Chapter 2: Linear Equations and Inequalities Section 2.1: Introduction to Algebra Part A: Definitions List all of the coefficients and variable parts of the following expressions. 1. PROBLEM: 4 1 x − SOLUTION: ANSWER: Coefficients: { } 1, 4 − ; Variable Parts: { } x 2. PROBLEM: 2 –7 2 1 x x − + SOLUTION: ANSWER: Coefficients: { } 7, 2, 1 − − ; Variable Parts: { } 2 , x x 3. PROBLEM: 2 5 3 x x − + − SOLUTION: ANSWER: Coefficients: { } 3, 1, 5 − − ; Variable Parts: { } 2 , x x Terms Coefficient Variable Part 4x 4 x −1 −1 Terms Coefficient Variable Part −7x 2 −7 x 2 −2x −2 x 1 1 Terms Coefficient Variable Part −x 2 −1 x 2 5x 5 x −3 −3
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Chapter2:LinearEquationsandInequalities
Section2.1:IntroductiontoAlgebra Part A: Definitions List all of the coefficients and variable parts of the following expressions. 1. PROBLEM:
4 1x − SOLUTION: ANSWER: Coefficients: { }1, 4− ; Variable Parts: { }x
2. PROBLEM:
2–7 2 1x x− + SOLUTION: ANSWER: Coefficients: { }7, 2, 1− − ; Variable Parts: { }2 , x x
3. PROBLEM:
2 5 3x x− + − SOLUTION: ANSWER: Coefficients: { }3, 1, 5− − ; Variable Parts: { }2 , x x
Terms Coefficient Variable Part 4x 4 x −1 −1
Terms Coefficient Variable Part −7x2 −7 x2 −2x −2 x
1 1
Terms Coefficient Variable Part −x2 −1 x2 5x 5 x −3 −3
4. PROBLEM: 2 2 23 7
3x y xy− +
SOLUTION:
ANSWER: Coefficients: 2 , 3, 73
⎧ ⎫−⎨ ⎬⎩ ⎭
; Variable Parts: { }2 2 , x y xy
5. PROBLEM:
21 1 53 2 7
y y− +
SOLUTION:
ANSWER: Coefficients: 1 1 5, ,2 3 7
⎧ ⎫−⎨ ⎬⎩ ⎭
; Variable Parts: { }2 , y y
6. PROBLEM:
2 24 5 1a b ab ab− + − + SOLUTION: ANSWER: Coefficients: { }4, 1, 1, 5− − ; Variable Parts: { }2 2, ,a b ab ab
Terms Coefficient Variable Part 3x2y2 3 x2y2 −2/3xy −2/3 xy
7 7
Terms Coefficient Variable Part 1/3y2 1/3 y2 −1/2y −1/2 y 5/7 5/7
Terms Coefficient Variable Part −4a2b −4 a2b 5ab2 5 ab2 −ab −1 ab
1 1
7. PROBLEM: ( ) ( )3 52 3a b a b+ − +
SOLUTION: ANSWER: Coefficients: { }3, 2− ; Variable Parts: ( ) ( ){ }3 5,a b a b+ +
8. PROBLEM:
( ) ( )25 2 2 2 7x x+ − + − SOLUTION: ANSWER: Coefficients: { }7, 2, 5− − ; Variable Parts: ( ){ }22 , ( 2)x x+ +
9. PROBLEM:
2 2 10 27m n mn mn− + − SOLUTION: ANSWER: Coefficients: { }27, 1, 1 , 10− − ; Variable Parts: { }2 2, ,m n mn mn
Terms Coefficient Variable Part 2(a + b)3 2 (a + b)3 −3(a + b)5 −3 (a + b)5
Terms Coefficient Variable Part 5(x + 2)2 5 (x + 2)2 −2(x + 2) −2 (x + 2)
−7 −7
Terms Coefficient Variable Part m2n 1 m2n −mn2 −1 mn2 10mn 10 mn −27 −27
10. PROBLEM: 4 3 22 3 4 1x x x x− − − −
SOLUTION: ANSWER: Coefficients: { }4, 3, 2, 1, 1− − − − ; Variable Parts: { }4 3 2, , ,x x x x
Part B: Evaluating Algebraic Expressions Evaluate. 11. PROBLEM:
3x + , where 4x = −
SOLUTION: ( )3 4 31
x + = − +
= −
ANSWER: 1−
12. PROBLEM:
2 3x − , where 3x = −
SOLUTION: ( )2 3 2 3 36 39
x − = − −
= − −= −
ANSWER: 9−
13. PROBLEM:
5 20x− + , where 4x =
SOLUTION: ( )5 20 5 4 20
20 200
x− + = − +
= − +=
ANSWER: 0
Terms Coefficient Variable Part x4 1 x4 −2x3 −2 x3 −3x2 −3 x2 −4x −4 x −1 −1
14. PROBLEM: 5y− , where 1y = −
SOLUTION: ( )5 5 15
y− = − −
=
ANSWER: 5 15. PROBLEM:
34
a , where 32a =
SOLUTION:
( )3 3 324 4
3 824
a =
= ⋅=
ANSWER: 24
16. PROBLEM:
2( 4)a − , where 1a = −
SOLUTION:
( )( )( )
2( 4) 2 1 4
2 510
a − = − −
= −
= −
ANSWER: 10−
17. PROBLEM:
10(5 )z− − , where 14z =
SOLUTION: ( )( )
10(5 ) 10 5 14
10 990
z− − = − −
= − −
=
ANSWER: 90
18. PROBLEM:
5 1y − , where 15
y = −
SOLUTION:
15 1 5 15
1 12
y ⎛ ⎞− = − −⎜ ⎟⎝ ⎠
= − −= −
ANSWER: 2−
19. PROBLEM:
2 1a− + , where 13
a = −
SOLUTION:
12 1 2 13
2 132 1 33 32 3
353
a ⎛ ⎞− + = − − +⎜ ⎟⎝ ⎠
= +
⋅= +
+=
=
ANSWER: 53
20. PROBLEM:
4 3x + , where 316
x =
SOLUTION:
34 3 4 316
3 343 3 44 43 12
4154
x ⎛ ⎞+ = +⎜ ⎟⎝ ⎠
= +
⋅= +
+=
=
ANSWER: 154
21. PROBLEM:
12
x− + , where 2x = −
SOLUTION:
( )1 122 2
122
2 2 12
5
24 1
2
2
x− + = − − +
= +
⋅= +
+=
=
ANSWER: 52
22. PROBLEM: 2 13 2
x − , where 14
x = −
SOLUTION:
2 1 2 1 13 2 3 4 2
1 16 21 1 36 61 36
4623
x ⎛ ⎞− = − −⎜ ⎟⎝ ⎠
= − −
⋅= − −
− −=
−=
= −
ANSWER: 32
−
For each problem below, evaluate 2 4b ac− , given the following values for a, b, and c. 23. PROBLEM:
1 a= , 2b= , 3c=
SOLUTION: ( ) ( )( )22 4 2 4 1 34 12
8
b ac− = −
= −= −
ANSWER: 8−
24. PROBLEM:
3a= , –4b= , –1c=
SOLUTION:
( ) ( )( )22 4 4 4 3 116 4 3 116 1228
b ac− = − − −
= + ⋅ ⋅= +=
ANSWER: 28
25. PROBLEM: –6a= , 0b= , –2c=
SOLUTION: ( ) ( )( )22 4 0 4 6 20 48
48
b ac− = − − −
= −= −
ANSWER: 48−
26. PROBLEM:
12
a = , 1b = , 23
c =
SOLUTION:
( )22 1 24 1 42 3
413
1 3 43 3
3 4313
b ac ⎛ ⎞⎛ ⎞− = − ⎜ ⎟⎜ ⎟⎝ ⎠⎝ ⎠
= −
⋅= −
−=
= −
ANSWER: 13
−
27. PROBLEM:
3a = − , 12
b = − , 19
c =
SOLUTION:
( )2
2 1 14 4 32 9
1 44 31 3 4 412 123 16
121912
b ac ⎛ ⎞ ⎛ ⎞− = − − −⎜ ⎟ ⎜ ⎟⎝ ⎠ ⎝ ⎠
= +
⋅ ⋅= +
+=
=
ANSWER: 1912
28. PROBLEM:
13a = − , 23
b = − , 0c =
SOLUTION:
( )( )2
2 24 4 13 03
4 4 13 094 0949
b ac ⎛ ⎞− = − − −⎜ ⎟⎝ ⎠
= + ⋅ ⋅
= +
=
ANSWER: 49
Evaluate. 29. PROBLEM:
24xy− , where 3x = − and 2y =
SOLUTION: ( )( )( )
2 24 3 24
4 3 448
xy = − −
= − −
=
−
ANSWER: 48
30. PROBLEM:
258
x y , where 1x = − and 16y =
SOLUTION:
( ) ( )225 5 1 168 8
5 16810
x y = −
= ⋅
=
ANSWER: 10
31. PROBLEM:
2 2a b− , where 2a = and 3b =
SOLUTION:
2 2 2 22 34 9
5
a b == −
− −
= −
ANSWER: 5−
32. PROBLEM:
2 2a b− , where 1a = − and 2b = −
SOLUTION: ( ) ( )2 22 2 1 21 4
3
a b− − − −
= −= −
=
ANSWER: 3−
33. PROBLEM: 2 2x y− , where 1
2x = and 1
2y = −
SOLUTION:
2 22 21 1
2 21 14 40
x y ⎛ ⎞ ⎛ ⎞= − −⎜ ⎟ ⎜ ⎟⎝ ⎠ ⎝
= −
−⎠
=
ANSWER: 0 34. PROBLEM:
23 5 1x x− + , where 3x = −
SOLUTION:
( ) ( )( )
223 5 1 3 3 5 3 1
3 9 5 3 127 15 143
x x− + = − − − +
= ⋅ − − +
= + +=
ANSWER: 43 35. PROBLEM:
2 6 y y− − , where 0y =
SOLUTION:
2 26 0 0 60 0 6
6
y y− − = − −= − −= −
ANSWER: 6−
36. PROBLEM: 21 y− , where 1
2y = −
SOLUTION:
22
114
1 4 14 4
4
11 1
14
34
2y =
= −
⋅= −
−=
⎛ ⎞−⎜ ⎟⎝ ⎠
=
− −
ANSWER: 34
37. PROBLEM:
( )( )3 2x x+ − , where 4x = −
SOLUTION:
( )( ) ( )( ) ( )( )( )( )
3 2 4 3 4 2
1 66
x x+ − = − + − −
= − −
=
ANSWER: 6 38. PROBLEM:
( ) ( )5 6y y− + , where 5y =
SOLUTION: ( )( ) ( )( )5 6 5 5 5 6
0 110
y y− + = − +
= ⋅=
ANSWER: 0
39. PROBLEM: ( )3 4α β− + , where 1α = − and 6β =
SOLUTION:
( ) ( )( )( )
3 4 3 1 6 4
3 7 421 417
α β− + = − − +
= − +
= − += −
ANSWER: 17− 40. PROBLEM:
2 23α β− , where 2α = and 3β = −
SOLUTION:
( ) ( )2 22 2
3 4 912 93
3 3 2 3α β =
= ⋅ −=
−
−
−
=
−
ANSWER: 3 41. PROBLEM:
Evaluate ( )4 x h+ , given 5x = and 0.01h = .
SOLUTION: ( ) ( )
( ).014 4 5 0
4 .52 040
01.
x h+ = +
=
=
ANSWER: 20.04 42. PROBLEM:
Evaluate ( )2 3x h− + + , given 3x = and 0.1h = .
SOLUTION:
( ) ( )( )
.12 3 2 3 0 3
2 3 36 3
.1.2.23
x h− + + = − + +
= − +
= − += −
ANSWER: 23.−
43. PROBLEM: Evaluate ( ) ( )22 5 3x h x h+ − + + , given 2x = and 0.1h = .
SOLUTION:
( ) ( ) ( ) ( )( ) ( )
2 2
2
2 5 3 2 2 0 5 2 0 3
2 2 5 2 32 4
.1 .1
.1 .1
.41 5 2.1 38.82 10.5 31.32
x h x h
− ⋅ += −
+ − + + = + − + +
= − +
=+
=
⋅
ANSWER: 1.32 44. PROBLEM:
Evaluate ( ) ( )23 2 1x h x h+ + + − , given 1x = and 0.01h = .
SOLUTION:
( ) ( ) ( ) ( )( ) ( )
2 2
2
.01 .01
.01 .0
3 2 1 3 1 0 2 1 0 1
3 1 2 1 13 1 0201 2 1
1. .01 1
3.0603 2.02 14.0803
x h x h
−=
+ +
+ −
+ − = + + + −
= + −
= ⋅ +
=
⋅
ANSWER: 4.0803
Part C: Using Formulas Convert the following temperatures to degrees Celsius given 5
9 ( 32)C F= − , where F represents degrees Fahrenheit. 45. PROBLEM:
86°F
SOLUTION: ( )
5 (86 32)95 5495 630 C
C = −
=
= ⋅= °
ANSWER: 30 C°
46. PROBLEM: 95°F
SOLUTION: ( )
5 (95 32)95 6395 735 C
C = −
=
= ⋅= °
ANSWER: 35 C° 47. PROBLEM:
−13°F
SOLUTION:
( )( )
( )
( )
5 13 3295 4595 5
25 C
C = − −
= −
= −
= − °
ANSWER: 25 C− °
48. PROBLEM:
14°F
SOLUTION: ( )
( )
5 (14 32)95 1895 2
10 C
C = −
= −
= ⋅ −
= − °
ANSWER: 10 C− °
49. PROBLEM: 32°F
SOLUTION:
( )
( )
5 32 3295 090 C
C = −
=
= °
ANSWER: 0 C° 50. PROBLEM:
0°F
SOLUTION: ( )
5 (0 32)95 32917 777. ...
.17 78 C
C = −
= −
≈ −≈ − °
ANSWER: 17 C.78− °
Given the base and height of a triangle, calculate the area ( 1
2A bh= ). 51. PROBLEM:
25b = cm and 10h = cm
SOLUTION: ( )( )
121 25
square cm
102125
A bh=
=
=
ANSWER: squar125 e cm
52. PROBLEM: 40b = in and 6h = in
SOLUTION: ( )( )
121 40 621 s20 quare in
A bh=
=
=
ANSWER: squar120 e in
53. PROBLEM:
12
b = ft and 2h = ft
SOLUTION: ( )
square
121 1 22 21 f2
t
A bh=
⎛ ⎞= ⎜ ⎟⎝ ⎠
=
ANSWER: squa1 2
re ft
54. PROBLEM:
34
b = in and 58
h = in
SOLUTION:
121 3 52 4 8
square 15 64
in
A bh=
⎛ ⎞⎛ ⎞= ⎜ ⎟⎜ ⎟⎝ ⎠⎝ ⎠
=
ANSWER: squar15 64
e in
55. PROBLEM: A certain cellular phone plan charges $23.00 per month plus $0.09 for each minute of usage. The monthly charge is given by the formula
0.09 23monthly charge x= + , where x represents the number of minutes of usage per month. What is the charge for a month with 5 hours of usage? SOLUTION: Usage 5 hours = 5 60 minutes = 300 minutes= ⋅
( ) 0.09 23
0.09 300 2327 23$50
monthly charge x= +
= +
= +=
ANSWER: Hence, the charge for a month with 5 hours of usage is $50. 56. PROBLEM:
A taxi service charges $3.75 plus $1.15 per mile given by the formula, 1.15 3.75charge x= + , where x represents the number of miles driven. What is the
charge for a 17-mile ride?
SOLUTION: ( )1.15 3.751.15 17 3.7519.55 3.75$23.30
charge x= +
= +
= +=
ANSWER: Hence, the charge for a 17-mile ride is $23.30.
57. PROBLEM:
If a calculator is sold for $14.95, then the revenue in dollars, R, generated by this item is given by the formula 14.95R q= , where q represents the number of calculators sold. Use the formula to determine the revenue generated by this item if 35 calculators are sold.
SOLUTION: ( )14.9514.95 35$523.25
R q=
=
=
ANSWER: Hence, the revenue generated by selling 35 calculators is $523.25.
58. PROBLEM: Yearly subscriptions to a tutoring Web site can be sold for $49.95. The revenue in dollars, R, generated by subscription sales is given by the formula 49.95R q= , where q represents the number of yearly subscriptions sold. Use the formula to calculate the revenue generated by selling 250 subscriptions.
SOLUTION: 49.9549.95 250$12, 487.50
R q== ⋅=
ANSWER: Hence, the revenue generated by selling 250 subscriptions is $12,487.50.
59. PROBLEM:
The cost of producing pens with the company logo printed on them consists of a onetime set-up fee of $175 plus $0.85 for each pen produced. This cost can be calculated using the formula 175 0.85C q= + , where q represents the number of pens produced. Use the formula to calculate the cost of producing 2,000 pens.
SOLUTION: ( )
175 0.85175 0.85 2,000175 1,700$1,875.00
C q= +
= +
= +=
ANSWER: Hence, the cost of producing 2,000 pens is $1,875.00. 60. PROBLEM:
The cost of producing a subscription Web site consists of an initial programming and set-up fee of $4,500 plus a monthly Web hosting fee of $29.95. The cost of creating and hosting the Web site can be calculated using the formula
4500 29.95C n= + , where n represents the number of months the Web site is hosted. How much will it cost to set-up and host the Web site for one year? SOLUTION: One year = 12 months
( )4500 29.954500 29.95 124500 359.4$4,859.40
C n= +
= +
= +=
ANSWER: Hence, it will cost $4,859.40 to set-up and host the Web site for one year.
61. PROBLEM: The perimeter of a rectangle is given by the formula 2 2P l w= + , where l represents the length and w represents the width. What is the perimeter of a fenced-in rectangular yard measuring 70 feet by 100 feet?
SOLUTION:
2 22 100 2 70200 140340 ft
P l w= += ⋅ + ⋅= +=
ANSWER: Hence, the perimeter of a fenced-in rectangular yard is 340 ft. 62. PROBLEM:
Calculate the perimeter of an 8x10 inch picture.
SOLUTION:
2 22 10 2 820 1636 in
P l w= += ⋅ + ⋅= +=
ANSWER: Hence, the perimeter of an 8x10 inch picture is 36 in.
63. PROBLEM:
Calculate the perimeter of a room that measures 12 feet by 18 feet.
SOLUTION:
2 22 18 2 1236 2460 ft
P l w= += ⋅ + ⋅= +=
ANSWER: Hence, the perimeter of the room is 60 ft. 64. PROBLEM:
A computer monitor measures 57.3 cm in length and 40.9 cm high. Calculate the perimeter.
SOLUTION:
2 22 57 3 2 40 9. .
.114 6 81 819 .6
.4 cm
P l h= += ⋅ + ⋅= +=
ANSWER: Hence, the perimeter of the computer monitor is 196.4 cm.
65. PROBLEM: The formula for the area of a rectangle, in square units, is given by A l w= ⋅ , where l represents the length and w represents the width. Use this formula to calculate the area of a rectangle with length 12 cm and width 3 cm.
SOLUTION: 12 336 s m q c
A l w= ⋅= ⋅=
ANSWER: Hence, the area of the rectangle is 36 sq cm. 66. PROBLEM:
Calculate the area of an 8x12 inch picture.
SOLUTION: 12 896 s n q i
A l w= ⋅= ⋅=
ANSWER: Hence, the area of an 8x12 inch picture is 96 sq in.
67. PROBLEM:
Calculate the area of a room that measures 12 feet by 18 feet.
SOLUTION: 18.36216
A l w⋅===
ANSWER: Hence, the area of the room is 216 sq ft. 68. PROBLEM:
A computer monitor measures 57.3 cm in length and 40.9 cm in height. Calculate the total area of the screen.
SOLUTION: 57 402343 57
.3 .9.. sq 2,343 cm6
A l h= ⋅= ⋅=≈
ANSWER: Hence, the total area of the screen is 2,343.6 sq cm.
69. PROBLEM: A concrete slab is poured in the shape of a rectangle for a shed measuring 8 feet by 10 feet. Determine the area and perimeter of the slab.
SOLUTION: 10 880 s t q f
A l w= ⋅= ⋅=
2 22 10 2 820 1636 ft
P l w= += ⋅ + ⋅= +=
ANSWER: Hence, the area of the concrete slab is 80 sq ft and its perimeter is 36 ft.
70. PROBLEM:
Each side of a square deck measures 8 feet. Determine the area and perimeter of the deck.
SOLUTION:
2
288 864 sq ft
A s=
== ⋅=
44 832 ft
P s== ⋅=
ANSWER: Hence, the area of the square deck is 64 sq ft and its perimeter is 32 ft.
71. PROBLEM: The volume of a rectangular solid is given by V lwh= , where l represents the length, w represents the width, and h is the height of the solid. Find the volume of a rectangular solid if the length is 2 in, the width is 3 in, and the height is 4 in.
SOLUTION: 2 3 424 cu in
V lwh== ⋅ ⋅=
ANSWER: Hence, the volume of a rectangular solid is 24 cu in.
72. PROBLEM:
If a trunk measures 3 ft by 2ft and is 2½ ft tall, then what is the volume of the trunk?
SOLUTION:
13 2 22
53 22
15 cu ft
V lwh=
= ⋅ ⋅
= ⋅ ⋅
=
ANSWER: Hence, the volume of the trunk is 15 cu ft.
