Section 2.1 Addition and Subtraction Properties of Equality
1. This is an expression. There is no equal sign.
2. This is an expression. There is no equal sign.
3. This is an equation. There is an equal sign.
4. This is an equation. There is an equal sign.
5. This is an expression. There is no equal sign.
6. This is an expression. There is no equal sign.
7. This is an equation. There is an equal sign.
8. This is an equation. There is an equal sign.
9. This is an equation. There is an equal sign.
10. This is an equation. There is an equal sign.
11. This is an expression. There is no equal sign.
12. This is an expression. There is no equal sign.
13. a. Using the given equation 10.75 Sh = replace h with 20 and solve for S 10.75 10.7520 215 Sh S S = =⋅ =
If Joe works 20 hours in a week, his salary will be $215.00.
b. Evaluate each side of the equation and see if the two sides are equal.
? 10.75 Original equation.
d. Evaluate each side of the equation and see if the two sides are equal.
? 10.75 Original equation.
Sh Sh = = =
43010.75(40) Substitute values for and . 430430 This statement is true.
Because the final statement is true, 40 h = and 430 S = is a solution to the equation 10.75 Sh = . If Joe works 40 hours, then he will earn a salary of $430 for the week.
14. a. Using the given equation 7 tg = replace g with 45 and solve for t
= It will take Danny 315 minutes, or 5.25 hours, to assemble 45 golf clubs.
b. Evaluate each side of the equation and see if the two sides are equal.
? ? 7 Original equation.
3407(50) Substitute values for and . 340350 The statement is not true.
340350 tg tg = = = ≠
376.2510.75(35) Substitute values for and . 376.25376.25 This statement is true.
Sh Sh = = =
Because the final statement is true, 35 h = and
376.25 S = is a solution to the equation 10.75 Sh = If Joe works 35 hours, then he will earn a salary of $376.25 for the week.
c. Evaluate each side of the equation and see if the two sides are equal.
? 10.75 Original equation.
Because the final results are not equal, 50 g = and 340 t = do not give a solution to the equation 7.tg = These values state that if Danny assembles 50 golf clubs, then he spent 340 minutes assembling those clubs. These values do not make the equation true.
c. Evaluate each side of the equation and see if the two sides are equal.
? 7 Original equation.
Sh Sh = = ≠
26010.75(25) Substitute values for and. 260268.75 The statement is not true.
Because the final results are not equal, 25 h = and 260 S = do not give a solution to the equation 10.75 Sh = . These values state that if Joe works 25 hours in a week, then his salary would be $260. These values do not make the equation true.
tg tg = = =
3297(47) Substitute values for and . 329329 This statement is true.
Because the final statement is true, 47 g = and 329 t = is a solution to the equation 7 tg = . If Danny assembles 47 golf clubs, then he spent 329 minutes assembling those clubs.
15. a. Evaluate each side of the equation and see if the two sides are equal.
? 60 Original equation.
24060(4) Substitute values for and . 240240 This statement is true.
dt dt = = =
Because the final statement is true, 4 t = and 240 d = is a solution to the equation 60 dt =
b. These values represent that 240 miles is the distance traveled when driving for 4 hours.
16. a. Evaluate each side of the equation and see if the two sides are equal.
? 65 Original equation.
b. These values represent that a circle with a radius of 26 has a circumference of 163.4 (rounded to one decimal place).
19. a. Evaluate each side of the equation and see if the two sides are equal.
? 15
Original equation.
FL FL = = =
450015(300) Substitute values for and. 45004500 This statement is true.
Because the final statement is true, 300 L = and 4500 F = is a solution to the equation 15 FL = .
b. These values represent that a loan of $300,000.00 would have an origination fee of $4,500.00.
195065(30) Substitute values for and . 19501950 This statement is true.
dt dt = = =
Because the final statement is true, 30 t = and 1950 d = is a solution to the equation 65 dt =
b. These values represent that 1950 miles is the distance traveled when driving for 30 hours.
17 a. Evaluate each side of the equation and see if the two sides are equal.
? 2 Original equation.
20. a. Evaluate each side of the equation and see if the two sides are equal.
? 20 Original equation.
FL FL = = =
900020(450) Substitute values for and. 90009000 This statement is true.
Because the final statement is true, 450 L = and 9000 F = is a solution to the equation 20 FL = .
b. These values represent that a loan of $450,000.00 would have an origination fee of $9,000.00.
37.72(3.14)(6) Substitute values for and . Substitue 3.14 for and round answer to one decimal place.
37.737.7 This statement is true.
π π = = =
Because the final statement is true, 6 r = and 37.7 C = is a solution to the equation 2 Cr π =
b. These values represent that a circle with a radius of 6 has a circumference of 37.7 (rounded to one decimal place).
18. a. Evaluate each side of the equation and see if the two sides are equal.
21. a. Evaluate each side of the equation and see if the two sides are equal.
? ? 10 Original equation.
200010(225) Substitute values for and. 20002250 This statement is not true. 20002250 FL FL = ≠
Because the final results are not equal, 225 L = and 2000 F = do not give a solution to the equation 10 FL =
? ? 2 Original equation.
163.42()(26) Substitute values for and . Use the button on your calculator and multiply w/o approximating.
163.4163.3628 Round to one decimal place.
163 Cr Cr π π π = = = .4163.4 This statement is true. =
Because the final statement is true, 26 r = and
163.4 C = is a solution to the equation 2 Cr π = .
b. These values state that a loan of $225,000.00 would have an origination fee of $2,000.00. These values do not make the equation true, so they do not make sense in this situation.
22. a. Evaluate each side of the equation and see if the two sides are equal.
25. Evaluate each side of the equation and see if the two sides are equal.
? ? 12.5 Original equation.
45005000 This statement is not true.
45005000 FL FL = ≠
450012.5(400) Substitute values for and .
Because the final results are not equal, 400 L = and 4500 F = do not give a solution to the equation
12.5 FL =
b. These values state that a loan of $400,000.00 would have an origination fee of $4,500.00. These values do not make the equation true, so they do not make sense in this situation.
23. a. Evaluate each side of the equation and see if the two sides are equal.
77.8 Original equation.
?
155.677.8(2) Substitute values for and . 155.6155.6 This statement is true.
Because the final statement is true, 2 f = and 155.6 S = is a solution to the equation 77.8 Sf = .
b. These values represent that it would require 155.6 cubic yards of sand to fill a volleyball court 2 feet deep.
24. a. Evaluate each side of the equation and see if the two sides are equal.
? ? 2830 Original equation.
2(11)830 Substitute 11. 22830 Simplify both sides
3030 This statement is true. x x += +== += =
Because the final statement is true, 11 x = is a solution to the equation 2830 x +=
26. Evaluate each side of the equation and see if the two sides are equal.
20077.8(3)
? ? Substitute values for and . 77.8 Original equation.
200233.4 This statement is not true.
200233.4
Sf = = = ≠
Because the final results are not equal, 3 f = and 200 S = do not give a solution to the equation 77.8 Sf =
b. These values state that it would require 200 cubic yards of sand to fill a volleyball court 3 feet deep. These values do not make the equation true, so they do not make sense in this situation.
? ? ? 42380 Original equation.
4(10)2380 Substitute 10. 402380 Simplify both sides
6380 This statement is not true.
6380 g g += +== += = ≠
Because the final results are not equal, 10 g = is not a solution to the equation 42380 g += .
27. Evaluate each side of the equation and see if the two sides are equal.
71510 Original equation.
? ? ?
7(8)1510(8) Substitute 8. 561580 Simplify both sides
7180 This statement is not true.
7180 mm m += +== += = ≠
Because the final results are not equal, 8 m = is not a solution to the equation 71510mm +=
28. Evaluate each side of the equation and see if the two sides are equal. ? ? ? 4725 Original equation.
4(6)725 Substitute 6.
24725 Simplify both sides
3125 This statement isnot true.
3125 y y −+=− −−+=−=− +=− =− ≠−
Because the final results are not equal, 6 y =− is not a solution to the equation 4725 y −+=−
29. Evaluate each side of the equation and see if the two sides are equal. ? 3 138 Original equation. 4 3 (28)138 Substitute 28. 4 x x −= −== ? 21138 Simplify both sides
88 This statement is true. −= = Because the final statement is true, 28 x = is a solution to the equation 3 138 4 x −= .
30. Evaluate each side of the equation and see if the two sides are equal.
? ? ? 51 7 Original equation. 33
7(4)51 Substitute 4. 33 201 7 Simplify both sides 33 21 7 3 77 This statement is true. y y =−+ =−−+=− =+ = =
Because the final statement is true, 4 y =− is a solution to the equation 51 7 33 y =−+
31. Evaluate each side of the equation and see if the two sides are equal. ? ? ? . 115 10 Original equation. 22 115 (5)10 Substitute 5. 22 515 10 Simplify both sides 22 52015 222 1515 This statement is true. 22 x x −=− −=−=
Because the final statement is true, 5 x = is a solution to the equation 115 10 22 x −=−
32. Evaluate each side of the equation and see if the two sides are equal. ? ? . 2 11 Original equation. 3 2 1(3)1 Substitute 3. 3
−=−+ −=−
121 Simplify both sides 11 This statement is true. x x −=+ −=−+=−
Because the final statement is true, 3 x =− is a solution to the equation 2 11 3 x −=+
33. Evaluate each side of the equation and see if the two sides are equal. ?
0.151.21.35 Original equation.
0.15(1)1.21.35 Substitute 1. x x += +==
? 0.151.21.35 Simplify both sides 1.351.35 This statement is true. += =
Because the final statement is true, 1 x = is a solution to the equation 0.151.21.35 x += .
34. Evaluate each side of the equation and see if the two sides are equal.
1.4614.4 Original equation.
1.4(5)614.4 Substitute 5.
1314.4 This statement is not true. 1314.4 x x −−=− −−=−=
7614.4 Simplify both sides
? ? ? .
Because the final results are not equal, 5 x = is not a solution to the equation 1.4614.4 x −−=− .
35. Evaluate each side of the equation and see if the two sides are equal.
0.53.21.7 Original equation.
0.5(3)3.21.7 Substitute 3. 1.53.21.7 Simplify both sides
4.71.7 This statement is not true. 4.71.7 t t += +== += = ≠
? ? ? .
Because the final results are not equal, 3 t = is not a solution to the equation 0.53.21.7 t +=
36. Evaluate each side of the equation and see if the two sides are equal. ? ? ? This statement is not true.
1.910.3 1.910.3 y y −+=− −+=−= −+=− −=−
6.1310.3 Original equation.
6.13(1.4)10.3 Substitute 1.4.
6.14.210.3 Simplify both sides
Because the final results are not equal, 1.4 y = is not a solution to the equation 6.1310.3 y −+=−
37. Evaluate each side of the equation and see if the two sides are equal. ? ?
3520 Original equation.
41. Evaluate each side of the equation and see if the two sides are equal. ? 25151040 Original equation.
25(3)1510(5)40 Substitute 3 and 5. 751550+40 Simplify both sides. 9090 This statement is true. mn mn +=+ +=+== + = =
Because the final statement is true, 3 m = and 5 n = is a solution to the equation 25151040 mn+=+
3(5)5(1)20 Substitute 5 and 1. 15520 Simplify both sides
2020 This statement is true.
xy xy += +=== += =
Because the final statement is true, 5 x = and 1 y = is a solution to the equation 3520 xy+=
38. Evaluate each side of the equation and see if the two sides are equal. ? 4930 Original equation.
4(3)9(2)30 Substitute 3 and 2. xy xy −= −−===− ? . 121830 Simplify both sides
3030 This statement is true. += = Because the final statement is true, 3 x = and 2 y =− is a solution to the equation 4930 xy−=
39. Evaluate each side of the equation and see if the two sides are equal.
42. Evaluate each side of the equation and see if the two sides are equal. ? ? ? 3749 Original equation.
3(0)74(2)9 Substitute 0 and 2. 0789 Simplify both sides. 71 This statement is not true.
Because the final results are not equal, 0 x = and 2 y =− is not a solution to the equation
3749 xy −+=+ .
43. Evaluate each side of the equation and see if the two sides are equal.
? ? ? 5315 Original equation.
1515 This statement is not true. 1515 xy xy
5(0)3(5)15 Substitute 0 and 5. 01515 Simplify both sides.
? ? 24 Original equation.
404 Simplify both sides
xy xy −+= −−+==−= += =
(4)2(0)4 Substitute 4 and 0.
44 This statement is true.
Because the final statement is true, 4 x =− and 0 y = is a solution to the equation 24xy −+=
40. Evaluate each side of the equation and see if the two sides are equal. ? ? 3412 Original equation.
3(0)4(3)12 Substitute 0 and 3. 01212 Simplify both sides. 1212 This statement is true. xy xy −+=− −+−−==− −=− = =
Because the final statement is true, 0 x = and 3 y =− is a solution to the equation 3412 xy −+=−
Because the final results are not equal, 0 x = and 5 y = is not a solution to the equation 5315 xy −+=− .
44. Evaluate each side of the equation and see if the two sides are equal.
? ? ? 4312 Original equation.
1312 This statement is not true. 1312 xy xy +=− −+−=−= −+−
4(4)3(1)12 Substitute 4 and 1. 16312 Simplify both sides.
Because the final results are not equal, 4 x =− and 1 y = is not a solution to the equation 4312 xy+=− .
45. a. Using the equation 220 slc++= substitute
75 s = for the weight of the cargo storage system and 100 l = for the weight of the luggage, then solve for c to find the number of pounds of other cargo you can store on the roof of the car.
220 Given equation.
(75)(100)220 Substitute 75 and 100. 175220 Simplify on left side.
175 175 Subtract 175 from both sides.
The solution is 45 c = which represents that an additional 45 pounds of other cargo could be stored on the roof of the car.
b. Using the equation 220 slc++= substitute
75 s = for the weight of the cargo storage system and 60 c = for the weight of the luggage, then solve for c to find the number of pounds of other cargo you can store on the roof of the car.
220 Given equation.
then solve for e to find how much he has left for entertainment.
50 Erbe e e e =++ =++ =+ =
Given equation.
(1000)(600)(350) Substitute value of variables. 1000950 Simplify on right side. 950 950 Subtract 950 from both sides.
The solution is 50 e = which represents that Alex has $50 left for entertainment.
47. Using the equation PRC =− substitute 47000 R = for the revenue, 39000 C = for costs, then solve for P to find the company’s profit.
Given equation.
(47000)(39000) Substitute value of variables. 8000 Simplify on right side.
The solution is 8000 P = which represents that the company has a monthly profit of $8,000.
48. Using the equation PRC =− substitute 180000 R = for the revenue, 165000 C = for costs, then solve for P to find the company’s profit.
Given equation.
(75)60220 Substitute 75 and 60. 135220 Simplify on left side.
135 135 Subtract 135 from both sides. 85 slc lsc l l
The solution is 85 l = which represents that 85 pounds of luggage could be stored on the roof of the car.
46. a. Using the equation Erbe =++ substitute 1200 E = for the amount Alex has this month for expenses, 600 r = for rent, and 230 b = for bills, then solve for e to find how much he has left for entertainment.
(180000)(165000) Substitute value of variables. 15000 Simplify on right side.
The solution is 15000 P = which represents that the company has monthly profit of $15,000.
49. Using the equation PRC =− substitute 38000 R = for the revenue, 41000 C = for costs, then solve for P to find the company’s profit.
PRC
Given equation.
(38000)(41000) Substitute value of variables. 3000 Simplify on right side.
The solution is 3000 P =− which represents that the company has monthly profit of $3,000 , representing that the company is operating at a loss.
370 Erbe
Given equation.
(1200)(600)(230) Substitute value of variables. 1200830 Simplify on right side. 830 830 Subtract 830 from both sides.
The solution is 370 e = which represents that Alex has $370 left for entertainment.
b. Using the equation Erbe =++ substitute
1000 E = for the amount Alex has this month for expenses, 600 r = for rent, and 350 b = for bills,
50. Using the equation PRC =− substitute 430000 R = for the revenue, 500000 C = for costs, then solve for P to find the company’s profit.
Given equation.
(430000)(500000) Substitute value of variables. 70000 Simplify on right side.
The solution is 70000 P =− which represents that the company has monthly profit of $70,000 , representing that the company is operating at a loss.
51. Using the equation PRC =− substitute 4000 P = for the revenue, 25000 C = for costs, then solve for R to find the company’s revenue.
29000 PRC R R =− =− + =
Given equation.
(4000)(25000) Substitute value of variables. 25000 +25000 Add 25000 to both sides.
The solution is 29000 R = which represents that the company would need to generate $29,000 in revenue to earn the desired profit.
52. Using the equation PRC =− substitute 11000 P = for the revenue, 156000 C = for costs, then solve for R to find the company’s revenue.
167000 PRC R R =− =− + =
Given equation.
(11000)(156000) Substitute value of variables. 156000 +156000 Add 156000 to both sides.
The solution is 167000 R = which represents that the company would need to generate $167,000 in revenue to earn the desired profit.
Exercises53-82:Answersaretobechecked.
53. ?
(25)530 Check the answer. 3030 The answer works. x x += = += =
530 Identify the variable term. 5 5 Subtract 5 from both sides.
25
54. ? 1228 Identify the variable term. 12 12 Subtract 5 from both sides. 16 (16)1228 Check the answer. 2828 The answer works. w w += = += = 55. ? 20.545 Identify the variable term.
(24.5)20.545 Check the answer. 4545 The answer works. x x += = += =
20.5 20.5 Subtract 20.5 from both sides. 24.5
4.27.3 Identify the variable term. 4.2 4.2 Subtract 4.2 from both sides. 11.5
4.2(11.5)7.3 Check the answer. 7.37.3 The answer works. t t +=− =− +−=− −=−
57. ? 145 Identify the variable term. +14 +14 Add 14 to both sides.
19
(19)145 Check the answer. 55 The answer works. m m −= = −= =
58. ? 822 Identify the variable term. +8 +8 Add 8 to both sides.
30 (30)822 Check the answer. 2222 The answer works. k k −= = −= =
59. ? 423 Identify the variable term. +4 +4 Add 4 to both sides.
27
4(27)23 Check the answer. 2323 The answer works. p p −+= = −+= = 60. ?
4(13)9 Check the answer. 99 The answer works. y y −+= = −+= =
49 Identify the variable term. +4 +4 Add 4 to both sides.
13
61. ? 116 Identify the variable term. +1 +1 Add 1 to both sides.
15
1(15)16 Check the answer. 1616 The answer works. k k −+=− =− −+−=− −=−
62. ? 39 Identify the variable term. +3 +3 Add 3 to both sides.
6
3(6)9 Check the answer. 99 The answer works. w w −+=− =− −+−=− −=−
53 Identify the variable term. +3 +3 Add 3 to both sides. 8 x x =−+
53(8) Check the answer. 55 The answer works. =−+
82 Identify the variable term. +2 +2 Add 2 to both sides.
82(6) Check the answer.
0.252.75 Identify the variable term. 0.25 0.25 Subtract 0.25 from both sides. 2.5
(2.5)0.252.75 Check the answer. 2.752.75 The answer works.
1.357.60 Identify the variable term. +1.35 +1.35 Add 1.35 to both sides. 8.95
1.35(8.95)7.60 Check the answer. 7.607.60 The answer works.
? 145 Check the answer.
55 The answer works.
73. ? 514 Identify the variable term. 5 5 Subtract 5 from both sides. 9
5(9)14 Check the answer. 1414 The answer works. x x
= 74. ? 75 Identify the variable term. 7 7 Subtract 7 from both sides. 2
Identify the variable term. 57 222 Subtract from both sides. 555 32 Find LCD35. 75 1514 Rewrite over LCD. 3535 1 35 x
7(2)5 Check the answer.
