ ]1,8 ( ],8=−∞
22.
Graph () ,6−∞ :
Graph [ ] 2,9 :
To find the union, take the portion of the number line representing the total collection of numbers in the two graphs.
Numbers in either () ,6 −∞ or [ ] 2,9 or both:
Thus, () ,6−∞ ∪ [ ] 2,9 ( ] ,9=−∞
23. Graph [ ) 3, ∞ :
Graph () 6, ∞ :
To find the intersection, take the portion of the number line that the two graphs have in common.
Numbers in both
[ ) 3, ∞ and () 6, ∞ :
Thus, [ ) 3, ∞ ∩ () 6, ∞ () 6, =∞
24. Graph [ ) 2, ∞ :
Graph () 4, ∞ :
To find the intersection, take the portion of the number line that the two graphs have in common.
Numbers in both
[ ) 2, ∞ and () 4, ∞ :
Thus, [ ) 2, ∞ ∩ () 4, ∞ () 4, =∞ .
25.
Graph [ ) 3, ∞ :
Graph () 6, ∞ :
To find the union, take the portion of the number line representing the total collection of numbers in the two graphs.
Numbers in either [ ) 3, ∞ or () 6, ∞ or both:
Thus, [ ) 3, ∞ ∪ () 6, ∞ [ ) 3, =∞
26. Graph [ ) 2, ∞ :
Graph () 4, ∞ :
To find the union, take the portion of the number line representing the total collection of numbers in the two graphs.
Numbers in either [ ) 2, ∞ or () 4, ∞ or both:
Thus, [ ) 2, ∞ ∪ () 4, ∞ [ ) 2, =∞
27. 5x + 11 < 26
5x < 15
x < 3
The solution set is {}3, xx < or (–∞, 3).
28. 2x + 5 < 17
2x < 12 x < 6
The solution set is {}6or xx < (–∞, 6).
29. 3x – 7 ≥ 13
3x ≥ 20 20 3 x ≥
The solution set is 20 3, xx ⎧⎫ > ⎨⎬ ⎩⎭ or 20 3, ⎡ ⎞ ∞ ⎟ ⎢ ⎣ ⎠ .
30. 8x – 2 ≥ 14
8x ≥ 16 x ≥ 2
The solution set is {}2or xx > [ ) 2, ∞
31. –9x ≥ 36 x ≤ –4
The solution set is {}4, xx ≤− or ( ] ,4∞−
32. –5x ≤ 30 x ≥ –6
The solution set is {}6or xx ≥− [ ) 6, ∞ .
33. 8x – 11 ≤ 3x – 13
8x – 3x ≤ –13 + 11 5x ≤ –2 2 5 x ≤−
The solution set is 2 5, xx ⎧⎫ ≤− ⎨⎬ ⎩⎭ or 2 ,5 ⎛⎤ ∞− ⎜ ⎥ ⎝⎦ .
34. 18x + 45 ≤ 12x – 8
18x – 12x ≤ –8 – 45 6x ≤ –53 53 6 x ≤
The solution set is 53 or 6 xx ⎧⎫ ≤− ⎨⎬ ⎩⎭
35. 4(x + 1) + 2 ≥ 3x + 6
4x + 4 + 2 ≥ 3x + 6
4x + 6 ≥ 3x + 6
4x – 3x ≥ 6 – 6
x ≥ 0
The solution set is {}0, xx > or [0, ∞).
36. 8x + 3 > 3(2x + 1) + x + 5
8x + 3 > 6x + 3 + x + 5
8x + 3 > 7x + 8
8x – 7x > 8 – 3 x > 5
The solution set is {}5or xx > (5, ∞ ).
37. 2x – 11 < – 3(x + 2)
2x – 11 < – 3x – 6
5x < 5
x < 1
The solution set is {}1, xx < or (–∞, 1).
38. –4(x + 2) > 3x + 20 –4x – 8 > 3x + 20
–7x > 28
x < –4
The solution set is {}4or xx <− (–∞, –4).
39. 1 – (x + 3) ≥ 4 – 2x 1 – x – 3 ≥ 4 – 2x –x – 2 ≥ 4 – 2x
x ≥ 6
The solution set is {}6, xx ≥ or [6, ∞).
40. 5(3 – x) ≤ 3x – 1 15 – 5x ≤ 3x – 1 –8x ≤ –16 x ≥ 2
The solution set is {}2or xx ≥ [2, ∞).
41. 3 1 422 xx −≤+ 4434 41 422 624 10 10 xx xx x x −≤+⋅ −≤+ −≤ ≥−
The solution set is {}10, xx ≥− or [ ) 10, −∞
42. 31 1 10510 xx +≥− 31 10110 10510 3102 48 2 xx xx x x ⎛⎞⎛⎞ +≥− ⎜⎟⎜⎟ ⎝⎠⎝⎠ +≥− ≥− ≥−
The solution set is {}2or xx ≥− [ ) 2,.−∞
43. 14 2 x −> 3 2 x −> x < –6
The solution set is {},6,xx or () ,6−∞−
44. 43 7 55 x −< 432 55 x −<− x > 8
The solution set is {}8or xx > (8, ∞).
45. 425 6918 3(4)2(2)5 312245 13 xx xx xx x ≥+ ≥−+ −≥−+ ≥
The solution set is { } 13, xx ≥ or [ ) 13, ∞
46. 4321 2 612
2(43)2421 862421 6181 619 19 6 xx xx xx x x x +≥ +≥− +≥− +≥− ≥− ≥−
The solution set is 19-19 or,. 66 xx ⎧−⎫
47. 4(3x – 2) – 3x < 3(1 + 3x) – 7 12x – 8 – 3x < 3 + 9x – 7 9x – 8 < – 4 + 9x –8 < –4
True for all x
The solution set is { } isanyrealnumber, xx or ( ) , ∞∞
48. 3(x – 8) – 2(10 – x) > 5(x – 1) 3x – 24 – 20 + 2x > 5x – 5 5x – 44 > 5x – 5 –44 > –5
Not true for any x.
The solution set is the empty set, .∅
49. 5(2)3(4)220 xxx−+≥− 510312220 222220 2220 xxx xx −−≥− ≥− −≥−
Not true for any x.
The solution set is the empty set, ∅
50. 6(1)(4)78 xxx −−−≥− 66478 71078 108 xxx xx −−+≥− −≥− −≥−
Not true for any x.
The solution set is the empty set, .∅
51. 6 < x + 3 < 8
6 – 3 < x + 3 – 3 < 8 – 3 3 < x < 5
The solution set is {} 35,xx<< or (3, 5).
52. 7 < x + 5 < 11
7 – 5 < x + 5 – 5 < 11 – 5 2 < x < 6
The solution set is {}26orxx<< (2, 6).
53. –3 ≤ x – 2 < 1 –1 ≤ x < 3
The solution set is {} 13,xx−≤< or [–1, 3).