73. PROBLEM:
The interior of an industrial freezer measures 3 feet wide by 3 feet deep and 4 feet high. What is the volume of the freezer?
SOLUTION: 3 3 436 cubic feet
V lwh== ⋅ ⋅=
ANSWER: Hence, the volume of the freezer is 36 cubic feet.
74. PROBLEM: A laptop case measures 1 ft 2 in by 10 in. by 2 in. What is the volume of the case? SOLUTION: 1 ft 2 in = 12 in 2 in = 14 in+
14 10 2280 cu in
V lwh== ⋅ ⋅=
ANSWER: Hence, the volume of the case is 280 cu in.
75. PROBLEM: If the trip from Fresno to Sacramento can be made by car in 2½ hours at an average speed of 67 miles per hour, then how far is Sacramento from Fresno?
SOLUTION:
167 22
5672
167.5 mi
D r t= ⋅
= ⋅
= ⋅
=
ANSWER: Hence, Sacramento is 167.5 mi from Fresno.
76. PROBLEM:
A high-speed train averages 170 miles per hour. How far can it travel in 1½ hours?
SOLUTION:
1170 12
31702
255 mi
D r t= ⋅
= ⋅
= ⋅
=
ANSWER: Hence, the high-speed train can travel 255 mi in 1½ hours.
77. PROBLEM: A jumbo jet can cruise at an average speed of 550 mph. How far can it travel in 4 hours?
SOLUTION: 550 42, 200 mi
D r t= ⋅= ⋅=
ANSWER: Hence, the jumbo jet can travel 2,200 mi in 4 hours.
78. PROBLEM:
A fighter jet reaches a top speed of 1,316 mph. How far will the jet travel if it can sustain this speed for 15 minutes? SOLUTION: 1 hour = 60 minutes
Therefore, 11 minute = hours60
Hence, 1515 minutes = hours60
151,31660
329 mi
D r t= ⋅
= ⋅
=
ANSWER: Hence, the jet will travel 329 mi if it can sustain a speed of 1,316 mph for 15 minutes.
79. PROBLEM:
The Hubble Space Telescope is in low earth orbit traveling at an average speed of 16,950 mph. What distance does it travel in 1½ hours?
SOLUTION:
116,950 12
316,9502
25, 425 mi
D r t= ⋅
= ⋅
= ⋅
=
ANSWER: Hence, the Hubble Space Telescope travels 25,425 mi in 1½ hours.
80. PROBLEM: The Earth orbits the Sun a speed of about 66,600 mph. How far does the Earth travel around the Sun in one day? SOLUTION: 1 day 24 hours=
66,600 241,598, 400 mi
D r t= ⋅= ⋅=
ANSWER: Hence, the Earth travels a distance of 1,598,400 mi around the Sun in one day.
81. PROBLEM:
Calculate the simple interest earned on a $2,500 investment at 3% annual interest rate for 4 years.
SOLUTION: 33% 0.03100
r = = =
2,500 0.03 4$300
I prt⋅ ⋅
=
==
ANSWER: Hence, the
simple interest earned is $300. 82. PROBLEM:
Calculate the simple interest earned on a $1,000 investment at 5% annual interest rate for 20 years.
SOLUTION: 55% 0.05100
r = = =
1,000 0.05 20$1,000
I prt⋅ ⋅
=
==
ANSWER: Hence, the
simple interest earned is $1,000.
83. PROBLEM: How much simple interest is earned on a $3,200 investment at a 2.4% annual interest for 1 year?
SOLUTION: 2.42.4% 0.024100
r = = =
3, 200 0.024 1$76.80
I prt⋅ ⋅
=
==
ANSWER: Hence, the
simple interest earned is $76.80. 84. PROBLEM:
How much simple interest is earned on a $500 investment at a 5.9% annual interest rate for 3 years?
SOLUTION: 5.95.9% 0.059100
r = = =
500 0.059 3$88.50
I prt⋅ ⋅
=
==
ANSWER: Hence, the
simple interest earned is $88.50. 85. PROBLEM:
Calculate the simple interest earned on a $10,500 investment at a 144 % annual
interest rate for 4 years.
SOLUTION: 1 17 4.254 % % 0.04254 4 100
r = = = =
10,500 0.0425 4$1,785
I prt⋅
=
=⋅=
ANSWER: Hence, the
simple interest earned is $1,785.
86. PROBLEM: Calculate the simple interest earned on a $6,250 investment at a 3
46 % annual interest rate for 1 year.
SOLUTION: 3 27 6.756 % % 6.75% 0.06754 4 100
r = = = = =
6,250 0.0675 1$421.875$421.88
I prt⋅ ⋅
=≈
==
ANSWER: Hence, the
simple interest earned is $421.88. Part D: Discussion Board Topics 87. PROBLEM:
Research and discuss the history of the symbols for addition (+) and subtraction (−). ANSWER: In Europe in the early 15th century the letters “P” and “M” are generally used. The symbols (P with stroke for piu, i.e. plus, and M with stroke for meno, i.e. minus) appeared for the first time in Luca Pacioli’s mathematics compendium, Summa de Arithmetica, Geometria, Proportioni et Proportionalita, first printed and published in Venice in 1494. The + is a simplification of the Latin “et.” The − may be derived from a tilde written over m when used to indicate subtraction; or it may come from a shorthand version of the letter m itself. (Source: http://en.wikipedia.org/wiki/Plus_and_minus_signs)
88. PROBLEM:
What are mathematical models and why are they useful in everyday life? ANSWER: A mathematical model is a description of a system using mathematical language. The process of developing a mathematical model is termed mathematical modelling (also written modeling). Mathematical models are used not only in the natural sciences (such as physics, biology, earth science, meteorology) and engineering disciplines (e.g. computer science, artificial intelligence), but also in the social sciences (such as economics, psychology, sociology and political science); physicists, engineers, statisticians, operations research analysts and economists use mathematical models most extensively. (Source: http://en.wikipedia.org/wiki/Mathematical_model)
89. PROBLEM: Find and post a useful formula. Demonstrate its use with some values. SOLUTION: (Students answers may vary.) Volume of a cylinder = πr2h The volume of a cylinder with a base radius of 4 cm and height 10 cm, is V = πr2h = π(4 cm)2(10 cm) = 502.65 cm2.
ANSWER: 502.65 cm2
90. PROBLEM:
Discuss the history and importance of the variable. How can you denote a variable when you run out of letters? ANSWER: In mathematics, a variable is a value that may change within the scope of a given problem or set of operations. In contrast, a constant is a value that remains unchanged, though often unknown or undetermined. The concepts of constants and variables are fundamental to many areas of mathematics and its applications. A “constant” in this context should not be confused with a mathematical constant which is a specific number independent of the scope of the given problem. Greek alphabets in lower case can be used to denote a variable. (Source: http://en.wikipedia.org/wiki/Variable_%28mathematics%29)
91. PROBLEM:
Find and post a useful resource describing the Greek alphabet. ANSWER: (Students answers may vary.) Source: http://www.greek-language.com/Alphabet.html
Section2.2:SimplifyingAlgebraicExpressionsPart A: Distributive Property Multiply. 1. PROBLEM:
( )3 3 2x −
SOLUTION: ( )3 3 2 3 3 3 29 6
x xx
− = ⋅ − ⋅
= −
ANSWER: 9 6x −
2. PROBLEM: 12( 5 1)y− +
SOLUTION: ( )12( 5 1) 12 5 12 1
60 12y y
y− + = ⋅ − + ⋅
= − +
ANSWER: 60 12y− +
3. PROBLEM:( )2 1x− +
SOLUTION: ( ) ( )2 1 2 2 12 2
x xx
− + = − ⋅ + −
= − −
ANSWER: 2 2x− −
4. PROBLEM:
5( )a b−
SOLUTION: 5( ) 5 55 5
a b a ba b
− = ⋅ − ⋅= −
ANSWER: 5 5a b− 5. PROBLEM:
( )5 8 168
x −
SOLUTION: ( )5 5 58 16 8 168 8 8
5 10
x x
x
− = ⋅ − ⋅
= −
ANSWER: 5 10x −
6. PROBLEM:
( )3 10 55
x− −
SOLUTION:
( )3 3 310 5 10 55 5 5
3 2 36 3
x x
xx
⎛ ⎞− − = − ⋅ − −⎜ ⎟⎝ ⎠
= − ⋅ += − +
ANSWER: 6 3x− +
7. PROBLEM:
(2 3) 2x + ⋅
SOLUTION: (2 3) 2 2 2 3 24 6
x xx
+ ⋅ = ⋅ + ⋅= +
ANSWER: 4 6x +
8. PROBLEM:
(5 1) 5x − ⋅
SOLUTION: (5 1) 5 5 5 1 525 5
x xx
− ⋅ = ⋅ − ⋅= −
ANSWER: 25 5x −
9. PROBLEM:
( )7 ( 3)x− + −
SOLUTION: ( ) ( ) ( )7 ( 3) 3 7 33 21
x xx
− + − = − − + ⋅ −
= −
ANSWER: 3 21x −
10. PROBLEM:
( )8 1 ( 2)x− + −
SOLUTION: ( )8 1 ( 2) 8 ( 2) 1 ( 2)16 2
x xx
− + − = − ⋅ − + ⋅ −
= −
ANSWER: 16 2x −
11. PROBLEM:(2 3 )a b− −
SOLUTION: ( )(2 3 ) 2 32 3
a b a ba b
− − = − − −
= − +
ANSWER: 2 3a b− +
12. PROBLEM:
( 1)x− −
SOLUTION: ( ) ( )( 1) 1 1 11
x xx
− − = − − − ⋅
= − +
ANSWER: 1x− +
13. PROBLEM:
1 (2 5)3
x +
SOLUTION:
1 1 1(2 5) 2 53 3 3
2 53 3
x x
x
+ = ⋅ + ⋅
= +
ANSWER: 2 53 3
x +
14. PROBLEM:
3 ( 2)4
y− −
SOLUTION:
3 3 3( 2) 24 4 4
3 34 2
y y
y
⎛ ⎞− − = − ⋅ − − ⋅⎜ ⎟⎝ ⎠
= − +
ANSWER: 3 34 2
y− +
15. PROBLEM:( )3 2 5a b c− + −
SOLUTION: ( ) ( ) ( )3 2 5 3 2 3 5 36 15 3
a b c a b ca b c
− + − = − ⋅ + − − −
= − − +
ANSWER: 6 15 3a b c− − + 16. PROBLEM:
( )22 5 7y y− − + SOLUTION: ( )2 22 5 7 2 5 7y y y y− − + = − + − ANSWER: 22 5 7y y− + −
17. PROBLEM:
( )25 6 9y y− −
SOLUTION: ( )2 2
2
5 6 5 9
5 30 4
6 5
5
5 9y y
y y
y y− − ⋅ − ⋅
= −
−
−
= ⋅
ANSWER: 25 30 45y y− −
18. PROBLEM:
( )26 5 2 1x x− + −
SOLUTION: ( ) ( ) ( )2 2
2
6 5 2 1 6 5 6 2 6 1
30 12 6x
x x
x
xx + − ⋅− + − = − ⋅ − − ⋅
= − − +
ANSWER: 230 12 6x x− − +
19. PROBLEM:
( )27 3 11x x− −
SOLUTION: ( ) ( )2 2
2
7 3 11 7 3 11
7 3 11
x x x x
x x
− − = − − −
= − +
ANSWER: 27 3 11x x− +
20. PROBLEM: ( )2 3a b c− − +
SOLUTION: ( )2 3 2 3a b c a b c− − + = − + + ANSWER: 2 3a b c− + +
21. PROBLEM:( )23 7 2 3x x− −
SOLUTION: ( )2
2
2 3 2 3
21 6 3
3 7 2 3 3 7x x x x
x x
− − ⋅ −
= −
= ⋅
−
−
ANSWER: 221 6 3x x− −
22. PROBLEM:
( )21 4 6 42
a a− +
SOLUTION: ( )2 2
2
1 1 1 14 6 4 42 2 2
6 4
2 3 22
a a a
a
a
a
⎛ ⎞ ⎛ ⎞− + ⋅⎜ ⎟ ⎜ ⎟⎝ ⎠ ⎝ ⎠
= −
+ = ⋅
+
−
ANSWER: 22 3 2a a− +
23. PROBLEM:
( )21 9 3 273
y y− − +
SOLUTION:
( ) ( )2 2
2
2
1 1 1 19 3 27 9 3 273 3 3 3
9 33
273 3
3 9
y y y y
y
y
y
y
⎛ ⎞ ⎛ ⎞− − + = − − − + −⎜ ⎟ ⎜ ⎟⎝ ⎠ ⎝ ⎠
+ −
= − +
=
−
−
ANSWER: 23 9yy− + −
24. PROBLEM: ( )25 7 9 ( 5)x x− + −
SOLUTION: ( )2 2
2
5 7 9 ( 5) 5 ( 5) 7 ( 5) 9 ( 5)
35 4525 x
x x x x
x
− + − = − − ⋅ − + ⋅ −
− + −=
⋅
ANSWER: 2 3525 45x x+− −
25. PROBLEM:
21 1 163 6 2
x x⎛ ⎞− +⎜ ⎟⎝ ⎠
SOLUTION:
2 2
2
2
1 1 1 1 1 16 6 6 63 6 2 3 6 2
63
6 66 2
2 3
x
x x x x
x
x x
⎛ ⎞ ⎛ ⎞ ⎛ ⎞− +
− +
= − + ⋅⎜ ⎟ ⎜ ⎟ ⎜ ⎟⎝ ⎠ ⎝ ⎠ ⎝
=
= − +
⎠
ANSWER: 22 3x x− + 26. PROBLEM:
( )3 22 3 2 3x x x− − + −
SOLUTION: ( ) ( ) ( ) ( )3 2 3 2
3 2
2 3 2 3 2 3 2 2
6 4 2 6
2 2 3x x x x x x
xx x
− − ⋅ + − ⋅ − − ⋅
= − + −
− + − = − ⋅
+
−
ANSWER: 3 26 4 2 6x x x− + − +
27. PROBLEM:
20 30 1010
x y z+ −
SOLUTION: 20 30 10 20 30 10
10 10 10 102 3
x y z x y z
x y z
+ −= + −
= + −
ANSWER: 2 3x y z+ −
28. PROBLEM: 4 20 8
4a b c− + −
SOLUTION: 4 20 8 4 20 8
4 4 4 45 2
a b c a b c
a b c
− + − −= + −
= − + −
ANSWER: 5 2a b c− + −
29. PROBLEM:
23 9 813
x x− +−
SOLUTION:
2 2
2
3 9 81 3 9 8
3 27
13 3 3 3
x x x x
xx
− += − +
−− +
−= −− −
ANSWER: 2 3 27x x+− −
30. PROBLEM:
215 20 55
y y− + −
SOLUTION:
2 2
2
15 20 5 15 20 55 5 5 5
13 4
y y y y
y y
− + − −= + −
= + −−
ANSWER: 2 13 4yy + −−
Translate the following sentences into algebraic expressions and then simplify. 31. PROBLEM:
Simplify two times the expression 225 9x − .
SOLUTION: ( )2 2
2
2 25 9 2 2 9
50 18
25x x
x
− ⋅
=
⋅
−
− =
ANSWER: 250 18x −
32. PROBLEM: Simplify the opposite of the expression 26 5 1x x+ − . SOLUTION: ( )2 2 6 5 1 6 5 1x x x x− + − =− − + ANSWER: 26 5 1x x− − +
33. PROBLEM: Simplify the product of 5 and 2 8x − .
SOLUTION: ( )2 2
2
5 8
5
5
40
5 8x x
x
− ⋅
=
⋅
−
− =
ANSWER: 25 40x −
34. PROBLEM:
Simplify the product of −3 and 22 8x x− + − .