55 The answer works. z
(8.25)1.756.5 Check the answer. 6.56.5 The answer works. d d −= = −= = 76.
1.756.5 Identify the variable term.
+1.75 +1.75 Add 1.75 to both sides. 8.25
2.812.3
+2.8 +2.8 Add 2.8 to both sides. 15.1 k k −+= =
Identify the variable term.
2.8(15.1)12.3 Check the answer. 12.312.3 The answer works. −+= = 77.
1.3(3.4)4.7 Check the answer. 4.74.7 The answer works. x x −+=− =− −+−=−
Identify the variable term. +1.3 +1.3 Add 1.3 to both sides. 3.4
78. ?
13.57.8(21.3) Check the answer. 13.513.5 The answer works. y y =−+ = =−+ =
13.57.8 Identify the variable term.
+7.8 +7.8 Add 7.8 to both sides. 21.3
? ? ? 213 Check the answer. 5357 1413 Find LCD35. 35357 153 Reduce on left side. 357 33 The answer works. 77
Identify the variable term. 23 111 Subtract from both sides. 222 21 Find LCD6.
112 Check the answer.
312 Find LCD6. 663 42 Reduce on left side. 63 22 The answer works. 33
the variable term
both sides
LCD10.
the answer.
Find LCD10. The answer works. 413 5210 853 101010 33 1010 =
Identify the variable term. 1122 666 Subtract from both sides. 111111 16 Find LCD22. 2211 112 Rewrite over LCD. 2222 11 Reduce. 22 1 2 611 11222
? Check the answer. 12111 Find LCD22. 222222 11 The answer works. 2222 −== = 83.
Identify the variable term to isolate. +C Add to both sides. or PRC CC CPRRCP =− + +==+ 84.
Identify the variable term to isolate. +D Add to both sides. or SPD DD DSPPDS =− + +==+ 85.
220 slc sss lcs ccc lcs ++= +=− =−− 86.
220 Identify the variable term to isolate. Subtract from both sides.
220 Subtract from both sides.
90 AB BBB AB AB ++= +=− =−
90180 Identify the variable term to isolate. Subtract from both sides.
90180 90 90 Subtract 90 from both sides.
88.
360 ABCD AAA BCDA BBB CDAB D +++= ++=− +=−− Subtract from both sides.
360 Identify the variable term to isolate. Subtract from both sides.
360 Subtract from both sides.
360 DD CABD =−−−
89.
45 or 45 TM TMMT =+ −==−
45 Identify the variable term to isolate. 45 45 Subtract 45 from both sides.
90.
3
3 or 3 gfh fff gfhhgf =+ −==− 91.
Identify the variable term to isolate. 3 3 Subtract 3 from both sides.
4 Identify the variable term to isolate. Subtract from both sides. 4 xy xxx yx +=− =−−
92.
Identify the variable term to isolate. Subtract from both sides. or RCM MMM RMCCRM =+ −==− 87.
27 Identify the variable term to isolate. 2 2 Subtract from both sides. 27 xy xxx yx += =−+ 93.
330 or 330 DFG FFF DFG DFGGDF +=+− −+=− + −+==−+
10320 Identify variable term to isolate. 3 3 Subtract 3 from both sides.
31020 20 +20 Add 20 to both sides.
5223 or 5223 abc ccc acbbac −=++ −−==−− 95.
58215 Identify variable term to isolate.
215 215 Subtract 2 and 15 from both sides.
3 or 3 tuw uuu tuwwtu =−+ ++ +==+
3 Identify variable term to isolate. 3 3 Add 3 to both sides.
96.
420 or 420 yxz xxx xyxxxy =−+− ++++ ++==++ 97.
420 Identify variable term to isolate. 420 4 20 Add 4 and 20 to both sides.
Identify variable term to isolate. Subtract and from both sides. or Pabc acacac PacbbPac =++ −−==−−
98.
Identify variable term to isolate. Subtract , and from both sides. or Pabcd abcabc abc PabcddPabc =+++ −−−==−−− 99.
35 or 35 yx yxxy =− ++ +==+
35 Identify variable term to isolate. 5 5 Add 5 to both sides.
49 or 49 yx yxxy −=+ −−==−−
49 Identify variable term to isolate. 9 9 Subtract 9 from both sides.
101. The statement has an equal sign, therefore it is an equation.
239 Identify variable term to isolate. 9 9 Subtract 9 from both sides. 14 14 x x x =+ = =
102. The statement has an equal sign, therefore it is an equation.
1445 Identify variable term to isolate. 14 14 Subtract 14 from both sides.
31 m m += =
103. The statement does not have an equal sign, therefore it is an expression. Simplify:
573 Combine like terms
27 xx x +− +
104. The statement does not have an equal sign, therefore it is an expression. Simplify:
83142 Combine like terms 714 hhh h −++ +
105. The statement does not have an equal sign, therefore it is an expression. Simplify: 22 2 206310 Combine like terms 17610 aaa aa +−+ ++
106. The statement does not have an equal sign, therefore it is an expression. Simplify: 22 2 11857 Combine like terms 1678 Order in conventional form nnn nn −+−+ −++
107. The statement has an equal sign, therefore it is an equation.
1410 Identify variable term to isolate. 14 14 Add 14 to both sides. 4 c c −+=− ++ =
108. The statement has an equal sign, therefore it is an equation.
164 Identify variable term to isolate. 16 16 Add 16 to both sides. 12 y y −+=− ++ =
109. The statement has an equal sign, therefore it is an equation.
1420 Identify variable term to isolate. 14 14 Add 14 to both sides. 34 p p −= ++ =
110. The statement has an equal sign, therefore it is an equation.
68 r r −= ++ =
2741 Identify variable term to isolate. 27 27 Add 27 to both sides.
111. The statement does not have an equal sign, therefore it is an expression. Simplify:
6 p p −+ +
1420 Combine like terms.
112. The statement does not have an equal sign, therefore it is an expression. Simplify:
14 r r −+ +
2741 Combine like terms.
113. P represents the population of South America, in millions, t years since 1960. Therefore, when t = 35, we have 1960351995 += P = 321.05 millions means 321.051000000321050000 ⋅= . These values mean that in the year 1995, the population of South America was 321,050,000.
114. P represents the population of South America, in millions, t years since 1960. Therefore, when t = 47, we have 1960472007 += P = 380.93 millions means 380.931000000380930000 ⋅= . These values mean that in the year 2007, the population of South America was 380,930,000.
115. P represents the population of South America, in millions, t years since 1960. Therefore, when t = 58, we have 1960582018 += . P = 435.82 millions means 435.821000000435820000 ⋅= . These values mean that in the year 2018, the population of South America was 435,820,000.
116. P represents the population of South America, in millions, t years since 1960. Therefore, when t = 25, we have 1960251985 += P = 271.15 millions
means 271.151000000271150000 ⋅= . These values mean that in the year 1985, the population of South America was 271,150,000.
117. a. To find the number of years since 1985, subtract the year from 1960: 1985196025 −=
The year 1985 corresponds to 25 years since 1960.
b. The variable t represents the number of years since 1960. Therefore, since 1985 is 25 years since 1960, t = 25.
c. The value of P represents the population of South America, in millions. To determine the population in the year 1985, substitute t = 25 into the equation
4.99146.4Pt=+
124.75146.4 271.15 P P P =+ =+ =
() 4.9925146.4
The population of South America in the year 1985 was about 271.15 million.
118. a. To find the number of years since 1998, subtract the year from 1960: 1998196038 −=
The year 1998 corresponds to 38 years since 1960. b. The variable t represents the number of years since 1960. Therefore, since 1998 is 38 years since 1960, t = 38.
c. The value of P represents the population of South America, in millions. To determine the population in the year 1998, substitute t = 38 into the equation 4.99146.4Pt=+ .
189.62146.4 336.02 P P P =+ =+ =
() 4.9938146.4
The population of South America in the year 1998 was about 336.02 million.
119. a. To find the number of years since 2016, subtract the year from 1960: 2016196056 −=
The year 2016 corresponds to 56 years since 1960.
b. The variable t represents the number of years since 1960. Therefore, since 2016 is 56 years since 1960, t = 56.
c. The value of P represents the population of South America, in millions. To determine the population in the year 2016, substitute t = 56 into the equation
4.99146.4Pt=+
The population of South America in the year 2016 was about 425.84 million.
120. a. To find the number of years since 2012, subtract the year from 1960: 2012196052 −=
The year 2012 corresponds to 52years since 1960.
b. The variable t represents the number of years since 1960. Therefore, since 2012 is 52 years since 1960, t = 52.
c. The value of P represents the population of South America, in millions. To determine the population in the year 2012, substitute t = 52 into the equation 4.99146.4Pt=+ .
The population of South America in the year 2012 was about 405.88 million.
Section 2.2 Multiplication and Division Properties of Equality
Exercises1-22:Answersaretobechecked. Exercises 1-5 are completed as examples.
1. Divide both sides of the equation by 3.
381 Variable term is isolated. 381 Divide both sides by 3.
3(27)81 Check the answer. 8181 The answer works.
2. Divide both sides of the equation by 4.
448 Variable term is isolated.
Divide both sides by 4.
4(12)48 Check the answer. 4848 The answer works.
3. Divide both sides of the equation by 10 1055 Variable term is isolated.
Divide both sides by10.
4. Divide both sides of the equation by 6 . 627 Variable term is isolated. 627 Divide both sides by6. 66 6 y y
The answer works.
5. Divide both sides of the equation by 7. 791 Variable term is isolated. 791 Divide both sides by 7.
7(13)91 Check the answer. 9191 The answer works.
6. Divide both sides of the equation by 14. 1484 Variable term is isolated. 1484 Divide both sides by 14. 1414
7. Divide both sides of the equation by 1 .
Variable term is isolated.
Divide both sides by1.
8. Divide both sides of the equation by 1 . 17 Variable term is isolated. 171 Divide both sides by1.
9. Divide both sides of the equation by 2.8.
422.8 Variable term is isolated.
422.8 Divide both sides by 2.8.
10. Divide both sides of the equation by 3.7.
3.731.45 Variable term is isolated.
3.731.45 Divide both sides by 3.7.
3.73.7 3.7
11. Divide both sides of the equation by 6.1.
15.256.1 Variable term is isolated.
15.256.1 Divide both sides by 6.1. 6.16.1 6.1 2.5 n n
12. Divide both sides of the equation by 5.4.
49.685.4 Variable term is isolated.
49.685.4 Divide both sides by 5.4.
13 Divide both sides of the equation by 6 642 Variable term is isolated.
14 Divide both sides of the equation by 4
18. Multiply both sides by the reciprocal.
21. Multiply both sides of the equation by 3.5 .
22. Multiply both sides of the equation by 7 ()
23. a.
17 Given equation.
17(80) Substitute 80. 1360 Multiply and solve for . Sh Sh SS = == = If Victoria works 80 hours, her salary will be $1360. b.
17 Given equation.
204017 Substitute 2040 and solve for 204017 Divide both sides by 17. 1717
17 120 . Sh hSh h = == = = 17 h
120 h =
Victoria has to work 120 hours to make $2040 a month. c.
17 Given equation.
70017 Substitute 700 and solve for 70017
Divide both sides by 17. 1717
17 41.18 Sh hSh h = == = ≈ 17 h 41.18 h ≈
Victoria has to work approximately 41.18 hours to earn $700.
24 a. and
60 Given equation.
60(12) Substitute 12. 720 Multiply solve for . Dt Dt DD = == =
Emily will travel 720 miles in a day if she drives for 12 hours. b.
60 Given equation.
45060 Substitute 450. 45060
Divide both sides by 60. 6060
60 7.5 Dt tD t = == = = 60 t 7.5 t =
Emily will have to drive for 7.5 hours to travel 450 miles a day.
c.
60 Given equation.
245060 Substitute 2450. 245060 Divide both sides by 60. 6060
60 40.83 Dt tD t = == = = 60 t 40.83 t ≈
It will take Emily approximately 40.83 hours to drive the approximate 2450 miles.
25. a. and
0.02 Given equation.
0.02(250000) Substitute 250000. 5000 Multiply solve for . FL FL FF = == =
The loan fees will be $5000 for a loan of $250,000.
b.
0.02 Given equation.
40000.02 Substitute 4000.
40000.02
Divide both sides by 0.02. 0.020.02
0.02
200000 FL LF L = == = = 0.02 L
200000 L =
Pablo can suggest a maximum loan amount of approximately $200,000 to keep the fees down to only $4000.
26 a.
18 Given equation. 2160018 Substitute 21600. 2160018
Divide both sides by 18. 1818
LEGO produced 1200 million (or 1.2 billion) bricks if there are 21,600 bricks that do not meet the company’s quality standards.
28 a.
0.025 Given equation.
0.025(300000) Substitute 300000. 7500 Multiply and solve for .
Rachel will pay $7500 in fees if her home loan is for $300,000.
b.
429 Given equation.
429(1) Substitute 1. 429 Multiply and solve for . Bm Bm BB = ==
The machine can make 429 bricks in 1 minute.
0.025 Given equation.
50000.025 Substitute 5000.
50000.025
0.0250.025
0.025
200000 FL LF L = == = = 0.025 L
200000 L =
Divide both sides by 0.025.
The most Rachel can borrow through this broker is $200,000.
27 a.
18 Given equation.
18(500) Substitute 500. 9000 Multiply and solve for .
There will be 9000 bricks that do not meet the quality standards.
b.
18 Given equation.
18(40.8) Substitute 40.8.
734.4 Multiply and solve for. Db Db DD = == =
Rounding to the nearest whole number, approximately 734 bricks will not meet the quality standards of the 40.8 million bricks produced in a day.
b. Convert hours to minutes (1 hour = 60 minutes) then substitute 60 m = into the equation 429 Bm = and solve for B
429 Given equation.
429(60) Substitute 60. 25740 Multiply and solve for . Bm Bm BB = == =
The machine can make 25,740 bricks in 1 hour.
c.
429 34965 Bm mB m
429 Given equation.
15000000429 Substitute 15,000,000. 15000000429 Divide both sides by 429. 429429
== = ≈ 429 m 34965 m ≈
It will take this machine approximately 34,965 minutes, which is approximately 24.28 days.
29. a.
300 Substitute 300. Solve for . 60 5 P B BPB B = == =
Given equation.
60
A total of 5 buses would be needed to transport 300 people.
Multiply both sides by 60 Given equation.
Public transportation can handle 21,000 people at a time with 350 city buses.
The smallest width would be 52 inches. d. First find the maximum capacity per seat.
Substitute 39 w = into the equation 13 Cw = and solve for C. We know from part a that the solution is 3 C = . Now multiply this by the total number of seats in the bus, which is 24, to find the total maximum capacity for the bus.
The city can transport 36,000 people at a time with 600 school and city buses. 30. a.
The total maximum capacity is 72 people for a bus with 24 seats in it that are 39 inches wide. 31. a. 2018 is 8 years since 2010, so t = 6.
75270.9712058.6 Given equation.
75270.98712058.6 Substitute 8.
602167.2712058.6 Multiply and solve for . 1314225.8 Mt Mt MM M
The total amount of mortgage debt for multifamily homes in the United States is $1,314,225.8 million. b.
75270.9712058.6 Given equation. 1464767.675270.9712058.6 Substitute 1464767.6. 712058.6 712058.6 Isolate variable term.
75270975270.9
75270975270.9
75270.9 Mt t M t
=
= 75270.9 Divide both sides by75270.9. 10 t t =
In the year 2018 the total amount of mortgage debt for multifamily homes in the United States was $1,464,767.6 million.
c.
2.990.99 Given equation.
2.990.99(5) Substitute 5.
2.994.95 Multiply & solve for . 7.94 Cn Cn CC C =+ =+= =+ =
The cost to ship 5 items via USPS is $7.94. b.
2.990.99 Given equation.
5.962.990.99 Substitute 5.96. 2.99 2.99 Isolate variable term.
2.970.99
2.970.99 0.99 Cn nC n =+ =+= = = 0.99 n Divide both sides by 0.99.
3 n =
Three items were shipped.
c.
2.990.99 Given equation.
12.892.990.99 Substitute 12.89.
2.99 2.99 Isolate variable term.
9.90.99
9.90.99 0.99 Cn nC n =+ =+= = = 0.99 n Divide both sides by 0.99.
10 n =
Ten items were shipped.
33 a.
0.0820 Given equation.
ff
0.08(17900)20 Substitute 17900. 143220 Multiply & solve for. 1452
The fee will be $1,452.
b.
0.0820 Given equation.
1040.0820 Substitute 104. 20 20 Isolate variable term.
840.08 840.08 0.08 fp pf p =+ =+= = = 0.08 p Divide both sides by 0.08. 1050 p =
The final price of the log splitter was $1,050.
0.0820 Given equation.
11000.0820 Substitute 1100. 20 20 Isolate variable term.
10800.08
10800.08 0.08 fp pf p =+ =+= = = 0.08 p Divide both sides by 0.08. 60000 p =
The final price of the tractor was $60,000.
34 a.
4.950.65 Given equation.
4.950.65(20) Substitute 20. 4.9513 Multiply and solve for. 17.95 Cn Cn CC C =+
The commission charged will be $17.95. b.
4.950.65 Given equation.
102.454.950.65 Substitute 102.45. 4.95 4.95 Isolate variable term.
97.50.65
97.50.65 0.65 Cn nC n =+ =+= = = 0.65 n Divide both sides by 0.65.
150 n =
The customer ordered 150 contracts.
c.
4.950.65 Given equation.
66.704.950.65 Substitute 66.70. 4.95 4.95 Isolate variable term.
61.750.65
61.750.65 0.65 Cn nC n =+ =+= = = 0.65 n Divide both sides by 0.65.
95 n =
The customer ordered 95 contracts.
35 a.
0.300.029 Given equation.
0.300.029(100) Substitute 100. 0.302.9 Multiply and solve for . 3.2 Fp Fp FF F =+ =+= =+ =
The fee for a $100 transaction is $3.20.
b.
0.300.029 Given equation.
10.740.300.029 Substitute 10.74. 0.30 0.30 Isolate variable term.
10.440.029
10.440.029
0.029 Fp pF p =+ =+= = = 0.029 p Divide both sides by 0.029. 360 p =
The total transaction was $360 if the fee charged was $10.74.
c.
0.300.029 Given equation.
43.800.300.029 Substitute 43.80. 0.30 0.30 Isolate variable term.
43.50.029
43.50.029
0.029 Fp pF p =+ =+= = = 0.029 p Divide both sides by 0.029. 1500 p =
The total transaction was $1500.
36. a.
15000.015 Given equation.
15000.015(325000) Substitute 325000. 15004875 Multiply and solve for . 6375 Cp Cp CC C
The commission would be $6,375.
b.
37. Step 1 Reason: Isolate the variable term by subtracting 15 from both sides.
Step 2 Algebraic Step:
38. Step 1 Algebraic Step: 6.56493.25 1515 h −−=−
Step 1 Reason: Isolate the variable by using the division property of equality.
Exercises39-64:Answersaretobechecked.
Exercises 39-42 are completed as examples. 39.
Identify variable term to isolate. Subtract 20 from both sides. 52060 20 20 540 5 x x += = 5 x 40 Divide both sides by 5. 5 8 x = = ? ? Check the answer. The answer works 5(8)2060 402060 6060 . += += = 40.
15000.015 Given equation.
562515000.015 Substitute 5625. 1500 1500 Isolate variable term.
41250.015
41250.015
0.015 Cp pC p =+ =+= = = 0.015 p
Divide both sides by 0.015.