54. –6 < x – 4 ≤ 1 –2 < x ≤ 5
The solution set is {|25} xx−<≤ or (–2, 5].
55. –11 < 2x –1 ≤ –5 –10 < 2x ≤ –4 –5 < x ≤ –2
The solution set is {} 52,xx−<≤− or (–5, –2].
56. 3 ≤ 4x – 3 < 19 6 ≤ 4x < 22 622 44 311 22 x x ≤< ≤<
The solution set is 311311 or,. 2222 xx ⎧⎫ ⎡ ⎞ ≤< ⎨⎬ ⎟ ⎢ ⎣
57. 2 351 3 x −≤−<− 2 24 3 x ≤<
3 ≤ x < 6
The solution set is {} 36,xx≤< or [3, 6).
58. 1 643 2 x ≤−<− 1 21 2 42 x x ≤< −≤<
The solution set is {} [ ) 42or4,2.xx−≥<−
59. |x| < 3 –3 < x < 3
The solution set is {} 33,xx−<< or (–3, 3).
60. |x| < 5 –5 < x < 5
The solution set is {} 55xx−<< or (–5, 5).
61. |x – 1| ≤ 2
–2 ≤ x – 1 ≤ 2
–1 ≤ x ≤ 3
The solution set is {} 13,xx−≤≤ or [–1, 3].
62. |x + 3| ≤ 4
–4 ≤ x + 3 ≤ 4
–7 ≤ x ≤ 1
The solution set is {} 71xx−≤≤ or [–7, 1].
63. |2x – 6| < 8
–8 < 2x – 6 < 8
–2 < 2x < 14
–1 < x < 7
The solution set is {} 17,xx−<< or (–1, 7).
64. 3517 x + <
–17 < 3x + 5 < 17
–22 < 3x < 12
The solution set is 22 4 3 xx ⎧⎫ −<< ⎨⎬ ⎩⎭ or 22 ,4 3 ⎛⎞ ⎜⎟ ⎝⎠ .
65. |2(x – 1) + 4| ≤ 8
–8 ≤ 2(x – 1) + 4 ≤ 8
–8 ≤ 2x – 2 + 4 ≤ 8
–8 ≤ 2x + 2 ≤ 8
–10 ≤ 2x ≤ 6
–5 ≤ x ≤ 3
The solution set is {} 53,xx−≤≤ or [–5, 3].
66. |3(x – 1) + 2| ≤ 20 –20 ≤ 3(x – 1) + 2 ≤ 20 –20 ≤ 3x – 1 ≤ 20 –19 ≤ 3x ≤ 21 19 7 3 x −≤≤
The solution set is 1919 7or,7. 33 xx ⎧⎫ ⎨⎬−≤≤−⎡⎤ ⎢⎥ ⎣⎦ ⎩⎭
67. 26 2 3 y + < 26 22 3 y + −<<
–6 < 2y + 6 < 6 –12 < 2y < 0 –6 < y < 0
The solution set is {} 60,xy−<< or (–6, 0).
68. ()31 6 4 x < ()31 66 4 x −<<
–24 < 3x – 3 < 24 –21 < 3x < 27 –7 < x < 9
The solution set is {} 79xx−<< or (–7, 9).
69. |x| > 3
x > 3 or x < –3
The solution set is {} 3or3,xxx><− that is, ()() ,3or3, −∞−∞
70 |x| > 5
x > 5 or x < –5
The solution set is {} ()() 5or5,thatis, allin,5or5,. xxx x <−> −∞−∞
71. |x – 1| ≥ 2 12or12 31 xx xx −≥−≤− ≥≤−
The solution set is {} 1or3,xxx≤−≥ that is, ( ] [ ) ,1or3,.−∞−∞
72. |x + 3| ≥ 4 34or34 17 xx xx + ≥+≤− ≥≤−
The solution set is { } ()() 7or1,thatis, allin,7or1,. xxx x ≤−≥ −∞−∞
73. |3x – 8| > 7 387or387 31531 1 5 3 xx xx xx >−<− >< ><
The solution set is 1 or5, 3 xxx ⎧ ⎫ <> ⎨ ⎬ ⎩⎭ that is, () 1 ,or5, 3 ⎛⎞ ⎜⎟∞∞ ⎝⎠ .
74. |5x – 2| > 13 5213or5213 515511 11 3 5 xx xx xx >−<− ><− ><−
The solution set is 11 or3, 5 xxx ⎧ −⎫ <> ⎨ ⎬ ⎩⎭
that is, all x in () 11 ,or3, 5 ⎛⎞ ⎜⎟∞∞ ⎝⎠
75. 22 2 4 x + ≥ 2222 2or2 44 228228 26210 35 xx xx xx xx + + ≥≤− + ≥+≤− ≥≤− ≥≤−
The solution set is { } 5or3,xxx≤−≥ that is, ( ] [ ) ,5or3, ∞−∞
76. 33 1 9 x ≥ 3333 1or1 99 339339 31236 42 xx xx xx xx ≥≤− ≥−≤− ≥≤− ≥≤−
The solution set is { } 2or4,xxx≤−≥ or ( ] [ ) ,2or4,.∞−∞
77. 2 35 3 x −> 22 35or35 33 22 28 33 312 xx xx xx −>−<− −>−<− <−>
The solution set is {} 3or12,xxx<−> that is, ()() ,3or12, −∞−∞
78. 3 39 4 x −> 33 39or39 44 33 612 44 816 xx xx xx −>−<− −>−<− <−> {}8or16,thatisallinxxxx <−> ()() ,8or16,.−∞−∞
79. 3|x – 1| + 2 ≥ 8 3|x – 1| ≥ 6 |x – 1| ≥ 2 12or12 31 xx xx −≥−≤− ≥≤−
The solution set is {} 1or3,xxx≤≥ that is, ( ] [ ) ,1or3, −∞−∞
80. 52139 x +−≥ 52112 12 21 5 x x +≥ +≥ 1212 2121 55 717 2or2 55 717 1010 xx xx xx +≥+≤− ≥≤− ≥≤−
The solution set is 177 or. 1010 xxx ⎧⎫ ≤−≥ ⎨⎬
81. 244 x −≥− 244 22 42 242 26 x x x x ≤ −≤ ≤−≤ ≤≤
The solution set is {} 26.xx≤≤
82. 3727 x +≥− 3727 33 79 979 162 x x x x −+ ≤ +≤ ≤+≤ −≤≤
The solution set is {} 162.xx−≤≤
83. 4116 x −<− 4116 44 14 x x > −> 1414 3or5 35 xx xx xx >−<− >−<− <−>
The solution set is {} 3or5.xxx<−>
84. 256 x −<− 256 256 22 53 x x x −<− > −> 5353 2or8 28 xx xx xx >−<− >−−<− <>
The solution set is {} 2or8.xxx<>
85. 321 x ≤ 213213 24or22 21 xx xx xx ≥−≤− ≥≤− ≥≤−
The solution set is {} 1or2.xxx≤−≥
86. 947 x ≤+ 479or479 42416 24 4 1 2 x x x x xx x +≥ +≤− ≥ ≤− ≤− ≥ ≥
The solution set is 1 4or. 2 xxx ⎧⎫ ≤−≥ ⎨⎬ ⎩⎭