SOLUTION: ( ) ( )( ) ( ) ( )2 2
2
3 2 8 3 2 3 3
2
8
3 46
x x x x
x x
− − + − = − − + −
= −
− ⋅
+
⋅ −
ANSWER: 2 3 246x x− +
Part B: Combining Like Terms Simplify. 35. PROBLEM:
2 3x x−
SOLUTION: ( )( )
2 3 2 3
1
x x x
xx
− = −
= −
= −
ANSWER: x−
36. PROBLEM: 2 5 12a a a− + −
SOLUTION: 2 5 12 ( 2 5 12)9
a a a aa
− + − = − + −= −
ANSWER: 9a−
37. PROBLEM: 10 30 15y y− −
SOLUTION: 10 30 15 10 15 30
(10 15) 305 30
y y y yy
y
− − = − −= − −= − −
ANSWER: 5 30y− −
38. PROBLEM:
1 53 12
x x+
SOLUTION:
1 5 1 4 53 12 12 12
4 512
91234
x x x
x
x
x
⋅⎛ ⎞+ = +⎜ ⎟⎝ ⎠
+⎛ ⎞= ⎜ ⎟⎝ ⎠
=
=
ANSWER: 34
x
39. PROBLEM: 1 4 34 5 8
x x− + +
SOLUTION:
1 4 3 1 3 44 5 8 4 8 5
1 2 3 48 8 52 3 48 5
1 48 5
x x x x
x
x
x
− + + = − + +
− ⋅⎛ ⎞= + +⎜ ⎟⎝ ⎠− +⎛ ⎞= +⎜ ⎟
⎝ ⎠
= +
ANSWER: 1 48 5
x +
40. PROBLEM:
2 4 7x x x x− + −
SOLUTION: 2 4 7 (2 4 7 1)4
x x x x xx
− + − = − + −=
ANSWER: 4x
41. PROBLEM:
3 2 10 4y y y y− − + −
SOLUTION: 3 2 10 4 ( 3 2 10 4)y y y y y
y− − + − = − − + −
=
ANSWER: y 42. PROBLEM:
5 7 8 2x x y y− + +
SOLUTION: 5 7 8 2 (5 7) (8 2)
2 10x x y y x y
x y− + + = − + +
= − +
ANSWER: 2 10x y− +
43. PROBLEM: 8 2 5 6α β α β− + − −
SOLUTION: 8 2 5 6 8 5 2 6
( 8 5) (2 6)13 4
α β α β α α β βα β
α β
− + − − = − − + −= − − + −= − −
ANSWER: 13 4α β− − 44. PROBLEM:
6 7 2α β α β− + − +
SOLUTION: 6 7 2 6 2 7
( 6 2) (7 1)8 8
α β α β α α β βα β
α β
− + − + = − − + += − − + += − +
ANSWER: 8 8α β− +
45. PROBLEM:
3 5 2 7 5 3x y x y+ − + − +
SOLUTION: 3 5 2 7 5 3 3 5 2 3 5 7
(3 5) ( 2 3) (5 7)2 12
x y x y x x y yx y
x y
+ − + − + = − − + + += − + − + + += − + +
ANSWER: 2 12x y− + + 46. PROBLEM:
– 8 3 14 1y x x y+ − + + −
SOLUTION: – 8 3 14 1– 8 14 – – 3 1
(8 14) ( 1 1) ( 3 1)22 2 2
y x x y x x y yx y
x y
+ − + + = + − += + + − − + − += − −
ANSWER: 22 2 2x y− −
47. PROBLEM:
4 6 2 8xy xy− + +
SOLUTION: 4 6 2 8 4 2 6 8
(4 2) ( 6 8)6 2
xy xy xy xyxy
xy
− + + = + − += + + − += +
ANSWER: 6 2xy +
48. PROBLEM: 12 3 4 20ab ab− − + −
SOLUTION: 12 3 4 20 12 4 3 20
( 12 4) ( 3 20)8 23
ab ab ab abab
ab
− − + − = − + − −= − + + − −= − −
ANSWER: 8 23ab− −
49. PROBLEM:
1 2 2 33 5 3 5
x y x y− + −
SOLUTION:
1 2 2 3 1 2 2 33 5 3 5 3 3 5 5
1 2 2 33 5
3 53 5
x y x y x x y y
x y
x y
x y
− + − = + − −
+ − −⎛ ⎞ ⎛ ⎞= +⎜ ⎟ ⎜ ⎟⎝ ⎠ ⎝ ⎠
−⎛ ⎞= + ⎜ ⎟⎝ ⎠
= −
ANSWER: x y− 50. PROBLEM:
3 2 1 38 7 4 14
a b a b− − +
SOLUTION:
3 2 1 3 3 1 2 38 7 4 14 8 4 7 14
3 1 2 2 2 38 8 14 143 2 4 3
8 141 18 14
a b a b a a b b
a b
a b
a b
− − + = − − +
⋅ − ⋅⎛ ⎞ ⎛ ⎞= − + +⎜ ⎟ ⎜ ⎟⎝ ⎠ ⎝ ⎠
− − +⎛ ⎞ ⎛ ⎞= +⎜ ⎟ ⎜ ⎟⎝ ⎠ ⎝ ⎠
= −
ANSWER: 1 18 14
a b−
51. PROBLEM: 2 24 3 7 4 5 3x xy x xy− − + + − −
SOLUTION:
2 2 2 2
2
4 3 7 4 5 3 4 4 3 5 7 3( 4 4) ( 3 5) (7 3)0 8 4
8 4
x xy x xy x x xy xyx xy
xyxy
− − + + − − = − + − − + −
= − + + − − + −= − += − +
ANSWER: 8 4xy− + 52. PROBLEM:
2 2 2 22 5x y xy x xy y+ − − + −
SOLUTION:
2 2 2 2 2 2 2 2
2 2(1 1) ( 2 5) (1 1)0 3
2 5 2 5
03
x y xy x xy y x x xy xyx xy y
xy
y y
xy
=
= − + − + + −= + +
+ − − + − − − + + −
=
ANSWER: 3xy
53. PROBLEM:
2 2 22 3x y x y− + −
SOLUTION:
2 2 2 2 2 2
2 2
2 2
2 3 2 3(1 2) 33 3
x y x y x x y yx y y
x y y
− + − = + − −
= + − −
= − −
ANSWER: 2 23 3x y y− −
54. PROBLEM: 2 2 2 21 2 1 1
2 3 8 5x y x y− − +
SOLUTION:
2 2 2 2 2 2 2 2
2 2
2 2
2 2
1 2 1 1 1 1 2 12 3 8 5 2 8 3 5
1
3 78 15
4 1 2 5 1 38 8 15 15
4 1 10 38 15
x y x y x x y y
x y
x y
x y
− − + − − +
⋅ − ⋅
=
⎛ ⎞= + ⎜ ⎟⎝ ⎠
⎛ ⎞ ⎛
⋅⎛ ⎞− +⎜ ⎟⎝ ⎠
⎞= +⎜ ⎟ ⎜ ⎟⎝ ⎠ ⎝ ⎠
= −
− − +
ANSWER: 2 23 78 15
x y−
55. PROBLEM:
2 23 4 1 116 5 4 4
a a− + −
SOLUTION:
2 2 2 2
2
2
2
3 4 1 1 3 1 4 116 5 4 4 16 4 5 4
4 4 1 520 20
16 520
7 2116 20
3 1 416 163 416
a a a a
a
a
a
− ⋅
− + − = + − −
⋅⎛ ⎞= +⎜⋅⎛ ⎞+ −⎜ ⎟
⎝ ⎠− −⎛ ⎞ ⎛ ⎞= +⎜ ⎟ ⎜ ⎟
⎝ ⎠ ⎝+
⎠
= −
⎟⎝ ⎠
ANSWER: 27 2116 20
a −
56. PROBLEM: 2 21 3 7 1
5 4 10 2y y− + −
SOLUTION:
2 2 2 2
2
2
2
1 3 7 1 1 7 3 15 4 10 2 5 10 4 2
1 2 7 3 1 210 10 4 42 710
3 24
9 510 4
y y y y
y
y
y
− + − = + − −
⋅ − ⋅⎛ ⎞ ⎛ ⎞= + −⎜ ⎟ ⎜ ⎟+
− −⎛ ⎞+ ⎜ ⎟
⎝ ⎠ ⎝ ⎠+⎛ ⎞= ⎜ ⎟
⎠⎝ ⎠ ⎝
= −
ANSWER: 29 510 4
y −
57. PROBLEM:
2 2 2 26 3 2 5x y xy x y xy− + −
SOLUTION:
2 2 2 2 2 2 2 2
2
2
2
2(6 2) ( 3 5)8
6 3 2 5 6 2 3 5
8
x y xy x y xy x y x y xy xy
x y xyx y xy
− + − + − −
− −
= −
=
= + +
ANSWER: 2 28 8x y xy− 58. PROBLEM:
2 2 2 212 3 13 10x y xy x y xy+ − +
SOLUTION:
2 2 2 2 2 2 2 2
2 2
2 2
12 3 13 10 12 13 3 10(12 13) (3 10)
13
x y xy x y xy x y x y xy xyx y xy
x y xy
+ − + = − + +
= − + +
= − +
ANSWER: 2 2 13x y xy− +
59. PROBLEM: 2 2 2 22 5ab a b ab a b− + − +
SOLUTION:
2 2 2 2 2 2 2 2
2
2
2
2
2 5 5 2(1 5)
3( 1 2)
6a
ab a b ab a b a b a b ab ab
a b abb ab
− + − + = + − −
= −
= + + − −
ANSWER: 2 26 3a b ab− 60. PROBLEM:
2 2 2 2 23 4m n mn mn m n m n− + − +
SOLUTION:
2 2 2 2 2 2 2 2 2 2
2 2 2
2 2 2
2 2 2
3 4 4 3(1 4) 3 05 3 05 3
m n mn mn m n m n m n m n m n mn mnm n m n
m n m nm n m n
− + − + + − − +
= + − +
= − +
= −
=
ANSWER: 2 2 25 3m n m n−
61. PROBLEM:
( ) ( )2 22 3x y x y+ + +
SOLUTION: ( ) ( ) ( )( )
( )
2 2 2
2
2 33
5
2 x y x y x y
x y
= +
=
+ + + +
+
ANSWER: ( )25 x y+
62. PROBLEM:
( ) ( )3 31 22 25 3
x x+ − +
SOLUTION:
( ) ( ) ( )
( )
( )
( )
3 3 3
3
3
3
1 3 2 5
1 2 1 22 2 25 3 5 3
2
2
15 153 10
157
152
x x x
x
x
x
⎛ ⎞= ⎜ ⎟⎝ ⎠⋅ ⋅⎛ ⎞= −⎜
+ − + − +
+⎟⎝ ⎠
−⎛ ⎞= ⎜ ⎟⎝ ⎠
= −
+
+
ANSWER: ( )32715
x− +
63. PROBLEM:
( ) ( )2 23 1 5 1x x x x− − + −
SOLUTION: ( ) ( )
( )
2 2 2
2
3 1 5 1 ( 3 5) ( 1)
2 1
x x x x x x
x x
− − + − = − + −
= −
ANSWER: ( )22 1x x − 64. PROBLEM:
( ) ( )5 3 8 3x x− − −
SOLUTION: ( ) ( ) ( )( )
( )5 3 8 3 5 8 3
3 3
x x x
x
− − − = − −
= − −
ANSWER: ( )3 3x− −
65. PROBLEM:
( ) ( )14 2 7 6 2 7x x− + + +
SOLUTION: ( ) ( ) ( )( )
( )14 2 7 6 2 7 14 6 2 7
8 2 7
x x x
x
− + + + = − + +
= − +
ANSWER: ( )8 2 7x− +
66. PROBLEM: ( ) ( ) ( )2 2 24 2 9 2 2xy x xy x xy x+ − + + +
SOLUTION: ( ) ( ) ( ) ( ) ( )
( )
2 2 2 2
2
4 9 14 2 9 2 2 2
4 2
xy
xy
xy x xy x xy x x
x
= −+ − + + + ++
+= −
ANSWER: ( )224xy x− +
Part C: Mixed Practice Simplify. 67. PROBLEM:
( )5 2 3 7x − +
SOLUTION: ( )5 2 3 7 5 2 5 3 7
10 15 710 8
x xxx
− + = ⋅ − ⋅ +
= − += −
ANSWER: 10 8x − 68. PROBLEM:
( )2 4 2 3y y− + −
SOLUTION: ( ) ( ) ( )2 4 2 3 2 4 2 2 3
8 4 311 4
y y y yy yy
− + − = − ⋅ + − ⋅ −
= − − −= − −
ANSWER: 11 4y− −
69. PROBLEM:
5 2(4 5)x x− −
SOLUTION: ( ) ( )5 2(4 5) 5 2 4 2 5
5 8 103 10
x x x xx x
x
− − = + − − −
= − += − +
ANSWER: 3 10x− +
70. PROBLEM: 3 (2 7)x− +
SOLUTION: 3 (2 7) 3 2 72 4
x xx
− + = − −= − −
ANSWER: 2 4x− −
71. PROBLEM: ( )2 3 4 1x x y− − −
SOLUTION: ( )2 3 4 1 2 3 4 1
(2 3) 4 14 1
x x y x x yx y
x y
− − − = − + +
= − + += − + +
ANSWER: 4 1x y− + +
72. PROBLEM: ( ) ( )10 8 40 20 7y x y− − + −
SOLUTION:
( ) ( )10 8 40 20 7 10 8 40 20 740 10 20 8 740 (10 20) ( 8 7)40 10 1
y x y y x yx y yx yx y
− − + − = − − − +
= − + − − += − + − + − += − − −
ANSWER: 40 10 1x y− − −
73. PROBLEM: 1 3 2 12 4 3 5
y x y x⎛ ⎞− − −⎜ ⎟⎝ ⎠
SOLUTION:
1 3 2 1 1 3 2 12 4 3 5 2 4 3 5
3 1 1 24 5 2 33 1 1 24 5 2 33 5 1 4 1 3 2 220 20 6 615 4 3 420 6
11 120 611 120 6
y x y x y x y x
x x y y
x y
x y
x y
x y
x y
⎛ ⎞− − − = − − +⎜ ⎟⎝ ⎠
= − + + −
⎛ ⎞ ⎛ ⎞= − + + −⎜ ⎟ ⎜ ⎟⎝ ⎠ ⎝ ⎠− ⋅ ⋅ ⋅ ⋅⎛ ⎞ ⎛ ⎞= + + −⎜ ⎟ ⎜ ⎟
⎝ ⎠ ⎝ ⎠− + −⎛ ⎞ ⎛ ⎞= +⎜ ⎟ ⎜ ⎟
⎝ ⎠ ⎝ ⎠− −⎛ ⎞ ⎛ ⎞= +⎜ ⎟ ⎜ ⎟
⎝ ⎠ ⎝ ⎠
= − −
ANSWER: 11 120 6
x y− −
74. PROBLEM:
1 3 3 15 4 15 2
a b a b− + −
SOLUTION:
1 3 3 1 1 3 3 15 4 15 2 5 15 4 2
1 3 3 3 1 215 15 4 43 3 3 215 4
6 515 42 55 4
a b a b a a b b
a b
a b
a b
a b
− + − = + − −
⋅ ⋅⎛ ⎞ ⎛ ⎞= + + − −⎜ ⎟ ⎜ ⎟⎝ ⎠ ⎝ ⎠
+ − −⎛ ⎞ ⎛ ⎞= +⎜ ⎟ ⎜ ⎟⎝ ⎠ ⎝ ⎠
= −
= −
ANSWER: 2 55 4
a b−
75. PROBLEM:
( )2 23
x y x y− + −
SOLUTION:
( )2 2 22 23 3 3
2 2 23 32 1 3 2 2 33 3 3 32 3 2 6
3 35 83 3
x y x y x y x y
x x y y
x y
x y
x y
− + − = − + −
= + − −
⋅ ⋅⎛ ⎞ ⎛ ⎞= + + − −⎜ ⎟ ⎜ ⎟⎝ ⎠ ⎝ ⎠
+ − −⎛ ⎞ ⎛ ⎞= +⎜ ⎟ ⎜ ⎟⎝ ⎠ ⎝ ⎠
= −
ANSWER: 5 83 3
x y−
76. PROBLEM:
( ) ( )1 1 16 1 4 1 2 23 2 6
x y x y⎛ ⎞− − + − − − + −⎜ ⎟⎝ ⎠
SOLUTION:
( ) ( )1 1 1 1 1 1 1 16 1 4 1 2 2 6 1 4 1 2 23 2 6 3 3 2 2 6
1 1 12 2 2 23 2 6
1 1 12 2 2 23 2 6
1 2 1 3 10 06 6 6
2 3 16
060
x y x y x y x y
x y x y
x x y y
⎛ ⎞ ⎛ ⎞− − + − − − + − = − ⋅ − − ⋅ + ⋅ − ⋅ + − +⎜ ⎟ ⎜ ⎟⎝ ⎠ ⎝ ⎠
= − + + − + − +
= − + + − + − +
⋅ ⋅= + + − +
− +=
=
= ANSWER: 0
77. PROBLEM: ( ) ( )2 22 7 1 7 5x x x x− + + + −
SOLUTION:
( ) ( )2 2 2 2
2 2
2
2
2 7 1 7 5 2 7 1 7 5
2 7 7 1 533
0 44
x x x x x x x x
x x x xxx
− + + + − = − + + + −
+ −
=
= + + + −
=
−
−
ANSWER: 23 4x − 78. PROBLEM:
( )2 26 2 3 1 10 5x x x x− + − + −
SOLUTION:
( ) ( ) ( )2 2 2 2
2 2
2
2
1
6 2 3 1 10 5 6 2 6 3 6 1 10 5
12 10 5( 12 10) (1
8 68 )
2 15 6
3 6
x
x x x x x x x x
x x xx x
x x
− + − + − = ⋅ − + ⋅ − ⋅ + −
= − + −
= − + + −
−
+ −
+= −
−
ANSWER: 2 1 62 3x x+− −
79. PROBLEM:
( )2 23 8 12x x x− − + + −
SOLUTION:
( )2 2 2 2
2 2
3 8 12 3 8 12
3 8 120 3 203 20
x x x x x x
x x xx
x
− − + + − = − + − + −
= − + + − −= + −= −
ANSWER: 3 20x − 80. PROBLEM:
( ) ( )2 3 4 4 2 3a b a b− + − +
SOLUTION: ( ) ( )2 3 4 4 2 3 2 3 2 4 4 2 4 3
6 8 8 122 4
a b a b a b a ba b a b
a b
− + − + = ⋅ − ⋅ − ⋅ + ⋅
= − − += − +
ANSWER: 2 4a b− +
81. PROBLEM: ( ) ( )7 10 7 6 8 4x y x y− − − +
SOLUTION:
( ) ( ) ( ) ( ) ( )7 10 7 6 8 4 7 10 7 6 8 6 470 49 48 2470 48 49 24
( 70 48) (49 24)118 25
x y x y x y x yx y x yx x y y
x yx y
− − − + = − ⋅ − − 7 + − ⋅ + − ⋅
= − + − −= − − + −= − − + −= − +
ANSWER: 118 25x y− + 82. PROBLEM:
( ) ( )10 6 9 80 35x x− − −
SOLUTION:
( ) ( )10 6 9 80 35 10 6 10 9 80 3560 90 80 35(60 80) ( 90 35)
20 55
x x x xx x
xx
− − − = ⋅ − ⋅ − +
= − − += − + − += − −
ANSWER: 20 55x− −
83. PROBLEM: ( )210 5 3 1x x− − −
SOLUTION:
( ) ( ) ( ) ( )2 2
2
2
2
15 515 5 10
10 5 3 1 10 5 5 3 5 1
1
1 15
0 555 5
x x x x
x
x
xx x
x
− − − = + − − − − −
= + +
= + +
= +
−
+
+
−
−
ANSWER: 2 155 15x x+− + 84. PROBLEM:
( )24 6 9y+ −
SOLUTION:
( )2 2
2
2
6 9
6 54 4
6
6 0
9
5
4 4 6y y
yy
− ⋅
= − +
=
+ − ⋅
−
= +
ANSWER: 26 50y −
85. PROBLEM: 23 1 2 7
4 2 3 5x x x⎛ ⎞− + −⎜ ⎟
⎝ ⎠
SOLUTION:
2 2
2
2
2
2
3
3 1 2 7 3 1 2 74 2 3 5 4 2 3 5
1
3 2 4
3 2 72 4 3 51 72 51 72 51
12 129 812
1 712 2 5
x x x x x x
x x x
x
x
x
x
x
x
−
+ −
⋅ ⋅⎛ ⎞+ −⎜ ⎟⎝ ⎠
−⎛ ⎞
⎛ ⎞− + − = − +⎜ ⎟⎝ ⎠
= − +
= − +
= +− +
+=
⎜ ⎟⎝
+
⎠
−
ANSWER: 2 112
1 72 5
x x− ++
86. PROBLEM:
2 27 1 7 13 6
x x x⎛ ⎞− + − + −⎜ ⎟⎝ ⎠
SOLUTION:
2 2 2 2
2
2
2
2
2
7 1
7
7 1 7 17 13 6 3 6
16
7 137 2 7 16
14 1 7 16
15 7 165 7 12
16
x x x x x
x
x
x
x
x
x
x
x
x
x
x
− + −
⎛ ⎞= − − + −⎜ ⎟⎝ ⎠− ⋅
=
⎛ ⎞− + − + − = −
− + −
− −= + −
−= + −
⎜ ⎟⎝ ⎠
⎛ ⎞⎜ ⎟
= − +
⎠
−
⎝
ANSWER: 25 7 12
x x− + −
87. PROBLEM: ( ) ( )2 22 3 1 5 10 7y y y y− + − + −
SOLUTION:
( ) ( )2 2 2 2
2 2
2
2
2 3 1 5 10 7 2 3 1 5 10 7
2 5 3 10 1 7(2 5) (
13 83 10) 8
3
y y y y y y y y
y y yy
y y
yy
− + − + − = − + − − +
= − − − +
−
+
= − + − +
= − +
−
ANSWER: 2 1 83 3y y−− + 88. PROBLEM:
( ) ( )2 2 2 210 12 4a b c a b c− − + + + −
SOLUTION:
( ) ( )2 2 2 2 2 2 2 2
2 2 2 2
2
2
10 12 4 10 12 4
10 12 0 3
2 3
42
a b c a b c a b c a b c
a a b b c caa
cc
− − + + + − = − − + + + −
= − + − + +
+
=
−
−
= −
ANSWER: 22 3a c−
89. PROBLEM:
( ) ( )2 24 2 3 2 5 4 1x x x x− + − + − − SOLUTION:
( ) ( ) ( ) ( ) ( )2 2 2 2
2 2
2
2
4 2 3 2 5 4 1 4 2 4 3 4 2 5 5 4 5 1
8 12 8 5 20 5( 8 5) ( 12 20) 8 5
3 32 3
x x x x x x x x
x x x xx x
x x
− + − + − − = − + − − − + ⋅ − ⋅ − ⋅
= − − + + − −
= − + + − − + −
= − − +
ANSWER: 23 32 3x x− − +
90. PROBLEM: ( ) ( )2 22 3 7 1 3 5 1x x x x− + − + −
SOLUTION: ( ) ( ) ( ) ( )2 2 2 2
2 2
2
2
2 7 2 1 3 3 5 3 1
6 14 2 3 15 3(6 3) ( 14 15
2 3 7 1 3 5
) (3 2)3 29
2 3
5
1x x x x x x
x x
x
x x
x xx x
x
− ⋅ + ⋅ − ⋅ + − ⋅ − − ⋅
=
− + − + − =
− + − − +
= − + − − + +
= − +
⋅
ANSWER: 23 29 5x x− +
91. PROBLEM:
( )2 2 2 23 2x y xy x y xy+ − −
SOLUTION:
( )2 2 2 2 2 2 2 2
2 2 2
2 2
2
2 2
3 2 3 2
2(1 2) (3 1)
3
4
x y xy x y xy x y xy x y xy
x y x y xy xyx y xy
x y xy
+ − − = + − +
− + +
= − +
=
= − + +
ANSWER: 2 24x y xy− + 92. PROBLEM:
( ) ( )2 2 2 23 12 7 20 18x y xy x y xy− − − + SOLUTION: ( ) ( )2 2 2 2 2 2 2 2
2 2 2 2
2 2
2 2
3 12 7 20 18 3 3 12 7 20 18
3 36 7 20 18(3 7) ( 36 20) 18
4 16 18
x y xy x y xy x y xy x y xy
x y xy x y xyx y xy
x y xy
− − − + = − ⋅ − + −
= − − + −
= − + − + −
= − − −
ANSWER: 2 24 16 18x y xy− − −
93. PROBLEM: ( ) ( )3 5 3 2 4ab ba− − + −
SOLUTION:
( ) ( )3 5 3 2 4 3 5 15 2 85 2 3 15 8
( 5 2) (3 15 8)3 10
ab ba ab baab ab
abab
− − + − = − + + −
= − + + + −= − + + + −= − +
ANSWER: 3 10ab− + 94. PROBLEM:
( ) ( )9 2 7 1xy yx− − + − −
SOLUTION:
( ) ( ) ( ) ( )9 2 7 1 9 2 2 7 19 2 14 1
( 2 1) ( 9 14 1)3 22
xy yx xy yxxy yxxy
xy
− − + − − = − + − + − ⋅ − +
= − − − − += − − + − − += − −
ANSWER: 3 22xy− −
95. PROBLEM:
( ) ( )5 4 2 1 10 3 2α β α β− − + + − + SOLUTION:
( ) ( ) ( ) ( )5 4 2 1 10 3 2 5 4 5 2 5 1 10 10 3 10 220 10 5 10 30 2020 10 10 30 5 20
( 20 10) (10 30) ( 5 20)10 20 15
α β α β α β α βα β α βα α β β
α βα β
− − + + − + = − ⋅ − − + − + ⋅ − ⋅ + ⋅
= − + − + − += − + + − − += − + + − + − += − − +
ANSWER: 10 20 15α β− − +
96. PROBLEM:
( ) ( )2 2 2 21 1100 50 2 50 10 52 5
α αβ β α αβ β− + − + −
SOLUTION:
( ) ( )2 2 2 2
2 2 2 2
2
2
2 2
2 2 2
2
1 1100 50 2 50 10 52 5
1 1 1 1 1 1100 50 2 50 10
50 25 10 2(50 10) ( 25 2) (1 1)40 2
52 2 2
7
5 5 5
2
α αβ β α αβ β
α αβ β α αβ β
α αβ
α
β α αβ β
α αβ
αβ β
β
⎛ ⎞ ⎛ ⎞ ⎛ ⎞− + − + − ⋅ − ⋅⎜ ⎟ ⎜ ⎟ ⎜ ⎟⎝ ⎠ ⎝
− + − + −
= ⋅ ⋅ + ⋅ ⋅ −
+ +⎠ ⎝ ⎠
= − − −
= − + − − + +
= − +
ANSWER: 2 240 27 2α αβ β− +
Translate the following sentences into algebraic expressions and then simplify. 97. PROBLEM:
What is the difference of 3 4x − and 2 5x− + ?