275000 p =
The sales price of the home was $275,000.
c.
15000.015 Given equation.
787515000.015 Substitute 7875. 1500 1500 Isolate variable term.
63750.015
63750.015
0.015 Cp pC p =+ =+= = = 0.015 p Divide both sides by 0.015.
425000 p =
The sales price of the home was $425,000.
834154 34 34 8120 8 m m += = 8 m 120 Divide both sides by 8. 8 15 m = = ? ? Check the answer.
Identify variable term to isolate. Subtract 34 from both sides.
The answer works 8(15)34154 12034154 154154 . += += = 41.
Identify variable term to isolate. Subtract 50 from both sides. 35085 50 50 3135 3 p p −+=− −=− 3 p Divide both sides by3 135 3 45 p = =
42.
answer
Identify variable term to isolate. Subtract 25 from both sides. 22545 25 25
43.
44.
Identify variable term to isolate. Subtract 20 from both sides. 2.32026.9 20 20 2.36.9 2.3 t t
Identify variable term to isolate. Subtract 50 from both sides
Identify variable term to isolate. Subtract 17 from both sides 20.117 8 17 17 3.1 8 3.18
Identify variable term to isolate. Add 14 to both sides 8.7514 2 14 14 5.25 2 5.252 2 b b b −=−− ++ = = 2 () Mulitply both sides by 2 10.5 b = 47.
Identify variable term to isolate. Add 30 to both sides 481230 30 30 7812 7812 12 m m =− ++ = = 12 m Divide both sides by 12 6.5 m = 48.
Identify variable term to isolate. Add 56 to both sides. 20856 56 56 368 368 8 d d −=− ++ = = 8 d Divide both sides by 8 4.5 d =
251.88.614.8 14.8 14.8 266.68.6 266.68.6 8.6 h h −=−+ −=− = 8.6 h Divide both sides by8.6. 31 h = 50.
Identify variable term to isolate. Subtract 14.8 from both sides.
Identify variable term to isolate. Add 12.3 to both sides
46.82.312.3 12.3 12.3 34.52.3 34.52.3 2.3 y y −=−− ++ −=− = 2.3 y Divide both sides by 2.3. 15 y =
134.04.716.5 16.5 16.5 117.54.7 117.54.7 4.7 x x −=−− ++ −=− = 4.7 x Divide both sides by 4.7. 25 x =
Identify variable term to isolate. Add 16.5 to both sides
53.
Identify
term
59.
Identify variable term to isolate. Subtract 3.8 from both sides
54.
55.
Identify variable term to isolate. Add 5 to both sides
Identify variable term to isolate. Subtract 8 from both sides.
56.
Identify variable term to isolate. Add 56 to both sides.
57.
Identify
term
both
Identify variable term to isolate. Subtract 6 from both sides.
Identify variable term to isolate. Subtract 14 from both sides.
Identify variable term to isolate. Subtract 5 from both sides.
Identify variable term to isolate. Add 6 to both sides.
Identify variable term to isolate. Subtract 16 from both sides. 51632 16 16
Identify variable term to isolate.
Multiply by reciprocal of coefficient. 1 Subtract from both sides.
Identify variable term to isolate.
65 a. Let n = the number of items to be shipped. Let C = the total cost in dollars for shipping.
6.991.99 Cn =+
b. Substitute 5 n = into the equation from part a, then solve for C.
6.991.99 Given equation.
6.991.99(5) Substitute 5. 6.999.95 Simplify. 16.94
It costs $16.94 to ship 5 items UPS Ground.
c. Substitute 12.96 C = into the equation from part a, then solve for n
6.991.99
12.966.991.99 6.99 6.99
5.971.99
5.971.99 1.99 CnC n n = =+ =+ = = 1.99 n 3 n =
Substitute 12.96. Isolate variable term. Subtract 6.99 from both sides. Divide both sides by 1.99.
A total of 3 items can be shipped UPS Ground for $12.96.
66. a. Let n = the number of option contracts. Let C = the commissions charged in dollars. 71 Cn =+
b. Substitute 20 n = into the equation from part a, then solve for C
Given equation.
Substitute 20. Simplify.
A Silver-level customer will be charged $27.00 for an order of 20 option contracts.
c. Substitute 92.00 C = into the equation from part a, then solve for n
Substitute 92.00. Isolate variable term. Subtract 7 from both sides.
This customer has ordered 85 option contracts.
67 a. Let c = the number of cranks to be produced. Let T = the total time in minutes it will take the machine shop to produce the cranks.
3605 Tc =+
b. Substitute 50 c = into the equation from part a, then solve for T.
Substitute 50. Simplify.
3605 Given equation. 3605(50) 360250 610 n Tc T T T = =+ =+ =+
It will take 610 minutes, or 10 hours and 10 minutes, to produce 50 cranks.
c. Substitute 1175 T = into the equation from part a, then solve for c.
3605 11753605 360 360 8155 8155 5 TcT c c = =+ =+ = = 5 c 163 c =
Substitute 1175.
Isolate variable term. Subtract 360 from both sides. Divide both sides by 5.
The machine shop made 163 cranks.
68. a. Let c = the number of EPs requested. Let T = the time it takes to burn the EPs.
12 Tc =
b. Substitute 250 c = into the equation from part a, then solve for T
12 Given equation. 12(250) 3000 n Tc T T = = = =
Substitute 250. Simplify.
It will take Richard 3000 minutes, or 50 hours.
c. Substitute 240 T = (4 hours = 240 minutes) into the equation from part a, then solve for c.
Substitute 240.
12 Given equation. 24012 24012 12 T Tc c = = = = 12 c Divide both sides by 12. 20 c =
He can create 20 EPs.
d. Substitute 1800 T = (6 hours = 360 minutes times 5 days) into the equation from part a, then solve for c
Substitute 1800.
12 Given equation. 180012 180012 12 T Tc c = = = = 12 c Divide both sides by 12. 150 c =
Richard can create 150 EPs in a week if he works 6 hours a day, 5 days a week.
Exercises69-86:Answersaretobechecked.
Exercises 69-72 are completed as examples.
69. Let x = a number. “Times” means to multiply so we have 3x , and “is” translates to “equals” so the equation is
345 x =
Solve the equation as follows
345 Divide both sides by 3. 3 x = 3 x 45 3 15 x = = ? Check the answer. The answer works 3(15)45 4545 . = = The number is 15. The answer is reasonable.
70. Let x = a number. “Sum” means to add and we are adding 20 to 4 times a number. “Is equal to” translates to the equal sign so the equation is 20444 x +=
Solve the equation as follows
20444 20 20 424 4 x x += = 4 x 24 4 6 x = = ? ? Check the answer. The answer works 204(6)44 202444 4444 . += += = The number is 6. The answer is reasonable.
Isolate variable term
Subtract 20 from both sides. Divide both sides by 4.
71. Recall that the perimeter of a rectangle can be calculated using the formula 22 Plw =+ where P is the perimeter of the rectangle with length, l, and width, w. Substitute 56 P = and 8 w= . Solve the equation for l.
. Substitute in known values
5622(8) 56216 16 16 402 402 2 l l l =+ =+ = = 2 l 20 l = ? ? Check the answer. The answer works 562(20)16 564016 5656 . =+ =+ = The length of the garden is 20 feet. The answer is reasonable.
Simplify & isolate variable term. Subtract 16 from both sides. Divide both sides by 2.
72. Recall that the perimeter of a rectangle can be calculated using the formula 22 Plw =+ where P is the perimeter of the rectangle with length, l, and width, w
Substitute 180 P = and 30 w = into the equation to find the perimeter of a rectangle. Solve the equation for l.
18022(30) 180260 60 60 1202 1202 2 l l l =+ =+ = = 2 l 60 l =
Substitute in known values
Simplify & isolate variable term. Subtract 60 from both sides. Divide both sides by 2.
May not be scanned, copied or duplicated, or posted to a publicly accessible website, in whole or in part.
? ? Check the answer.
The answer works
1802(60)60 18012060 180180 . =+ =+ = The length of the volleyball court is 60 feet. The answer is reasonable.
73. Recall that the area of a triangle can be calculated using the formula 1 2 Abh = where A is the area of the triangle with base, b, and height, h
Substitute 40 A = and 8 b = . Solve the equation for h
Substitute in known values
Simplify & isolate variable term. 1 40(8) 2 404 404 4 h h = = = 4 h Divide both sides by 4. 10 h =
The height of the triangle is 10 inches. The answer is reasonable.
74. Recall that the area of a triangle can be calculated using the formula 1 2 Abh = where A is the area of the triangle with base, b, and height, h.
Substitute 75 A = and 15 h = into the equation to find the area of a triangle. After substituting in these values, solve the equation for b.
Substitute in known values
Multiply by reciprocal of coefficient. 1 75(15) 2 15 75 2 2 15 b b = = 5 75 15 = 2 2 b
The base of the triangle is 10 inches. The answer is reasonable.
75. Let x = a number. “Quotient” represents division and “is” represents the equals sign. Remember that the order of division is important. Read from left to right. Write the quotient in the same order as it is read. The equation translates as
Solve the equation as follows
Multiply both sides by 7.
The number is 91. The answer is reasonable.
76. Let x = a number. The equation translates as 1219.2 5 x +=
Solve the equation as follows
1219.2 Isolate variable term. 5 12 12 Subtract 12 from both sides. 7.2 5 5 x x += = 5 x 7.25 Multiply both sides by 5. 36 x =⋅ =
The number is 36. The answer is reasonable.
77. Let x = a number. “Plus” means to add and “product” tells us to multiply. We are adding to 7 times a number. “Is equal to” translates to the equal sign so the equation is 6762 x +=
Solve the equation as follows
Isolate variable term Subtract 6 from both sides. Divide both sides by 7. 6762 6 6 756 7 x x += = 7 x 56 7 8 x = =
The number is 8. The answer is reasonable.
78. Let x = a number. “Times” means to multiply and “times the sum” means to multiply an addition that has already been done. To show the addition, include a set of parentheses around the sum. 4(3)68 x +=
Solve the equation as follows
4(3)68 41268 12 12 456 4 x x x += += = 4 x 56 4 14 x = =
. Distribute 4 on left side
Now isolate variable term
Subtract 12 from both sides. Divide both sides by 7.
The number is 14. The answer is reasonable.
79. Let x = a number. “Of” means to multiply so we will multiply 1 3 times x, subtract 8 then set it equal to 28. 1 828 3 x −=
Solve the equation as follows
Isolate variable term
Add 8 to both sides.
Multiply both sides by reciprocal. 1 828 3 +8 +8 1 36 3 3 x x −= = 1 1 3 3 36 1 108 x x =⋅ =
The number is 108. The answer is reasonable.
80. Let x = a number. “Half a number” means to multiply x by 1 2, we then add 18 and set it equal to 15.
Solve the equation as follows
Isolate variable term
Subtract 18 from both sides. Multiply both sides by reciprocal.
The number is –6. The answer is reasonable.
81. Let x = a number. “Difference” represents subtraction, “twice a number” means to multiply x by 2, and “is” represents the equal sign. The equation translates as
21540 x −=
Solve the equation as follows
21540 Isolate variable term. 15 15 Add 15 to both sides 255 2 x x −= ++ = 2 x 55 Divide both sides by 2. 2
27.5 x = =
The number is 27.5. The answer is reasonable.
82. Let x = a number. “Difference” represents subtraction, “three times a number” means to multiply x by 3, and “is” represents the equal sign. The equation translates as
3717 x −=
Solve the equation as follows
3717 Isolate variable term. 7 7 Add 7 to both sides 324 3 x x −= ++ = 3 x 24 Divide both sides by 3. 3 8 x = =
The number is 8. The answer is reasonable.
83. Let x = an unknown number. We will subtract the unknown number from 8 then set it equal to 19.
819 x −=
Solve the equation as follows
819 Isolate variable term.
8 8 Subtract 8 from both sides 11
1111 Multiply both sides by1. 11 x x x x −= −= −⋅−=⋅−− =−
The unknown number is –11. The answer is reasonable.
84. Let x = an unknown number. We will subtract the unknown number from 36 then set it equal to 5 .
365 x −=−
Solve the equation as follows
365 Isolate variable term. 36 36 Subtract 36 from both sides 41
1411 Multiply both sides by1. 41 x x x x −=−
The unknown number is 41. The answer is reasonable.
85. Let x = a number. “Less than” translates to subtraction so we will subtract 7 from twice (2 times) a number then set it equal to 11.
2711 x −=
Solve the equation as follows 2711 Isolate variable term. 7 7 Add 7 to both sides
218 2 x x −= ++ = 2 x 18 Divide both sides by 2. 2 9 x = =
The number is 9. The answer is reasonable.
86. Let x = a number. “Less than” translates to subtraction so we will subtract 8 from 3 times a number then set it equal to 23 .
3823 x −=−
Solve the equation as follows 3823 Isolate variable term. 8 8 Add 8 to both sides.
The number is –5. The answer is reasonable.
Identify variable term to isolate Subtract 2 from both sides. Identify variable to isolate 22 .
both sides by 2.
on
to
Identify
Identify variable term to isolate Add
Identify variable to isolate
Identify
Identify
Identify variable term to isolate Subtract 2 from both sides.
Identify variable to isolate
Identify
to
Identify variable term to isolate Subtract 5 from both sides.
Identify variable to isolate
Identify variable term to isolate Add 3 to both sides.
to isolate
100.
Identify variable term to isolate
Subtract from both sides.
Identify variable to isolate
Identify variable term to isolate
Multiply each term by 3 to clear fraction.
Subtract from both sides.
Identify variable to isolate
Distribute on right side
Subtract 21 from both sides.
Identify variable to isolate
Distribute on right side
Subtract 10 from both sides.
Identify variable to isolate
on right side
Distribute on right side
Subtract 6 from both sides.
Identify variable to isolate
= 9 x Divide both sides by 9.
Identify
on right side
Distribute on right side
Subtract 1 from both sides.
Identify variable to isolate .
IPrt IP Pt = = rt Pt Divide both sides by Pt I r Pt = 106.
Identify variable to isolate
Identify variable to isolate. IPrt IP Pr = = rt Pr Divide both sides by Pr I t Pr = 107. The statement does not have an equal sign; therefore, it is an expression.
terms
108. The statement does not have an equal sign; therefore, it is an expression.
2 Identify like terms
Combine like terms.
109. The statement has an equal sign; therefore, it is an equation.
Isolate variable term
Subtract 7 from both sides.
Divide both sides by 2.
110. The statement has an equal sign; therefore, it is an equation.
Isolate variable term
Subtract 14 from both sides. Divide both sides by6.
111. The statement has an equal sign; therefore, it is an equation.
Combine like terms on left side
Rewrite
Subtract
both sides. Multiply both sides by reciprocal.
112. The statement has an equal sign; therefore, it is an equation.
Combine like terms
Now isolate variable term
Subtract 15 from both sides.
Multiply both sides by reciprocal.
113 The statement does not have an equal sign; therefore, it is an expression.
Identify like terms
Combine like terms.
578204 . 20311 nnm mn +−++ −+
Arrange in conventional form.
114. The statement does not have an equal sign; therefore, it is an expression.
Identify like terms
Combine like terms.
8641423 . 4617 wvw vw −+−+
Arrange in conventional form.
115. The statement has an equal sign; therefore, it is an equation.
Divide both sides by 1.3. 3.442.111.28 1.3411.28 4 4 1.37.28 1.37.28 1.31.3 5. yy y y y y +−= += = = = 6
Combine like terms on left side.
Isolate variable term
Subtract 4 from both sides.
116. The statement has an equal sign; therefore, it is an equation.
Divide bot 4.68.71.232.84 3.48.732.84 8.7 8.7
Combine like terms on left side
Isolate variable term
Add 8.7 to both sides.
3.424.14 hh h h −−+=− −−=− ++ −=− h sides by3.4. 3.424.14 3.43.4 7.1 h h = =
Section 2.3 Solving Equations with Variables on Both Sides
Exercises1-16:Answersaretobechecked.Exercises 1-5arecompletedasexamples. 1 433
433 Identify variable terms. Subtract from both sides
Identify variable terms. Subtract from both sides
710418 Identify like terms. 4 4 Subtract 4 from both sides. 31018 10 10 Subtract 10 from both sides. 328 Divide both sides by 3. 3 xx xxx x x +=− +=− =−
25(6) xx
25(6) Distribute on the right side. 2530 Identify like terms. 5 5 Subtract 5 from both sides.
on the right side. 7210 Identify like terms. 2 2 Subtract 2 from both sides.
7. 203422 xx −=−
203422 Identify like terms. 4 4 Subtract 4 from both sides. 20722
20 20 Subtract 20 from both sides. 742 Divide both sides by7.
2(6)3(2) 21236 3 3 126 12 12 18 x xx xx xx x x
Distribute on the right side.
Identify like terms.
Subtract 3 from both sides.
Add 12 to both sides. M
ultiply both sides by1. 1181 18 x
Identify like terms.
Subtract 2 from both sides.
Multiply both sides by1. (2)5(2)
10. 1 57 3 xx+=−
Multiply each term by 3 to clear fraction.
(3 1 ) 3
(3)5(3)7(3) 15321 3 3 21521 15 15 236 2 x xx
Identify like terms.
Subtract 3 from both sides.
Subtract 15 from both sides.
Multiply each term by 2 to clear fraction (2)5(2 + ) 2 x (2)7(2) x =− ()()
Identify like terms.
Subtract 2 from both sides.
Subtract 10 from both sides. Multiply both sides by1. 10214 2 2 1014 10 10 24 1241 24 x xx xx x x x x +=− −=−
= 12 528 3 x x =+
Multiply each term by 3 to clear fraction (3 ) 3 x
Identify like terms.
Subtract 15 from both sides. (3)528(3) 1584 15 15 1484 14 x x xx xx x =+ =+ −= 14 x Divide both sides by14. 84 14 6 x = =− 13 854xx−=+ 854 Identify like terms.
5 5 Subtract 5 from both sides. 484 8 8 Add 8 to both sides. 412 Divide both sides by7. 4 xx xxx x x −=+ −−= ++ −=− 4 x 12 4 3 x = =− 14. 2927 x =− 2927
Identify like terms. 7 7 Add 7 to both sides. 362 362 2 x x =− ++ = = 2 x Divide both sides by 2. 18 x =
Multiply each term by 4 to clear fraction (4 1 ) 4 (4)8(4)229(4)xx−=−
Identify like terms.
Subtract 8 from both sides.
32 to both sides.
7 x Divide both sides by7. 84 7 12 x
= 16. 1 2035 2 xx+=+
Multiply each term by 2 to clear fraction (2 1 ) 2
Identify like terms.
Step 2 Reason: Combine like variable terms by subtracting 5x from both sides.
Step 3 Reason: Isolate the variable by adding 60 to both sides.
Exercises19-30:Answersaretobechecked. Exercises19-22arecompletedasexamples. 19.
Combine like terms across = sign Subtract 7 from both sides.
Subtract 12 from both sides. Divide both sides by4.
Subtract 6 from both sides.
Subtract 40 from both sides. (2)20(2)35(2) 40610 6 6 54010 40 40 530 5 x xx xx xx x
sides by5.
17. Step 1 Reason: Distribute to simplify.
Step 2 Algebraic Step: 211518 xx−=−−
Step 3 Reason: Combine like variable terms by adding 5x to both sides.
Step 4 Algebraic Step: 1111 71118 x ++ −=−
Step 5 Reason: Divide both sides by 7 to isolate the variable.