87. 54 x >− is equivalent to 45 x −< 545 91 91 111 91 19 x x x x x −<−< −<−< >> >>− −<<
The solution set is {} 19.xx−<<
88. 211 x >− is equivalent to 112 x −< . 2112 139 139 111 139 913 x x x x x −<−< −<−<− >> >> <<
The solution set is {} 913.xx<<
89. 123 x <− is equivalent to 231 x −> . 231231 3133 31or33 3333 11 3 x x x x x x xx −> −<− −>− −<− < > > <
The solution set is 1 or1. 3 xxx ⎧⎫ <> ⎨⎬ ⎩⎭
90. 42 x < is equivalent to 24 x −> or 2424 26 26 1111 26 xx xx xx xx >−<− >−<− <> <−>
The solution set is { } 2or6.xxx<−>
91. 63 122 77 x < −++ 816 2 77 x < −+ 681681 2or2 7777 7587 22 77 7587 1414 xx xx xx +>−+<− −>−<− <−>
The solution set is 7587 or, 1414 xxx ⎧ ⎫ <−> ⎨ ⎬ ⎩⎭ that is, 7587 ,or, 1414 ⎛⎞⎛⎞ ∞−∞ ⎜⎟⎜⎟ ⎝⎠⎝⎠
92. 117 1 33 x < −+ {} 411 33 114 Sinceistrueforall, 33 thesolutionsetisisanyrealnumber or(–,). x xx xx −<− −>− ∞∞
93. 439 3 x + −≥ 35 3 x ≥ 35or35 33 28 33 624 xx xx xx ≥−≤− ≥−≤− ≤−≥
The solution set is { } 6or24,xxx≤−≥ that is, ( ] [ ) ,6or24, ∞−∞
94. 211 2 x −−≤ 22 2 222 2 40 2 80 x x x x −≤ −≤−≤ −≤−≤ ≥≥
The solution set is {} 08xx≤≤ or [ ] 0,8.
95. 12yy ≤ 5 3 232 5 636 232 666(5) 6(3) 232 318215 3 xx xx xx xx x +≤+ ⎛⎞⎛⎞ +≤+ ⎜⎟⎜⎟ ⎝⎠⎝⎠ +≤+ +≤+ ≤−
The solution set is ( ] ,3.−∞−
96. 12yy > () 2 3(69)451 2 3(69)4351 3 2(69)12153 121812153 126153 39 39 33 3 xx xx xx xx xx x x x −+>+ ⎛⎞ ⎜⎟−+>+ ⎝⎠ −+>+ −+>+ −>+ −> < <−
The solution set is (),3.−∞−
97. 4 y ≥ 1(3)24 1324 24 6 xx xx x x −++≥ −−+≥ −≥ ≥
The solution set is [ ) 6,. ∞
98. 0 y ≤ 2113(2)0 211360 550 55 1 xx xx x x x ++≤ ++≤ ≤ ≤ ≤
The solution set is ( ] ,1.−∞
99. 8 y < 3428 346 6346 2310 2310 333 210 33 x x x x x x −+< −< <−< −<< << <<
The solution set is 210 ,.33 ⎛⎞ ⎜⎟ ⎝⎠
100. 9 y > 2519 258 x x +> > 258or258 21323 133 22 xx xx xx >−<− ><− ><
The solution set is 313 ,,. 22 ⎛⎞⎛⎞ −∞−∞ ⎜⎟⎜⎟ ⎝⎠⎝⎠ ∪
101. 4 y ≤ 724 2 23 2 23 2 x x x −+≤ +≤− +≥ 2323 or 22 4646 210 xx xx xx + ≥+≤− + ≥+≤− ≥≤−
The solution set is ( ] [ ) ,102,.−∞−∞ ∪
102. 6 y ≥ () () 8536 532 532 532 2532 551 551 555 1 1 5 x x x x x x x x −+≥ −+≥− −−+≤−− +≤ −≤+≤ −≤≤− ≤≤ −≤≤−
The solution set is 1 1,. 5 ⎡⎤ ⎢⎥ ⎣⎦
103. The graph’s height is below 5 on the interval ()1,9.
104. The graph’s height is at or above 5 on the interval ( ] [ ) ,19,.−∞−∞ ∪
105. Thesolutionsetis {} |12xx−≤< or [ ) 1,2
106. Thesolutionsetis {} |14xx<≤ or ( ] 1,4
107. Let x bethenumber. |43|5or|34|5 xx −≥−≥
39 3 x x x −≥ ≥ ≥ or 345 31 1 3 x x x −≤− ≤− ≤−
Thesolutionsetis 1 |or3 3 xxx ⎧⎫ ≤−≥ ⎨⎬ ⎩⎭ or [ ) 1 ,3, 3 ⎛⎤ −∞−∪∞ ⎜ ⎥ ⎝⎦
108. Let x bethenumber. |54|13or|45|13 xx≤−≤ 134513 8418 9 2 2 x x x ≤−≤ −≤≤ −≤≤
Thesolutionsetis9 |22xx ⎧ ⎫ −≤≤ ⎨ ⎬ ⎩⎭ or 2 2, 9 ⎡⎤ ⎢⎥ ⎣⎦ .
109. ( ) 0,4
110. [ ]0,5
111. passion ≤ intimacyorintimacy ≥ passion
112. commitment ≥ intimacyor intimacy ≤ commitment
113. passion<commitmentor commitment>passion
114. commitment>passionor passion<commitment
115. 9,after3years
116 . Afterapproximately152years
117. 3.125.863 3.137.2 12 x x x + > > >
Sincexisthenumberofyearsafter1994,we calculate1994+12=2006.63%ofvoterswilluse electronicsystemsafter2006.
118. 2.563.138.1 2.525 10 x x x +< −< > 1994+10=2004 Inyearsafter2004,fewerthan38.1%ofU.S.voters willusepunchcardsorlevermachines.
119. ( ) 28200.406040 28200.402440
280.40440 320.4044 80110 x x x x x ≤ +−≤ ≤+−≤ ≤−≤ ≤≤ ≤≤ Between80and110tenminutes,inclusive.
120.
273263 5995 F F ≤−≤ ≤≤
TherangeforFahrenheittemperaturesis59F ° to 95F ° ,inclusiveor [ ] 59F,95F. °°
121. 50 1.645 5 h ≥
The number of outcomes would be 59 or more, or 41 or less.
122. 50 + 0.20x < 20 + 0.50x 30 < 0.3x 100< x Basic Rental is a better deal when driving more than 100 miles per day.
123. 150.083.12 120.04
300 xx x x +<+ < < Plan A is a better deal when driving more than 300 miles a month.