SOLUTION: ( )3 4 2 5 3 4 2 55 9
x x x xx
− − − + = − + −
= −
ANSWER: 5 9x − 98. PROBLEM:
Subtract 2 3x − from 5 7x + .
SOLUTION: ( ) ( )5 7 2 3 5 7 2 33 10
x x x xx
+ − − = + − +
= +
ANSWER: 3 10x +
99. PROBLEM:
Subtract 4 3x + from twice the quantity 2x − .
SOLUTION: ( ) ( )2 2 4 3 2 4 4 32 7
x x x xx
− − + = − − −
= − −
ANSWER: 2 7x− −
100. PROBLEM: Subtract three times the quantity 8x− + from 10 9x − .
SOLUTION: ( )10 9 3 8 10 9 3 2413 33
x x x xx
− − − + = − + −
= −
ANSWER: 13 33x −
Part D: Discussion Board Topics 101. PROBLEM:
Do we need a distributive property for division, ( )a b c+ ÷ ? Explain. ANSWER: The distributive property is used when dividing grouped algebraic
expressions, ( ) a ba b cc c
+ ÷ = + .
102. PROBLEM:
Do we need a separate distributive property for three terms, ( )a b c d+ + ? Explain. ANSWER: No. The same distributive property can be used for three terms.
( )a b c d a b a c a d+ + = ⋅ + ⋅ + ⋅ 103. PROBLEM:
Explain how to subtract one expression from another. Give some examples and demonstrate the importance of the order in which subtraction is performed. ANSWER: The traditional names for the parts of the formula c − b = a are minuend (c) − subtrahend (b) = difference (a). The words “minuend” and “subtrahend” are uncommon in modern usage. In mathematics, it is often useful to view or even define subtraction as a kind of addition, the addition of the additive inverse. We can view 7 − 3 = 4 as the sum of two terms: 7 and -3. This perspective allows us to apply to subtraction all of the familiar rules and nomenclature of addition. Subtraction is not associative or commutative—in fact, it is anticommutative and left-associative—but addition of signed numbers is both. (Source: http://en.wikipedia.org/wiki/Subtraction)
104. PROBLEM: Given the algebraic expression 8 5(3 4)x− + , explain why subtracting 8 5− is not the first step. ANSWER: The term inside the brackets should be multiplied with 5 first. Always consider the order of operations when simplifying.
105. PROBLEM:
Can you apply the distributive property to the expression 5( )abc ? Explain why or why not and give some examples. ANSWER: Distributive property cannot be applied in the expression 5( )abc , because the expression does not contain any binary operator “+” or “–”. 5(abc) = 5abc
106. PROBLEM:
How can you check to see if you have simplified an expression correctly? Give some examples. ANSWER:
For expressions containing variables, use a constant term instead of the variable And find the solution. If the solution satisfies the expression the simplification is correct. For e.g.: 2x + 4 = 6 when x = 1 2(1) + 4 =6 2 + 4 = 6 6 = 6
Section2.3:SolvingLinearEquations:PartIPart A: Solutions to Linear Equations Is the given value a solution to the linear equation? 1. PROBLEM:
6 20; 26x x− = = SOLUTION:
6 20
26 6 2020 20
x − =− =
=
ANSWER: Yes, 26x = is a solution to the linear equation 6 20x − = .
2. PROBLEM: 7 6; 13y y+ = − = −
SOLUTION:
7 6
13 7 66 6
y + = −− + = −
− = −
ANSWER: Yes, 13y = − is a solution to the linear equation 7 6y + = − .
3. PROBLEM:
5 17; 12x x− + = = SOLUTION:
5 12 5 7 17x− + = − + = − ≠
ANSWER: No, 12x = is not a solution to the linear equation 5 17x− + = .
4. PROBLEM:
2 44; 11y y− = = SOLUTION: 2 2 11 22 44y− = − ⋅ = − ≠
ANSWER: No, 11y = is not a solution to the linear equation 2 44y− = .
5. PROBLEM:
4 24; 6x x= − = − SOLUTION:
( )4 24
4 6 2424 24
x = −− = −
− = −
ANSWER: Yes, 6x = − is a solution to the linear equation 4 24x = − .
6. PROBLEM:
5 1 34; 7x x− = = − SOLUTION: ( )5 1 5 7 1 35 1 36 34x − = ⋅ − − = − − = − ≠ ANSWER: No, 7x = − is not a solution to the linear equation 5 1 34x − = .
7. PROBLEM: 2 7 7; 0a a− − = − =
SOLUTION:
2 7 72 0 7 7
0 7 77 7
a− − = −− ⋅ − = −
− = −− = −
ANSWER: Yes, 0a =
is a solution to the linear equation 2 7 7a− − = − . 8. PROBLEM:
1 4 5; 33
x x− − = − = −
SOLUTION: ( )1 14 3 4 1 4 3 53 3
x− − = − ⋅ − − = − = − ≠ −
ANSWER: No, 3x = − is not a solution to the linear equation 1 4 53
x− − = − .
9. PROBLEM:
1 2 1 11; 2 3 4 6
x x− + = − =
SOLUTION:
1 2 12 3 4
1 11 2 12 6 3 4
11 2 112 3 4
11 2 4 112 12 4
11 8 112 4
3 112 41 14 4
x− + = −
⎛ ⎞− + = −⎜ ⎟⎝ ⎠
− + = −
− ⋅+ = −
− += −
− = −
− = −
ANSWER: Yes, 116
x = is a solution to the linear equation 1 2 12 3 4
x− + = − .
10. PROBLEM: 8 33 3 ; 3x x x− − = =
SOLUTION:
8 33 38 3 33 3 324 33 9
57 9
x x− − =− ⋅ − = ⋅− − =
− ≠
ANSWER: No,
3x = is not a solution to the linear equation 8 33 3x x− − = .
11. PROBLEM: 3 5 2 15; 2y y y− = − − = − SOLUTION:
( ) ( )3 5 2 15
3 2 5 2 2 156 5 4 15
11 11
y y− = − −
− − = − − −
− − = −− = −
ANSWER: Yes, 2y = − is a solution to the linear equation 3 5 2 15y y− = − − .
12. PROBLEM:
( ) 13 2 1 4 3; 2
x x x+ = − − = −
SOLUTION:
( )
( )( )
3 2 1 4 3
1 13 2 1 4 32 2
3 1 1 2 3
3 0 10 1
x x+ = − −
⎛ ⎞⎛ ⎞ ⎛ ⎞⋅ − + = − ⋅ − −⎜ ⎟ ⎜ ⎟⎜ ⎟⎝ ⎠ ⎝ ⎠⎝ ⎠− + = −
= −
≠ −
ANSWER: No, 1 2
x = −
is not a solution to the linear equation
( )3 2 1 4 3x x+ = − − .
13. PROBLEM: 1 1 1 12 3 3 6
y y− = + ; 3y =
SOLUTION:
1 1 1 12 3 3 61 1 1 13 32 3 3 6
3 1 112 3 6
3 3 1 2 1 6 16 6 6 6
9 2 6 16 6
7 76 6
y y− = +
⋅ − = ⋅ +
− = +
⋅ ⋅ ⋅− = +
− +=
=
ANSWER: Yes, 3y = is a solution to the linear equation 1 1 1 12 3 3 6
y y− = + .
14. PROBLEM:
4 1 2 13 9 3 9
y y− + = − − ; 13
y =
SOLUTION: 4 1 2 13 9 3 9
4 1 1 2 1 13 3 9 3 3 9
4 1 2 19 9 9 94 1 2 19 9
3 39 91 13 3
y y− + = − −
− ⋅ + = − ⋅ −
− + = − −
− + − −=
− −=
− = −
ANSWER: Yes, 13
y = is a solution to the linear equation 4 1 2 13 9 3 9
y y− + = − − .
Part B: Solving Basic Linear Equations Solve. 15. PROBLEM:
3 13x + = SOLUTION:
3 133 3 13 3
0 1010
xx
xx
+ =+ − = −
+ ==
ANSWER: 10x = 16. PROBLEM:
4 22y − = SOLUTION: 4 22
4 4 22 40 26
26
yy
yy
− =− + = +
+ ==
ANSWER: 26y =
17. PROBLEM:
6 12x− + = SOLUTION:
6 126 6 12 6
0 1818
xxx
x
− + =− + + = +
+ ==
ANSWER: 18x =
18. PROBLEM: 9 4y+ = − SOLUTION: 9 4
9 9 4 90 13
13
yy
yy
+ = −+ − = − −
+ = −= −
ANSWER: 13y = −
19. PROBLEM:
1 12 3
x − =
SOLUTION:
1 12 3
1 1 1 12 2 3 2
1 2 1 306 6
2 36
56
x
x
x
x
x
− =
− + = +
⋅ ⋅+ = +
+=
=
ANSWER: 56
x =
20. PROBLEM: 2 13 5
x + = −
SOLUTION: 2 13 52 2 1 23 3 5 3
1 3 2 5015 15
3 1015
1315
x
x
x
x
x
+ = −
+ − = − −
− ⋅ ⋅+ = −
− −=
= −
ANSWER: 1315
x = −
21. PROBLEM:
1 12 32 3
x + =
SOLUTION:
1 12 32 35 102 3
5 5 10 52 2 3 2
10 2 5 306 6
20 156
56
x
x
x
x
x
x
+ =
+ =
+ − = −
⋅ ⋅+ = −
−=
=
ANSWER: 56
x =
22. PROBLEM: 37 37x− + = −
SOLUTION: 37 37
37 37 37 370 0
0
xxxx
− + = −− + + = − +
+ ==
ANSWER: 0x =
23. PROBLEM:
4 44x = − SOLUTION:
4 444 444 4
1 1111
xx
xx
= −−
=
⋅ = −= −
ANSWER: 11x = −
24. PROBLEM:
9 63x− = SOLUTION:
9 639 639 9
1 77
xx
xx
− =−
=− −⋅ = −
= −
ANSWER: 7x = −
25. PROBLEM:
13y− = SOLUTION: 13
13 1 1
13
yy
y
− =−
=− −= −
ANSWER: y = −13
26. PROBLEM: 10x− = −
SOLUTION: 10
101 1
10
xx
x
− = −− −
=− −=
ANSWER: 10x =
27. PROBLEM:
9 0x− = SOLUTION: 9 0
9 09 9
1 00
xx
xx
− =−
=− −⋅ =
=
ANSWER: 0x =
28. PROBLEM:
3 33a− = − SOLUTION:
3 333 333 3
1 1111
aa
aa
− = −− −
=− −⋅ =
=
ANSWER: 11a =
29. PROBLEM: 27 18y= SOLUTION: 27 18
27 1818 18
3 1232
32
yy
y
y
y
=
=
= ⋅
=
=
ANSWER: 32
y =
30. PROBLEM:
14 7x= − SOLUTION:
14 714 7
7 72 1
2
xx
xx
= −−
=− −− = ⋅= −
ANSWER: 2x = −
31. PROBLEM:
5.6 39.2a = − SOLUTION:
5.6 39.25.6 39.25.6 5.61 7
7
aa
aa
= −−
=
⋅ = −= −
ANSWER: 7a = −
32. PROBLEM: 1.2 3.72y− =
SOLUTION: 1.2 3.72
1.2 3.721.2 1.21 3.1
3.1
yy
yy
− =−
=− −
⋅ = −= −
ANSWER: 3.1y = −
33. PROBLEM:
1 13 2
x = −
SOLUTION: 1 1
3 21 3 1 33 1 2 1
31232
x
x
x
x
= −
⋅ = − ⋅
⋅ = −
= −
ANSWER: 32
x = −
34. PROBLEM:
112 4t
− =
SOLUTION:
( ) ( )
112 4
112 1212 4
1 33
t
t
tt
− =
− ⋅ − = ⋅ −
⋅ = −= −
ANSWER: 3t = −
35. PROBLEM: 7 13 2
x− =
SOLUTION: 7 1
3 27 3 1 33 7 2 7
31143
14
x
x
x
x
− =
⎛ ⎞ ⎛ ⎞− ⋅ − = −⎜ ⎟ ⎜ ⎟⎝ ⎠ ⎝ ⎠
⋅ = −
= −
ANSWER: 314
x = −
36. PROBLEM:
35x= −
SOLUTION: 35
5 3 55
1 1515
x
x
xx
= −
⋅ = − ⋅
⋅ = −= −
ANSWER: 15x = −
37. PROBLEM: 4 29 3
y = −
SOLUTION: 4 2
9 34 9 2 99 4 3 4
31232
y
y
y
y
= −
⋅ = − ⋅
⋅ = −
= −
ANSWER: 32
y = −
38. PROBLEM:
5 58 2
y− = −
SOLUTION: 5 58 2
5 8 5 88 5 2 5
1 44
y
y
yy
− = −
⎛ ⎞ ⎛ ⎞− ⋅ − = − ⋅ −⎜ ⎟ ⎜ ⎟⎝ ⎠ ⎝ ⎠
⋅ ==
ANSWER: 4y =
Part C: Solving Linear Equations Solve. 39. PROBLEM:
5 7 32x + = SOLUTION:
5 7 32
5 7 7 32 75 0 25
5 255 5
1 55
xx
xx
xx
+ =+ − = −
+ =
=
⋅ ==
ANSWER: 5x = 40. PROBLEM:
4 3 21x − = SOLUTION:
4 3 21
4 3 3 21 34 0 24
4 244 4
1 66
xx
xx
xx
− =− + = +
+ =
=
⋅ ==
ANSWER: 6x =
41. PROBLEM:
3 7 23a − = SOLUTION: 3 7 23
3 7 7 23 73 0 30
3 303 3
1 1010
aa
aa
aa
− =− + = +
+ =
=
⋅ ==
ANSWER: 10a =
42. PROBLEM: 12 1 1y + = SOLUTION:
12 1 1
12 1 1 1 112 0 0
12 012 121 0
0
yy
yy
yy
+ =+ − = −+ =
=
⋅ ==
ANSWER: 0y =
43. PROBLEM: 21 7 0x − = SOLUTION:
21 7 0
21 7 7 0 721 0 7
21 721 21
11313
xx
xx
x
x
− =− + = +
+ =
=
⋅ =
=
ANSWER: 13
x =
44. PROBLEM:
3 2 13y− + = − SOLUTION:
3 2 13
3 2 2 13 23 0 15
3 153 3
1 55
yy
yy
yy
− + = −− + − = − −
− + = −− −
=− −⋅ =
=
ANSWER: 5y =
45. PROBLEM: 5 9 8x− + =
SOLUTION:
5 9 8
5 9 9 8 95 0 1
5 15 5
11515
xx
xx
x
x
− + =− + − = −
− + = −− −
=− −
⋅ =
=
ANSWER: 15
x =
46. PROBLEM:
22 55 22x − = − SOLUTION:
22 55 2222 55 55 22 55
22 0 3322 3322 22
31232
xx
xx
x
x
− = −− + = − +
+ =
=
⋅ =
=
ANSWER: 32
x =
47. PROBLEM: 4.5 2.3 6.7x − = SOLUTION: 4.5 2.3 6.7
4.5 2.3 2.3 6.7 2.34.5 0 9.0
4.5 9.04.5 4.51 2
2
xx
xx
xx
− =− + = +
+ =
=
⋅ ==
ANSWER: 2x = 48. PROBLEM:
1.4 3.2 3x− = SOLUTION:
1.4 3.2 3
1.4 1.4 3.2 3 1.40 3.2 1.6
3.2 1.63.2 3.21 0.5
0.5
xxxx
xx
− =− − = −
− =−
=− −
⋅ = −= −
ANSWER: 0.5x = −
49. PROBLEM:
9.6 1.4 10.28y− = − SOLUTION: 9.6 1.4 10.28
9.6 1.4 9.6 10.28 9.61.4 0 19.88
1.4 19.881.4 1.41 14.2
14.2
yy
yy
yy
− = −− − = − −− + = −
− −=
− −⋅ =
=
ANSWER: 14.2y =
50. PROBLEM: 4.2 3.71 8.89y − = SOLUTION: 4.2 3.71 8.89
4.2 3.71 3.71 8.89 3.714.2 0 12.6
4.2 12.64.2 4.21 3
3
yy
yy
yy
− =− + = +
+ =
=
⋅ ==
ANSWER: 3y =
51. PROBLEM: 3 2 11y− = − SOLUTION: 3 2 11
3 2 3 11 32 142 142 2
1 77
yy
yy
yy
− = −− − = − −
− = −− −
=− −⋅ =
=
ANSWER: 7y = 52. PROBLEM:
4 7 24a− − = SOLUTION: 4 7 24
4 4 7 24 40 7 28
7 287 7
1 44
aaaa
aa
− − =− + − = +
− =−
=− −⋅ = −
= −
ANSWER: 4a = −
53. PROBLEM: 10 2 5x− = −
SOLUTION:
10 2 5
10 5 2 5 55 2 05 2
2 25 12
52
xxxx
x
x
− = −− + = − +
− = +−
=
− = ⋅
= −
ANSWER: 52
x = −
54. PROBLEM:
24 6 12y= − SOLUTION:
24 6 12
24 6 6 6 1218 0 12
18 1212 123 12
32
yy
yy
y
y
= −− = − −
= −−
=− −
− = ⋅
= −
ANSWER: 32
y = −
55. PROBLEM: 5 1 26 2 3
x − =
SOLUTION: 5 1 26 2 3
5 1 1 2 16 2 2 3 2
5 2 2 1 306 6 6
5 4 36 65 76 6
5 6 7 66 5 6 5
71575
x
x
x
x
x
x
x
x
− =
− + = +
⋅ ⋅+ = +
+=
=
⋅ = ⋅
⋅ =
=
ANSWER: 75
x =
56. PROBLEM: 1 1 22 3 5
x + =
SOLUTION:
1 1 22 3 5
1 1 1 2 12 3 3 5 3
1 2 3 1 502 15 15
1 6 52 151 12 15
1 12 22 15
21152
15
x
x
x
x
x
x
x
x
+ =
+ − = −
⋅ ⋅+ = −
−=
=
⋅ = ⋅
⋅ =
=
ANSWER: 215
x =
57. PROBLEM: 2 143 6
a − = −
SOLUTION:
2 143 6
2 2 1 243 3 6 3
1 2 24 06 61 446
3461421
4 24 4
1 112 418
a
a
a
a
a
a
a
a
a
− = −
− + = − +
− ⋅+ = +
− +=
=
=
=
⋅ = ⋅
=
ANSWER: 18
a =
58. PROBLEM: 3 1 15 2 10
x − =
SOLUTION:
3 1 15 2 10
3 1 1 1 15 2 2 10 2
3 1 1 505 10 10
3 1 55 103 65 10
3 5 6 55 3 10 3
1 11
x
x
x
x
x
x
xx
− =
− + = +
⋅+ = +
+=
=
⋅ = ⋅
⋅ ==
ANSWER: 1x =
59. PROBLEM: 4 1 15 3 15
y− + =
SOLUTION: 4 1 1
5 3 154 1 1 1 15 3 3 15 3
4 1 1 505 15 15
4 1 55 154 45 15
4 5 4 55 4 15 4
11313
y
y
y
y
y
y
y
y
− + =
− + − = −
⋅− + = −
−− =
−− =
−⎛ ⎞ ⎛ ⎞− ⋅ − = ⋅ −⎜ ⎟ ⎜ ⎟⎝ ⎠ ⎝ ⎠
⋅ =
=
ANSWER: 13
y =
60. PROBLEM:
9 4 416 3 3
x− + =
SOLUTION: 9 4 416 3 3
9 4 4 4 416 3 3 3 3
9 0 016
9 169
160( )16 9
1 00
x
x
x
x
xx
− + =
− + − = −
− + =
⎛ ⎞− ⋅ − = −⎜ ⎟⎝ ⎠
⋅ ==
ANSWER: 0x =
61. PROBLEM: 5 14x− + =
SOLUTION: 5 14
5 5 14 50 9
91 1
1 99
xx
xx
xx
− + =− + − = −
− + =−
=− −⋅ = −
= −
ANSWER: 9x = − 62. PROBLEM:
7 12y− − = − SOLUTION: 7 12
7 7 12 70 5
51 1
5
yy
yy
y
− − = −− − + = − +
− + = −− −
=− −
=
ANSWER: 5y =
63. PROBLEM:
75 200a− = SOLUTION: 75 200
75 75 200 750 125
1251 1
1 125125
aa
aa
aa
− =− − = −
− =−
=− −⋅ = −
= −
ANSWER: 125a = −
64. PROBLEM: 15 5 x= − SOLUTION: 15 5
15 5 5 510 010
1 110
10
xx
xx
xx
= −− = − −
= −−
=− −− =
= −
ANSWER: 10x = −
65. PROBLEM:
8 4 2x− = − SOLUTION: 8 4 2
8 4 4 2 412 0 212 22 26
6
xxx
x
xx
− = −− − = − −− = −− −
=− −
==
ANSWER: 6x = 66. PROBLEM:
33 33x− = SOLUTION: 33 33
33 33 33 330 0
01 1
0
xxxx
x
− =− − = −
− =−
=− −
=
ANSWER: 0x =
67. PROBLEM: 18 6 y= − SOLUTION: 18 6
18 6 6 612 012
1 112
12
yyy
y
yy
= −− = − −
= −−
=− −− =
= −
ANSWER: 12y = −
68. PROBLEM:
12 2 3x− = − + SOLUTION: 12 2 3
12 3 2 3 315 2 015 22 215 12
152
xxxx
x
x
− = − +− − = − + −
− = − +− −
=− −
= ⋅
=
ANSWER: 152
x =
69. PROBLEM:
3 3.36 1.2a− = − SOLUTION: 3 3.36 1.2
3 3.36 3.36 1.2 3.366.36 0 1.26.36 1.21.2 1.25.3
5.3
aa
aa
aa
− = −− − = − −
− = −− −
=− −
==
ANSWER: 5.3a =
70. PROBLEM: 0 3.1 32.55a= − + SOLUTION:
0 3.1 32.55
0 32.55 3.1 32.55 32.5532.55 3.1 032.55 3.1
3.1 3.110.5 1
10.5
aaaa
aa
= − +− = − + −− = − +− −
=− −
= ⋅=
ANSWER: 10.5a =
71. PROBLEM: 1 3 104 8
x= − +
SOLUTION:
1 3 104 8
1 3 3 3104 8 8 8
1 2 3 0 108 8
2 3 108
5 1085
10810 10
5 1 18 10
116
116
x
x
x
x
x
x
x
x
x
= − +
+ = − + +
⋅+ = +
+=
=
=
⋅ = ⋅
=
=
ANSWER: 116
x =
72. PROBLEM: 170 502
y= −
SOLUTION:
( ) ( )
170 502
170 50 50 502
120 02
120 2 22
40 140
y
y
y
y
yy
= −
− = − −
= −
⋅ − = − ⋅ −
− = ⋅= −
ANSWER: 40y = −
Translate the following sentences into linear equations and then solve. 73. PROBLEM:
The sum of 2x and 5 is equal to 15. SOLUTION:
The linear equation is 2 5 15x + = . 2 5 15
2 5 5 15 52 0 10
2 102 2
1 55
xx
xx
xx
+ =+ − = −
+ =
=
⋅ ==
ANSWER: 5x =
74. PROBLEM: The sum of −3x and 7 is equal to 14. SOLUTION:
The linear equation is 3 7 14x− + = . 3 7 14
3 7 7 14 73 0 7
3 73 3
71373
xx
xx
x
x
− + =− + − = −
− + =−
=− −
⋅ = −
= −
ANSWER: 73
x = −
75. PROBLEM:
The difference of 5x and 6 is equal to 4. SOLUTION: The linear equation is 5 6 4x − = .