18. Step 1 Algebraic Step: () () 12 152154 35 12 1515215154 35 530660 xx xx xx ⎛⎞⎛⎞ +=−
312728 7 7 41228 12 12 440 4 x xx xx x x +=− −+=− −=− 4 x 40 4 10 x = = ? ? Check the answer. The answer works 3(10)127(10)28 30127028 4242 . +=− +=− = 20.
Combine like terms across = sign Subtract 10 from both sides.
Subtract 20 from both sides. Divide both sides by4 6201036 10 10 42036 20 20 456 x xx xx x x +=− −+=− −=− 4 4 x 56 4 14 x = = ? ? Check the answer. The answer works 6(14)2010(14)36 842014036 104104 . +=− +=− = 21.
Combine like terms across = sign Subtract 4 from both sides. Add 50 to both sides. Divide both sides by 8. 1250414 4 4 85014 50 50 864 8 t tt tt t t −=+ −= ++ = 8 t 64 8 8 t = = ? ? Check the answer. The answer works
12(8)504(8)14 96503214 4646 . −=+ −=+ =
Combine like terms across = sign
Subtract 3 from both sides.
Add 30 to both sides.
Divide both sides by 4.
answer.
Combine like terms across = sign.
Subtract 9 from both sides.
Subtract 36 from both sides.
Divide both sides by5.
22.
Combine like terms across = sign
Subtract 4 from both sides.
Add 20 to both sides.
Divide both sides by 11.
Combine like terms on left side
Combine like terms across = sign.
Subtract 3 from both sides.
This is a false statement.
This equation has no solution.
Combine like terms on right side
Combine like terms across = sign .
Subtract 5 from both sides.
This is a false stateme
()() Distribute to simplify.
Combine like terms on left side
Combine like terms across = sign
Subtract 8 from both sides.
Isolate the variable term by adding 2 toboth sides.
Divide both sides by 2 to isolate the variable.
()() Distribute to simplify.
Combine like terms on both sides
Combine like terms across = sign
Subtract 12 from both side .
Isolate the variable term by adding 24 to both sides.
Divide both sides by 7 to isolate the variable.
Multiply by 3 to remove the fraction.
Combine like terms across = sign.
Add 66 to both sides.
Divide both sides by 13.
Multiply by the LCD, 28, to remove the fraction.
Substitute two random numbers to check. ? ? ? ? Check the answer using7. The answer works Check the answer using 0.
Combine like terms across = sign.
Add 140 to both sides.
Exercises31-52:Answersaretobechecked.
Exercises31and32arecompletedasexamples.
31.
Combine like terms on right side
Combine like terms across = sign .
Subtract 4 from both sides.
This is a true statement
The equation is an identity therefore the solution is all real numbers or
Substitute two random numbers to check.
4(11)53(11)5(11) 44533511 4949 −+=−++−
? ? ? ? Check the answer using4.
The answer works
4(4)53(4)5(4) 1651254 1111 .
Check the answer using 11. Th
Subtract 3 from both sides. This is a true statement. 28538
Combine like terms on left side
Combine like terms across = sign
The equation is an identity therefore the solution is all real numbers or .
2(7)85(7)3(7)8 14835218 2929 .
The answer works
2(0)85(0)3(0)8 08008 88 . −+−−=−−+ −++=+ = +−=−+ +−=+ =
Combine like terms across = sign. Subtract 2 from both sides. Divide both sides by 1.5. 3.5212 2 2 1.512 1.5 h hh hh h =− =− 1.5 h 12 1.5 8 h = =−
Combine like terms across = sign Subtract 4.8 from both sides. Divide both sides by 2.5. 4.8207.3 4.8 4.8 202.5 202.5 2.5 d dd dd d += = = 2.5 d 8 d =
Isolate variable.
Subtract 15 from both sides. Multiply both sides by1. 158 15 15 7 171 7 x x x x −= −=− −⋅−=−⋅− =
()()
85920 520 5 5 15 1151 15 xx x x x x +−= −+= −=
Combine like terms on left side. Isolate variable.
Subtract 5 from both sides.
()()
Multiply both sides by1.
Combine like terms on left side Isolate variable.
Subtract 8 from both sides.
Multiply both sides by1. 38413 813 8 8 5 151 5 hh h h h h +−= −+= −=
()()
39.
40.
41.
Multiply both sides by1. 14586 148 8 8 22 1221 22 mm m m m m
Combine like terms on right side
Isolate variable.
Add 8 to both sides.
Distribute to simplify on the left.
Combine like terms across = sign
Subtract 4 from both sides.
2(35)422 610422 4 4 21022 10 10 x xx xx xx x +=+ +=+ += th sides.
Subtract 10 from bo
Divide both sides by 2. 212 2 x = 2 x 12 2 6 x = =
3(89)3099 24273099 30 30 62799 27 +27 Add 27 to x xx xx xx x
Simplify on left - distribute
Combine like terms across = sign
+ sides.
Subtract 30 from both sides. both
Divide both sides by6. 672 6 x −=− 6 x 72 6 12 x = =
5(23)83(42)2 101581262 ccc ccc +−=+− +−=+− ?
Distribute on both sides
Simplify on both sides
This is a false statement 107106 10 10 76 c cc cc +=+ = This equation has no solution.
Combine like terms across = sign
Subtract 10 from both sides.
44.
42. ? Distribute on both sides
2(65)3(42)7 12101267 12101213 12 12 1013 w ww ww ww ww −=−−
Simplify on right side
−=− −=− This is a false statement
Combine like terms across = sign
Subtract 12 from both sides.
This equation has no solution.
Subtract 20 from both 7203950 420950 9 9 52050 20 20 d ddd dd dd d +−=+ +=+ −+= sides. Divide both sides by5. 530 5 d −= 5 d 30 5 6 d = =−
Combine like terms on left side
Combine like terms across = sign
Subtract 9 from both sides.
Subtract 3 from both sides. both si 12155351 157351 3 3 15451 15 15 Add 15 to r rrr rr rr r −−=− −+=− −+=− ++ des.
Combine like terms on left side
Combine like terms across = sign
Divide both sides by 4. 436 4 r =− 4 r 36 4 9 r = =− 45.
Multiply by 2 to remove fraction 1 5345 2 (2 xx+=− 1 ) 2 (2)5(2)345(2)xx+=−
Subtract 10 from both sides. Divide both sides by5 10690 6 6 51090 10 10 5100 x xx xx x x +=− −+=− −=− . 5 5 x 100 5 20 x = = 46.
Combine like terms across = sign
Subtract 6 from both sides.
Multiply by 2 to remove fraction. 1 4531 2 (2 zz−=− 1 ) 2
Add 8 to both sides. Divi (2)4(2)531(2) 81062 10 10 9862 8 8 954 z zz zz zz z z −=− −=− −−=− ++ −=− de both sides by5. 9 9 x 54 9 6 x = =
Combine like terms across = sign
Subtract 10 from both sides.
48.
Multiply by 3 to remove fraction 1 20220 3 (3 xx+=+ 1 ) 3
(3)202(3)20(3) 60660 60560 60 60 05 x xx xx xx x x +=+
Combine like terms across = sign
Subtract from both sides.
Subtract 60 from both sides. Div
sides by
3510.259.9 5 5 3010.29.9 10.2 10.2 300.3 t tt tt t y +=+ += =− Divide both sides by 4. 30 30 y 0.3 30 0.01 y = =−
Combine like terms across = sign
Subtract 5 from both sides.
Subtract 10.2 from both sides.
Multiply by 7 to remove fraction. 1 88 7 (7 h += 1 ) 7
(7)88(7) 5656 56 56 0 h h h += += =
Isolate variable.
Subtract 56 from both sides.
49.
Combine like terms across = sign.
Subtract 5 from both sides. both sides. Divide both sides by 4
50.
Combine like terms across = sign.
Subtract 8 from both sides.
Subtract 3.6 from both sides.
both sides by
52.
Combine like terms across = sign. Add 4 to both sides. Add 7.9 to both sides. Divide both sides 37.94.754 4 4 77.94.75 7.9 7.9 73.15 w ww ww w w −=−− ++ −=− ++ = by 7.
7 7 w 3.15 7 0.45 w = =
53. Add the measures of the given angles and set them equal to 180° then solve for x 25180 Combine like terms. 8180 Divide both sides by 8. 8 xxx x ++= = 8 x 180 8 22.5 x = =
Substitute 22.5 x = into the expressions that represent the. The measures of the 3 angles are 22.5,45,and112.5. °°°
To check the solution, the measures of the angles should add up to 180. ? Check the answer. The answer works
22.545112.5180 180180 . °+°+°=° °=°
54. Add the measures of the given angles and set them equal to 180° then solve for y.
36180 Combine like terms.
10180 Divide both sides by 10. 10 yyy y ++= = 10 y 180 10 18 y = =
Substitute 18 x = into the expressions that represent the angles. The measures of the 3 angles are 18,54,and108. °°°
To check the solution, the measures of the angles should add up to 180.
?
Check the answer. The answer works 1854108180 180180 . °+°+°=° °=°
55. Substitute 66 P = , 31lx=+ , and 5 wx = into the formula for perimeter and solve for x
22
662(31)2(5) 666210 66162 2 2 Plw xx xx x =+ =++ =++ =+
Formula for perimeter.
Substitute in values
Distribute/mulitply on right side
Simplify & isolate variable term.
Subtract 2 from both sides. Divide both sides by 16. 6416 6416 16 x = = 16 x 4 x =
Now substitute 4 x = into the expressions for length and width.
3(4)15(4) 12120 13 lwx lw l = =+=
Substitute 4. Simplify each equation.
The length and width of the yard measure 13 ft and 20 ft.
To check the solution, substitute these values into the formula for perimeter and set it equal to 66.
Now substitute 6 x = into the expressions for length and width.
Substitute 6. Simplify each equation. 4(6)23(6) 24218 26 lwx lw l = =+= =+= =
The length and width of the yard measure 26 ft and 18 ft.
To check the solution, substitute these values into the formula for perimeter and set it equal to 88.
? ? Check the answer. The answer works 22
662(13)2(20) 662640 6666 . Plw =+ =+ =+ =
56. Substitute 88 P = , 42lx=+ , and 3 wx = into the formula for perimeter and solve for x
22
882(42)2(3) 88846 88144 4 4 Plw xx xx x =+ =++ =++ =+
Formula for perimeter.
Substitute in values
Distribute/mulitply on right side
Simplify & isolate variable term.
Subtract 4 from both sides. Divide both sides by 14. 8414 8414 14 x = = 14 x 6 x =
? ? Check the answer.
The answer works 22
882(26)2(18) 885236 8888 . Plw =+ =+ =+ =
57. Substitute 83 P = , and 32sx=− into the formula for perimeter and solve for x
Formula for perimeter.
Substitute in value for Distribute/mulitply on right side Add 8 to both sides. 4 834(32) 83128 8 8 9112 s Ps x x x = =− =− ++ = Divide both sides by 12. 9112 12 = 12 x 791 7 or 1212 xx ==
Now substitute 91 12 x = into the expression for the length of each side.
The length of each side of the yard measures 20.75 feet.
To check the solution, substitute this value into the formula for perimeter and set it equal to 83.
? Check the answer. The answer works 4
834(20.75) 8383 . Ps = = =
58. Substitute 60 P = , and 63sx=− into the formula for perimeter and solve for x.
Add 8 to both sides. 4 604(63) 602412 12 12
Formula for perimeter.
Substitute in value for
7224 s Ps x x x = =− =− ++ = Divide both sides by 12. 7224 24 = 24 x 3 x =
Distribute/mulitply on right side
Now substitute 3 x = into the expression for the length of each side.
60. Substitute 58 P = , 7 lx=− , and 3 wx = into the formula for perimeter and solve for x . Formula for perimeter.
582(7)2(3) 582146 58814 14 14 Plw xx xx x =+ =−+ =−+ =− ++ Add 14 to both sides.
728 728 8 x = = 8 x 9 x =
Substitute in values
Distribute/mulitply on right side
Simplify and isolate variable term.
Divide both sides by 10.
Now substitute 9 x = into the expressions for length and width.
Substitute . Simplify
6(3)3 3 183 15 sx s s = =− =− =
The length of each side of the yard measures 15 feet.
To check the solution, substitute this value into the formula for perimeter and set it equal to 60.
Substitute 9. Simplify each equation (9)73(9) 227 lwx lw = =−= ==
The length and width of the lawn measure 2 feet and 27 feet.
? Check the answer. The answer works 4
604(15) 6060 . Ps = = =
59. Substitute 100 P = , 5 lx=− , and 4 wx = into the formula for perimeter and solve for x
22
1001010 10 Plw xx xx x
1002(5)2(4) 1002108
Formula for perimeter. Substitute in values
Distribute/mulitply on right side
=− + Add 10 to both sides. Divide both sides by 10. 10 11010 11010 10 x + = = 10 x 11 x =
Simplify and isolate variable term.
To check the solution, substitute these values into the formula for perimeter and set it equal to 58. ? ? Check the answer. The answer works 22 582(2)2(27) 58454 5858 . Plw =+ =+ =+ = 61.
22180 Combine like terms. 5180 Divide both sides by 5. 5 xxx x ++= = 5 x 180 5 36 x = =
Now substitute 11 x = into the expressions for length and width.
Substitute 11. Simplify each equation (11)54(11) 644 lwx lw = =−= ==
The length and width of the lawn measure 6 feet and 44 feet.
To check the solution, substitute these values into the formula for perimeter and set it equal to 100.
22
Now substitute 36 x = into the expressions that represent the angles to find the measure of each angle. The measures of the 3 angles are 36,72,and72. °°°
To check the solution, the measures of the angles should add up to 180. ? Check the answer. The answer works
367272180 180180 . °+°+°=° °=°
1002(6)2(44) 1001288 100100 . Plw =+ =+ =+ =
? ? Check the answer. The answer works
62.
6 xxx x ++= = 6 x 180 6
23180 Combine like terms.
6180 Divide both sides by 6.
30 x = =
Now substitute 30 x = into the expressions that represent the angles to find the measure of each angle. The measures of the 3 angles are 30,60,and90. °°°
To check the solution, the measures of the angles should add up to 180.
? Check the answer. The answer works 306090180 180180 . °+°+°=° °=°
Exercises63-76:Answersaretobechecked. 63.
3(0.5)130.5
31.5130.5 30.530.5 3 3 x xx xx xx xx −+=− −+=− −=− ides. This is a true statement. 0.50.5−=−
Distribute to simplify on the left.
Combine like terms on left side
Subtract 3 from both s .
Combine like terms across = sign
The equation is an identity therefore the solution is all real numbers or .
To check this, we will randomly choose two real numbers 2 and 2, and substitute these into the equation.
[] ? ? ?
3(20.5)13(2)0.5
3(2.5)160.5 7.516.5 6.56.5 . −−+=−− −+=−− −+=− −=−
Check the answer using2.
The answer works
3(20.5)13(2)0.5
[] ? ? ? Check the answer using 2.
The answer works
2.1(2)32.11.2
2.14.232.11.2
Distribute to simplify on the left.
Combine like terms on left side
Combine like terms across = sign
2.11.22.11.2 2.1 2.1 xx xx xx xx +−=+ +−=+ +=+ from both sides.
Subtract 2.1
This is a true statement. 1.21.2 x =
The equation is an identity therefore the solution is all real numbers or
To check this, we will randomly choose two real numbers 9 and 8, and substitute these into the equation.
65.
3(1.5)160.5 4.515.5 5.55.5 . −+=− +=− += = 64.
2.1(92)32.1(9)1.2
−+−=−+
2.1(7)318.91.2 14.7317.7
66.
17.717.7 . 2.1(82)32.1(8)1.2
2.1(10)316.81.2 213
−−=−+ −−=− −=− +−=+ −=+ −= ? The answer works 18 1818 . =
The answer works
Check the answer using 8.
? ? ? ? ? Check the answer using9.
0.30.30.50.10.2
0.30.30.30.1
0.3 0.3
0.30.1 x xxx xx xx +=+− +=+ = ent
? Simplify on right side
Combine like terms across = sign 0.3 . Subtract from both sides.
This is a false statem
This equation has no solution.
0.90.20.5(31)0.6
0.90.21.50.50.6
0.90.20.90.5
0.9 0.9
0.20.5 x xxx xxx xx xx −=+− −=+− −=+
? Simplify on right side
Combine like terms across = sign
−= his is a false statement
Subtract 0.9 from both sides. T
This equation has no solution.
68.
Multiply by 10 (52) to clear fractions 17 55 25 17 (10)(10)5(10)5(10) 25 (10 rr rr +=+ +=+ 1 ) 2 2 50(10 r += 7 ) 5
5
Subtract 50 from both sides. Divide b 50 5501450 14 14 95050 50 50 90 r r rr rr r r + +=+ −+= −= oth sides by9. 9 9 r 0 9 0 r = = ? ? Check the answer. The answer works 17(0)5(0)5 25 0505 55 . +=+ +=+ =
Combine like terms across = sign.
Subtract 14 from both sides.
The equation is an identity therefore the solution is all real numbers or To check this, we will randomly choose two real numbers 20 and 10, and substitute these into the equation.
31(20)1(62010) 510 1 121(12010) 10 1 13(130) 10 1313 . 31 510(10)1(61010) 1 61(6010) 10 1 5( 10
Check the answer using20. The answer works
Check the answer using 10.
answer works 50) 55 .
Multiply by 4 to clear fraction 1 50350 4 (4 ww+=+ 1 ) 4
Multiply by 3 to clear fractions. 21 336(218) (3 nn+=+ 2 ) 3 (3)6(3 n += 1 ) 3
Subtract 200 from both sides. (4)50(4)350(4) 20012200 12 12 11200200 200 200 110 w ww ww ww w w +=+ +=+ −+= −= Divide both sides by11. 11 11 w 0 11 0 w = = ? ? Check the answer. The answer works 1 (0)503(0)50 4 050050 5050 . +=+ +=+ = 69. 2
Combine like terms across = sign
Subtract 12 from both sides.
Multiply by 10 to clear fractions 31 5101(610) (10 dd−=− 3 ) 5 (10)1(10 d −= 1 ) 10
Subtract 6 from both sides. This is a true statement. (610) 610610 6 6 1010 d d dd dd
Combine like terms across = sign
Combine like terms across = sign Subtract 2 from both sides. This is a true statement. (218) 218218 2 2 1818 n n nn nn + +=+ =
The equation is an identity therefore the solution is all real numbers or To check this, we will randomly choose two real numbers 12 and 6, and substitute these into the equation.
Check the answer using 6. 21 33(12)621218 1 86(2418) 3 1 2(6) 3 22 . 21 33(6)62618 1 46(1218) 3 1 10(30) 3 1010
? ? Check the answer using12. The answer works
= The answer works.
May not be scanned, copied or duplicated, or posted to a publicly accessible website, in whole or in part.
Multiply by 5 to clear fractions 12 72 55 (5 gg +=+ 1 ) 5 2 (5)7(5)2 5 g +=+ (5g
Multiply both sides by1. ) 35102 2 2 3510 35 35 25 1 g gg gg g g +=+
Combine like terms across = sign.
Subtract 2 from both sides.
Subtract 35 from both sides.
= ? ? Check the answer.
The answer works 12 55(25)72(25) 57210 1212 . +=+ +=+ = 72.
Multiply by 7 to clear fractions. 23 515 77 (7 tt −=− 2 ) 7 3 (7)5(7)15 7 t −=− (7 t
++
Divide both sides by 5. ) 2351053 3 3 Add 535105 35 35 5140 5 t tt tt
t
Combine like terms across = sign 3 to both sides.
++ = 5 t 140 5 28 t = = ? ? Check the answer. The answer works 23 77(28)515(28) 851512 33 .
Add 35 to both sides.