124. 1800 + 0.03x < 200 + 0.08x 1600 < 0.05x 32000 < x A home assessment of greater than $32,000 would make the first bill a better deal.
125. 20.0880.05 0.03<6 200 xx x x +<+ < The credit union is a better deal when writing less than 200 checks.
126. 210,0000.40 1.610,000 1.610,000 1.61.6
6250 xx x x x >+ > > > More than 6250 tapes need to be sold a week to make a profit.
127. 300035.5 30002.5 1200 xx x x + < < <
More then 1200 packets of stationary need to be sold each week to make a profit.
128. 265652800 652535 39 x x x + ≤ ≤ ≤ 39 bags or fewer can be lifted safely.
129. 245953000 952755 29 x x x + ≤ ≤ ≤ 29 bags or less can be lifted safely.
130. Let x = the grade on the final exam. 86889284 90 6 86889284540 2350540 2190 95 xx xx x x x + ++++ ≥ +++++≥ +≥
You must receive at least a 95% to earn an A.
131. a. 86 + 88 + x 3 ≥ 90
You must get at least a 96. b. 86 + 88 + x 3
+ x < 240 x < 66
This will happen if you get a grade less than 66.
132. Let x =thenumberofhoursthemechanic worksonthecar. 22617534294 5134119 1.53.5 x x x ≤ +≤ ≤≤ ≤≤ Themanwillbeworkingonthejobatleast 1.5andatmost3.5hours.
133. Let x =thenumberoftimesthebridgeis crossedperthreemonthperiod
Thecostwiththe3-monthpassis 37.500.50. Cx =+
Thecostwiththe6-monthpassis630. C =
Becauseweneedtobuytwo3-monthpasses per6-monthpass,wemultiplythecostwiththe 3-monthpassby2.
145 a. The cost of Plan A is 4 + 0.10x; The cost of Plan B is 2 + 0.15x.
142.
143.
()
1530 15 x x x +< +< <
27.500.5030
Wealsomustconsiderthecostwithout purchasingapass.Weneedthiscosttobeless thanthecostwitha3-monthpass.
37.500.50
2.507.50 3 xx x x >+ > >
The3-monthpassisthebestdealwhenmaking morethan3butlessthan15crossingsper3monthperiod.
4 x <
3 x <−
144. Verifyexercise142.
c. 41 or more checks make Plan A better.
d. 40.1020.15 20.05 40 xx x x + <+ < > The solution set is {}40or(40,). xx >∞
146. a. False; |2x – 3| > –7 is true for any x because the absolute value is 0 or positive.
b. False; 2x > 6, x > 3 3.1 is a real number that satisfies the inequality.
c. True; 40 x > is not satisfied only when x = 4. Since 4 is rational, all irrational numbers satisfy the inequality.
d. False (c) is true.
147. Because x > y, y – x represents a negative number. When both sides are multiplied by (y – x) the inequality must be reversed.
148.a. |4|3 x <
b. |4|3 x ≥
Verifyexercise143.
149. Model 1: 577
5064 T T T −< −<−< <<
7577
Model 2: 5022
2872 T T T −< −<−< <<
225022
Model 1 describes a city with monthly temperature averages ranging from 50 degrees to 64 degrees Fahrenheit. Model 2 describes a city with monthly temperature averages ranging from 28 degrees to 72 degrees Fahrenheit.
Model 1 describes San Francisco and model 2 describes Albany.
Chapter 1 Review Exercises
1.
x = –3, y = –8
x = –2, y = –6
x = –1, y = –4
x = 0, y = –2
x = 1, y = 0
x = 2, y = 2
x = 3, y = 4
2.
x = –3, y = 6
x = –2, y = 1
x = –1, y = –2
x = 0, y = –3
x = 1, y = –2
x = 2, y = 1
x = 3, y = 6
3.
x = –3, y = –3
x = –2, y = –2
x = –1, y = –1
x = 0, y = 0
x = 1, y = 1
x = 2, y = 2
x = 3, y = 3 4. 3,1 2,0 1,1 0,2 1,1 2,0 3,1 xy xy xy xy xy xy xy
5. A portion of Cartesian coordinate plane with minimum x-value equal to –20, maximum
x-value equal to 40, x-scale equal to 10 and with minimum y-value equal to –5, maximum y-value equal to 5, and y-scale equal to 1.
6. x-intercept:–2;Thegraphintersectsthe x-axisat (–2,0).
y-intercept:2;Thegraphintersectsthe y-axisat (0,2).
7. x-intercepts:2,–2;Thegraphintersectsthe x-axisat(–2,0)and(2,0).
y-intercept:–4;Thegraphinterceptsthe y-axis at(0,–4).
8. x-intercept:5;Thegraphintersectsthe x-axisat (5,0).
y-intercept:None;Thegraphdoesnotintersect the y-axis.
9. Point A is(91,125).Thismeansthatin1991, 125,000acreswereusedforcultivation
10. Opiumcultivationwas150,000acresin1997.
11. Opiumcultivationwasataminimumin2001 whenapproximately25,000acreswereused.
12. Opiumcultivationwasatamaximumin2004 whenapproximately300,000acreswereused.
13. Opiumcultivationdidnotchangebetween1991 and1992.
14. Opiumcultivationincreasedatthegreatestrate between2001and2002.Theincreaseinacres usedforopiumcultivationinthistimeperiodwas approximately180,000–25,000=155,000acres.
15. 2x –5=7 2x =12 x =6
Thesolutionsetis{6}. Thisisaconditionalequation.
16. 5x +20=3x 2x =–20 x =–10
Thesolutionsetis{–10}. Thisisaconditionalequation.
17. 7(x –4)= x +2
7x –28= x +2 6x =30 x =5
Thesolutionsetis{5}. Thisisaconditionalequation.
18. 1–2(6– x)=3x +2 1–12+2x =3x +2
–11– x =2 –x =13 x =–13
Thesolutionsetis{–13}. Thisisaconditionalequation.
19. 2(4)3(5)22 2831522 5722 39 3 xxx xxx xx x x ++=− ++=− + =− =− =−
Thesolutionsetis{–3}. Thisisaconditionalequation.
20. 2x –4(5x +1)=3x +17 2x –20x –4=3x +17 –18x –4=3x +17 –21x =21 x =–1
Thesolutionsetis{–1}. Thisisaconditionalequation.
21. 755(3)2 xxx + =++ 755152 75715 515 xxx xx + =++ +=+ = Thesolutionsetis. ∅ Thisisaninconsistentequation.
22. 7132(25)323 xxx + =−++ 7132(25)323 713410323 713713 1313 xxx xxx xx + =−++ +=−++ +=+ = Thesolutionsetisallrealnumbers. Thisisanidentity.
23. 2 1 36 2(2)6 46 36 2 xx xx xx x x = + = + = + = =
The solution set is {2}. This is a conditional equation.