5 6 45 6 6 4 6
5 0 105 105 5
1 22
xx
xx
xx
− =− + = +
+ =
=
⋅ ==
ANSWER: 2x =
76. PROBLEM: Twelve times x is equal to 36. SOLUTION: The linear equation is 12 36x = . 12 3612 3612 121 3
3
xx
xx
=
=
⋅ ==
ANSWER: 3x =
77. PROBLEM:
A number n divided by 8 is 5.
SOLUTION: The linear equation is 58n= .
588 5 8
81 40
40
n
n
nn
=
⋅ = ⋅
⋅ ==
ANSWER: 40n = 78. PROBLEM:
Six subtracted from two times a number x is 12. SOLUTION: The linear equation is 2 6 12x − = .
2 6 122 6 6 12 6
2 0 182 182 2
1 99
xx
xx
xx
− =− + = +
+ =
=
⋅ ==
ANSWER: 9x =
79. PROBLEM: Four added to three times a number n is 25. SOLUTION: The linear equation is 3 4 25n + = .
3 4 253 4 4 25 4
3 0 213 213 3
1 77
nn
nn
nn
+ =+ − = −
+ =
=
⋅ ==
ANSWER: 7n = 80. PROBLEM:
Three-fourths of a number x is 9.
SOLUTION: The linear equation is 3 94
x = .
3 94
3 4 494 3 3
1 3 412
x
x
xx
=
⋅ = ⋅
⋅ = ⋅=
ANSWER: 12x =
81. PROBLEM:
Negative two-thirds times a number x is equal to 20.
SOLUTION: The linear equation is 2 203
n− = .
2 2032 3 3203 2 2
1 10 330
n
n
nn
− =
⎛ ⎞ ⎛ ⎞− ⋅ − = ⋅ −⎜ ⎟ ⎜ ⎟⎝ ⎠ ⎝ ⎠
⋅ = − ⋅= −
ANSWER: 30n = −
82. PROBLEM: One-half of a number x plus 3 is equal to 10.
SOLUTION: The linear equation is 1 3 102
x + = .
1 3 102
1 3 3 10 32
1 0 721 2 7 22
1 1414
x
x
x
x
xx
+ =
+ − = −
+ =
⋅ = ⋅
⋅ ==
ANSWER: 14x =
Find a linear equation of the form 0ax b+ = with the given solution, where a and b are integers. (Answersmayvary) 83. PROBLEM:
2x = SOLUTION: The linear equation is 2 0x − = where 1a = and 2b = − . The solution 2x = can be arrived at as follows:
2 02 2 0 2
0 22
xx
xx
− =− + = +
+ ==
ANSWER: 2x =
Find a linear equation of the form 0ax b+ = with the given solution, where a and b are integers. (Solutionsmayvary) 84. PROBLEM:
3x = − SOLUTION: The linear equation is 3 0x + = where 1a = and 3b = . The solution 3x = − can be arrived at as follows:
3 03 3 0 3
0 33
xx
xx
+ =+ − = −
+ = −= −
ANSWER: 3x = −
85. PROBLEM:
12
x = −
SOLUTION: The linear equation is 2 1 0x + = where 2a = and 1b = . The
solution 12
x = − can be arrived at as follows:
2 1 02 1 1 0 12 0 12 12 2
112
12
xxxx
x
x
+ =+ − = −+ = −−
=
⋅ = −
= −
ANSWER: 12
x = −
86. PROBLEM: 23
x =
SOLUTION: The linear equation is 3 2 0x − = where 3a = and 2b = − . The
solution 2 3
x = can be arrived at as follows:
3 2 03 2 2 0 2
3 0 23 23 3
21323
xx
xx
x
x
− =− + = +
+ =
=
⋅ =
=
ANSWER: 23
x =
Part D: Discussion Board Topics 87. PROBLEM:
How many steps are needed to solve any equation of the form ax b c+ = ? Explain. SOLUTION: 3 steps.
ax b cax b b c bax c ba c bxa a
c bxa
+ =+ − = −= −
−=
−=
ANSWER: c bxa−
=
88. PROBLEM: Instead of dividing by 6 when 6 12x = , could you multiply by the reciprocal of 6? Does this always work? ANSWER: Multiplying 12 with the reciprocal of 6 gives the same SOLUTION as dividing 12 with 6. Yes. This method works always
Section2.4:SolvingLinearEquations:PartIIPart A: Checking for Solutions Is the given value a solution to the linear equation? 1. PROBLEM:
( )2 3 5 6 3 8; 4x x x+ − = − = − SOLUTION: On substituting 4x = − in the linear equation, we have:
( )( )( ) ( )( )
( )
2 3 5 6 3 8
2 3 4 5 6 3 4 8
2 12 5 6 12 8
2 7 6 2014 6 20
20 20
x x+ − = −
− + − = − −
− + − = − −
− − = −
− − = −− = −
ANSWER: Substituting 4x = − in the given linear equation leads to a true statement; therefore, the value is a solution to the linear equation.
2. PROBLEM:
17 8 9 ; 1x x x x− + − = − = − SOLUTION: On substituting 1x = − in the linear equation, we have:
( ) ( ) ( )17 8 9
1 17 8 1 9 11 17 8 9 1
26 10
x x x− + − = −
− − + − − = − −
+ + = +≠
ANSWER: Substituting 1x = − in the given linear equation leads to a false statement; therefore, the value is not a solution to the linear equation.
3. PROBLEM:
( ) ( ) 14 3 7 3 2 1; 3
x x x− − + = − =
SOLUTION: On substituting 13
x = in the linear equation, we have:
( ) ( )
( )
( )
4 3 7 3 2 1
1 14 3 7 3 2 13 3
1 2 34 1 7 3 13 3
1 64 6 3 13
24 7 131 1
x x− − + = −
⎛ ⎞⎛ ⎞ ⎛ ⎞− − + = −⎜ ⎟ ⎜ ⎟⎜ ⎟⎝ ⎠ ⎝ ⎠⎝ ⎠⋅⎛ ⎞− − + = −⎜ ⎟
⎝ ⎠+⎛ ⎞− − = −⎜ ⎟
⎝ ⎠− − = −
− = −
ANSWER: Substituting 13
x = in the given linear equation leads to a false
statement; therefore, the value is not a solution to the linear equation. 4. PROBLEM:
( ) ( )5 2 5 3 ; 8x x x− − − = − + = − SOLUTION: On substituting 8x = − in the linear equation, we have:
( ) ( )( )( ) ( )( )
( ) ( )
5 2 5 3
5 2 8 5 8 3
5 2 13 55 26 5
21 5
x x− − − = − +
− − − − = − − +
− − − = − −
− + =≠
ANSWER: Substituting 8x = − in the given linear equation leads to a false statement; therefore, the value is not a solution to the linear equation.
5. PROBLEM: 17 2 6 1; 102
x x x⎛ ⎞− − = − =⎜ ⎟⎝ ⎠
SOLUTION: On substituting 10x = in the linear equation, we have:
( )
( )( )
17 2 6 12
17 2 10 6 10 12
7 2 5 6 9
7 2 1 97 2 9
9 9
x x⎛ ⎞− − = −⎜ ⎟⎝ ⎠
⎛ ⎞− − = −⎜ ⎟⎝ ⎠− − =
− − =
+ ==
ANSWER: Substituting 10x = in the given linear equation leads to a true statement; therefore, the value is a solution to the linear equation.
6. PROBLEM:
( )2 43 9 2 0; 3 9
x x x− − = =
SOLUTION: On substituting 49
x = in the linear equation, we have:
( )
4 2 43 9 2 09 3 9
4 2 4 2 03 3
4 2 2 03 3
4 4 03 3
0 0
⎛ ⎞⎛ ⎞ ⎛ ⎞− − =⎜ ⎟ ⎜ ⎟⎜ ⎟⎝ ⎠ ⎝ ⎠⎝ ⎠
− − =
− ⋅ =
− =
=
ANSWER: Substituting 49
x = in the given linear equation leads to a true
statement; therefore, the value is a solution to the linear equation.
Part B: Solving Linear Equations Solve. 7. PROBLEM:
4 7 7 5x x− = +
SOLUTION:
4 7 7 54 7 7 7 5 7
3 7 53 7 7 5 7
3 123 123 3
4
x xx x x x
xx
xx
x
− = +− − = + −− − =
− − + = +− =−
=− −
= −
ANSWER: 4x = − 8. PROBLEM:
5 3 8 9x x− + = − −
SOLUTION:
5 3 8 95 3 8 8 9 8
3 3 93 3 3 9 3
3 123 123 3
4
x xx x x x
xx
xx
x
− + = − −− + + = − − +
+ = −+ − = − −
= −−
=
= −
ANSWER: 4x = −
9. PROBLEM:
3 5 2 17x x− = −
SOLUTION:
3 5 2 173 5 2 2 17 2
5 175 5 17 5
12
x xx x x x
xx
x
− = −− − = − −
− = −− + = − +
= −
ANSWER: 12x = −
10. PROBLEM: 2 52 3 13y y− − = +
SOLUTION:
2 52 3 132 52 3 3 13 3
5 52 135 52 52 13 52
5 655 655 5
13
y yy y y y
yy
yy
y
− − = +− − − = + −
− − =− − + = +
− =−
=− −
= −
ANSWER: 13y = −
11. PROBLEM:
4 2 7 20x x− + = −
SOLUTION:
4 2 7 204 2 7 7 20 7
11 2 2011 2 2 20 2
11 2211 2211 11
2
x xx x x x
xx
xx
x
− + = −− + − = − −
− + = −− + − = − −
− = −− −
=− −
=
ANSWER: 2x = 12. PROBLEM:
4 3 6 15x x− = −
SOLUTION:
4 3 6 154 3 6 6 15 6
2 3 152 3 3 15 3
2 122 122 2
6
x xx x x x
xx
xx
x
− = −− − = − −− − = −
− − + = − +− = −− −
=− −
=
ANSWER: 6x =
13. PROBLEM: 9 25 12 25x x− = −
SOLUTION:
9 25 12 259 25 12 12 25 12
3 25 253 25 25 25 25
3 03 03 3
0
x xx x x x
xx
xx
x
− = −− − = − −− − = −
− − + = − +− =−
=− −
=
ANSWER: 0x = 14. PROBLEM:
12 15 6 23y y+ = − +
SOLUTION:
12 15 6 2312 15 6 6 23 6
18 15 2318 15 15 23 15
18 818 818 18
49
y yy y y y
yy
yy
y
+ = − ++ + = − + +
+ =+ − = −
=
=
=
ANSWER: 49
y = 15. PROBLEM:
1.2 0.7 3 4.7x x− = +
SOLUTION:
1.2 0.7 3 4.71.2 0.7 3 3 4.7 3
1.8 0.7 4.71.8 0.7 0.7 4.7 0.7
1.8 5.41.8 5.41.8 1.8
3
x xx x x x
xx
xx
x
− = +− − = + −
− − =− − + = +
− =−
=− −
= −
ANSWER: 3x = −
16. PROBLEM: 2.1 6.1 1.3 4.4x x+ = − +
SOLUTION:
2.1 6.1 1.3 4.42.1 6.1 1.3 1.3 4.4 1.3
3.4 6.1 4.43.4 6.1 6.1 4.4 6.1
3.4 1.73.4 1.73.4 3.4
0.5
x xx x x x
xx
xx
x
+ = − ++ + = − + +
+ =+ − = −
= −−
=
= −
ANSWER: 0.5x = −
17. PROBLEM:
2.02 4.8 1 4.782 1.2 x x+ = −
SOLUTION:
2.02 4.8 1 4.782 1.2 2.02 4.8 1.2 14.782 1.2 1.2
3.22 4.8 14.7823.22 4.8 4.8 14.782 4.8
3.22 9.9823.22 9.9823.22 3.22
3.1
x xx x x x
xx
xx
x
+ = −+ + = − +
+ =+ − = −
=
=
=
ANSWER: 3.1x =
18. PROBLEM: 3.6 5.5 8.2 6.5 4.6 x x x− + + = +
SOLUTION:
3.6 5.5 8.2 6.5 4.6 4.6 5.5 6.5 4.6
4.6 5.5 4.6 6.5 4.6 4.6 5.5 6.5
x x xx x
x x x x
− + + = ++ = +
+ − = + −≠
ANSWER: ∅ . Solving leads to a false statement; therefore, the equation is a contradiction and there is no solution.
19. PROBLEM:
1 2 12 3 5
x x− = +
SOLUTION:
( ) ( )
1 2 12 3 51 2 12 3 51 1 2 2 12 2 3 51 2 12 3 51 2 2 1 22 3 3 5 31 1 3 2 52 15 151 3 102 151 132 22 15
2615
x x
x x x x
x
x
x
x
x
x
x
− = +
− − = + −
⋅⎛ ⎞− − =⎜ ⎟⎝ ⎠
− − =
− − + = +
⋅ ⋅− = +
+− =
− ⋅ − = ⋅ −
= −
ANSWER: 2615
x = −
20. PROBLEM: 1 1 1 13 2 4 3
x x− = − −
SOLUTION:
1 1 1 13 2 4 3
1 1 1 1 1 13 2 4 4 3 4
1 4 1 3 1 112 12 2 3
7 1 1 1 112 2 2 3 2
7 1 112 6 67 2 3
12 67 12 1 12
12 7 6 727
x x
x x x x
x
x
x
x
x
x
− = − −
− + = − − +
⋅ ⋅⎛ ⎞+ − = −⎜ ⎟⎝ ⎠
− + = − +
− ⋅2 ⋅3= +
− +=
⋅ = ⋅
=
ANSWER: 27
x =
21. PROBLEM: 1 2 1 3
10 5 5 10y y− + = +
SOLUTION: 1 2 1 310 5 5 10
1 2 1 1 3 110 5 5 5 10 51 1 2 2 3
10 10 5 101 2 2 310 5 103 2 2 3 2
10 5 5 10 53 3 2 2
10 10 103 3 4
10 103 10 1 10
10 3 10 313
y y
y y y y
y
y
y
y
y
y
y
− + = +
− + − = + −
− ⋅⎛ ⎞− + =⎜ ⎟⎝ ⎠
− −⎛ ⎞ + =⎜ ⎟⎝ ⎠
− + − = −
⋅− = −
−− =
⎛ ⎞ ⎛ ⎞− ⋅ − = − −⎜ ⎟ ⎜ ⎟⎝ ⎠ ⎝ ⎠
=
ANSWER: 13
y =
22. PROBLEM: 20 5 53 2 6
x x− = +
SOLUTION:
20 5 53 2 6
20 5 5 5 53 2 2 6 2
1 2 5 20 52 2 3 6
2 5 20 52 3 6
3 20 52 3 6
3 20 20 5 202 3 3 6 3
3 5 20 22 6 63 5 402 63 452 6
3 2 45 22 3 6 3
5
x x
x x x x
x
x
x
x
x
x
x
x
x
− = +
− − = + −
⋅⎛ ⎞− − =⎜ ⎟⎝ ⎠
−⎛ ⎞ − =⎜ ⎟⎝ ⎠
− − =
− − + = +
⋅− = +
+− =
− =
⎛ ⎞ ⎛ ⎞− − = −⎜ ⎟ ⎜ ⎟⎝ ⎠ ⎝ ⎠
= −
ANSWER: 5x = −
23. PROBLEM: 2 1 5 373 2 8 24
y y+ = +
SOLUTION:
( )
2 1 5 373 2 8 24
2 1 5 5 37 53 2 8 8 24 8
2 8 5 3 1 3724 24 2 24
16 15 1 3724 2 24
1 1 3724 2 24
1 1 1 37 124 2 2 24 2
1 37 1 1224 24 241 37 1224 24
1 2524 2424 24
25
y y
y y y y
y
y
y
y
y
y
y
y
+ = +
+ − = + −
⋅ ⋅⎛ ⎞− + =⎜ ⎟⎝ ⎠
−⎛ ⎞ + =⎜ ⎟⎝ ⎠
+ =
+ − = −
⋅= −
−=
= ⋅
=
ANSWER: 25y =
24. PROBLEM: 1 4 10 1 23 3 7 3 21
x x x+ = + −
SOLUTION: 1 4 10 1 23 3 7 3 211 4 10 3 2 13 3 21 21 31 4 30 2 13 3 21 31 4 4 13 3 3 3
1 4 4 4 1 43 3 3 3 3 3
1 13 3
x x x
x x
x x
x x
x x x x
+ = + −
⋅⎛ ⎞+ = − +⎜ ⎟⎝ ⎠
−⎛ ⎞+ = +⎜ ⎟⎝ ⎠
+ = +
+ − = + −
=
ANSWER: . Solving leads to a true statement; therefore, the equation is an identity and any real number is a solution.
25. PROBLEM: 8 11 7 19 18 6 2
x x− = −
SOLUTION:
8 11 7 19 18 6 2
8 11 1 7 1 19 18 2 6 2 2
8 11 1 9 79 18 18 6
8 11 9 79 18 6
8 2 79 18 6
8 2 8 7 89 18 9 6 9
2 7 8 218 18 182 21 16
18 182 18 5 18
18 2 18 252
x x
x x x x
x
x
x
x
x
x
x
x
− = −
− + = − +
⋅⎛ ⎞+ − + =⎜ ⎟⎝ ⎠
− +⎛ ⎞+ =⎜ ⎟⎝ ⎠
− =
− − = −
⋅3 ⋅− = −
−− =
⎛ ⎞ ⎛ ⎞− ⋅ − = −⎜ ⎟ ⎜ ⎟⎝ ⎠ ⎝ ⎠
= −
ANSWER: 52
x = −
26. PROBLEM: 1 4 19
3 9 2x x− = +
SOLUTION:
1 4 193 9 2
1 1 4 1 193 2 9 2 2
9 2 1 1 42 2 3 9
18 1 1 42 3 9
19 1 42 3 9
19 1 1 4 12 3 3 9 3
19 4 1 32 9 9
19 4 32 9
19 12 9
19 2 1 22 19 9 19
2171
x x
x x x x
x
x
x
x
x
x
x
x
x
− = +
− − = + −
− ⋅⎛ ⎞− + =⎜ ⎟⎝ ⎠
− −⎛ ⎞ + =⎜ ⎟⎝ ⎠
− + =
− + − = −
⋅− = −
−− =
− =
⎛ ⎞ ⎛ ⎞− − = −⎜ ⎟ ⎜ ⎟⎝ ⎠ ⎝ ⎠
= −
ANSWER: 2171
x = −
27. PROBLEM: 12 5 9 44x x− + =
SOLUTION:
12 5 9 4421 5 44
21 5 5 44 521 4921 4921 21
73
x xx
xxx
x
− + =− =
− + = +=
=
=
ANSWER: 73
x =
28. PROBLEM:
10 6 13 12x− − = SOLUTION:
10 6 13 126 3 12
6 3 3 12 36 156 156 6
52
xx
xxx
x
− − =− − =
− − + = +− =−
=− −
= −
ANSWER: 52
x = −
29. PROBLEM: 2 4 9 7 8 2x x x− + + = + −
SOLUTION: 2 4 9 7 8 2
4 7 5 84 7 5 5 8 5
7 87 7 8 7
11
1 11
x x xx x
x x x xx
xxx
x
− + + = + −+ = +
+ − = + −− + =
− + − = −− =−
=− −
= −
ANSWER: 1x = − 30. PROBLEM:
20 5 12 6 7x x x− + = − +
SOLUTION: 20 5 12 6 7
32 5 1332 5 13
33 5 1333 5 5 13 5
33 1833 1833 33
18 / 336 /11
x x xx x
x x x xx
xxx
xx
− + = − +− = −
− + = − +− =
− + = +=
=
==
ANSWER: 6 /11x =
31. PROBLEM: 3 5 2 7a a a+ − = +
SOLUTION:
3 5 2 72 5 2 7
2 5 2 2 7 25 7
a a aa a
a a a a
+ − = ++ = +
+ − = + −≠
ANSWER: ∅ . Solving leads to a false statement; therefore, the equation is a contradiction and there is no solution.