Multiply by 10 (52) to clear fractions 12 79 25 (10 xx+=+ 1 ) 2 2 (10)7(10 x += 2 ) 5
Combine like terms across = sign. Subtract 4 from both sides.
Subtract 70 from both sides. 9(10) 570490 4 4 7090 70 70 20 x x xx xx x x + +=+ += = ? ? Check the answer. The answer works 12 25(20)7(20)9 10789 1717 . +=+ +=+ = 74. 4
Multiply by 12 (LCD) to clear fractions 119 4 342 (12 vv+=+ 1 ) 3 3 (12)4(12 v += 1 ) 4 9 2 v + 6 (12
Subtract 48 from both sides. ) 448354 3 3 4854 48 48 6 v vv vv v v +=+ += = ? ? ? Check the answer. The answer works 119 (6)4(6) 342 39 24 22 12 6 2 66 . +=+ +=+ = = 75.
Combine like terms across = sign Subtract 3 from both sides.
Multiply by 2 to clear fractions 11 53 22 (2 xx+=+ 1 ) 2 (2)5(2 x += 1 ) 2 ?
This is a false statement 3(2) 106 106 x x xx xx + +=+ =
Combine like terms across = sign.
Subtract from both sides.
This equation has no solution. 76.
Simplify on each side - distribute 11 32(36)24 xx ⎛⎞ +=+⎜⎟ ⎝⎠ ?
This is a false statement. 28 28 x xx xx +=+ = This equation has no solution.
Combine like terms across = sign
Subtract from both sides.
Combine like terms
75(5)180 . 270180 70 70 2110 2 xx x x ++−= += = 2 x Divide both sides by 2 110 2 55 x = =
Subtract 70 from both sides.
Now substitute 63 x = into the expressions that represent the angles. The measures of the three angles are 63,68,and49. °°°
To check the solution, the measures of the angles should add up to 180.
Now substitute 55 x = into the expressions that represent the angles. The measures of the three angles are 75,55,and50. °°°
To check the solution, the measures of the angles should add up to 180.
78.
80.
636849180 180180 . °+°+°=° °=°
? Check the answer. The answer works
Combine like terms
Subtract 20 from both sides.
? Check the answer. The answer works
755550180 180180 . °+°+°=° °=°
2(30)(10)180 . 420180 20 20 4160 4 xxx x x +++−= += = 4 x Divide both sides by 4 160 4 40 x = =
Combine like terms
Subtract 70 from both sides.
80(10)(20)180 . 270180 70 70 2110 2 xx x x +++−= += = 2 x Divide both sides by 2 110 2 55 x = =
Now substitute 55 x = into the expressions that represent the angles to find the measure of each angle. The measures of the three angles are 80,65,and35. °°°
To check the solution, the measures of the angles should add up to 180.
? Check the answer. The answer works
806535180 180180 . °+°+°=° °=° 79.
Combine like terms
Add 9 to both sides. (5)(14)180 . 39180 9 9 3189 3 xxx x x +++−= −= ++ = 3 x Divide both sides by 3 189 3 63 x = =
Now substitute 40 x = into the expressions that represent the angles. The measures of the three angles are 80,70,and30. °°°
To check the solution, the measures of the angles should add up to 180. ? Check the answer. The answer works 807030180 180180 . °+°+°=° °=°
81. The sum of the measures of the angles in a quadrilateral must equal 360° . Add the measures of the given angles and set them equal to 360° then solve for x
(25)65(20)360 360360 . 60 60 3300 3 xxx x x +−+++= += = 3 x Divide both sides by 3 300 . 3 100 x = =
Combine like terms
Subtract 60 from both sides.
Now substitute 100 x = into the expressions that represent the angles. The measures of the four angles are 100,75,65,and120. °°°°
To check the solution, the measures of the angles should add up to 360.
? Check the answer. The answer works 1007565120360 360360 . °+°+°+°=° °=° 82.
(210)(25)(315)360 8360 . 8 xxxx x −+−+++= = 8 x Divide both sides by 8 360 8 45 x = =
Combine like terms
Now substitute 45 x = into the expressions that represent the angles. The measures of the four angles are 80,85,150,and45. °°°°
To check the solution, the measures of the angles should add up to 360.
? Check the answer. The answer works 808515045360 360360 . °+°+°+°=° °=°
83. Add the given expressions for side lengths, set them equal to 342, and solve for p
(3)(20)(5)342 422342 22 22 4320 4 pppp p p −+++++= += = 4 p 320 4 80 p = =
Combine like terms on left side. Subtract 22 from both sides. Divide both sides by 4.
Now substitute 80 p = into the expressions that represent the lengths of each side. The lengths of the sides are 77 ft, 100 ft, 80 ft, and 85 ft.
To check the solution, the sum of the lengths of the sides should equal 342.
84. Add the given expressions for side lengths, set them equal to 335, and solve for m
(5)(10)335 35335 5 5 3330 3 mmm m m −+++= += = 3 m 330 3 110 m = =
Combine like terms on left side. Subtract 5 from both sides.
Divide both sides by 3.
Now substitute 110 m = into the expressions that represent the lengths of each side. The lengths of the sides are 105 feet, 110 feet, and 120 feet.
To check the solution, the sum of the lengths of the sides should equal 335.
? Check the answer. The answer works 105110120335 335335 . ++= =
85. Add the given expressions for side lengths, set them equal to 345, and solve for s
2(5)(210)345 615345 15 15 6360 6 ssss s s ++−+−= −= ++ = 6 s 360 6 60 s = =
Combine like terms on left side. Add 15 to both sides.
Divide both sides by 6.
Now substitute 60 s = into the expressions that represent the lengths of each side. The lengths of the sides are 60 feet, 120 feet, 55 feet, and 110 feet.
To check the solution, the sum of the lengths of the sides should equal 345.
? Check the answer. The answer works 6012055110345 345345 . +++= =
771008085342 342342 . +++= =
? Check the answer. The answer works
86. Add the given expressions for side lengths, set them equal to 500, and solve for x
(20)(35)(310)(30)500 95500 5 5 9495 9 xxxxx x x +++++++−= += = 9 x 495 9 55 x = =
Combine like terms on left side. Subtract 5 from both sides.
Divide both sides by 9.
149
2019 Cengage Learning, Inc. May not be scanned, copied or duplicated, or posted to a publicly accessible website, in whole or in part.
Now substitute 55 x = into the expressions that represent the lengths of each side. The lengths of the sides are 75 feet, 170 feet, 55 feet, 175 feet, and 25 feet.
To check the solution, the sum of the lengths of the sides should equal 500.
? Check the answer. The answer works 751705517525500 500500 . ++++= =
87. Let s = the number of students going on the bike tour.
• 210(s) dollars for the bikes
• 2(2s) dollars for the water bottles
• 8.50(3s) dollars for the food sacks
We write out the following equation equal to Carlotta's budget of $2275.00
2102(2)8.50(3)7664 sss ++=
Now solve the equation for s Multiply on left side Combine like terms on left side.
2102(2)8.50(3)7664 . 210425.57664 239.57664
Divide both sides by 239.5.
Carlotta can supply 32 students for the tour.
88. Let c = the number of Alphabet Cars Bill is going to build.
Bill will need to buy the following:
• 4c wheels, 4 for each car.
• 2c axles, two for each car.
• 5c hubcaps, 5 for each car.
Now we multiply each of these expressions by how much they cost and then add those costs together to get the total of $14,700.
• 1.00(4c) dollars for the wheels
• 0.39(2c) dollars for the axles
• 0.06(5c) dollars for the hubcaps
We write out the following equation equal to Bill's budget of $12,192.00
1.00(4)0.39(2)0.06(5)12192 ccc++=
Now solve the equation for c.
1.00(4)0.39(2)0.06(5)12192 40.780.312192 5.0812192 5.08 ccc ccc c ++= ++= = 5.08 c Divide both sides by 5.08. 12192 5.08 2400 c = =
Combine like terms on left side.
Bill can build 2400 Alphabet Cars.
89 a. First consider how many of each item is needed for k kids.
• 8 k + pine 2x2, 1foot for each kid plus 8 extra feet.
• 1.510 k + pine 1x3, 1.5 feet for each kid plus 10 extra feet.
• 12 k + sets of nails, 1 set for each kid plus 12 extra sets.
Now we multiply each of these expressions by how much they cost.
• 1.22(8) k + dollars for pine 2x2
• 1.34(1.510) k + dollars for pine 1x3
• 0.05(12) k + dollars for nails
Skkk Skkk Sk =+++++ =+++++ =+
1.22(8)1.34(1.510)0.05(12) 1.229.762.0113.40.050.6 . 3.2823.76
Let S = the total amount in dollars needed for supplies. Distribute Combine like terms.
b. Substitute 75 k = into the equation and solve for S. Substitute 75 Multiply then add.
3.28(75)23.76 . 24623.76 269.76 Sk S S = =+ =+ = The cost would be $269.76.
c. Substitute 417.36 S = into the equation and solve for k
393.63.28
300.33.28 3.28 kS k = =+ = = 3.28 k
Substitute 417.36
417.363.2823.76 . 23.76 23.76
Subtract 23.76 from both sides. Divide both sides by 3.28.
Rounded to hundredths 91.55 k ≈ The manager can buy supplies for 91 projects. Round down as there would not be enough money for 92 projects.
150 © 2019 Cengage Learning, Inc. May not be scanned, copied or duplicated, or posted to a publicly accessible website, in whole or in part.
90. a. First consider how many of each item is needed for y yards.
• 1215 y + 10-ft PVC pipes, 12 for each yard plus 15 extra.
• 96 y + sprinkler heads, 9 for each yard plus 6 extra.
• 2 y + sets of misc. PVC parts, 1 set for each yard plus 2 extra sets. Now we multiply each of these expressions by how much they cost.
• 1.50(1215) y + dollars for 10-ft PVC pipes
• 2.50(96) y + dollars for sprinkler heads
• 15(2) y + dollars for misc. PVC parts
Let S = the total amount in dollars needed for supplies.
55.567.5 Syyy Syyy Sy =+++++ =+++++ =+
1.50(1215)2.50(96)15(2) 1822.522.5151530
Distribute Combine like terms.
b. Substitute 12 y = into the equation and solve for S.
733.5 St S S =+= =+ = It will cost $733.50 for supplies.
55.5(12)67.5 Substitute 12
66667.5
Multiply then add.
c. Substitute 6006 S = into the equation and solve for y.
Use the formula M = rP. ()0.3120 36 M M = =
Sales price, S, is the original price plus the markup. 12036 156 SPM S S =+ =+ =
The sales price of the book is $156.
92. In order to determine the sales price, the markup must be determined, then added to the wholesale cost.
Markup rate: r = 3% = 0.03
Wholesale cost is the original price: P = $1.50
Markup: M
Use the formula M = rP. ()0.031.50 0.045 M M = =
Sales price, S, is the original price plus the markup. 1.500.045 1.5451.55 SPM S S =+ =+ =≈
The sales price of a pound of grapes is $1.55.
93. The discount is found by using the formula D = rA, where D is the discount, or decrease, r is the percent decrease, and A is the original amount.
Decrease: D
Substitute 6006
600655.567.5 .
5938.555.5
5938.555.5
55.5 yS y = =+ = = 55.5 y Rounded to hundredths 107 y =
Subtract 67.5 from both sides Divide both sides by 55.5.
67.5 67.5 .
107 sprinkler systems can be installed.
91. In order to determine the sales price, the markup must be determined, then added to the wholesale cost.
Markup rate: r = 30% = 0.3
Wholesale cost is the original price: P = $120
Markup: M
Percent decrease: r = 35% = 0.35
Original amount: A = 20 ()0.3520 7 D D = =
The discount is $7.00.
In order to determine the sales price, S, the discount is subtracted from the original price. 207 13 SAD S S =− =− =
The sales price of a set of headphones is $13.
94. The discount is found by using the formula D = rA, where D is the decrease, or discount, r is the percent decrease, and A is the original amount.
Decrease: D
Percent decrease: r = 15% = 0.15
Original amount: A = 22 ()0.1522 3.3 D D =
The discount is $3.30.
In order to determine the sales price, S, the discount is subtracted from the original price.
The sales price of a storage container is $18.70.
95. To find the percent decrease, use the formula D = rA. After 1 year, the current value of the car is $26,158, whereas, originally the car was purchased for $31,900, so A = 31900.The decrease is found by subtracting the current value from the original value.
Current Amount
3190026158
Substitute D = 12942 and A = 31900 into the formula D = rA and solve for r. () 31900 1294231900
12942 31900 r = = 31900 r 0.4057 r ≈
The percent drease in the value of the car after 1 year is approximately 40.6%.
96. To find the percent decrease, use the formula D = rA. After 1 year, the current value of the car is $11,470, whereas, originally the car was purchased for $15,500, so A = 15500.The decrease is found by subtracting the current value from the original value.
Current Amount
Substitute D = 4030 and A = 15500 into the formula D = rA and solve for r () 15500 403015500 4030 15500 r = = 15500 r 0.26 r =
The percent drease in the value of the car after 1 year is 26%.
97. To find the percent increase, use the formula I = rA. The current value of the house is $512,000, whereas, originally the house was originally $456,000, so A = 456000.The increase is found by subtracting the original value from the current value.
Current Amount
512000456000 56000 IA I I
= Substitute I = 56000 and A = 456000 into the formula I = rA and solve for r () 456000
56000456000 56000 456000 r = = 456000 r 0.1228 r ≈
The percent increase in the value of the house is approximately 12.3%.
98. To find the percent increase, use the formula I = rA. The current value of the house is $462,000, whereas, originally the house was originally $345,000, so A = 345000.The increase is found by subtracting the original value from the current value.
Current Amount
462000345000
Substitute I = 117000 and A = 345000 into the formula I = rA and solve for r.
117000345000 117000 345000 r = = 345000 r 0.3391 r ≈
The percent increase in the value of the house is approximately 33.9%.
99. The statement does not have an equal sign; therefore, it is an expression.
2 2 Identify like terms. Arrange in conventional form. 564123 64173 abaaab aaab +−++ −++
100. The statement does not have an equal sign; therefore, it is an expression.
22 2 Identify like terms. Arrange in conventional form. 109174 395 mnnmn mnn ++−+ ++
101. The statement has an equal sign; therefore, it is an equation.
104. The statement has an equal sign; therefore, it is an equation.
Multiply each term by 16 to clear fractions. 312 1420 416 (16 dd+=− 3 ) 4 (16)14(16 d += 12 ) 16 ?
4
Identify like terms.
0.5427 Identify like terms. 2 2 Subtract 2 from both sides. 1.547 4 4 Add 4 to both sides. 1.511 Divide both sides by1.5.
1.5 xx xxx x x −=+ −−= ++ −=− 1.5 x 11 1.5 7.3 x = =−
102. The statement has an equal sign; therefore, it is an equation.
Identify like terms.
Subtract 12 from both sides. Divide bot 0.25124.515 4.5 4.5 4.751215 12 12 4.7527 . r rr rr r r −+=−
Subtract 4.5 from both sides
103. The statement has an equal sign; therefore, it is an equation. 3
Multiply each term by 24 to clear fractions. 13 122 824 (24 nn+=− 1 ) 8 (24)12(24 n += 3 ) 24 ? Identify like terms.
Subtract 3 from both sides
This is a false statement. 2(24) 3288348 3 3 28848 . n n nn nn +=− =−
This equation has no solution.
May not be
1222412320 12 12 224320 . d d dd dd +=− =−
This equation has no solution.
This is a false statement. 20(16)
Subtract 12 from both sides
105. The statement does not have an equal sign; therefore, it is an expression.
Identify like terms.
Arrange in conventional form. 21148 219 xx x +−+ −+
106. The statement does not have an equal sign; therefore, it is an expression. Simplify as follows 32 32
71032 738 yyyy yyy −++ +−
Identify like terms.
Arrange in conventional form.
107. The statement has an equal sign; therefore, it is an equation.
Multiply each term by 6 to clear fractions. 11 23(34)20 (6 xx+=+ 1 ) 2 2 (34)(6 x += 1 ) 3
3
or duplicated, or
Distribute on left side.
Identify like terms.
Subtract 2 from both sides
Subtract 12 from both sides. 20(6)
3(34)2120 9122120 2 2 712120 12 12 7108 . x x xx xx xx x x + +=+ +=+ += = Divide both sides by 7. 7 7 x 108 7 3 15 7 x = =
to a publicly accessible website, in whole or in part.
108. The statement has an equal sign; therefore, it is an equation.
Multiply each term by 36 to clear fractions. 15 (52)11 418 (36 ww +=− 1 ) 4 5 (52)(36)11 18 w +=− 2 (36w
9
Distribute on left side.
Subtract 18 from both sides. )
Identify like terms. Add 10 to both sides
9(52)39610 451839610 10 10 . 5518396 18 18 55 w ww ww ww w w +=− +=− ++ +=
Divide both sides by 55. 378 55 = 55 w 378 55 48 6 55 x = =
Section 2.4 Solving and Graphing Linear
Inequalities on a Number Line
Exercises1-22:Answersaretobechecked.Exercises 1-5arecompletedasexamples.
1.
Divide both sides by 3. 321 3 x ≥ 3 x 21 3 7 x ≥ ≥
Divide both sides by 8. 856 8 z < 8 z 56 8 7 z < < ? Check the answer. The answer works 8(7)56 5656 . = =
Check a number less than 7 ( 6 z = ) to test the direction of the inequality symbol.
? Check the answer. The answer works
3(7)21 2121 . = =
Check a number greater than 7 ( 8 x = ) to test the direction of the inequality symbol.
? Check the direction of the inequality. The direction of the inequality works 3(8)21 2421 . ≥ ≥ 2.
Divide both sides by 5. 5300 5 w ≥ 5 w 300 5 60 w ≥ ≥
? Check the direction of the inequality. The direction of the inequality works 8(6)56 4856 . < <
Multiplyboth sides by4.
Multiply by neg., reverse inequality. 1 48 4 1 4448 4 192 x x x −≥
() ? Check the answer. The answer works 1 19248 4 4848 . −−= =
? Check the answer. The answer works
5(60)300 300300 . = =
Check a number greater than 60 ( 61 w = ) to test the direction of the inequality symbol.
? Check the direction of the inequality. The direction of the inequality works 5(61)300 305300 . ≥ ≥ 3.
Divide both sides by 2. 215 2 d < 2 d 15 2 7.5 d < < ? Check the answer. The answer works 2(7.5)15 1515 . = =
Check a number less than 7.5 ( 7 d = ) to test the direction of the inequality symbol.
? Check the direction of the inequality. The direction of the inequality works 2(7)15 1415 . < <
Check a number less than 192 ( 200 x =− ) to test the direction of the inequality symbol. () ? Check the direction of the inequality. The direction of the inequality works 1 20048 4 5048 . −−≥ ≥
Multiply both sides by3.
Multiply by neg., reverse inequality. 1 15 3 1 3315 3 45 y y y −≤ ⎛⎞ −−≥− ⎜⎟ ⎝⎠ ≥− 7.
Divide both sides by1.
Divide by neg., reverse inequality. 61 61 11 61 a a a −<− > > 8
Divide both sides by1.
by neg., reverse inequality.
Subtract
Subtract
Subtract
Divide both sides by1. Divide by neg., rev
Subtract
Subtract 5 from both sides
Subtract 7 from both sides
Divide both sides by9.
Subtract
Subtract
20.
2(5)318
Distribute on left side.
23. a. Write the inequality as follows: 11275 h ≥
21.
22.
Subtract 7 from both sides 718 7 7 11 . kkk k k +< <
Subtract from both sides
4(3)8720
Distribute on left side.