24. 11 21052 5125 36 2 xx xx x x −=+ −=+ = =
The solution set is {2}. This is a conditional equation.
25. 2 6 34
4(2)12(6)3 8723 1172 72 11 xx xx xx x x =− =− =− = =
The solution set is 72 11 ⎧⎫ ⎨⎬ ⎩⎭
This is a conditional equation.
26. 3 2 43 xx =− 1212(3) 12(2) 43 324412 736 36 7 xx xx x x =− =−+ = =
The solution set is 36 7 ⎧⎫ ⎨⎬ ⎩⎭
This is a conditional equation.
27. 31131 324 xx +− −=
4(31)6(13)3(1) 1247833 127433 1577 77 15 xx xx xx x x +−=− +−=− −=− = =
The solution set is 77 15 ⎧⎫ ⎨⎬ ⎩⎭
This is a conditional equation.
28. 914 42 9216 918 2 xx x x x = = = =
The solution set is {2}. This is a conditional equation.
29. 72 2 55 72(5)2 72102 232 5 x xx xx xx xx x + += + −=+ + −=+ =+ =
5 does not check and must be rejected. The solution set is the empty set, ∅ This is an inconsistent equation.
30. 2 112 111xxx −= −+− 112 11(1)(1) 1(1)2 112 22 xxxx xx xx −= ++− +−−= +−+= =
The solution set is all real numbers except –1 and 1. This is a conditional equation.
31. 2 518 326xxxx += +− + 518 32(3)(2) 5(3)(2)(3)(2)8(3)(2) 32(3)(2) 5(2)1(3)8 51038 678 615 15 6 5 2 xxxx xxxxxx xxxx xx xx x x x x += +−+− + −+−+− += +−+− −++= −++= −= = = =
The solution set is 5 . 2 ⎧⎫ ⎨⎬ ⎩⎭ This is a conditional equation.
32. 1 0 5 x = + 1 (5)(5)(0) 5 10 xx x +=+ + =
The solution set is the empty set, ∅ This is an inconsistent equation.
33. 2 4310 22 xxxx += ++ 4310 2(2) 4(2)3(2)10(2) 2(2) 43(2)10 43610 7610 74 4 7 xxxx xxxxxx xxxx xx xx x x x += ++ ⋅+⋅+⋅+ +=
The solution set is 4 7 ⎧⎫ ⎨⎬ ⎩⎭
This is a conditional equation.
34. 35(21)2(4)0 xx −+−−= 35(21)2(4)0 3105280 1260 126 6 12 1 2 xx xx x x x x −+−−= −−−+=
The solution set is 1 2 ⎧⎫ ⎨⎬ ⎩⎭
This is a conditional equation.
35. 2 21 10 323 x xxx + + −= ++− 22 21 10 3(3)(1) 21 1 3(3)(1) (2)(3)(1)1(3)(1) 3 (2)(1)1(3)(1) 2123 123 2 2 x xxx x xxx xxx xx x xxxx xxxx xx x x + +−= ++− + += ++− ++− + =+− + + −+=+− +−+=+−
The solution set is { } 2. This is a conditional equation.
36 Let x =thenumberofcaloriesinBurger King’sChickenCaesar. 125 x + =thenumberofcaloriesinTacoBell’s ExpressTacoSalad. 95 x + =thenumberofcaloriesinWendy’s MandarinChickenSalad. ( ) ( ) 125951705 32201705 31485 495 xxx x x x ++++= += = = 125495125620 x + =+= 9549595590 x + =+=
Thereare495caloriesintheChickenCaesar, 620caloriesintheExpressTacoSalad,and590 caloriesintheMandarinChickenSalad.
37. Let x = the number of years after 1970. 0.537.4 18.40.537.4 190.5 190.5 0.50.5 38 Px x x x x = −+ =−+ −=− = = If the trend continues only 18.4% of U.S. adults will smoke cigarettes 38 years after 1970, or 2008.
38. 15.055.07 10.02 500 xx x x + =+ = = Both plans cost the same at 500 minutes.
39. Let x =theoriginalpriceofthephone
480.20
480.80 60 xx x x =− = = Theoriginalpriceis$60.
40. Let x =theamountsoldtoearn$800inoneweek 8003000.05
5000.05 10,000 x x x =+ = = Salesmustbe$10,000inoneweektoearn$800.
41. Let x =theamountinvestedat4% Let y =theamountinvestedat7%
0.040.07555 xy xy += +=
9000
Multiplythefirstequationby–0.04andadd.
0.040.04360
0.040.07555
0.03195 6500 xy xy y y −−=− += = =
Back-substitute6500for y inoneoftheoriginal equationstofind x 9000
65009000 2500 xy x x += += = Therewas$2500investedat4%and$6500invested at7%.
42. Let x = the amount invested at 2% Let 8000 x −= the amount invested at 5%.
43. Let w =thewidthoftheplayingfield, Let3w –6=thelengthoftheplayingfield ( )() () 2length2width 3402362 3406122 340812 3528 44 P ww ww w w w =+ =−+ =−+ =− = = Thedimensionsare44yardsby126yards.
44. a. Let x =thenumberofyears(after2007). CollegeA’senrollment:14,1001500 x + CollegeB’senrollment:41,700800 x 14,100150041,700800 xx + =−
b. Checksomepointstodeterminethat 114,1001500 yx = + and 241,700800 yx = .Since 1232,100yy = = when12 x = ,thetwo collegeswillhavethesameenrollmentinthe year2007122019 + = .Thatyearthe enrollmentswillbe32,100students.
45. 2 vtgts + = 2 2 22 2 gtsvt gtsvt tt svt g t = = =
0.05(8000)0.0285
4000.050.0285
0.050.0285400
0.07315 0.07315
=
0.070.07 4500 80003500 xx xx xx x x x x −=+ −=+
−= $4500 was invested at 2% and $3500 was invested at 5%.
46. Tgrgvt = + ( ) () Tgrvt Tgrvt rvtrvt T g rvt T g rvt =+ + = ++ = + = +
47. Pr AP T = () ()Pr 1 1 AP PrTPr
A P rT = =− += += = +
48. (8 – 3i) – (17 – 7i) = 8 – 3i – 17 + 7i = –9 + 4i
49. 2 4(32)(4)(3)(4)(2) 128 128 iiiii ii i −=+− =− =−−
50. (7)(23) 727(3)()(2)()(3) 142123 1719 ii iiii ii i −+ =⋅++−+− =+−+ =+
51. 222 (34)323(4)(4) 92416 724 iii i i −=+⋅−+− =−− =−−
52. 22 (78)(78)784964113 ii +−=+=+=
53. 665 555 i iii =⋅ ++− 306 251 306 26 153 13 153 1313 i i i i = + = = =−
54. 343442 424242 iii iii +++ =⋅ −−+ 2 2 126168 164 12228 164 422 20 111 510 iii i i i i +++ = +− = + + = =+
55. 32183218 ii−−−=− 16292 4232 (43)2 2 ii ii ii i =⋅−⋅ =− =− =
56. 22 2 2 (2100)(2100) (210) 440(10) 440100 9640 i i ii i i −+−=−+ =−+ =−+ =−− =−−
57. 4848422 22 222 ii i +−++ ===+ 58. 2 2 2158 21580 (21)(8)0 xx xx xx + = + −= += 2x – 1 = 0 x + 8 = 0 1 2 x = or x = –8
The solution set is 1 ,8. 2 ⎧ ⎫ ⎨ ⎬ ⎩⎭ 59. 2 5200 5(4)0 xx xx + = + = 5x = 0 x + 4 = 0 x = 0 or x = –4
The solution set is {0, –4}.