32. PROBLEM:
7 3 2 5 1 2b b b− + = − + −
SOLUTION:
7 3 2 5 1 27 3 7 3
7 3 7 7 3 73 3
b b bb b
b b b b
− + = − + −− + = − +
− + + = − + +=
ANSWER: . Solving leads to a true statement; therefore, the equation is an identity and any real number is a solution.
33. PROBLEM:
7 2 3 4 2 2x x x− + = + −
SOLUTION:
7 2 3 4 2 210 2 2 2
10 2 2 2 2 28 2 2
8 2 2 2 28 48 48 8
12
x x xx x
x x x xx
xxx
x
− + = + −− = +
− − = + −− =
− + = +=
=
=
ANSWER: 12
x =
34. PROBLEM: 3 8 4 2 10x x− + − + =
SOLUTION:
3 8 4 2 107 10 10
7 10 10 10 107 07 07 7
0
x xx
xxx
x
− + − + =− + =
− + − = −− =−
=− −
=
ANSWER: 0x =
35. PROBLEM:
6 2 3 2 13x x x+ − = − −
SOLUTION:
6 2 3 2 133 2 2 13
3 2 2 2 13 25 2 13
5 2 2 13 25 155 155 5
3
x x xx x
x x x xx
xxx
x
+ − = − −+ = − −
+ + = − − ++ = −
+ − = − −= −−
=
= −
ANSWER: 3x = − 36. PROBLEM:
3 0.75 0.21 1.24 7.13x x x− + = +
SOLUTION:
3 0.75 0.21 1.24 7.13
3.21 0.75 1.24 7.133.21 0.75 1.24 1.24 7.13 1.24
1.97 0.75 7.131.97 0.75 0.75 7.13 0.75
1.97 7.881.97 7.881.97 1.97
4
x x xx x
x x x xx
xxx
x
− + = +− = +
− − = + −− =
− + = +=
=
=
ANSWER: 4x =
37. PROBLEM: 2 4 5 3 7x x x− − + = + −
SOLUTION:
2 4 5 3 73 2 3 2
3 2 3 3 2 32 2
x x xx x
x x x x
− − + = + −− = −
− − = − −− = −
ANSWER: . Solving leads to a true statement; therefore, the equation is an identity and any real number is a solution.
38. PROBLEM:
2 5 8 6 10y y y− − = − −
SOLUTION:
2 5 8 6 102 5 2 6
2 5 2 2 6 25 6
y y yy y
y y y y
− − = − −− − = − −
− − + = − − +− ≠ −
ANSWER: ∅ . Solving leads to a false statement; therefore, the equation is a contradiction and there is no solution.
39. PROBLEM: 1 1 1 1 7
10 3 30 15 15x x− = − −
SOLUTION:
1 1 1 1 710 3 30 15 151 1 1 7 2 1
10 3 30 30 151 1 1 14 1
10 3 30 151 1 1 13 1 1
10 3 15 30 15 151 3 1 2 1 1330 30 3 30
3 2 1 1330 3 30
5 1 1330 3 30
1 1 1 13 16 3 3 30 3
1 13 1 106 30 301 13 106
x x
x x
x x
x x x x
x
x
x
x
x
x
− = − −
⋅⎛ ⎞− = − −⎜ ⎟⎝ ⎠−
− = −
−− + = − +
⋅ ⋅⎛ ⎞+ − = −⎜ ⎟⎝ ⎠
+⎛ ⎞ − = −⎜ ⎟⎝ ⎠
− = −
− + = − +
− ⋅= +
− +=
( )
301 36 30
1 36 66 30
35
x
x
x
−=
−= ⋅
= −
ANSWER: 35
x = −
40. PROBLEM: 5 4 1 3 1 18 3 3 9 4 3
x x x− + = − − +
SOLUTION:
5 4 1 3 1 18 3 3 9 4 3
4 5 3 1 8 1 1 13 24 24 3 3 4
4 15 8 103 24 4
4 23 13 24 4
4 23 23 1 233 24 24 4 24
4 1 6 233 24 244 6 233 244 293 244 293 24
4 3 29 33 4 24 4
293
x x x
x x
x
x
x
x
x
x
x
x
x
− + = − − +
⋅ ⋅⎛ ⎞ ⎛ ⎞− + + = − + −⎜ ⎟ ⎜ ⎟⎝ ⎠ ⎝ ⎠
+⎛ ⎞− + = −⎜ ⎟⎝ ⎠
− + = −
− + − = − −
− ⋅− = −
− −− =
−− =
− = −
⎛ ⎞ ⎛ ⎞− ⋅ − = − −⎜ ⎟ ⎜ ⎟⎝ ⎠ ⎝ ⎠
=2
ANSWER: 2932
x =
Part C: Solving Linear Equations Involving Parentheses Solve. 41. PROBLEM:
( )5 2 3 2 12y− − + =
SOLUTION:
( )( )
5 2 3 2 12
5 2 5 3 2 1210 15 2 12
10 17 1210 17 17 12 17
10 510 510 10
12
y
yy
yy
yy
y
− − + =
− ⋅ − − + =
− + + =− + =
− + − = −− = −− −
=− −
=
ANSWER: 12
y =
42. PROBLEM:
( )3 5 4 5 8x x+ + = −
SOLUTION:
( )3 5 4 5 8
3 5 3 4 5 815 12 5 8
20 12 820 12 12 8 12
20 2020 2020 20
1
x xx x
x xx
xxx
x
+ + = −
⋅ + ⋅ + = −+ + = −
+ = −+ − = − −
= −−
=
= −
ANSWER: 1x = −
43. PROBLEM: ( )4 2 5 2x− − = −
SOLUTION:
( )( )
4 2 5 2
4 2 2 5 24 2 10 2
2 14 22 14 14 2 14
2 162 162 2
8
x
xxx
xxx
x
− − = −
− ⋅ − − = −
− + = −− + = −
− + − = − −− = −− −
=− −
=
ANSWER: 8x = 44. PROBLEM:
( ) ( )10 5 3 1 5 4x x− + = − SOLUTION:
( ) ( )( )
10 5 3 1 5 4
10 5 3 5 1 5 5 410 15 5 5 20
15 5 5 2015 5 5 5 20 5
20 5 2020 5 5 20 5
20 2520 2520 20
54
x x
x xx xx x
x x x xx
xxx
x
− + = −
− ⋅ + − ⋅ = ⋅ − ⋅
− − = −− + = −
− + − = − −− + = −
− + − = − −− = −− −
=− −
=
ANSWER: 54
x =
45. PROBLEM: ( ) ( )9 7 2 1x x− + = −
SOLUTION:
( ) ( )9 7 2 19 7 2 2 1
2 2 22 2 2 2 23 2 2
3 2 2 2 23 43 43 3
43
x xx xx x
x x x xx
xxx
x
− + = −
− − = ⋅ − ⋅− + = −
− + − = − −− + = −
− + − = − −− = −− −
=− −
=
ANSWER: 43
x =
46. PROBLEM:
( )5 2 1 3 12x− − + = − SOLUTION:
( )( )
5 2 1 3 12
5 2 5 1 3 1210 5 3 12
10 8 1210 8 8 12 8
10 2010 2010 10
2
x
xx
xx
xx
x
− − + = −
− ⋅ − − ⋅ + = −
− + + = −− + = −
− + − = − −− = −− −
=− −
=
ANSWER: 2x =
47. PROBLEM: ( )3 2 1 5x x x− + = +
SOLUTION:
( )3 2 1 53 2 2 5
2 52 5
2 5
x x xx x x
x xx x x x
− + = +
− − = +− = +
− − = + −− ≠
ANSWER: ∅ . Solving leads to a false statement; therefore, the equation is a contradiction and there is no solution.
48. PROBLEM:
( ) ( )5 3 2 1 2 3x x x− − = − SOLUTION: ( ) ( )
( )5 3 2 1 2 3
5 3 2 3 1 2 2 35 6 3 2 6
3 2 63 2 2 6 23 3 6
3 3 3 6 33 93 93 3
3
x x x
x x xx x x
x xx x x x
xx
xx
x
− − = −
− ⋅ − − = ⋅ − ⋅
− + = −− + = −
− + − = − −− + = −
− + − = − −− = −− −
=− −
=
ANSWER: 3x =
49. PROBLEM: ( ) ( )6 1 3 3 8x x x− − − = +
SOLUTION:
( ) ( )( )
6 1 3 3 8
6 6 1 3 3 3 86 6 3 3 24
9 6 3 249 6 3 3 24 3
12 6 2412 6 6 24 6
12 1812 1812 12
32
x x x
x x xx x x
x xx x x x
xx
xx
x
− − − = +
− ⋅ − − ⋅ − = ⋅ + ⋅
− + − = +− + = +
− + − = + −− + =
− + − = −− =−
=− −
= −
ANSWER: 32
x = −
50. PROBLEM:
( ) ( )3 15 10 4 125 2
x x− + = −
SOLUTION:
( ) ( )3 15 10 4 125 2
3 3 1 15 10 4 125 5 2 2
3 6 2 63 6 2 2 6 2
5 6 65 6 6 6 6
5 05 05 5
0
x x
x x
x xx x x x
xx
xx
x
− + = −
⎛ ⎞− ⋅ + − ⋅ = ⋅ − ⋅⎜ ⎟⎝ ⎠− − = −
− − − = − −− − = −
− − + = − +− =−
=− −
=
ANSWER: 0x =
51. PROBLEM: ( )3.1 2 3 0.5 22.2x − + =
SOLUTION:
( )3.1 2 3 0.5 22.23.1 2 3.1 3 0.5 22.2
6.2 9.3 0.5 22.26.2 8.8 22.2
6.2 8.8 8.8 22.2 8.86.2 316.2 316.2 6.2
5
xx
xx
xxx
x
− + =
⋅ − ⋅ + =− + =
− =− + = +
=
=
=
ANSWER: 5x = 52. PROBLEM:
( ) ( )4.22 3.13 1 5.2 2 1 11.38x x− − = + − SOLUTION:
( ) ( )( )
4.22 3.13 1 5.2 2 1 11.38
4.22 3.13 3.13 1 5.2 2 5.2 1 11.384.22 3.13 3.13 10.4 5.2 11.38
7.35 3.13 10.4 6.187.35 3.13 10.4 10.4 6.18 10.4
7.35 13.53 6.187.35 13.53 7.35 6.18 7.35
13.53
x x
x xx x
x xx x x x
xx
x
− − = + −
− ⋅ − − ⋅ = ⋅ + ⋅ −
− + = + −− = −
− − = − −− = −
− − = − −− 13.5313.53 13.5313.53 13.53
1
x
x
= −− −
=− −
=
ANSWER: 1x =
53. PROBLEM: ( ) ( )6 2 7 12 14x x− − − =
SOLUTION:
( ) ( )6 2 7 12 146 6 2 7 12 14
6 12 7 12 141414
1 114
x xx x
x xxx
x
− − − =
⋅ − ⋅ − + =− − + =
− =−
=− −
= −
ANSWER: 14x = − 54. PROBLEM:
( ) ( )9 3 3 3 4 9x x x− − − = − +
SOLUTION:
( ) ( )( ) ( )
9 3 3 3 4 9
9 9 3 3 3 4 3 99 27 3 12 27
12 27 12 2712 27 12 12 27 12
27 27
x x x
x x xx x x
x xx x x x
− − − = − +
− ⋅ − − ⋅ − = − ⋅ + − ⋅
− + − = − −− + = − −
− + + = − − +≠ −
ANSWER: ∅ . Solving leads to a false statement; therefore, the equation is a contradiction and there is no solution.
55. PROBLEM: ( ) ( )3 2 4 3 4 5x x− + = − −
SOLUTION:
( ) ( )( )
3 2 4 3 4 5
3 2 2 4 3 4 3 53 2 8 12 15
2 5 12 152 5 12 12 15 12
10 5 1510 5 5 15 5
10 2010 2010 10
2
x x
x xx xx x
x x x xx
xxx
x
− + = − −
− ⋅ − ⋅ = − ⋅ − −
− − = − +− − = − +
− − + = − + +− =
− + = +=
=
=
ANSWER: 2x = 56. PROBLEM:
( ) ( )12 2 2 1 4 1x x− + = − SOLUTION: ( ) ( )
( )12 2 2 1 4 1
12 2 2 2 1 4 4 112 4 2 4 4
10 4 4 410 4 4 4 4 4
10 8 410 8 10 4 10
8 148 148 8
74
x x
x xx x
x xx x x x
xx
xx
x
− + = −
− ⋅ + − ⋅ = ⋅ − ⋅
− − = −− = −
− − = − −− = −
− − = − −− = −− −
=− −
=
ANSWER: 74
x =
57. PROBLEM: ( ) ( )3 5 2 2 3 7 9x x x+ − + = +
SOLUTION:
( ) ( )( )
3 5 2 2 3 7 9
3 3 5 2 2 2 3 7 93 15 4 6 7 9
9 7 99 7 7 9 78 9 9
8 9 9 9 98 08 08 8
0
x x x
x x xx x x
x xx x x x
xx
xx
x
+ − + = +
⋅ + ⋅ − ⋅ + − = +
+ − − = +− + = +
− + − = + −− + =
− + − = −− =−
=− −
=
ANSWER: 0x = 58. PROBLEM:
( ) ( )3 2 1 4 3 2 5 10x x x− − − = − + SOLUTION: ( ) ( )
( )3 2 1 4 3 2 5 10
3 2 3 1 4 3 4 2 5 106 3 12 8 5 10
6 5 5 106 5 5 5 10 5
5 105 5 10 5
55
1 15
x x x
x x xx x x
x xx x x x
xx
xx
x
− − − = − +
⋅ − ⋅ − ⋅ − − ⋅ = − +
− − + = − +− + = − +
− + + = − + +− + =
− + − = −− =−
=− −
= −
ANSWER: 5x = −
59. PROBLEM: ( ) ( )3 2 3 2 3 7a a− − + = +
SOLUTION:
( ) ( )( )
3 2 3 2 3 7
3 2 3 3 2 3 3 76 9 2 3 21
6 11 3 216 11 3 3 21 3
9 11 219 11 11 21 11
9 109 109 9
109
a a
a aa a
a aa a a a
aa
aa
a
− − + = +
− ⋅ − − + = ⋅ + ⋅
− + + = +− + = +
− + − = + −− + =
− + − = −− =−
=− −
= −
ANSWER: 109
a = −
60. PROBLEM:
( ) ( )2 5 3 1 5 2 1x x− − − = − +
SOLUTION:
( ) ( )( ) ( )
2 5 3 1 5 2 1
2 5 2 3 1 5 2 5 110 6 1 10 5
10 5 10 510 5 10 10 5 10
5 5
x x
x xx x
x xx x x x
− − − = − +
− ⋅ − − ⋅ − = ⋅ − + ⋅
− + − = − +− + = − +
− + + = − + +=
ANSWER: . Solving leads to a true statement; therefore, the equation is an identity and any real number is a solution.
61. PROBLEM:
( ) ( ) ( )1 12 1 8 2 3 42 4
x x x+ − + = −
SOLUTION:
( ) ( ) ( )1 12 1 8 2 3 42 4
1 1 1 12 1 8 2 3 3 42 2 4 4
1 12 3 122 2
3 123 3 12 34 124 124 4
3
x x x
x x x
x x x
x xx x x x
xx
x
+ − + = −
⎛ ⎞⋅ + ⋅ − ⋅ + − ⋅ = ⋅ − ⋅⎜ ⎟⎝ ⎠
+ − − = −
− = −− − = − −
− = −− −
=− −
=
ANSWER: 3x =
62. PROBLEM:
( ) ( )2 1 36 3 4 13 2 2
x x− − − = +
SOLUTION:
( ) ( )2 1 36 3 4 13 2 2
2 2 1 3 36 3 4 13 3 2 2 2
1 34 2 62 2
2 2 1 34 62 2 24 1 34 6
2 23 34 62 2
3 3 3 34 62 2 2 2
4 64 6 6 6
10 010 010 10
0
x x
x x
x x
x x
x x
x x
x x
x xx x x x
xx
x
− − − = +
⎛ ⎞− ⋅ − − ⋅ − = ⋅ + ⋅⎜ ⎟⎝ ⎠
− + − = +
⋅− + − = +
−⎛ ⎞− + = +⎜ ⎟⎝ ⎠
− + = +
− + − = + −
− =− − = −
− =−
=− −
=
ANSWER: 0x =
63. PROBLEM:
( ) ( )1 13 1 2 5 02 3
x x− + − =
SOLUTION: ( ) ( )1 13 1 2 5 0
2 31 1 1 13 1 2 5 02 2 3 3
3 1 2 5 02 2 3 3
3 3 2 2 1 3 5 2 06 6 6 6
9 4 3 10 06 6
13 13 06 6
13 13 13 1306 6 6 6
13 136 6
13 6 13 66 13 6 13
1
x x
x x
x x
x
x
x
x
x
x
x
− + − =
⋅ − ⋅ + ⋅ − ⋅ =
− + − =
⋅ ⋅ ⋅ ⋅⎛ ⎞ ⎛ ⎞+ − + =⎜ ⎟ ⎜ ⎟⎝ ⎠ ⎝ ⎠
+ +⎛ ⎞ ⎛ ⎞− =⎜ ⎟ ⎜ ⎟⎝ ⎠ ⎝ ⎠
− =
− + = +
=
⎛ ⎞ ⎛ ⎞⋅ =⎜ ⎟ ⎜ ⎟⎝ ⎠ ⎝ ⎠
=
ANSWER: 1x =
64. PROBLEM:
( ) ( )1 1 12 3 33 5 9
x x− + = +
SOLUTION:
( ) ( )1 1 12 3 33 5 9
1 1 1 1 12 3 33 3 5 9 9
2 1 13 3 5 3 3
2 1 3 13 15 15 3 3
10 3 13 15 3 3
7 13 15 3 37 1
3 15 3 3 3 37 1
15 3
x x
x x
x x
x x
x x
x x
x x x x
− + = +
⋅ − ⋅ + = ⋅ + ⋅
− + = +
− ⋅5 ⋅⎛ ⎞+ + = +⎜ ⎟⎝ ⎠
− +⎛ ⎞+ = +⎜ ⎟⎝ ⎠
− = +
− − = + −
− ≠
ANSWER: ∅ . Solving leads to a false statement; therefore, the equation is a contradiction and there is no solution.
65. PROBLEM: ( ) ( ) ( )2 2 7 3 6 1x x x− − − + = −
SOLUTION: ( ) ( ) ( )
( )2 2 7 3 6 1
2 2 2 7 3 6 6 14 14 3 6 6
5 11 6 65 11 6 6 6 6
11 11 611 11 11 6 11
11 1711 1711 11
1711
x x x
x x xx x x
x xx x x x
xx
xx
x
− − − + = −
− ⋅ − − − − = ⋅ − ⋅
− + − − = −− + = −
− + − = − −− + = −
− + − = − −− = −− −
=− −
=
ANSWER: 1711
x =
66. PROBLEM:
( ) ( ) ( )10 3 5 5 4 2 2 5 20x x x+ − + = +
SOLUTION:
( ) ( ) ( )( )
10 3 5 5 4 2 2 5 20
10 3 10 5 5 4 5 2 2 5 2 2030 50 20 10 10 40
10 40 10 4010 40 10 10 40 10
40 40
x x x
x x xx x x
x xx x x x
+ − + = +
⋅ + ⋅ − ⋅ + − ⋅ = ⋅ + ⋅
+ − − = ++ = +
+ − = + −=
ANSWER: . Solving leads to a true statement; therefore, the equation is an identity and any real number is a solution.
67. PROBLEM: ( ) ( ) ( )2 3 6 2 1 5 2 4x x x− − + = − −
SOLUTION:
( ) ( ) ( )( ) ( )
2 3 6 2 1 5 2 4
2 2 3 6 2 6 1 5 2 5 42 6 12 6 10 20
10 12 10 2010 12 10 10 20 10
12 20
x x x
x x xx x x
x xx x x x
− − + = − −
⋅ − ⋅ − ⋅ + − ⋅ = − ⋅ − −
− − − = − +− − = − +
− − + = − + +− ≠
ANSWER: ∅ . Solving leads to a false statement; therefore, the equation is a contradiction and there is no solution.