Subtract 7 from both sides
Add 4 to both sides
Divide both si
des by3. 3 3 z Divide by neg., reverse inequality 24 3 8 z ≤ ≤−
Distribute on left side.
Subtract 8 from both sides
Add 13 to both sides
Divide both s
ides by1.
Divide by neg., reverse inequality 7 11 7 x x < <−
9(2)31013 91831013
Distribute on left side.
9151013 10 10 1513 15 15 2 x
Subtract 10 from both sides
Subtract 15 from both sides
−<− Divide both sides by1.
Divide by neg., reverse inequality 2 11 2 x x > >
b. Divide both sides by 11. 11275 h ≥ 11 11 h 275 11 25 h ≥ ≥
Kati needs to work at least 25 hours this week.
c. Rewrite the inequality and solve as follows: Divide both sides by 11. 11400 11 h ≥ 11 h 400 11 36.36 h ≥ ≥
Kati must work at least 36.36 hours.
24 a. Write the inequality as follows: 11.25225 h < b. Divide both sides by 11.25. 11.25225 11.25 h < 11.25 h Answer rounded to hundredths. 225 11.25 20 h < <
Tyler must work less than 20 hours.
c. Rewrite the inequality and solve: Divide both sides by 11.25. 11.25180 11.25 h ≥ 11.25 h
Answer rounded to hundredths. 180 11.25 16 h ≥ ≥
Tyler must work at least 16 hours a week.
25 a. Pro Floors charges are given by the expression
253 s + . The inequality that represents Kevin keeping the refinishing charge to at most $5000 is: 2535000 s +≤
b.
Isolate variable term.
Subtract 25 from both sides 2535000 25 25 34975 3 s s +≤ ≤ 3 s Divide both sides by 3. 4975 3 1658.33 s ≤ ≤
Kevin can have no more that 1658 square feet of wood floors professionally refinished.
c.
2534000 25 25
Isolate variable term.
Subtract 25 from both sides
both sides by 3.
Kevin can have no more that 1325 square feet of wood floors professionally refinished.
26. a. Add the costs in an inequality less than or equal to 220.
10040220 n +≤
b.
Isolate variable term.
Subtract 100 from both sides 10040220
n Divide both sides by 40.
Ricardo can have up to a total of 6 rooms cleaned (the initial 3 rooms then 3 more).
27. a. The inequality that represents Eddie keeping his flat-bed truck rental cost to at most $119 is: 1920119 h +≤
b.
1920119 1920119 19 19
Distribute to simplify. Isolate variable term.
Subtract 19 from both sides.
Divide both sides by 20.
To stay within his budget, Eddie can rent a flat-bed truck for up to 5 hours, over the intial 1.5 hours.
c.
Distribute to simplify. Isolate variable term.
Subtract 19 from both sides. 192089 192089 1919
Divide both sides by 20.
To stay within his budget, Eddie can rent a flat-bed truck for up to 3.5 hours, over the initial 1.5 hours.
28 a. The inequality that represents Steven keeping his Bobcat rental cost to at most $224 is: 1925244 h +≤
1925244 1925244 19 19 25225
Distribute to simplify. Isolate variable term. Add 19 to both sides.
both sides by 25.
To stay within his budget, Steve can rent a Bobcat rental for up to 9 hours, over the initial 1.5 hours. c.
Distribute to simplify. Isolate variable term. Add 19 to both sides.
1925169 1925169 19 19 25150 25 h h h +≤ +≤ ≤ 25 h Divide both sides by 25. 150 25 6 h ≤ ≤
To stay within his budget, Steve can rent a Bobcat rental for up to 6 hours, over the initial 1.5 hours.
29 a. Add the costs in an inequality less than or equal to 115.
753115 m +≤ b.
Isolate variable term.
753115 75 75 340 3 m m +≤ ≤ 3 m Divide both sides by 3. 40 3 1 13 3 m ≤ ≤
Subtract 75 from both sides
Alicia can have her car towed up to 13 miles, over the initial 10 miles.
30. a. Add the costs in an inequality less than or equal to 95.
200.695 m +≤
b.
m m +≤ ≤
Isolate variable term.
Subtract 20 from both sides 200.695 20 20
m Divide both sides by 0.6. 75 0.6
Amy can drive the rental truck up to 125 miles on the day she rents it.
c.
Isolate variable term.
Subtract 20 from both sides 200.6125 20 20
Divide both sides by 0.6.
Amy can drive the rental truck up to 175 miles on the day she rents it.
31.
2.93 2.93
Isolate variable term. Subtract 2.93 from both sides
both sides by 0.01. 0.07 0.01
The population of Arkansas is projected to be 3 million or more after the year 2017.
32.
0.0919.4120.3 19.41 19.41 0.090.89
Isolate variable term. Subtract 19.41 from both sides
0.09 t Divide both sides by 0.09. Rounded to hundredths. 0.89 0.09 9.89 t
The population of New York is projected to be 20.3 million or more in the year 2019. 33.
0.2618.8421 18.84 18.84
0.262.16
The population of Florida is projected to be 21 million or more in the year 2018. 34.
Isolate variable term. Add 400 to both sides 24000 400 400 2400 2 . p p −> ++ > 2 p Divide both sides by 2. 400 2 200 p > >
Juanita must sell more than 200 pieces of jewelry in order to make a profit.
35. If 4 x > , then x must be less than 4. This means we can rewrite the inequality as 4 x <
36. If 5 y > , then y must be less than 5. This means we can rewrite the inequality as 5 y <
37. If 9 n −< , then n must be greater than 9 . This means we can rewrite the inequality as 9 n >−
38. If 16 m −> , then m must be less than 16 . This means we can rewrite the inequality as 16 m <−
39. If 0 t ≥ , then t must be less than or equal to 0. This means we can rewrite the inequality as 0 t ≤
40. If 10 t ≤ , then t must be greater than or equal to 10. This means we can rewrite the inequality as 10 t ≥
Isolate variable term. Subtract 18.84 from both sides
0.26 . t t +≥ ≥ 0.26 t Divide both sides by 0.26. Rounded to hundredths. 2.16 0.26 8.31 t ≥ ≥
41. If 62 x > , then 2x must be less than 6. This means we can rewrite the inequality as 26 x < To solve, divide both sides by 2. 2 2 x 6 2 3 x < <
42. If 48 x <− , then 8 x must be greater than 4. This means we can rewrite the inequality as 84 x −>
To solve, divide both sides by 8 . 8 8 x 4
43. If 521 x ≤+ , then 21 x + must be greater than or equal to 5. This means we can rewrite the inequality as 215 x +≥
To solve, isolate the variable as follows: Isolate variable term.
Subtract 1 from both sides
44. If 336 x −≥+ , then 36 x + must be less than or equal to 3 . This means we can rewrite the inequality as 363 x +≤−
To solve, isolate the variable as follows:
Isolate variable term.
Subtract 6 from both sides
by 3.
Exercises45-62:Answersaretobechecked.
Exercises45-47arecompletedasexamples.
45.
Divide both sides by 4.
Check a number greater than 5 ( 6 x = ) to test the direction of the inequality symbol. ?
Check the direction of the inequality. The direction of the inequality works 4(6)20 2420 . > >
Using interval notation, the answer is written as (5,)
Using a number line: 46. Divide both sides by 2.
Check the answer. The answer works 2(8)16 1616 . =
Check a number greater than 8 ( 9 x = ) to test the direction of the inequality symbol. ? Check the direction of the inequality. The direction of the inequality works 2(9)16 1816 . > >
Using interval notation, the answer is written as (8,) ∞
Using a number line:
answer.
Check a number less than 6 ( 9 m =− ) to test the direction of the inequality symbol. ? Check the direction of the inequality. The direction of the inequality works (9) 2 3 32 .
Using interval notation, the answer is written as (,6) −∞−
Using a number line:
48.
Multiply both sides by 4.
Using interval notation, the answer is written as (,6)
Using a number line:
49. Divide both sides by3.
Using interval notation, the answer is written as (,4]
Using a number line:
50.
sides by6.
Using interval notation, the answer is written as (,5]
Using a number line: 51. Multiply both sides by.
Using interval notation, the answer is written as
Using a number line:
52.
Multiply both sides by. 7 537 755 y
by neg., reverse inequality
Using interval notation, the answer is written as 21 , 25 ⎛⎞
Using a number line: 53.
Multiply both sides by. 4 3 3 3 a −≤− 3 a () Multiply by neg., reverse inequality. 43 12 a
Using interval notation, the answer is written as [12,) ∞ .
Using a number line: 54.
Multiply both sides by. 11 5 5 5 b −≥− 5 b ⋅ () Multiply by neg., reverse inequality 115 55 b ⋅− ⎛⎞ ≤− ⎜⎟ ⎝⎠
Using interval notation, the answer is written as (,55] −∞
Using a number line:
55.
Distribute on left side.
Subtract from both sides
Subtract 10 from both sides
Using interval notation, the answer is written as (,3] −∞
Using a number line:
56.
Distribute on left side. Simplify on left side.
Subtract 3 from both sides
Subtract 1 from both sides
4 4 p Divide by neg., reverse inequality. 8 4 2 p ≥ ≥−
Using interval notation, the answer is written as [2,) −∞
Using a number line:
57.
Subtract 3.5 from both sides
Subtract 8 from both sides
Divide both sides by1.
Divide by neg., rev
Using interval notation, the answer is written as (,12] −∞
Add to both sides
Divide both sides by 0.5. 0.5114 0.5114 11 11 0.515 0.515 0.50.5 30 x xx xx x x x x −+≥−− ++ +≥− ≥− ≥ ≥−
Subtract 11 from both sides
Using interval notation, the answer is written as [30,) −∞
Using a number line:
59.
Subtract 6 from both sides
Divide both sides by4. 27617 6 6 4717 7 7 410 4 c cc cc c c +≤+ −+≤ −≤ 4 c Divide by neg., reverse inequality 10 4 2.5 c ≥ ≥−
Subtract 7 from both sides
Using interval notation, the answer is written as [2.5,) −∞
Using a number line:
60.
Using a number line: 58.
Divide both sides by14. 516944 9 9 141644 16 16 1428 14 . . z zz zz z z −−≤− −−≤− ++ −≤− 14 z Divide by neg., reverse inequality 28 14 2 z ≥ ≥
Subtract 9 from both sides
Add 16 to both sides
Using interval notation, the answer is written as [2,) ∞
Using a number line:
61.
5122(510)30 512102030 5121050 10 10 51250 12 12 w ww ww ww ww w
Distribute on right side. Simplify on right side
Subtract 10 from both sides
Add 12 to both sides
Divide both sides by14. 538 5 . w −≥− 5 w Divide by neg., reverse inequality 38 5 7.6 w ≤ ≤
Using interval notation, the answer is written as (,7.6]
Using a number line:
62.
This is equivalent to the inequality 12 x <− 67.
This is equivalent to the inequality 69 x > 68.
This is equivalent to the inequality 42 x > 69.
This is equivalent to the inequality 16 x ≤− 70.
Subtract 32 fro 4(38)5(410)26 1232205026 12322024 20 20 83224 32 32 . g gg gg gg gg g +<−+ +<−+ +<− −+<− m both sides
Distribute on both sides. Simplify on right side
Subtract 20 from both sides
Divide both sides by8. 856 8 g −<− 8 g Divide by neg., reverse inequality 56 8 7 g > >
Using interval notation, the answer is written as (7,) ∞
Using a number line: 63.
This is equivalent to the inequality 1 x ≥− 64.
This is equivalent to the inequality 7 x ≥− . 65.
This is equivalent to the inequality 9 x <
This is equivalent to the inequality 0 x ≤ .
71. Let s = the number of students enrolled in the beginning algebra class.
1638 s ≤≤
72. Let p = the number of passengers that can be on the Boeing 717-200.
0106 p ≤≤
73. Let s = the number of students that can ride on school bus 12.
071 s ≤≤
74. Let p = the number of people the camp can accommodate.
0177 p ≤≤
75. Let p = the number of people the camp can accommodate.
200450 p ≤≤
76. Let c = the amount in dollars this accountant charges to prepare tax returns.
150500 c ≤≤
77. Let h = the height requirements in inches for children to ride on some attractions at LEGOLAND with an adult.
4252 h ≤≤
78. Let n = the course number.
100300 n ≤<
79. Let r = the average adult’s resting heart rate in beats per minute.
66100 r ≤≤
80. Let r = the average newborn baby’s resting heart rate in beats per minute.
100160 r ≤≤
81.
Isolate by dividing all sides by 2. 628 628 222 34 xx x x << << <<
Since both sides do not have “equal to” as part of the inequality, the solution written in interval notation will have parentheses on both sides. Always write intervals from lowest to highest numbers, so we get the interval () 3,4 .
Using a number line:
82.
Isolate by dividing all sides by 5. 25555 25555 555 511 xx x x << << <<
Since both sides do not have “equal to” as part of the inequality, the solution written in interval notation will have parentheses on both sides. Always write intervals from lowest to highest numbers, so we get the interval () 5,11
Using a number line:
83.
intervals from lowest to highest numbers, so we get the interval () 6,2
Using a number line: 84.
Isolate . Subtract 5 from all sides 456 5 5 5 91 . bb b −<+< −<<
Since both sides do not have “equal to” as part of the inequality, the solution written in interval notation will have parentheses on both sides. Always write intervals from lowest to highest numbers, so we get the interval () 9,1
Using a number line: 85.
Divide all sides by 3. 123321 3 3 3 9318 9318 333 36 . xx x x x <+< << << <<
Isolate . Subtract 3 from all sides
Since both sides do not have “equal to” as part of the inequality, the solution written in interval notation will have parentheses on both sides. Always write intervals from lowest to highest numbers, so we get the interval () 3,6 .
Using a number line: 86.
Isolate . Subtract 10 from all sides
Isolate . Add 1 to all sides
713 +1 1 +1 62 . yy y −<−<− + −<<−
Since both sides do not have “equal to” as part of the inequality, the solution written in interval notation will have parentheses on both sides. Always write
Divide all sides by 2. 2021036 10 10 10 10226 10226 222 513 . xx x x x <+< << << <<
Since both sides do not have “equal to” as part of the inequality, the solution written in interval notation will have parentheses on both sides. Always write
intervals from lowest to highest numbers, so we get the interval () 5,13
Using a number line:
87.
intervals from lowest to highest numbers, so we get the interval [] 6,10
Using a number line:
Isolate by multiplying all sides by 2. 35 2
(2)3(2)(2)5 2
610 k k k k << << <<
Since both sides do not have “equal to” as part of the inequality, the solution written in interval notation will have parentheses on both sides. Always write intervals from lowest to highest numbers, so we get the interval () 6,10 .
Using a number line:
88.
Isolate by multiplying all sides by 4. 23 4 n n −<<
2(4)(4)(4)3 4 812 n n −<< −<<
Since both sides do not have “equal to” as part of the inequality, the solution written in interval notation will have parentheses on both sides. Always write intervals from lowest to highest numbers, so we get the interval () 8,12
Using a number line:
89.
Isolate .
Divide all sides by 6. 406464 4 4 4 36660 36660 666 610 xx x x x ≤+≤ ≤≤ ≤≤ ≤≤
Subtract 4 from all sides
Since both sides have “equal to” as part of the inequality, the solution written in interval notation will have brackets on both sides. Always write
90. Isolate . Add 1 to all sides
Divide all sides by 8. 258115 1 1 1 24816 24816 888 32 . xx x x x −≤−≤ +++ −≤≤ ≤≤ −≤≤
Since both sides have “equal to” as part of the inequality, the solution written in interval notation will have brackets on both sides. Always write intervals from lowest to highest numbers, so we get the interval [] 3,2
Using a number line: 91.
Multiply all sides by 6. 1.50.5 6 1.5(6)(6)0.5(6) 6 93 m m m m −≤≤− −≤≤− −≤≤−
Isolate .
Since both sides have “equal to” as part of the inequality, the solution written in interval notation will have brackets on both sides. Always write intervals from lowest to highest numbers, so we get the interval [] 9,3
Using a number line: 92.
Isolate .
Multiply all sides by 4. 325 4 2 2 2 13 4 (4)1(4)(4)3 4 412 w w w w w ≤+≤ ≤≤ ≤≤ ≤≤
Subtract 2 from all sides
Since both sides have “equal to” as part of the inequality, the solution written in interval notation
May not be scanned, copied or duplicated, or posted to a publicly accessible website, in whole or in part.
will have brackets on both sides. Always write intervals from lowest to highest numbers, so we get the interval [] 4,12
Using a number line:
93. Isolate . Add 2 to all sides
Since both sides have “equal to” as part of the inequality, the solution written in interval notation will have brackets on both sides. Always write intervals from lowest to highest numbers, so we get the interval [] 20,4
Using a number line:
94.
Multiply
Since both sides have “equal to” as part of the inequality, the solution written in interval notation will have brackets on both sides. Always write intervals from lowest to highest numbers, so we get the interval [] 3,45 .
Using a number line:
Isolate by multiplying all sides by 2 430 2 (2)4(2)(2)30 2 860 x x x x << << <<
Since both sides do not have “equal to” as part of the inequality, the solution written in interval notation will have parentheses on both sides. Always write intervals from lowest to highest numbers, so we get the interval () 8,60 .
Using a number line:
96.
Isolate by multiplying all sides by 15. 525 15
(15)5(15)(15)25 15 75375 x x x x << << <<
Since both sides do not have “equal to” as part of the inequality, the solution written in interval notation will have parentheses on both sides. Always write intervals from lowest to highest numbers, so we get the interval () 75,375
Using a number line:
97. () () Multiply by Isolate 4 3 3 71 4 4434 71. 3343 284 33 x x x x −<<− ⎛⎞⎛⎞⎛⎞ −<<−⎜⎟⎜⎟⎜⎟ ⎝⎠⎝⎠⎝⎠ −<<−
Since both sides do not have “equal to” as part of the inequality, the solution written in interval notation will have parentheses on both sides. Always write intervals from lowest to highest numbers, so we get the interval 284 , 33 ⎛⎞ ⎜⎟ ⎝⎠
Using a number line:
98. () () Multiply by
Isolate 7 5 5 252 7 7757 252. 5575 14 35 5 x x x x −<<−
Since both sides do not have “equal to” as part of the inequality, the solution written in interval notation will have parentheses on both sides. Always write intervals from lowest to highest numbers, so we get the interval 14 35, 5 ⎛⎞
Using a number line:
99.
Using a number line:
101.
3.5228.5 2 2 2
5.5210.5 5.5210.5 222 2.755.25 . xx x x x ≤−≤
Isolate . Add 2 to all sides
Divide all sides by 2.
Since both sides have “equal to” as part of the inequality, the solution written in interval notation will have brackets on both sides. Always write intervals from lowest to highest numbers, so we get the interval [] 2.75,5.25
Using a number line: 100.
12.3646.6 4 4 4
Isolate . Subtract 4 from all sides
Divide all sides by 2.
16.362.6 16.362.6 666 2.720.43 . xx x x x −≤+≤
Answers rounded to hundredths.
Since both sides have “equal to” as part of the inequality, the solution written in interval notation will have brackets on both sides. Always write intervals from lowest to highest numbers, so we get the interval [] 2.72,0.43 .
Isolate . Subtract 6 from all sides
Divide all sides by 3.
9363 6 6 6 1533 . vx v −≤+< −≤<− 1533 333 51 v v ≤< −≤<−
In interval notation, the solution set will have a bracket on the left side, since it is less than or equal to, and a parenthesis on the right side, since it is less than. This gives us the interval [ )5,1 .
Using a number line: 102.