60. 2 2 2 23125 2128 64 8 x x x x −= = = = ±
The solution set is {8, –8}.
61. 2 53 2 x + =− 2 2 2 8 2 16 16 4 x x x xi =− =− = ±− =±
62. 2 (3)10 x + =− 2 (3)10 310 310 x xi xi +=±− +=± =−±
63. 2 (34)18 x −= 2 (34)18 3432 3432 3432 33 432 3 x x x x x −=± −=± =± ± = ± =
64. 2 2 2 22 20 20 10100 2 20100(10) xx xxx + ⎛⎞ == ⎜⎟ ⎝⎠ ++=+
65.
66. 2 2 2 1227 12362736 (6)9 63 63 9,3 xx xx x x x x −=− −+=−+ −= −=± =± = The solution set is {9, 3}.
67.
68. 2 2 24 240 xx xx = + −−= 2 2(2)4(1)(4) 2(1) 2416 2 220 2 225 2 15 x x x x x ± = ±+ = ± = ± = =±
The solution set is {} 15,15. +−
69. 2 2 2190 2(2)4(1)(19) 2(1) 2476 2 272 2 262 2 132 xx x x x i x xi −+= ±−− = ±− = ±− = ± = =±
The solution set is {} 132,132. ii+−
70. 2 2 234 2430 xx xx = +−= 2 444(2)(3) 2(2) 41624 4 440 4 4210 4 210 2 x x x x x ±−− = −±+ = −± = −± = −± =
−+−− ⎪⎪ ⎨⎬ ⎪⎪ ⎩⎭
The solution set is 210210 ,.22
71. 2 2 4130 (4)4(1)(13) xx−+= = 16 – 52 = –36; 2 complex imaginary solutions
72. 2 2 2 923 9320 34(9)(2) xx xx =− +−= = 9 + 72 = 81; 2 unequal real solutions
73. 2 21150 xx−+= (2x – 1)(x – 5) = 0 2x – 1 = 0 x – 5 = 0 1 2 x = or x = 5
The solution set is 1 5,. 2 ⎧⎫
74. 2 2 (35)(3)5 359155 34200 xx xxx xx +−= +−−= −−= 2 4(4)4(3)(20) 2(3) 416240 6 4256 6 416 6 2012 , 66 10 ,2 3 x x x x x x ±−−− = ±+ = ± = ± = = =−
The solution set is 10 2,. 3 ⎧⎫ ⎨⎬ ⎩⎭
75. 2 3710 xx−+= 2 7(7)4(3)(1) 2(3) 74912 6 737 6 x x x ±−− = ±− = ± = The solution set is 737737 ,.66 ⎧⎫ +− ⎪⎪
76. 2 2 90 9 3 x x x −= = = ± The solution set is {–3, 3}.
77. 2 2 (3)250 (3)25 35 35 8,2 x x x x x −−= −= −=± = ± =
The solution set is {8, –2}.
78. 2 320 xx += 2 1(1)4(3)(2) 2(3) 1124 6 123 6 123 6 x x x i x ±−− = ±− = ±− = ± =
The solution set is 123123 ,.66 ii ⎧ ⎫ +− ⎪ ⎪ ⎨ ⎬ ⎪ ⎪ ⎩⎭
79. 2 3108 xx = 2 31080 (32)(4)0 xx xx −= + −= 32040 or 324 2 3 xx xx x + =−= = −= =−
The solution set is 2 ,4. 3 ⎧ ⎫ ⎨ ⎬ ⎩⎭
80. 2 (2)40 x + += 2 2 (2)4 (2)4 22 22 x x xi xi +=− + =±− +=± = −±
The solution set is { } 22,22. ii−+−−
81. 51 2 14 x x += + 2 2 54(1)(1)4(1)24(1) 14 20(1)(1)8(1) 20188 8110 xxx x x xxx xx xx ⋅+−⋅+ +=⋅+ +
(8)(8)4(1)(11)
The solution set is {} 45,45. +−
82. () 2 2 2 3 5883 196 Wtt t t = = =
Applythesquarerootproperty. 2196 196 14 t t t = =± =±
Thesolutionsare–14and14.Wedisregard–14, becausewecannothaveanegativetime measurement.Thefetuswillweigh588grams after14weeks.
83. 2 0.0350.657.6Pxx =−++ 2 2 2 00.0350.657.6 4 2 (0.65)(0.65)4(0.035)(7.6) 2(0.035) 278(rejected) xx bbac x a x xx =−++ −±− = −±−− =
If this trend continues, corporations will pay no taxes 27 years after 1985, or 2012.
84. () 2 2 1527 1527 02715 0(23)(5) 5 273 Alw ll ll ll ll l l = =− =− =−− = +− = −=
The length is 5 yards, the width is 3 yards.
85. Let x = height of building 2x = shadow height 222 22 2 2 (2)300 490,000 590,000 18,000 134.164 xx xx x x x += += = = ≈± Discard negative height. The building is approximately 134 meters high.
86. 42250xx = 42 2500 xx = ( ) 22 2250 xx =
x = 0
x = ± 5
The solution set is {–5, 0, 5}.
87. 32 21890 xxx−+= ( ) ( ) 2219210 xxx−−= ( ) ( ) 29210xx −=
x = ± 3, 1 2 x =
The solution set is 1 3,,3. 2 ⎧⎫ ⎨⎬ ⎩⎭
88. 2 2 2 2 2 233 233 2396 8120 812 8161216 (4)4 42 42 xx xx xxx xx xx xx x x x −+= −=− −=−+ −+= −=− −+=−+ −= −=± =+ x = 6, 2 The solution set is {2}.
89. 415 451 425101(1) 426101 30101 31 91 8 xx xx xxx xxx x x x x −++=
The solution set is {8}.
90. 3 4 3240 x −= () 3 4 3 4 4 34 3 43 324 8 8 16 x x x x = = ⎛⎞ = ⎜⎟ ⎝⎠ = The solution set is {16}.
91. 2 3 (7)25 x −= () 3 23 2 32 3 22 3 (7)25 75 75 7125 132 x x x x x ⎡⎤ ⎢⎥−= ⎣⎦ −= −= −= = The solution set is {132}.