68. PROBLEM:
( ) ( ) ( )5 2 4 1 2 3x x x− − − = − − SOLUTION: ( ) ( ) ( )
( )5 2 4 1 2 3
5 5 2 4 1 2 3 25 10 4 1 6 2
9 6 29 2 6 2 2
9 69 9 6 9
33
1 13
x x x
x x xx x x
x xx x x x
xx
xx
x
− − − = − −
⋅ − ⋅ − + = − ⋅ − −
− − + = − +− = − +
− − = − + −− − = −
− − + = − +− =−
=− −
= −
ANSWER: 3x = −
69. PROBLEM: ( ) ( )6 3 2 12 1 4 0x x− − − + =
SOLUTION:
( ) ( )6 3 2 12 1 4 06 3 6 2 12 1 4 0
18 12 12 5 06 7 0
6 7 7 0 76 76 76 6
76
x xx x
x xx
xxx
x
− − − + =
⋅ − ⋅ − + + =− − + =
− =− + = +
=
=
=
ANSWER: 76
x = 70. PROBLEM:
( ) ( )3 4 2 9 3 6 0x x x− − − + − = SOLUTION:
( ) ( )
( )3 4 2 9 3 6 0
3 4 3 2 9 3 6 012 6 9 3 6 0
27 3 027 3 3 0 3
27 327 327 27
19
x x x
x x xx x x
xx
xx
x
− − − + − =
− ⋅ − − ⋅ − − − =
− + − − − =− + =
− + − = −− = −− −
=− −
=
ANSWER: 19
x =
Part D: Literal Equations Solve for the indicated variable. 71. PROBLEM:
Solve for w: A l w= ⋅ .
SOLUTION:
A l wA l wl lA wl
Awl
= ⋅⋅
=
=
=
ANSWER: Awl
=
72. PROBLEM:
Solve for a: F ma= .
SOLUTION:
F maF mam mF am
Fam
=
=
=
=
ANSWER: Fam
=
73. PROBLEM: Solve for w: 2 2P l w= + .
SOLUTION:
2 22 2 2 22 22 2
2 22
22
2
P l wP l l w lP l wP l w
P l w
P lw
= +− = + −− =−
=
−=
−=
ANSWER: 22
P lw −=
74. PROBLEM:
Solve for r: 2C rπ= .
SOLUTION:
22
2 2
2
2
C rC r
C r
Cr
ππ
π π
π
π
=
=
=
=
ANSWER: 2Crπ
=
75. PROBLEM:
Solve for b: P a b c= + + .
SOLUTION:
P a b cP a c a b c a cP a c b
b P a c
= + +− − = + + − −− − =
= − −
ANSWER: b P a c= − −
76. PROBLEM:
Solve for C: 9 325
F C= + .
SOLUTION:
( )
( )
( )
9 325932 32 3259325
5 9 5329 5 95329
5 329
F C
F C
F C
F C
F C
C F
= +
− = + −
− =
⎛ ⎞− ⋅ = ⋅⎜ ⎟⎝ ⎠
− ⋅ =
= −
ANSWER: ( )5 329
C F= −
77. PROBLEM:
Solve for h: 12
A bh= .
SOLUTION:
12
2 1 22
2
2
A bh
A bhb bA h
bAh
b
=
⎛ ⎞⋅ = ⋅⎜ ⎟⎝ ⎠
=
=
ANSWER: 2Ahb
=
78. PROBLEM: Solve for t: I Prt= .
SOLUTION:
I PrtI Prt
Pr PrI t
PrIt
Pr
=
=
=
=
ANSWER: ItPr
=
79. PROBLEM:
Solve for y: ax by c+ = .
SOLUTION:
ax by cax by ax c ax
by c axby c axb b
ax cyb
+ =+ − = −
= −−
=
− +=
ANSWER: ax cyb
− +=
80. PROBLEM: Solve for h: 22 2S r rhπ π= + . SOLUTION:
2
2 2 2
2
2
2
2
2 22 2 2 22 22 2
2 22
22
2
S r rhS r r rh rS r rhS r rh
r rS r
rS r
h
hr
π π
π π π π
π π
π ππ πππ
ππ
= +
− + −
−
−
=
=
=
=
=
−
−
ANSWER: 22
2S
rh rπ
π−
=
81. PROBLEM:
Solve for x: 25
x yz += .
SOLUTION:
25
25 55
5 25 25 25 2
2 25
25
2
x yz
x yz
z x yz y x y yz y xz y x
z y x
z yx
+=
+⋅ = ⋅
= +− = + −− =−
=
−=
−=
ANSWER: 52
z yx −=
82. PROBLEM:
Solve for c: 3 23
b ca −= .
SOLUTION:
3 23
3 23 33
3 3 23 3 3 2 33 3 23 3 2
2 23 3
23 3
2
b ca
b ca
a b ca b b c ba b ca b c
a b c
a bc
−=
−⋅ = ⋅
= −− = − −− = −− −
=− −−
− =
−= −
ANSWER: 3 32
a bc −= −
83. PROBLEM:
Solve for b: y mx b= + .
SOLUTION:
y mx by mx mx b mxy mx b
b y mx
= +− = + −− =
= −
ANSWER: b y mx= −
84. PROBLEM: Solve for m: y mx b= + . SOLUTION:
y mx by b mx b by b mxy b mx
x xy b m
xy bm
x
= +− = + −− =−
=
−=
−=
ANSWER: y bmx−
= 85. PROBLEM:
Solve for y: 3 2 6x y− = .
SOLUTION:
3 2 63 2 3 6 3
2 6 32 6 32 2
6 32
3 62
x yx y x x
y xy x
xy
xy
− =− − = −
− = −− −
=− −
−= −
−=
ANSWER: 3 62
xy −=
86. PROBLEM: Solve for y: 5 2 12x y− + = .
SOLUTION:
5 2 125 2 5 12 5
2 5 122 5 122 2
5 122
x yx y x x
y xy x
xy
− + =− + + = +
= ++
=
+=
ANSWER: 5 122
xy +=
87. PROBLEM:
Solve for y: 13 5x y− = .
SOLUTION:
( ) ( )
13 5
13 5 3 3
1 35 3 3
35 55 3
5 153
x y
x y x x
y x
y x
xy
− =
− − = −
⋅− = −
−⎛ ⎞− ⋅ − = −⎜ ⎟⎝ ⎠
−=
ANSWER: 5 153
xy −=
88. PROBLEM:
Solve for y: 3 1 14 5 2
x y− = .
SOLUTION:
( ) ( )
3 1 14 5 2
3 1 3 1 34 5 4 2 4
1 1 35 2 41 1 2 35 4 41 2 35 4
5 51 2 35 4
15 104
x y
x y x x
y x
y x
xy
xy
xy
− =
− − = −
− = −
⋅− = −
−− =
−− ⋅ − = −
−=
ANSWER: 15 104
xy −=
Translate the following sentences into linear equations and then solve. 89. PROBLEM:
The sum of 3x and 5 is equal to the sum of 2x and 7.
SOLUTION:
3 5 2 73 5 2 2 7 2
5 75 5 7 5
2
x xx x x x
xx
x
+ = ++ − = + −
+ =+ − = −
=
ANSWER: 2x =
90. PROBLEM: The sum of −5x and 6 is equal to the difference of 4x and 2.
SOLUTION:
5 6 4 25 6 4 4 2 4
9 6 29 6 6 2 6
9 89 89 9
89
x xx x x x
xx
xx
x
− + = −− + − = − −
− + = −− + − = − −
− = −− −
=− −
=
ANSWER: 89
x =
91. PROBLEM:
The difference of 5x and 25 is equal to the difference of 3x and 51. SOLUTION:
5 25 3 515 25 3 3 51 3
2 25 512 25 25 51 25
2 262 262 2
13
x xx x x x
xx
xx
x
− = −− − = − −
− = −− + = − +
= −−
=
= −
ANSWER: 13x = −
92. PROBLEM:
The sum of 12
x and 34
is equal to 23
x .
SOLUTION:
( ) ( )
1 3 22 4 3
1 3 2 2 22 4 3 3 3
1 3 2 2 3 06 6 4
3 4 3 06 4
1 3 06 4
1 3 3 306 4 4 4
1 36 4
1 36 66 4
92
x x
x x x x
x
x
x
x
x
x
x
+ =
+ − = −
⋅ ⋅⎛ ⎞− + =⎜ ⎟⎝ ⎠
−⎛ ⎞ + =⎜ ⎟⎝ ⎠
−+ =
− + − = −
− = −
− ⋅ − = − −
=
ANSWER: 92
x =
93. PROBLEM: A number n divided by 5 is equal to the sum of twice the number and 3. SOLUTION:
2 35
2 2 3 252 5 3
5 510 35
9 35
9 5 535 9 9
53
n nn n n n
n n
n n
n
n
n
= +
− = + −
⋅− =
−=
−=
− ⎛ ⎞ ⎛ ⎞⋅ − = −⎜ ⎟ ⎜ ⎟⎝ ⎠ ⎝ ⎠
= −
ANSWER: 53
n = −
94. PROBLEM:
Negative ten times a number n is equal to the sum of three times the number and 13.
SOLUTION:
10 3 1310 3 3 13 3
13 1313 1313 13
1
n nn n n n
nn
n
− = +− − = + −
− =−
=− −
= −
ANSWER: 1n = −
Part E: Discussion Board Topics 95. PROBLEM:
What is the origin of the word algebra? ANSWER: The word algebra is a Latin variant of the Arabic word al-jabr. This came from the title of a book, Hidab al-jabr wal-muqubala, written in Baghdad about 825 A.D. by the Arab mathematician Mohammed ibn-Musa al-Khowarizmi. (Source: http://www.und.nodak.edu/instruct/lgeller/algebra.html)
96. PROBLEM: What is regarded as the main business of algebra? ANSWER: Algebra is used in companies to figure out their annual budget which involves their annual expenditure. It, along with statistics is also used to predict their annual turnover. Algebra is used in schools by the teachers to prepare the annual reports of the children. (Source: http://wiki.answers.com/Q/What_is_the_importance_of_algebra_in_today%27s_world)
97. PROBLEM: Why is solving equations such an important algebra topic? ANSWER: Most of the equations that arise in real world contexts cannot be solved. Even if they can, it is often simpler and faster to use a computational method to find a numerical solution. The real power of equations is that they provide a very precise way to describe various features of the world. (Students answers may vary.)
98. PROBLEM:
Post some real-world linear formulas not presented in this section. ANSWER: In business economic analysis, the demand curve is a linear equation that illustrates the inverse relationship between two variables: the quantity of a particular good and its price. Modeling this relationship illustrates how price affects the willingness or ability of a consumer to buy a certain product. This demand schedule shows how much of a product or service consumers are willing to purchase at varying price levels. The foreign exchange rate tells you the price of one currency in terms of a foreign currency. Say for example that you want to convert U.S. dollars into British pounds. You use your dollars to buy pounds and receive the value of your dollars in British currency. Depending on the currency exchange rate of the British pound, you may be able to buy more or less British pounds with your U.S. dollars. This relationship between currencies may also be modeled with a linear equation.
(Students answers may vary.) (Source: http://www.ehow.com/list_5958699_real-world-uses-linear equations.html)
99. PROBLEM: Research and discuss the contributions of Diophantus of Alexandria. ANSWER: Diophantus of Alexandria, sometimes called "the father of algebra", was an Alexandrian Greek mathematician and the author of a series of books called Arithmetica. These texts deal with solving algebraic equations, many of which are now lost. In studying Arithmetica, Pierre de Fermat concluded that a certain equation considered by Diophantus had no solutions, and noted without elaboration that he had found “a truly marvelous proof of this proposition,” now referred to as Fermat’s Last Theorem. This led to tremendous advances in number theory, and the study of Diophantine equations (“Diophantine geometry”) and of Diophantine approximations remain important areas of mathematical research. Diophantus was the first Greek mathematician who recognized fractions as numbers; thus he allowed positive rational numbers for the coefficients and solutions. In modern use, Diophantine equations are usually algebraic equations with integer coefficients, for which integer solutions are sought. Diophantus also made advances in mathematical notation. (Source: http://en.wikipedia.org/wiki/Diophantus)
100. PROBLEM:
Create an identity or contradiction of your own and share on the discussion board. Provide a solution and explain how you found it. ANSWER:
Contradiction: The Top looks nice, But I Don't like it (Student answers may vary.)
(Source: http://wiki.answers.com/Q/What_are_some_examples_of_contradiction) Section2.5:ApplicationsofLinearEquationsPart A: Translate Translate the following into algebraic equations. 1. PROBLEM:
The sum of a number and 6 is 37. SOLUTION: Let x represent the number. The sum of the number and 6 is represented as 6x + . ANSWER: Therefore, the algebraic equation is 6 37x + = .
2. PROBLEM: When 12 is subtracted from twice some number the result is 6. SOLUTION: Let x represent the number. 12 subtracted from twice the number is represented as 2 12x − . ANSWER: Therefore, the algebraic equation is 2 12 6x − = .
3. PROBLEM: Fourteen less than 5 times a number is 1. SOLUTION: Let x represent the number. Fourteen less than 5 times the number is represented as 5 14x − . ANSWER: Therefore, the algebraic equation is 5 14 1x − = .
4. PROBLEM:
Twice some number is subtracted from 30 and the result is 50. SOLUTION: Let x represent the number. Twice the number subtracted from 30 is represented as 30 2x− . ANSWER: Therefore, the algebraic equation is 30 2 50x− = .
5. PROBLEM:
Five times the sum of 6 and some number is 20. SOLUTION: Let x represent the number. Five times the sum of 6 and the number is represented as ( )5 6x + . ANSWER: Therefore, the algebraic equation is ( )5 6 20x + = .
6. PROBLEM:
The sum of 5 times some number and 6 is 20. SOLUTION: Let x represent the number. The sum of 5 times the number and 6 is represented as 5 6x + . ANSWER: Therefore, the algebraic equation is 5 6 20x + = .
7. PROBLEM: When the sum of a number and 3 is subtracted from 10 the result is 5. SOLUTION: Let x represent the number. The sum of the number and 3 subtracted from 10 is represented as ( )10 3x− + . ANSWER: Therefore, the algebraic equation is ( )10 3 5x− + = .
8. PROBLEM:
The sum of three times a number and five times that same number is 24. SOLUTION: Let x represent the number. The sum of three times the number and five times the number is represented as 3 5x x+ . ANSWER: Therefore, the algebraic equation is 3 5 24x x+ = .
9. PROBLEM:
Ten is subtracted from twice some number and the result is the sum of the number and 2. SOLUTION: Let x represent the number. Ten subtracted from twice the number is represented as 2 10x − . The sum of the number and 2 is represented as 2x + . ANSWER: Therefore, the algebraic equation is 2 10 2x x− = + .
10. PROBLEM:
Six less than some number is ten times the sum of that number and 5. SOLUTION: Let x represent the number. Six less than the number is represented as 6x − . Ten times the sum of the number and 5 is represented as ( )10 5x + . ANSWER: Therefore, the algebraic equation is ( )6 10 5x x− = + .
Part B: Number Problems Set up an algebraic equation, and then solve. 11. PROBLEM:
A larger integer is 1 more than twice another integer. If the sum of the integers is 25, find the integers. SOLUTION: Let x represent the smaller integer. Let 2 1x + represent the larger integer. The algebraic equation is: ( )2 1 25x x+ + = Solve the equation to obtain x:
( )2 1 252 1 253 1 25
3 1 1 25 13 243 243 3
8
x xx x
xx
xx
x
+ + =+ + =
+ =+ − = −
=
=
=
Back substitute to obtain the larger integer: 2 1 2 8 1 16 1 17x + = ⋅ + = + = ANSWER: Hence, the two integers are 8 and 17.
12. PROBLEM:
If a larger integer is 2 more than 4 times another integer and their difference is 32, find the integers. SOLUTION: Let x represent the smaller integer. Let 4 2x + represent the larger integer. The algebraic equation is: ( )4 2 32x x+ − = Solve the equation to obtain x: ( )4 2 32
4 2 323 2 32
3 2 2 32 23 303 303 3
10
x xx x
xx
xx
x
+ − =+ − =
+ =+ − = −
=
=
=
Back substitute to obtain the larger integer: 4 2 4 10 2 40 2 42x + = ⋅ + = + = ANSWER: Hence, the two integers are 10 and 42.
13. PROBLEM: One integer is 30 more than another integer. If the difference between the larger and twice the smaller is 8, find the integers. SOLUTION: Let x represent the smaller integer. Let 30x + represent the larger integer. The algebraic equation is: ( )30 2 8x x+ − = Solve the equation to obtain x: ( )30 2 8
30 2 830 8
30 30 8 302222
1 122
x xx x
xx
xx
x
+ − =+ − =− + =
− + − = −− = −− −
=− −
=
Back substitute to obtain the larger integer: 30 22 30 52x + = + = ANSWER: Hence, the two integers are 22 and 52.
14. PROBLEM:
The quotient of some number and 4 is 22. Find the number. SOLUTION: Let x represent the number.
The algebraic equation is: 224x=
Solve the equation to obtain x:
2244 422
4 1 188
x
x
x
=
⋅ = ⋅
=
ANSWER: Hence, the number is 88.
15. PROBLEM: Eight times a number is decreased by three times the same number, giving a difference of 20. What is the number? SOLUTION: Let x represent the number. The algebraic equation is: 8 3 20x x− = Solve the equation to obtain x: 8 3 20
5 205 205 5
4
x xxx
x
− ==
=
=
ANSWER: Hence, the number is 4.
16. PROBLEM:
One integer is two units less than another. If their sum is −22, find the two integers. SOLUTION: Let x represent the larger integer. Let 2x − represent the smaller integer. The algebraic equation is: ( )2 22x x+ − = − Solve the equation to obtain x:
( )2 222 22
2 2 222 2 2 22 2
2 202 202 2
10
x xx x
xx
xx
x
+ − = −+ − = −
− = −− + = − +
= −−
=
= −
Back substitute to obtain the smaller integer: 2 10 2 12x − = − − = − ANSWER: Hence, the two integers are −12 and −10.
17. PROBLEM: The sum of two consecutive integers is 139. Find the integers. SOLUTION: Let x represent the smaller integer. Let 1x + represent the next integer. The algebraic equation is: ( )1 139x x+ + = Solve the equation to obtain x:
( )1 1391 139
2 1 1392 1 1 139 1
2 1382 1382 2
69
x xx x
xx
xx
x
+ + =+ + =
+ =+ − = −
=
=
=
Back substitute to obtain the larger integer: 1 69 1 70x + = + = . ANSWER: Hence, the two consecutive integers are 69 and 70.
18. PROBLEM:
The sum of three consecutive integers is 63. Find the integers. SOLUTION: Let x represent the smallest of the three integers. Let 1x + represent the next integer. Let 2x + represent the last integer. The algebraic equation is: ( ) ( )1 2 63x x x+ + + + = Solve the equation to obtain x:
( ) ( )1 2 631 2 63
3 3 633 3 3 63 3
3 603 603 3
20
x x xx x x
xx
xx
x
+ + + + =+ + + + =
+ =+ − = −
=
=
=
Back substitute to obtain the other integers: 1 20 1 21x + = + = and 2 20 2 22x + = + =
ANSWER: Hence, the three consecutive integers are 20, 21 and 22.
19. PROBLEM: The sum of three consecutive integers is 279. Find the integers. SOLUTION: Let x represent the smallest of the three integers. Let 1x + represent the next integer. Let 2x + represent the last integer. The algebraic equation is: ( ) ( )1 2 279x x x+ + + + = Solve the equation to obtain x:
( ) ( )1 2 2791 2 279
3 3 2793 3 3 279 3
3 2763 2763 3
92
x x xx x x
xx
xx
x
+ + + + =+ + + + =
+ =+ − = −
=
=
=
Back substitute to obtain the other integers: 1 92 1 93x + = + = and 2 92 2 94x + = + =
ANSWER: Hence, the three consecutive integers are 92, 93, and 94.
20. PROBLEM:
The difference of twice the smaller of two consecutive integers and the larger is 39. Find the integers. SOLUTION: Let x represent the smaller integer. Let 1x + represent the next integer. The algebraic equation is: ( )2 1 39x x− + = Solve the equation to obtain x:
( )2 1 392 1 39
1 391 1 39 1
40
x xx x
xx
x
− + =− − =
− =− + = +
=
Back substitute to obtain the larger integer: 1 40 1 41x + = + = ANSWER: Hence, the two consecutive integers are 40 and 41.
21. PROBLEM: If the smaller of two consecutive integers is subtracted from two times the larger, then the result is 17. Find the integers. SOLUTION: Let x represent the smaller integer. Let 1x + represent the next integer. The algebraic equation is: ( )2 1 17x x+ − = Solve the equation to obtain x: ( )2 1 172 2 17
2 172 2 17 2
15
x xx x
xx
x
+ − =+ − =
+ =+ − = −
=
Back substitute to obtain the larger integer: 1 115 1 16x + = + = ANSWER: Hence, the two consecutive integers are 15 and 16.
22. PROBLEM:
The sum of two consecutive even integers is 46. Find the integers. SOLUTION: Let x represent the smaller even integer. Let 2x + represent the next even integer. The algebraic equation is: ( )2 46x x+ + = Solve the equation to obtain x:
( )2 462 46
2 2 462 2 2 46 2
2 442 442 2
22
x xx x
xx
xx
x
+ + =+ + =
+ =+ − = −
=
=
=
Back substitute to obtain the larger integer: 2 22 2 24x + = + = ANSWER: Hence, the two consecutive even integers are 22 and 24.