Isolate . Add 4 to all sides
Divide all sides by 2. 5248 4 4 4 9212 9212 222 4.56 . xx x x x <−≤ +++ <≤ <≤ <≤
In interval notation, the solution set will have a parenthesis on the left side, since it is less than, and a bracket on the right side, since it is less than or equal to. This gives us the interval ( ] 4.5,6
Using a number line: 103.
Multiply all sides by. 5 5 783 3 8 8 8 5 111 3 . a a a −<−≤ +++ <≤ 3353 111 5535 333 55 a a ⎛⎞⎛⎞⎛⎞ <≤ ⎜⎟⎜⎟⎜⎟ ⎝⎠⎝⎠⎝⎠ <≤
Isolate . Add 8 to all sides 3
In interval notation, the solution set will have a parenthesis on the left side, since it is less than, and a
bracket on the right side, since it is less than or equal to. This gives us the interval 333 , 55 ⎛⎤ ⎜ ⎥
Using a number line: 104.
Isolate . Add 9 to all sides
Divide all sides by 4.
In interval notation, the solution set will have a parenthesis on the left side, since it is less than, and a bracket on the right side, since it is less than or equal to. This gives us the interval ( ] 1.5,6.5
Using a number line:
105 The statement has an inequality symbol; therefore, it is an inequality.
Isolate variable term.
Subtract 7 from both sides
by 2.
Using interval notation, the answer is written as (17,) ∞ . Use a parenthesis next to the 17 because the inequality is not also “equal to” 17. Always use a parenthesis next to the infinity symbol.
106. The statement has an inequality symbol; therefore, it is an inequality.
Isolate variable term. Add 10 to both sides 61032 10 10 642 6 c c −< ++ < 6 c Divide both sides by 6. 42 6 7 c < <
Using interval notation, the answer is written as () ,7−∞ . Use a parenthesis next to the 7 because the
inequality is not also “equal to” 7. Always use a parenthesis next to the infinity symbol.
107. The statement does not have an equal sign or an inequality symbol, therefore, it is an expression.
Combine like terms 4712 . 87 xx x +− −+
108. The statement does not have an equal sign or an inequality symbol, therefore, it is an expression. 22 2 Combine like terms Write in conventional form 2347 . 543 . mmm mm +−− −−+
109. The statement has an equal sign; therefore, it is an equation.
Isolate variable term.
Subtract 3 from both sides
Subtract 12 from both sides 512316 3 3 21216 12 12 228 h hh hh h
Divide both sides by 2. 28 2 14 h = =−
110. The statement has an equal sign; therefore, it is an equation.
Isolate variable term. Add 5 to both sides Add 14 to both sides 2(47)519
Distribute on left side.
111. The statement has an inequality symbol; therefore, it is an inequality.
Isolate variable term.
Subtract 7 from both sides
Subtract 4 from both sides Divide both sides by4.
Using interval notation, the answer is written as ( ] ,6
. Use a bracket next to the 6 because the inequality is also “equal to” 6. Always use a parenthesis next to the infinity symbol.
112. The statement has an inequality symbol; therefore, it is an inequality.
Isolate variable term.
Subtract 3 from both sides
Add 14 to both sides
Divide both sides by8.
Using interval notation, the answer is written as [ ) 2,
113. The statement has an equal sign; therefore, it is an equation. Multiply each term by 3 first to clear the denominators.
Multiply by 3.
115. The statement has an inequality symbol; therefore, it is an inequality.
Isolate .
Isolate variable term.
Subtract from both sides
Subtract 4 from both sides
114. The statement has an equal sign; therefore, it is an equation. Multiply each term by 7 first to clear the denominators. (7 4 ) 7 (7 d 2 ) 7 3 (7)2 7 d =+ (7
Isolate variable term.
Subtract 14 from both sides
Add 2 to both sides
Divide both sides by10. )
Subtract 2 from all sides
Divide all sides by 4. 204212 2 2 2 22410 22410 444 5.52.5 . aa a a a −<+< −<< << −<<
The answer in interval notation is () 5.5,2.5 .
116. The statement has an inequality symbol; therefore, it is an inequality. Multiply each term by 4 first to clear the denominator.
Multiply by 4.
Isolate .
Subtract 40 from all sides 81016 4
(4)8(4)(4)1016(4) 4 324064 40 40 40 824 . y y y y y ≤+< ≤+< ≤+<
The answer in interval notation is [ ) 8,24 .
117. The statement does not have an equal sign or an inequality symbol, therefore, it is an expression.
Combine like terms
Rewrite fractions with LCD9 Write in conventional form
118. The statement does not have an equal sign or an inequality symbol, therefore, it is an expression.
Combine like terms
Rewrite -term fractions with LCD10
Rewrite constant-term fractions with LCD2
119. The statement has an inequality symbol; therefore, it is an inequality.
Isolate .
120. The statement has an inequality symbol; therefore, it is an inequality.
Isolate .
Multiply all sides by.
Multiply by neg., reverse inequalities Reorder compound inequality from 47 2
(2)4(2)7(2) 2 2 814 . 148 g g g g g
smallest to greatest
The answer, in interval notation, is [ ) 14,8
Multiply all sides by . 84 3
(3)8(3)4(3) 3 3 2412 x x x x −≤≤ −≤≤
The answer in interval notation is [ ] 24,12
Chapter 2 Review
1 a. To check if the given values of the variables are a solution to the equation, evaluate each side of the equation and see if the two sides are equal.
262.575(3.5)
? Substitute values for and . 75 Original equation.
262.5262.5 This statement is true. Dt
Dt = = =
Because the final statement is true, 3.5 t = and 262.5 D = is a solution to the equation 75 Dt =
b. These values represent that 262.5 miles is the distance traveled when driving for 3.5 hours.
2 a. To check if the given values of the variables are a solution to the equation, evaluate each side of the equation and see if the two sides are equal.
4. To check if the given values of the variables are a solution to the equation, evaluate each side of the equation and see if the two sides are equal. ? ? Substitute 35 Original equation. (14)3(3)5 3 and 14 1495 Simplify both sides 1414 This statement is true. yx xy =−+ =−−+=−= =+ = Because the final statement is true, 3 x =− and 14 y = is a solution to the equation 35yx=−+ .
Given equation. (49000)(26000) Substitute value of variables
FL = = =
? Substitute values for and . 11.5 Original equation.
690011.5(600) 69006900 This statement is true. FL
Because the final statement is true, 600 L = and 6900 F = is a solution to the equation 11.5 FL = .
b. These values represent that a loan of $600,000.00 would have an origination fee of $6,900.00.
3. To check if the given values of the variables are a solution to the equation, evaluate each side of the equation and see if the two sides are equal.
? ? Substitute 248 Original equation. 2(2)4(3)8 2 and 3.
4128 Simplify left side 168 This statement isnot true.
Because the final results are not equal, 2 x =− and 3 y = is not a solution to the equation 248. xy −+=−
5. Using the equation PRC =− substitute 49000 R = for the revenue, 26000 C = for costs, then solve for P to find the company's profit. .
Simplify on right side . 23000 P PRC P =− =− = The solution is 23000 P = which represents that the company has monthly profit of $23,000.
6. Using the equation PRC =− substitute 225000 R = for the revenue, 186000 C = for costs, then solve for P to find the company's profit.
Given equation. (225000)(186000) Substitute value of variables Simplify on right side 39000 P PRC P =− =− = The solution is 39000 P = which represents that the company has monthly profit of $39,000. Exercises7-14:Answersaretobechecked.
7. ? Add 3 to both sides
Check the answer. The answer works. 39 +3 +3 6 (6)39 99 x x −=− =− −−=− −=− 8. ? Add 16 to both sides
Check the answer. The answer works. 169 +16 +16
Subtract 0.02 from both
Check the answer. The answer works.
15. To solve for q means to isolate q on one side of the equal sign.
Identify the variable term to isolate. Subtract 2 from both sides 29 2 2
16. To solve for x means to isolate x on one side of the equal sign.
Identify the variable term to isolate. Add to both sides 2 2 y xy yy xy −= ++
17. To solve for y means to isolate y on one side of the equal sign.
Identify the variable term to isolate. Subtract 8 from both sides 82448 8 8 24848 x xy xx yx −= −=−+
Divided both sides by24. 24848 2424 848
18. To solve for p means to isolate p on one side of the equal sign.
Identify the variable term to isolate. Add 5 to both sides
Divided both sides by 3.
Exercises19-28:Answersaretobechecked. Exercises19-22arecompletedasexamples. 19.
Divide both sides by 7 Check the answer. The answer works.
20.
Divide both sides by 4. Check the answer. The answer works.
answer.
6.28.4 6.28.4
Divide both sides by 6.2 Answer rounded to hundredths.
Divide both sides by0.5 0.516.5 0.516.5 0.50.5 33 k k k
29. a. To find the number of airplanes that are needed to transport 520 people, substitute 520 P = into the equation 126 P J = and solve for J
hundredths.
Since you cannot have 4.13 airplanes, we round the final answer up to the nearest whole number. To transport 520 people, 5 airplanes would be needed.
b. To find the number of people that the carrier can transport at a time with 15 planes, substitute 15 J = into the equation 126 P J = and solve for P
Substitute 15.
Multiply both sides by 126. 126 15 126 12615126 126 1890 J P J P P P = = = ⋅=⋅ =
If the airline carrier has 15 planes, they can transport 1890 people at a time.
30 a. To fine Devora's monthly salary if she works 140 hours, substitute 140 h = into the equation 14 Sh = and solve for S
Substitute 140. 14 14(140) 1960 Shh S S = = = =
Devora's monthly salary will be $1960 if she works 140 hours.
b. To find the number of hours Devora needs to work a month to make $2100 a month, substitute 2100 S = into the equation 14 Sh = and solve for h.
Substitute 2100.
Divide both sides by 14 14 210014 210014 1414 150 ShS h h h = = = = =
Devora has to work 150 hours a month to make $2100 a month.
c. To find the number of hours Devora needs to work a month to earn $1330 a month, substitute 1330 S = into the equation 14 Sh = and solve for h.
Substitute 1330.
Divide both sides by 14 14 133014 133014 1414 95 ShS h h h = = = = =
Devora has to work 95 hours a month to earn $1330 a month.
Exercises31-38:Answersaretobechecked. Exercises31and32arecompletedasexamples. 31.
Add 9 to both sides 6963 9 9 654 x x −=− ++ =−
6(9)963 54963
Divide both sides by 6.
Check the answer.
The answer works.
174
Add 17 to both sides
The answer works. 51733 17 17 550 550 55 10 5(10)1733 501733 3333 y y
Divide both sides by 5.
Check the answer.
Subtract 9 from both sides
Divide both sides by2. .
Add 19 to both sides Divide both sides by8.
Add 1 to both sides
Multiply both sides by 2.
2019 Cengage Learning, Inc. May not be scanned, copied or duplicated, or posted to a publicly accessible website, in whole or in part.
Subtract 2 from both sides
Multiply both sides by3. 24 3 2 2 6 3 363 3 18 x x x x −+=−
37.
Add 8.6 to both sides
Divide both sides by1.2. . 8.61.210.6 8.6 8.6 1.219.2 1.219.2 1.21.2 16 x x x x −−= ++
38.
Add 1 to both sides
Divide both sides by 0.25. . 0.2517 1 1 0.256 0.256 0.250.25 24 y y y y −=− ++ =−
Exercises39-46:Answersaretobechecked.
Exercises39and40arecompletedasexamples.
39 Let x = a number. The sentence translates as follows ? ? Divide both sides by 9. Check the answer. The answer works. 976 : 976 99 4 8 9 4 9876 9 76 976 9 7676 x Solve
40. Let x = a number. The sentence translates as follows Divide both sides by2.
Check the answer. The answer works.
41. Let x = a number. The sentence translates as follows
Subtract 16 from both sides
Divide both sides by 2.
42. Let x = a number. The sentence translates as follows
Subtract 4 from both sides
Multiply both sides by 2.
43. Let x = a number. The sentence translates as follows Multiply both sides by 4. 0.25 4 : 40.254 4 1 x Solve x x = ⋅=⋅ =
44. Let x = a number. The sentence translates as follows
both sides by3.
45. Let x = a number. The sentence translates as follows
Subtract 8 from both sides
Divide both sides by3.
46 Let x = a number. The sentence translates as follows
47. Identify the variable to isolate. Dcma Dc ca = = ma ca Divide both sides by .ca
Identify the variable to isolate.
sides by .lw
Identify the variable to isolate.
51.
Identify the variable to isolate.
= (2bh Multiply both sides by 2. ) 2 Abh h ⋅ = h Divide both sides by . 2 h A b h =
Identify the variable to isolate.
Subtract 9 from both sides
both sides by3.
Identify the variable to isolate. Add 8 to both sides
both sides by3. .
Exercises53-62:Answersaretobechecked.
53. Let x = a number. The sentence translates as follows
Subtract 3 from both sides
Subtract 1 from both sides
Multiply both sides by1.
54. Let x = a number. The sentence translates as follows 1 104 2 xx
Multiply each term by 2.
Subtract 2 from both sides
Subtract 20 from both sides.
answer.
answer works.
Add 8 to both sides
sides by
Subtract 7 from both sides
Subtract 12 from both sides
Divide both sides by 2. 91276 7 7 2126 12 12 26 26 22 3 y yy yy y y y y +=+ += =− = =− ? ? Check the answer. The answer works. 9(3)127(3)6 2712216 1515 −+=−+ −+=−+ −=− 57.
Distribute on left side.
Subtract 12 from both sides
+=−⎜⎟ ⎜⎟ ⎢⎥ ⎝⎠⎣⎦⎝⎠
Subtract 25 from both sides . . 5(5)1213 5251213 12 12 72513 25 25 738 738 77 a aa aa aa a a a +=− +=− −+=− −=− = ? Divide both sides by7.
Check the answer. The answer works. 383 or 5 77 3838 551213 77 38353891 512 7777 7345691 5 777 365365 77 aa==
58.
Simplify on left side.
Divide both 117752 71852 5 5 2182 18 18 220 220 22 x xx xx xx x x x ++=− +=− +=− =− = ? ? sides by 2. Check the answer.
Subtract 5 from both sides
Subtract 18 from both sides
117(10)75(10)2 1870502 5252 x =− +−+=−−
59. ?
The answer works. 10
4153(31) 41593 4143 4 4 13 k kkk kkk kk kk +=−+− +=−+− +=− =− nt.
Distribute on right side. Simplify on right side.
Subtract 4 from both sides
This is a false stateme
This equation has no solution. 60. ? Distribute on left side. Simplify on left side.
97(1)29 97729 2729 2 2 79 h hhh hhh hh hh −+=+ −−=+ −=+ −=
Subtract 2 from both sides
This is a false statement.
This equation has no solution. 61.
53135(2) 5313510 5353
5 5 33 x xx xx xx xx −+=−+ −+=−− −+=−+ ++ =
Distribute on right side. Simplify on right side.
Add 5 to both sides
This is a true statement.
The equation is an identity therefore the solution is all real numbers or
To check this, we will randomly choose two real numbers 3 and 1, and substitute these into the equation.
()
62.
() ? ? ? ? Check the answer using3.
The answer works
5(3)313532 153135(1) 1818 . 5(0)313502 03135(2) 33 −−+=−−+ +=−− = −+=−+ +=− = s.
Check the answer using 0.
The answer work
7102(32)6 710646 66 66 y yyy yyy yy yy −+−+=− −−+=− −=− ++ −=−
Distribute-left side. Simplify on left side.
Add to both sides
This is a true statement.
The equation is an identity therefore the solution is all real numbers or
To check this, we will randomly choose two real numbers 1 and 5, and substitute these into the equation. [] () ? ? ?
Check the answer using1. 7(1)102312(1)6
7102(32)7 172(5)7 77 . −−+−−+=−− −−++=− −+=− −=−
[] () ? ? ?
The answer works
Check the answer using 5. 7(5)102352(5)6 35102(152)1 252(13)1 11 . −+−+=− −+−+=− +−=− −=−
The answer works
63. The sum of the measures of the angles in a triangle must equal 180° . Add the measures of the given angles and set them equal to 180° then solve for x
Combine like terms
(6)(12)180 .
Add 6 to both sides.
Divide both sides by 3
Now substitute 62 x = into the expressions that represent the angles to find the measure of each angle. The measures of the three angles are 62,68,and50. °°°
To check the solution, the measures of the angles should add up to 180.
? Check the answer. The answer works 626850180 180180 . °+°+°=° °=°
64.The sum of the measures of the angles in a triangle must equal 180° . Add the measures of the given angles and set them equal to 180° then solve for x
3(20)(10)180
Combine like terms
Subtract 10 from both sides.
Divide both sides by 5
Now substitute 34 x = into the expressions that represent the angles to find the measure of each angle. The measures of the three angles are 102,54,and24. °°°
To check the solution, the measures of the angles should add up to 180.
? Check the answer. The answer works 1025424180 180180 . °+°+°=° °=°
Exercises65-78:Answersaretobechecked. Exercises65and66arecompletedasexamples.
65.
Divide both sides by 4. 4149 1 1 448 x x +> >
Subtract 1 from both sides
448 44 12 x x > > ? ? Check the answer.
4(12)149 48149 4949 . += += =
The answer works
Check a number greater than 12 ( 13 x = ) to test the direction of the inequality symbol.
? ? Check the direction of the inequality. The direction of the inequality works
4(13)149 52149 5349 . +> +> >
Using interval notation, the answer is written as () 12, ∞ . Use a parenthesis next to the 12 because x is not also “equal to” 12. Always use a parenthesis next to the infinity symbol.
Using a number line:
66.
Add 2 to both sides Divide both sides by 3. 3219 2 2
.
? ? Check the answer.
3(7)219 21219 1919 . −= −= =
The answer works
Check a number less than 7 ( 6 x = ) to test the direction of the inequality symbol. ? ? Check the direction of the inequality.
3(6)219 18219 1619 . −≤ −≤ ≤
The direction of the inequality works
Using interval notation, the answer is written as ( ] ,7−∞ . Use a bracket next to the 7 because x is also “equal to” 7. This means that 7 is part of the solution set. Always use a parenthesis next to the infinity symbol.
Using a number line:
Multiply both sides by1.
()()
Multiply by neg., reverse inequality. 7 171 7 x x x −≤
Using interval notation, the answer is written as [ ) 7, −∞ . Use a bracket next to the 7 because x is also “equal to” 7 . This means that 7 is part of the solution set. Always use a parenthesis next to the infinity symbol.
Using a number line:
68.
Multiply both sides by1.
()()
Multiply by neg., reverse inequality. 3 131 3 y y y −≥
Using interval notation, the answer is written as ( ],3−∞− . Use a bracket next to the 3 because y is also “equal to” 3 . This means that 3 is part of the solution set. Always use a parenthesis next to the infinity symbol.
Using a number line:
69.
Subtract 1 from both sides
Divide both sides by5.
Divide by neg., reverse inequality. 5116 1 1 517 517 55 17 5 . y y y y
Using interval notation, the answer is written as 17 , 5 ⎛⎤
. Use a bracket next to the 17 5 because y is also “equal to” 17 5 . This means that 17 5 is part of the solution set. Always use a parenthesis next to the infinity symbol.
Using a number line:
Add 7 to both sides
Divide both sides by6.
Divide by neg., reverse
Using interval notation, the answer is written as 2 , 3
. Use a parenthesis next to the 2 3 because
x is not also “equal to” 2 3 . Always use a parenthesis next to the infinity symbol.
Using a number line: 71.
Subtract 3 from both sides
Multiply both sides by4.
Multiply by neg., reverse inequality.