92. x 4 5 x 2 + 4 = 0 Let t = x 2 t 2 5t + 4 = 0 22 41 or 41 21 tt xx xx = = = = = ±=±
The solution set is {–2, –1, 1, 2}.
93. x 1 / 2 + 3 x 1/ 4 10 = 0 Let t = x 1/ 4 23100 (5)(2)0 tt tt + −= + −= () () 11 44 4411 4444 52 or 52 52 62516 tt xx xx xx =−= =−=
625 does not check and must be rejected. The solution set is {16}.
94. 2x + 1 = 7 217or217 2628 38 xx xx xx + =+=− = =− = =−
The solution set is {–4, 3}.
95. 2 x 3 6 = 10 2 x 3 = 16 x 3 = 8
38or38 115 xx xx =−=− = =−
The solution set is {–5, 11}.
96. 3 x 4 / 3 5 x 2 / 3 + 2 = 0
Let 2 3 tx = 2 3520 (32)(1)0 tt tt −+= −−= () 2 3 23 2332 32 33 222 3 3 2 32010 or 321 2 31 2 31 21 3 2 3 22 33 223 333 26 9 tt tt tx x x x x x x x x −= −= = = = =
=± =±⋅⋅ =± The solution set is 2626 ,,1,199 ⎧⎫
97. 21xx −= 2 2 2 2 4(1) 44 440 (2)0 2 xx xx xx x x −= −= −+= −= =
The solution set is {2}.
98. 2530 x −−= 253or253 2822 41 xx xx xx −=−=− == == The solution set is {4, 1}.
99. 3229180xxx+−−= 2 2 (2)9(2)0 (2)(9)0 (2)(3)(3)0 xxx xx xxx +−+= +−= ++−=
The solution set is {–3, –2, 3}.
100. 820 xx−= () () 22 2 2 82 82 82 028 0(4)(2) xx xx xx xx xx −= −= −= =+− = +− 40or20 42 xx xx + =−= = −= –4 does not check. Thesolutionsetis { } 2.
101. 323260xxx + −−= ( ) ( ) () () 2 2 3230 320 xxx xx + −+= + −= 2 2 30or20 32 2 xx xx x −= += =−= =±
Thesolutionsetis {} 3,2,2.
102. 41120 x ++= 4112 13 x x +=− += 13or13 24 xx xx + =+=− = =−
Thesolutionsetis {}4,2.
103. Weneedtosolve4.30.33.4 x =+ for x () 2 2 4.30.33.4 0.90.3 3 3 9 x x x x x =+ = = = =
ThemodelindicatesthatthenumberofHIV infectionsinIndiawillreach4.3millionin2007 (9 x = yearsafter1998).
104. { }35xx≤<
105. { }2 xx >−
106. {} 0 xx ≤
107. Graph ( ]2,1 :
Graph [ ) 1,3 :
To find the intersection, take the portion of the number line that the two graphs have in common.
Numbers in both ( ]2,1 and [ ) 1,3 :
Thus, ( ]2,1 ∩ [ ) 1,3 [ ]1,1=−
108. Graph ( ]2,1 :
Graph [ ) 1,3 :
To find the union, take the portion of the number line representing the total collection of numbers in the two graphs.
Numbers in either ( ]2,1 or [ ) 1,3 or both:
Thus, ( ]2,1 ∪ [ ) 1,3 () 2,3=−
109. Graph [ ) 1,3 :
Graph () 0,4 :
To find the intersection, take the portion of the number line that the two graphs have in common.
Numbers in both [ ) 1,3 and () 0,4 :
Thus, [ ) 1,3 ∩ () 0,4 [ ) 1,3 = .
110. Graph [ ) 1,3 :
Graph () 0,4 :
To find the union, take the portion of the number line representing the total collection of numbers in the two graphs.
Numbers in either [ ) 1,3 or () 0,4 or both:
Thus, [ ) 1,3 ∪ () 0,4 () 0,4 =
111. –6x +3 ≤ 15 –6x ≤ 12 x ≥ 2 Thesolutionsetis [ ) 2,.
113. 3 1 342 xx −−> 3 12112 342 xx ⎛⎞⎛⎞ −−> ⎜⎟⎜⎟ ⎝⎠⎝⎠
4x –9–12>6x –21>2x 21 2 x −> Thesolutionsetis 21 ,. 2 ⎛⎞ −∞− ⎜⎟ ⎝⎠
114. 6x +5>–2(x –3)–25
6x +5>–2x +6–25
8x +5>–19 8x >–24 x >–3
Thesolutionsetis ()3,.−∞
115. 3(2x –1)–2(x –4) ≥ 7+2(3+4x) 6x –3–2x +8 ≥ 7+6+8x 4x +5 ≥ 8x +13 –4x ≥ 8 x ≤ –2
Thesolutionsetis [ ) ,2.−∞−
116. 5(2)3(4)220 xxx −−+≥− 510312220 222220 2220 xxx xx −−−≥− −≥− −≥−
Thesolutionsetis ∅
117. 7 < 2x + 3 ≤ 9 4 < 2x ≤ 6 2 < x ≤ 3 (2, 3]
The solution set is [ ) 2,3.
118. 2315 x +≤ –15 ≤ 2x + 3 ≤ 15 –18 ≤ 2x ≤ 12 –9 ≤ x ≤ 6
The solution set is [ ] 9,6.
119. 26 2 3 x + > 26 2 3 266 20 0 x x x x + > + > > > 26 –2 3 26–6 212 6 x x x x + < +< <− <−
The solution set is ()() ,6or0,.−∞−∞
120. 2576 251 x x + −≥− +≥
2x + 5 ≥ 1 or 2x + 5 ≤ –1 2x ≥ –4 2x ≤ –6 x ≥ –2 or x ≤ –3
The solution set is ( ] [ ) ,3or2,.−∞−−∞
121. 4257 x ++≤− 4212 23 x x +≤− +≥ 2323 or 15 xx xx + ≥+≤− ≥≤−
Thesolutionsetis ( ] [ ) ,51,.−∞−∞ ∪
122. 12yy > 103(21)81 106381 61381 1414 1414 1414 1 xx xx xx x x x −+>+ −−>+ −>+ −> < <−
Thesolutionsetis ( ) ,1.−∞−
Chapter 1
123. 3256 x −−≥− 259 259 11 259 9259 4214 27 x x x x x x −−≥− ≤ −≤ −≤−≤ −≤≤ −≤≤
Thesolutionsetis [ ] 2,7.
124. 0.202440 x +≤ 0.2016 0.2016 0.200.20 80 x x x ≤ ≤ ≤
A customer can drive no more than 80 miles.
125. 95799186 8090 5 x ++++ ≤< 40095799186450 400351450 4999 x x x ≤++++< ≤+< ≤<
A grade of at least 49% but less than 99% will result in a B.
126. 0.0759000 x ≥ 0.0759000 0.0750.075 120,000 x x ≥ ≥
The investment must be at least $120,000.