Using interval notation, the answer is written as [ ) 20, ∞ . Use a bracket next to the 20 because y is also “equal to” 20. This means that 20 is part of the solution set. Always use a parenthesis next to the infinity symbol.
Using a number line: 72.
Subtract 4 from both sides 5
Multiply both sides by.
Multiply by neg., reverse inequality.
Using interval notation, the answer is written as () ,30−∞ . Use a parenthesis next to the 30 because t is not also “equal to” 30. Always use a parenthesis next to the infinity symbol.
Using a number line:
73.
Add 2 to both sides
Add 3 to both sides
Multiply both sides by1. Multiply by neg., reverse inequa
lity. 12 x >−
Using interval notation, the answer is written as () 12, −∞ . Use a parenthesis next to the 12 because x is not also “equal to” 12 . Always use a parenthesis next to the infinity symbol.
Using a number line:
74.
Add 3 to both sides
Divide both sides by. 8139 3 3 519 1 1 510 5 510 55 y yy yy y y y −−>−+ ++ −−> ++ −> < y. 2 y <−
Add 1 to both sides
Divide by neg., reverse inequalit
Using interval notation, the answer is written as () ,2−∞−
Using a number line:
75.
The solution in interval notation is () 3,4
Using a number line: 76. Isolate . Add 6 to all sides
Divide all sides by 5. 15624 6 6 6 5530 . xx x −<−< +++ << 5530 555 16 x x << <<
The solution in interval notation is () 1,6 .
Using a number line: 77. Isolate .
Multiply each side by 3 29 3 (3)2(3)9(3) 3 627 . x x x x −<< −<< −<< ?? (3) 29 Check direction using a number 3 between6 and 27. 219 The direction of the inequality symbols works. −<< −<<
The solution in interval notation is () 6,27 .
Using a number line:
78.
Multiply each side by 2 8115 2 11 11 11 1916 2 (2)19(2)16(2) 2 3832 . x y y y y −<+<− −<<− −<<− −<<−
() Isolate . Subtract 11 from all sides
Isolate .
Subtract 3 from all sides
Divide all sides by 2. 32311 3 3 3 628 628 222 34 . xx x x x −<+< −<< << −<<
The solution in interval notation is () 38,32
Using a number line:
79 The graph of the interval [3,) −∞ is below:
The values that are shown start at 3 , include 3 , and all numbers greater than 3 .
This is equivalent to the inequality 3 x ≥−
80. The graph of the interval (5,) ∞ is below:
The values shown start at 5, do not include 5, and include all numbers greater than 5. This is equivalent to the inequality 5 x > .
81. The graph of the interval (,6) −∞ is below:
The values that are shown start at 6, do not include 6, and include all numbers less than 6. This is equivalent to the inequality 6 x <
82. The graph of the interval (,12] −∞− is below:
The values that are shown start at 12 , include 12 , and all numbers less than 12
This is equivalent to the inequality 12 x ≤− .
83. If 6 x −> , then x must be less than 6 . This means we can rewrite the inequality as 6 x <− .
84. If 0 y ≤ , then y must be greater than or equal to 0. This means we can rewrite the inequality as 0 y ≥
86
Isolate variable term. 32.4921.65119.12
Subtract 21.65 from both sides
Jessie can rent a car for 3 days or less.
87. a. The inequality that represents Karyn keeping the bartending bill at the wedding to at most $235 is: 13540235 h +≤ , where h is the number of hours after 3 hours that the bartender is working at the wedding.
b. Isolate variable term.
Subtract 135 from both sides
13540235 135 135 40100 40 . h h +≤ ≤ 40 h Divide both sides by 40. 100 40 2.5 h ≤ ≤
Karyn can have the bartender work no more than 2.5 hours past the first 3 hours.
88. a. Add the costs in an inequality less than or equal to 65.
501.565 m +≤
b.
501.565 m +≤
Isolate variable term.
8024000 n −>
85. Isolate variable term.
Subtract 50 from both sides 50 50 1.515 1.5 m ≤ 1.5 m Divide both sides by 1.5. 15 1.5 10 m ≤ ≤
Add 2400 to both sides 2400 2400
802400
80 n
2400
80 n ++ >
80 30 n > >
Divide both sides by 80.
Kiano must sell more than 30 stained-glass windows in order to make a profit.
Trey can have the TV delivered and stay within his budget for the delivery charge if he lives 10 miles or less from the store.
Chapter 2 Test
1. To check if the given value of the variable is a solution to the equation, evaluate each side of the equation and see if the two sides are equal.
? ? ? Substitute . 582 Original equation. 5(2)82 2 1082 Simplify both sides
22 This statement is nottrue.
Because the final results are not equal, 2 b =− is not a solution to the equation 582 b +=
2. To check if the given values of the variables are a solution to the equation, evaluate each side of the equation and see if the two sides are equal.
5450 Variable term is isolated. 5450 Divide both sides by5. 55 90 w w w −= =−
5(90)450 Check the answer. 450450 The answer works. −−= =
Identify variable term to isolate. Subtract 25 from both sides.
Divide both sides by4
? ? Check the answer.
? Substitute values for and . 70 Original equation.
29470(4.2) 294294 This statement is true. dt dt = = =
Because the final statement is true, 4.2 t = and 294 d = is a solution to the equation 70 dt = 3.
Identify the variable term Check the answer. The answer works. 517 +5 +5 12 5(12)17 1717 x
? Add 5 to both sides
4. To solve for c means to isolate c on one side of the equal sign.
abab
Identify the variable term to isolate. ubtract and from both sides. S
ab
5. Using the equation PRC =− substitute 250000 R = for the revenue, 88000 C = for costs, then solve for P to find the company's profit.
Given equation. (250000)(88000) . Simplify on right side 162000 P PRC P =− =− = The solution is 162000 P = which represents that the company has monthly profit of $162,000.
Substitute value of variables
The answer works 4(36)25119 14425119 119119 . −+=−
Identify variable term to isolate. Subtract 7 from both sides. Multiply both sides by 3 75 3 7 7 12 3 3123 . 3 36 x x x x +=−
⋅=−⋅ =− ? ? Check the answer. The answer works (36) 75 3 1275 55 . +=− −+=−
Distribute-left side. Simplify on left side.
This is a true statement. 4(4)5411 4165411 411411 4 4 1111 x
Subtract 4 from both sides
The equation is an identity therefore the solution is all real numbers or
To check this, we will randomly choose two real numbers 6 and 10, and substitute these into the equation.
Check the answer using6. 46454(6)11
13 Let x = a number. The sentence translates as follows
4(2)52411 8513 1313 .
The answer works
Check the answer using 10. 410454(10)11
Subtract 7 from both sides
Divide both sides by 4. .
The answer works
4(14)54011 56551 5151 .
320203(5)
Distribute on right side.
Simplify on right side.
Subtract 3 from both sides
This is a false statement
This equation has no solution.
11. Recall that the perimeter of a rectangle can be calculated using the formula 22 Plw =+ where P is the perimeter of the rectangle with length, l, and width, w. We are given the perimeter and width of the rectangular lot, so substitute 330 P = and 40 w = . After substituting in these values, solve the equation for l
? ? Check the answer.
The answer works.
Subtract 7 from both sides
Add 15 to both sides
Divide both sides by 3. . Check the
330280 80 80 2502 2502 22 125 l
33022(40)
Substitute in known values
Simplify and isolate variable term.
Subtract 80 from both sides. Divide both sides by 2.
The answer works. 2015149 55
3302(125)80
? ? Check the answer.
The answer works
33025080 330330 . =+ =+ = The length of the building lot is 125 feet.
12 To solve for x means to isolate x on one side of the equal sign.
Distribute-right side. Simplify on right side.
Subtract 4 from both sides
This is a true statement.
Divide both sides by . . b m ymxb bb ybmx ybm m =+ −= = x m yb x m =
Identify the variable to isolate.
Subtract from both sides
434(1)7 43447 4343 4 4 33 x xx xx xx xx +=−+ +=−+ +=+ = The equation is an identity therefore the solution is all real numbers or To check this, we will randomly choose two real numbers 2 and 7, and substitute these into the equation.
Check the answer using2. 4(2)34217 834(3)7 5127 55 .
Check the answer using 7. 4(7)34717
? ? ? ? ? ? The answer works
Subtract 4 from both sides
Multiply by neg., reverse ine 2143 4 4 213 1 1 24 24 22 . x xx xx x x x +>− −+>− −>− < quality.
Subtract 1 from both sides
Divide both sides by2.
2 x <
The answer works
2834(6)7 31247 3131 .
=
16. The sum of the measures of the angles in a triangle must equal 180° . Add the measures of the given angles and set them equal to 180 ° then solve for x
Combine like terms 116180 . 18180 xxx x ++= =
Divide both sides by 18 18180 1818 10 x x = =
Now substitute 10 x = into the expressions that represent the angles to find the measure of each angle. The measures of the three angles are 10,110,and60. °°°
To check the solution, the measures of the angles should add up to 180.
? Check the answer. The answer works 1011060180 180180 . °+°+°=° °=°
17. If 16 x −≤ , then x must be greater than or equal to 16 . This means we can rewrite the inequality as 16 x ≥−
Graphing the solution on a number line:
19. Isolate . Add 7 to all sides
Divide all sides by 2. 52715 7 7 7 12222 12222
20 a. Aiden's hourly wage, $10.25, times the number of hours h he works in a week must be greater than $287 in order to meet his expenses.
Write the inequality as follows: 10.25287 h > b.
10.25287 10.25287 10.2510.25 28 h h h > > >
Divide both sides by 10.25.
Aiden needs to work more than 28 hours a week to meet his expenses.
Cumulative Review for Chapters 1 and 2
1. The base is 9 4 and we multiply 9 4 by itself 6 times. So the exponent is 6. The exponential form is 6 9 4 ⎛⎞
. 6 9999999 4444444
2. The base is 4 , and the exponent of 3 tells us we are multiplying 4 by itself 3 times. The expanded form is (4)(4)(4) −⋅−⋅− 3 (4)(4)(4)(4) −=−⋅−⋅−
3. The base is 1 and we multiply 1 by itself 5 times. ()()()()() 111111 −⋅−⋅−⋅−⋅−=−
4. The exponent on 10 is 8. Move the decimal point 8 places to the right.
8 00,000, 3.04510 000 3.0451304,500,000×=⋅= Therefore, 8 3.04510 × has the standard form 304,500,000.
7 (16)(37)(37)(16) −+−=−+−
This is an example of the commutative property of addition, because the order of the terms has been changed.
8. In this problem, there are 2 quantities that can change. One quantity is the amount of money (revenue) the soccer club makes, and the second is the number of candy bars sold. Let R = the amount of revenue in dollars the soccer club makes and c = the number of candy bars sold.
84
⋅−−−+÷⋅−
32(5)204(3) 106
=⋅−−−+÷⋅−
84 32(5)204(3) 106 84 32(5)204(3) 4 2132(5)204(3) 2134(5)204(3)
Outline the terms.
5. Evaluating the expression using the order of operations would be as follows: 2 2 2 2
Simplify fraction first.
=⋅−−−+÷⋅− =⋅−−−+÷⋅− =⋅−−−+÷⋅− each term. Multiply in 1st term, divide in last. Multiply in last term.
Evaluate exponent. Simplify within
634(5)5(3) 634515 49 =−−−+⋅− =−+− =
Add/subtract from left to right.
6. Evaluating the expression using the order of operations would be as follows:
9. Let x = a number. “Less than” translates as subtraction, but it is done in reverse order. So, this translates as something minus 7. "8 times a number" means to multiply a number by 8 so this translates as 8 x and the sentence translates as 87. x 10. Convert 95 feet to meters. First, form the unity fraction. We want to convert to the units of meters, so meters will go in the numerator. We want to convert from the units of feet, so feet will go in the denominator. There are 3.28 ft in 1 m, so the unity fraction is 1 m 3.28 ft . Multiply 95 ft by this unity fraction and simplify the expression as follows. 1 m
95 ft 3.28 ft
95 ft ⋅ = 1 m 1 3.28 ft
95 m 28.96 m 3.28 =≈
Therefore, 95 feet is equal to approximately 28.96 m.
18(2)67(2)(3) 18(2)614(3) 18(2)20(3) 18(2)209 9209 20 ÷−+−⋅−+− =÷−+++− =÷−++− =÷−++ =−++
2 2 2 Outline the terms. Evaluate absolute value.
Evaluate exponent. Divide.
= Add/subtract from left to right.
11. Evaluate 7 3 x x −+ for 6 x =− 7 Substitute in 6. 3 (6) 7(6) Simplify. 3
. Number of Biscuits for Scout Number of Biscuits for Shadow
0 50 – 0 = 50
5 50 – 5 = 45
10 50 – 10 = 40
15 50 – 15 = 35 n 50 – n
To find number of dog biscuits Lisa will give to Shadow, subtract the number of biscuits given to Scout from 50 biscuits. The general variable expression is 50 n
13. Identify like terms first. Then combine like terms and arrange in conventional form.
22 2 73698 12515 xyxxxxyx xxxy −+−+− =−++
14. Evaluate parentheses first. Then identify like terms, combine like terms, and arrange in conventional form.
Evaluate parentheses Identify like terms. Combine like terms 6(35)8 . 6358 313 . xx xx x −−+ =−++ =+
15. Evaluate parentheses first. Then identify like terms, combine like terms, and arrange in conventional form.
Evaluate parentheses Identify like terms. Combine like terms (3)(7) . 37 48 . bcbc bcbc bc −−−+ =−+− =−
16. To determine the coordinates of the points we need to find the values of the x- and y-coordinates. Beginning in the top right quadrant with the point on the x-axis, then following counterclockwise for the other points, the first point is located 4 units to the right of the y-axis, so the x-coordinate is 4 x = . The point is located on the x-axis, so the y-coordinate is 0 y = .
The coordinates of this point are (4,0)
The next point is located 1 unit to the right of the y-axis, so the x-coordinate is 1 x = . The point is located 2 units above the x-axis, so the y-coordinate is 2 y =
The coordinates of this point are (1,2)
The next point is located 3 units to the left of the yaxis, so the x-coordinate is 3 x =− . The point is located 4 units above the x-axis, so the y-coordinate is 4 y = .
The coordinates of this point are (3,4)
The next point is located 3 units to the left of the yaxis, so the x-coordinate is 3 x =− . The point is located 5 units below the x-axis, so the y-coordinate is 5 y =−
The coordinates of this point are (3,5) .
The next point is located on the y-axis, so the xcoordinate is 0 x = . The point is located 3 units below the x-axis, so the y-coordinate is 3 y =− .
The coordinates of this point are (0,3)
The next point is located 6 units to the right of the yaxis, so the x-coordinate is 6 x = . The point is located 2 units below the x-axis, so the y-coordinate is 2 y =−
The coordinates of this point are (6,2) .
17. The input values range from 3 to 7. It is reasonable to scale the horizontal axis by 2's. The output values range from 3 to 6 so it will be reasonable to scale the y-axis by 2's. The points are plotted on the graph below.
18. A reasonable vertical axis for this data would range between 0 and 9.2 with a scale of 0.1 and a break from 0 to 8. The horizontal axis will start at 0 and end at 1990 with a scale of 3 and a break from 0 to 1960. For a scatterplot, instead of using a bar to represent the height, use a large dot. This scatterplot should have a title and would be as follows:
19. To check if the given values of the variables are a solution to the equation, evaluate each side of the equation and see if the two sides are equal. ? ? . Substitute 1.5 and 0 476 Original equation. 4(1.5)7(0)6 606 Simplify both sides 66 This statement is true. xy xy =−=
Because the final statement is true, 1.5 x =− and 0 y = are a solution to the equation 476 xy−=− . 20 Add 2 to both sides
Identify the variable term 27 +2 +2 5 y y
Identify the variable term Find LCD6. 15
1 Add to both sides 3
Identify the variable term Add 0.051 to both sides 0.0514.006 +0.051 +0.051 4.057 n n −+= =
23. We are solving for y so we want to isolate the variable y on one side of the equation. Identify variable term to isolate Subtract 3 from both sides. Identify variable to isolate Divide both sides by Simp 3530 . 3 3
25. Multiply both sides by 2 8 3 3233 8 . 2322 12 n n n =− ⋅=−⋅ =− 26. Divide both sides by
27. a. Substitute 120 h = into the equation 16 Sh = and solve for S.
16 Given equation.
16(120) Substitute 120. 1920 Multiply and solve for . Sh Sh SS = == =
If Corinthia works 120 hours, her salary will be $1920.
b. Substitute 2240 S = into the equation 16 Sh = and solve for h.
16 Given equation.
31. Let x = a number. The sentence translates as follows:
224016 Substitute 2240 and solve for.
224016 Divide both sides by 16. 1616 140 Sh hSh h h = == = =
Corinthia needs to work 140 hours in a month to make $2240 a month.
c. Substitute 1600 S = into the equation 16 Sh = and solve for h.
16 Given equation.
160016 Substitute 1600 and solve for. 160016 Divide both sides by 16. 1616 100 Sh hSh h h = == =
Corinthia needs to work 100 hours in a month if she only wants to earn $1600 a month. 28
Subtract 20 from both sides. Divide both sides by
32. Isolate the variable m
Identify variable to isolate. Prgm Pr rg = = gm rg Divide both sides by .lh
33. Isolate the variable y.
Identify variable term to isolate Subtract 6 from both sides.
Identify variable to isolate Divide both sides by
lify on right side. 210yx=+
34. Let x = a number. The sentence translates as follows:
Subtract 4 from both sides
Subtract 8 from both sides Divide both sides by
36.
37.
38.
Divide both sides by 841220 12 12 4420 4 4 416 416 4. 44 4 y yy yy y y y y +=+ −+= −= =− =−
Subtract 12 from both sides
Subtract 4 from both sides
7832(412) 783824 78524 5 5 2824 8 8 n nnn nnn nn nn n +=−+− +=−+− +=− +=−
Distribute on right side Simplify on right side.
Subtract 5 from both sides
Subtract 8 from both sides
Divide both sides by 232 232 2. 22 16 n n n =− = =−
Subtract 21 from both sides 563(49)5 5612275 7215 8215 21 21 8 w www www ww ww w
Distribute on left side Simplify on left side.
Subtract from both sides
Divide both sides by 16 816 8. 88 2 w w w =− =− =
Subtract 1 from both sides Divide both sides by 6. 6149 1 1 648 648 66 8 . x x x x +> > > > 40.
Multiply by neg., reverse inequality. 1 59 7 5 5 1 4 7 1 747 7 28 . r r r r −+≤ −≤
Subtract 5 from both sides
Multiply both sides by7.
41.
Isolate .
Subtract 5 from all sides
Divide all sides by 3. 143526 5 5 5 9321 9321 333 37 . xx x x x <+< << << << 42.
Subtract 10 from both sides
Subtract 15 from both sides
Combine like terms Divide both sides by 3 (18)(3)180 . 315180 15 15 3165 33 55
Now substitute 55 x = into the expressions that represent the angles to find the measure of each angle. The measures of the three angles are 55,73,and52. °°°
To check the solution, the measures of the angles should add up to 180. ?
Check the answer. The answer works 557352180 180180 . °+°+°=° °=°
Multiply both sides by. 3 104 8 10 10 38 14 83 838 14 383 1121 or 37 33 t t t tt
Multiply by neg., reverse inequality.
43. The English class can have between 15 students and 25 students enrolled. The maximum enrollment is 25, so we include 25 in the interval. The class must have at least 15 students, so we include 15 in the interval. Let s = the number of students enrolled in the class. 1525 s ≤≤
44. The upper-division courses are numbered between 200 and 400. The problem states that 200 is included in the interval, but 400 is not included. Let n = the course number.
200400 n ≤<