Chapter 1 Test
1. 7(2)4(1)21 xx−=+− 7144421 714417 33 1 xx xx x x −=+− −=− =− =−
The solution set is {–1}.
2. 103(21)810 xx −−+−−= 1063810 14140 1414 1 xx x x x −−−−−= −−= −= =−
The solution set is {–1}.
3. 2341 424 232(4)(1) 23281 239 6 xxx xxx xxx xx x −−+ =− =−−+ −=−−− −=− =−
The solution set is {–6}.
4. 248 33(3)(3)xxxx −= +−+
2(3)4(3)8 264128 2188 210 5 xx xx x x x + −−= +−+= −+= =− =
The solution set is {5}.
5. 2 2320 xx −= (2x + 1)(x – 2) = 0 210or20 1 or2 2 xx xx + =−= =−=
The solution set is 1 ,2. 2 ⎧ ⎫ ⎨ ⎬ ⎩⎭
6. () 2 3175 x −= 3175 3153 153 3 x x x −=± =± ± =
The solution set is 153153 ,.33 ⎧ ⎫ −+ ⎪ ⎪ ⎨ ⎬ ⎪ ⎪ ⎩⎭
7. () 2 3250 x + += () 2 325 325 35 x x xi +=− + =±− = −±
The solution set is { } 35,35. ii−+−−
8. x(x – 2) = 4 2240xx−−= ()()() 2 2 4 2 22414 2 bbac x a x −±− = ±−−− = 225 2 x ± = 15 x =±
The solution set is {} 15,15. −+
9. 2 485 xx=− 2 4850 xx−+= ()()() () 2 2 4 2 88445 24 bbac x a x −±− = ±−− = 816 8 x ±− = 84 8 i x ± = 1 1 2 xi =±
The solution set is 11 1,1. 22 ii ⎧⎫ +− ⎨⎬ ⎩⎭
10. 32440xxx−−+= ()() 24140 xxx−−−= () () 2140xx−−=
(x – 1)(x + 1)(x – 4) = 0 x =1 or x = –1 or x = 4
The solution set is {–1, 1, 4}.
11. 2 2 35 35 31025 11280 xx xx xxx xx −+= −=− =−+ −+= 2 11114(1)(28) 2(1) 11121112 2 119 2 113 2 7or4 x x x x xx ±− = ±− = ± = ± = == 4 does not check and must be rejected. The solution set is {7}.
12. 820 xx−= () () 22 2 2 82 82 82 028 0(4)(2) xx xx xx xx xx −= −= −= =+− = +− 40or20 42 xx xx + =−= = −= –4 does not check and must be rejected. The solution set is {2}.
13. 415xx + +−= 451 425101(1) 4251011 20101 21 41 5 xx xxx xxx x x x x +=−− + =−−+− + =−−+− −=−− =− =− = The solution set is {5}.
14. 3/2 5100 x = 3/2 510 x = 3/22 x = 2/3 2 x = 34 x =
The solution set is {}34.
15. 2/31/31/3 980let xxtx
16.
17. 347150
20. 3(x + 4) ≥ 5x – 12 3x + 12 ≥ 5x – 12 –2x ≥ –24 x ≤ 12
The solution set is (,12]. −∞
21. 13 6824 431218 821 21 8 xx xx x x +≤− +≤− −≤− ≥
The solution set is 21 8,. ⎡⎞ ∞ ⎟ ⎢ ⎣⎠
22. 25 36 3 x + −≤<
–9 ≤ 2x + 5 < 18 –14 ≤ 2x < 13 13 7 2 x −≤<
The solution set is 13 7,. 2 ⎡⎞ ⎟ ⎢ ⎣⎠
23. 323 x + ≥ 323or323 3135 15 33 xx xx xx + ≥+≤− ≥≤− ≥≤−
The solution set is 51 ,,. 33 ⎛⎤⎡⎞ −∞−∞ ⎜⎟ ⎥⎢ ⎝⎦⎣⎠
24. 37 y ≤≤ 3257 2212 16 x x x ≤−≤ ≤≤ ≤≤
The solution set is [ ] 1,6.
25. 1 y ≥ 2 1 4 x ≥ 22 or 11 44 2424 26 26 xx xx xx xx ≥≤− ≥−≤− ≥−≤− ≤−≥
The solution set is ( ] [ ) ,26,.−∞−∞ ∪
26. Graph [ ) 1,2 :
Graph ( ]0,5 :
To find the union, take the portion of the number line representing the total collection of numbers in the two graphs.
Numbers in either [ ) 1,2 or ( ]0,5 or both:
Thus, [ ) 1,2 ∪ ( ]0,5 [ ]1,5=−
27. Graph [ ) 1,2 :
Graph ( ]0,5 :
To find the intersection, take the portion of the number line that the two graphs have in common.
Numbers in both [ ) 1,2 and ( ]0,5 :
Thus, [ ) 1,2 ∩ ( ]0,5 () 0,2 = .
(67)(25)123014352
34. 2493642(7)3(8) 1424 38 ii ii i −+−=+ =+ =
35. 435751177 x + = 43602 14 x x = = The system’s income will be $1177 billion 14 years after 2004, or 2018.
36. 2 0.0747.4500Bxx =++
11770.0747.4500
(47.4)(47.4)4(0.07)(677) 2(0.07)
The system’s income will be $1177 billion 14 years after 2004, or 2018.
37. The formulas model the data quite well.
38. Let x = the number of books in 2002. Let 62 x + = the number of books in 2003. Let 190 x + = the number of books in 2004. ( ) ( ) ( ) 621902598 621902598 32522598 32346 782 62844 190972 xxx xxx x x x x x ++++= ++++= += = = += +=
The number of books in 2002, 2003, and 2004 were 782, 844, and 972 respectively.
39. 2970015050001100 24700950 26 xx x x + =+ = = In 26 years, the cost will be $33,600.
40. Let x = amount invested at 8% 10000 – x = amount invested at 10% () .08.110000940 .081000.1940 .0260 3000 100007000 xx xx x x x +−= +−= −=− = −= $3000 at 8%, $7000 at 10%
41. 2 2 2 24 48(24) 4824 02448 0224 0(6)(4) lw Alw ww ww ww ww ww =+ =
6040 64 242(4)412 ww ww w +=−= =−= +=+= width is 4 feet, length is 12 feet
42. 222 2 2 2426 576676 100 10 x x x x += += = =± The wire should be attached 10 feet up the pole.
43. Let x =theoriginalsellingprice 200.60 200.40 50 xx x x =− = = Theoriginalpriceis$50.
44. Let x =thenumberoflocalcalls
ThemonthlycostusingPlan A is25. A C = ThemonthlycostusingPlan B is130.06. B Cx = + ForPlan A tobebetterdeal,itmustcostlessthan Plan B
25130.06 120.06
200
200 ABCC x x x x < <+ < < > Plan A isabetterdealwhenmorethan200local callsaremadepermonth.