23. PROBLEM: The sum of two consecutive even integers is 238. Find the integers. SOLUTION: Let x represent the smaller even integer. Let 2x + represent the next even integer. The algebraic equation is: ( )2 238x x+ + = Solve the equation to obtain x:
( )2 2382 238
2 2 2382 2 2 238 2
2 2362 2362 2
118
x xx x
xx
xx
x
+ + =+ + =
+ =+ − = −
=
=
=
Back substitute to obtain the larger integer: 2 118 2 120x + = + = ANSWER: Hence, the two consecutive even integers are 118 and 120.
24. PROBLEM:
The sum of three consecutive even integers is 96. Find the integers. SOLUTION: Let x represent the smallest of the three integers. Let 2x + represent the next integer. Let 4x + represent the last integer. The algebraic equation is: ( ) ( )2 4 96x x x+ + + + = Solve the equation to obtain x:
( ) ( )2 4 962 4 96
3 6 963 6 6 96 6
3 903 903 3
30
x x xx x x
xx
xx
x
+ + + + =+ + + + =
+ =+ − = −
=
=
=
Back substitute to obtain the other integers: 2 30 2 32x + = + = and 4 30 4 34x + = + =
ANSWER: Hence, the three consecutive even integers are 30, 32 and 34.
25. PROBLEM: If the smaller of two consecutive even integers is subtracted from 3 times the larger the result is 42. Find the integers. SOLUTION: Let x represent the smaller even integer. Let 2x + represent the next even integer. The algebraic equation is: ( )3 2 42x x+ − = Solve the equation to obtain x: ( )3 2 423 6 42
2 6 422 6 6 42 6
2 362 362 2
18
x xx x
xx
xx
x
+ − =+ − =
+ =+ − = −
=
=
=
Back substitute to obtain the larger integer: 2 18 2 20x + = + = ANSWER: Hence, the two consecutive even integers are 18 and 20.
26. PROBLEM:
The sum of three consecutive even integers is 90. Find the integers. SOLUTION: Let x represent the smallest of the three integers. Let 2x + represent the next integer. Let 4x + represent the last integer. The algebraic equation is: ( ) ( )2 4 90x x x+ + + + = Solve the equation to obtain x:
( ) ( )2 4 902 4 90
3 6 903 6 6 90 6
3 843 843 3
28
x x xx x x
xx
xx
x
+ + + + =+ + + + =
+ =+ − = −
=
=
=
Back substitute to obtain the other integers: 2 28 2 30x + = + = and 4 28 4 32x + = + =
ANSWER: Hence, the three consecutive even integers are 28, 30 and 32.
27. PROBLEM: The sum of two consecutive odd integers is 68. Find the integers. SOLUTION: Let x represent the smaller odd integer. Let 2x + represent the next odd integer. The algebraic equation is: ( )2 68x x+ + = Solve the equation to obtain x:
( )2 682 68
2 2 682 2 2 68 2
2 662 662 2
33
x xx x
xx
xx
x
+ + =+ + =
+ =+ − = −
=
=
=
Back substitute to obtain the larger integer: 2 33 2 35x + = + = ANSWER: Hence, the two consecutive odd integers are 33 and 35.
28. PROBLEM:
The sum of two consecutive odd integers is 180. Find the integers. SOLUTION: Let x represent the smaller odd integer. Let 2x + represent the next odd integer. The algebraic equation is: ( )2 180x x+ + = Solve the equation to obtain x:
( )2 1802 180
2 2 1802 2 2 180 2
2 1782 1782 2
89
x xx x
xx
xx
x
+ + =+ + =
+ =+ − = −
=
=
=
Back substitute to obtain the larger integer: 2 89 2 91x + = + = ANSWER: Hence, the two consecutive odd integers are 89 and 91.
29. PROBLEM: The sum of three consecutive odd integers is 57. Find the integers. SOLUTION: Let x represent the smallest of the three integers. Let 2x + represent the next integer. Let 4x + represent the last integer. The algebraic equation is: ( ) ( )2 4 57x x x+ + + + = Solve the equation to obtain x:
( ) ( )2 4 572 4 57
3 6 573 6 6 57 6
3 513 513 3
17
x x xx x x
xx
xx
x
+ + + + =+ + + + =
+ =+ − = −
=
=
=
Back substitute to obtain the other integers: 2 17 2 19x + = + = and 4 17 4 21x + = + =
ANSWER: Hence, the three consecutive odd integers are 17, 19 and 21.
30. PROBLEM:
If the smaller of two consecutive odd integers is subtracted from twice the larger the result is 23. Find the integers. SOLUTION: Let x represent the smaller odd integer. Let 2x + represent the next odd integer. The algebraic equation is: ( )2 2 23x x+ − = Solve the equation to obtain x:
( )2 2 232 2 2 23
4 234 4 23 4
19
x xx x
xx
x
+ − =⋅ + ⋅ − =
+ =+ − = −
=
Back substitute to obtain the larger integer: 2 19 2 21x + = + = ANSWER: Hence, the two consecutive odd integers are 19 and 21.
31. PROBLEM: Twice the sum of two consecutive odd integers is 32. Find the integers. SOLUTION: Let x represent the smaller odd integer. Let 2x + represent the next odd integer. The algebraic equation is: ( )( )2 2 32x x+ + = Solve the equation to obtain x:
( )( )( )( )
2 2 322 2 32
2 2 2 324 4 32
4 4 4 32 44 284 284 4
7
x xx x
xx
xxx
x
+ + =+ + =
+ =+ =
+ − = −=
=
=
Back substitute to obtain the larger integer: 2 7 2 9x + = + = ANSWER: Hence, the two consecutive odd integers are 7 and 9.
32. PROBLEM:
The difference between twice the larger of two consecutive odd integers and the smaller is 59. Find the integers. SOLUTION: Let x represent the smaller odd integer. Let 2x + represent the next odd integer. The algebraic equation is: ( )2 2 59x x+ − = Solve the equation to obtain x:
( )2 2 592 2 2 59
4 594 4 59 4
55
x xx x
xx
x
+ − =⋅ + ⋅ − =
+ =+ − = −
=
Back substitute to obtain the larger integer: 2 55 2 57x + = + = ANSWER: Hence, the two consecutive odd integers are 55 and 57.
Part C: Geometry Problems Set up an algebraic equation, and then solve. 33. PROBLEM:
If the perimeter of a square is 48 inches, then find the length of each side. SOLUTION: Let s represent the side of the square. The formula for the perimeter of a square is 4P s= . The algebraic equation is: 48 4s= Solve the equation to obtain s: 48 448 44 412
12
ss
ss
=
=
==
ANSWER: Hence, the length of each side of the square is 12 inches.
34. PROBLEM:
The length of a rectangle is 2 inches longer than its width. If the perimeter is 36 inches, find the length and width. SOLUTION: Let l and w represent the length and width of the rectangle. Length, 2l w= + The formula for the perimeter of a rectangle is 2 2P l w= + . The algebraic equation is: ( )36 2 2 2w w= + + Solve the equation to obtain w:
( )36 2 2 236 2 2 2 236 4 4
36 4 4 4 432 432 44 48
8
w ww w
wwww
ww
= + += ⋅ + ⋅ += +
− = + −=
=
==
Back substitute to obtain the length: 2 8 2 10l w= + = + = ANSWER: Hence, the width of the rectangle is 8 inches and its length is 10 inches.
35. PROBLEM: The length of a rectangle is 2 feet less than twice its width. If the perimeter is 26 feet, find the length and width. SOLUTION: Let l and w represent the length and width of the rectangle. Length, 2 2l w= − The formula for the perimeter of a rectangle is 2 2P l w= + . The algebraic equation is: ( )26 2 2 2 2w w= − + Solve the equation to obtain w:
( )26 2 2 2 226 4 4 226 6 4
26 4 6 4 430 630 66 65
5
w ww wwwww
ww
= − += − += −
+ = − +=
=
==
Back substitute to obtain the length: 2 2 2 5 2 10 2 8l w= − = ⋅ − = − = ANSWER: Hence, the width of the rectangle is 5 feet and its length is 8 feet.
36. PROBLEM: The width of a rectangle is 2 centimeters less than one-half its length. If the perimeter is 56 centimeters, find the length and width. SOLUTION: Let l and w represent the length and width of the rectangle.
Width, 1 22
w l= −
The formula for the perimeter of a rectangle is 2 2P l w= + .
The algebraic equation is: 156 2 2 22
l l⎛ ⎞= + −⎜ ⎟⎝ ⎠
Solve the equation to obtain l: 156 2 2 22
56 2 2 2 22
56 2 456 3 4
56 4 3 4 460 360 33 320
20
l l
lll lllll
ll
⎛ ⎞= + −⎜ ⎟⎝ ⎠
= + ⋅ − ⋅
= + −= −
+ = − +=
=
==
Back substitute to obtain the width: 1 12 20 2 10 2 82 2
w l= − = ⋅ − = − =
ANSWER: Hence, the width of the rectangle is 8 cm and its length is 20 cm.
37. PROBLEM: The length of a rectangle is 3 feet less than twice its width. If the perimeter is 54 feet, find the dimensions of the rectangle. SOLUTION: Let l and w represent the length and width of the rectangle. Length, 2 3l w= − The formula for the perimeter of a rectangle is 2 2P l w= + . The algebraic equation is: ( )54 2 2 3 2w w= − + Solve the equation to obtain w:
( )54 2 2 3 254 4 6 254 6 6
54 6 6 6 660 660 66 610
10
w ww wwwww
ww
= − += − += −
+ = − +=
=
==
Back substitute to obtain the length: 2 3 2 10 3 20 3 17l w= − = ⋅ − = − = ANSWER: Hence, the width of the rectangle is 10 feet and its length is 17 feet.
38. PROBLEM:
If the length of a rectangle is twice as long as the width and its perimeter measures 72 inches, find the dimensions of the rectangle. SOLUTION: Let l and w represent the length and width of the rectangle. Length, 2l w= The formula for the perimeter of a rectangle is 2 2P l w= + . The algebraic equation is: ( )72 2 2 2w w= + Solve the equation to obtain w:
( )72 2 2 272 4 272 672 66 612
12
w ww www
ww
= += +=
=
==
Back substitute to obtain the length: 2 2 12 24l w= = ⋅ = ANSWER: Hence, the width of the rectangle is 12 inches and its length is 24 inches.
39. PROBLEM: The perimeter of an equilateral triangle measures 63 centimeters. Find the length of each side. SOLUTION: Let s represent the side of the equilateral triangle. For an equilateral triangle, the lengths of all three sides are equal. The formula for the perimeter of an equilateral triangle is 3P s s s s= + + = . The algebraic equation is 63 3s= . Solve the equation to obtain s: 63 363 33 321
21
ss
ss
=
=
==
ANSWER: Hence, the length of each side of the equilateral triangle is 21 centimeters.
40. PROBLEM: An isosceles triangle whose base is one-half as long as the other two equal sides has a perimeter of 25 centimeters. Find the length of each side. SOLUTION: Let a and c represent the two equal sides and b represent the base of the isosceles triangle.
Length of the base, 1 1 or 2 2
b a c= , as a c=
The formula for the perimeter of a triangle is P a b c= + + .
The algebraic equation is: 1252
a a a= + +
1252125 22
2 2 1252 2
4 1252
5252
2 5 2255 2 5
1010
a a a
a
a
a
a
aa
a
= + +
⎛ ⎞= +⎜ ⎟⎝ ⎠
⋅⎛ ⎞= +⎜ ⎟⎝ ⎠
+⎛ ⎞= ⎜ ⎟⎝ ⎠
=
⋅ = ⋅
==
Back substitute to obtain the base: 1 1 10 52 2
b a= = ⋅ =
ANSWER: Hence, the length of the base of the triangle is 5 centimeters and the length of each of the equal legs is 10 centimeters.
41. PROBLEM: Each of the two equal legs of an isosceles triangle are twice the length of the base. If the perimeter is 105 centimeters, then how long is each leg? SOLUTION: Let a and c represent the two equal legs and b represent the base of the isosceles triangle. Length of each of the equal legs of the isosceles triangle = 2b. The formula for the perimeter of a triangle is P a b c= + + . The algebraic equation is: 105 2 2b b b= + + Solve the equation to obtain b: 105 2 2105 5105 5
5 521
21
b b bbb
bb
= + +=
=
==
Back substitute to obtain the length of the equal leg: 2 2 21 42b = ⋅ = ANSWER: Hence, the length of the base of the triangle is 21 centimeters and the length of each of the equal legs is 42 centimeters.
42. PROBLEM: A triangle has sides whose measures are consecutive even integers. If the perimeter is 42 inches, find the measure of each side. SOLUTION: Let x represent the smallest side of the triangle. Let 2x + and 4x + represent the other two sides of the triangle. The formula for the perimeter of a triangle is P a b c= + + . The algebraic equation is: ( ) ( )42 2 4x x x= + + + + Solve the equation to obtain x:
( ) ( )42 2 442 2 442 3 6
42 6 3 6 636 336 33 312
12
x x xx x xxxxx
xx
= + + + += + + + += +
− = + −=
=
==
Back substitute to obtain the length of the remaining two sides: 2 12 2 14x + = + = and 4 12 4 16x + = + =
ANSWER: Hence, the measures of each side of the triangle are 12 inches, 14 inches and 16 inches.
43. PROBLEM: A triangle has sides whose measures are consecutive odd integers. If the perimeter is 21 inches, find the measure of each side. SOLUTION: Let x represent the smallest side of the triangle. Let 2x + and 4x + represent the other two sides of the triangle. The formula for the perimeter of a triangle is P a b c= + + . The algebraic equation is: ( ) ( )21 2 4x x x= + + + + Solve the equation to obtain x:
( ) ( )21 2 421 2 421 3 6
21 6 3 6 615 315 33 35
5
x x xx x xxxxx
xx
= + + + += + + + += +
− = + −=
=
==
Back substitute to obtain the length of the remaining two sides: 2 5 2 7x + = + = and 4 5 4 9x + = + =
ANSWER: Hence, the measures of each side of the triangle are 5 inches, 7 inches and 9 inches.
44. PROBLEM: A triangle has sides whose measures are consecutive integers. If the perimeter is 102 inches, then find the measure of each side. SOLUTION: Let x represent the smallest side of the triangle. Let 1x + and 2x + represent the other two sides of the triangle. The formula for the perimeter of a triangle is P a b c= + + . The algebraic equation is: ( ) ( )102 1 2x x x= + + + + Solve the equation to obtain x:
( ) ( )102 1 2102 1 2102 3 3
102 3 3 3 399 399 33 333
33
x x xx x xxxxx
xx
= + + + += + + + += +
− = + −=
=
==
Back substitute to obtain the length of the remaining two sides: 1 33 1 34x + = + = and 2 33 2 35x + = + =
ANSWER: Hence, the measures of each side of the triangle are 33 inches, 34 inches and 35 inches.
45. PROBLEM:
The circumference of a circle measures 50π units. Find the radius. SOLUTION: Let r represent the radius of the circle. The formula for the circumference of a circle is 2C rπ= . The algebraic equation is: 50 2 rπ π= Solve the equation to obtain r: 50 250 22 225
25
rr
rr
π ππ ππ π
=
=
==
ANSWER: Hence, the radius of the circle measures 25 units.
46. PROBLEM: The circumference of a circle measures 10π units. Find the radius. SOLUTION: Let r represent the radius of the circle. The formula for the circumference of a circle is 2C rπ= . The algebraic equation is: 10 2 rπ π= Solve the equation to obtain r: 10 210 22 2
55
rr
rr
π ππ ππ π
=
=
==
ANSWER: Hence, the radius of the circle measures 5 units.
47. PROBLEM: The circumference of a circle measures 100 centimeters. Determine the radius to the nearest tenth. SOLUTION: Let r represent the radius of the circle. The formula for the circumference of a circle is 2C rπ= . The algebraic equation is: 100 2 rπ= Solve the equation to obtain r:
100 2100 22 2
15.91...15.9
rr
rr
ππ
π π
=
=
=≈
ANSWER: Hence, the radius of the circle measures 15.9 centimeters.
48. PROBLEM: The circumference of a circle measures 20 centimeters. Find the diameter rounded off the nearest hundredth. SOLUTION: Let d represent the diameter of the circle.
Radius, d2
r =
The formula for the circumference of a circle is 2C rπ= .
The algebraic equation is: 20 22dπ= ⋅
Solve the equation to obtain d:
20 22
2020
6.366...6.37
d
dd
dd
πππ
π π
= ⋅
=
=
==
ANSWER: Hence, the diameter of the circle measures 6.37 centimeters.
49. PROBLEM:
The diameter of a circle measures 5 inches. Determine the circumference to the nearest tenth. SOLUTION: Let r represent the radius of the circle.
Radius, diameter 5 2.5 inches2 2
r = = =
The formula for the circumference of a circle is 2C rπ= . The algebraic equation is: ( )2 2.5C π= Solve the equation to obtain C:
( )2 2.515.707...15.7
C π=
==
ANSWER: Hence, the circumference of the circle measures 15.7 inches.
50. PROBLEM: The diameter of a circle is 13 feet. Calculate the exact value of the circumference. SOLUTION: Let r represent the radius of the circle.
Radius, diameter 13 6.5 feet2 2
r = = =
The formula for the circumference of a circle is 2C rπ= . The algebraic equation is: 2 6.5C π= ⋅ Solve the equation to obtain C:
2 6.513
CC
ππ
= ⋅=
ANSWER: Hence, the circumference of the circle measures 13π feet.
Part D: Percent and Money Problems Set up an algebraic equation, and then solve. 51. PROBLEM:
Calculate the simple interest earned on a two-year investment of $1,550 at a 8¾% annual interest rate? SOLUTION: Let I represent the simple interest earned on an interest rate of 8¾% 8.75% 0.0875= = . The formula for the simple interest is I prt= . The algebraic equation is: 1,550 0.0875 2I = ⋅ ⋅ Solve the equation to obtain I:
1,550 0.0875 2271.25
II= ⋅ ⋅=
ANSWER: Hence, the simple interest earned on a two-year investment of $1,550 at a 8¾% annual interest rate is $271.25.
52. PROBLEM: Calculate the simple interest earned on a one-year investment of $500 at a 6% annual interest rate? SOLUTION: Let I represent the simple interest earned on an interest rate of 6% 0.06= . The formula for the simple interest is I prt= . The algebraic equation is: 500 0.06 1I = ⋅ ⋅ Solve the equation to obtain I:
500 0.06 130
II= ⋅ ⋅=
ANSWER: Hence, the simple interest earned on a one-year investment of $500 at a 6% annual interest rate is $30.00.
53. PROBLEM: For how many years must $10,000 be invested at an 8½% annual interest rate to yield $4,250 in simple interest? SOLUTION: Let t represent the number of years required to yield a simple interest of $4,250 at 8½% 8.5% 0.085= = . The formula for the simple interest is I prt= . The algebraic equation is: 4, 250 10,000 0.085 t= ⋅ ⋅ Solve the equation to obtain t:
4, 250 10,000 0.0854,250 10,000 0.085
10,000 0.085 10,000 0.0855
5
tt
tt
= ⋅ ⋅⋅ ⋅
=⋅ ⋅
==
ANSWER: Hence, it will take 5 years for $10,000 invested at an 8½% annual interest rate to yield $4,250 in simple interest.
54. PROBLEM: For how many years must $1,000 be invested at a 7.75% annual interest rate to yield $503.75 in simple interest? SOLUTION: Let t represent the number of years required to yield a simple interest of $503.75 at 7.75% 0.0775= . The formula for the simple interest is I prt= . The algebraic equation is: 503.75 1,000 0.0775 t= ⋅ ⋅ Solve the equation to obtain t: 503.75 1,000 0.0775503.75 77.5503.75 77.577.5 77.5
6.56.5
ttt
tt
= ⋅ ⋅=
=
==
ANSWER: Hence, it will take 6.5 years for $1,000 invested at an 7.75% annual interest rate to yield $503.75 in simple interest.
55. PROBLEM: At what annual interest rate must $2,500 be invested for 3 years in order to yield $412.50 in simple interest? SOLUTION: Let r represent the annual interest rate at which $2,500 should be invested. The formula for the simple interest is I prt= . The algebraic equation is: 412.50 2,500 3r= ⋅ ⋅ Solve the equation to obtain r:
412.50 2,500 3412.50 2,500 3
2,500 3 2,500 30.055
5.51005.5%
rr
rrr
= ⋅ ⋅⋅ ⋅
=⋅ ⋅
=
=
=
ANSWER: Hence, $2,500 must be invested at 5.5% annual interest rate for 3 years in order to yield $412.50 in simple interest.
56. PROBLEM: At what annual interest rate must $500 be invested for 2 years in order to yield $93.50 in simple interest? SOLUTION: Let r represent the annual interest rate at which $500 should be invested. The formula for the simple interest is I prt= . The algebraic equation is: 93.50 500 2r= ⋅ ⋅ Solve the equation to obtain r:
93.50 500 293.50 1,00093.50 1,0001,000 1,000
0.09359.351009.35%
rrr
rrr
= ⋅ ⋅=
=
=
=
=
ANSWER: Hence, $500 must be invested at 9.35% annual interest rate for 2 years in order to yield $93.50 in simple interest.
57. PROBLEM: If the simple interest earned for 1 year was $47.25 and the annual rate was 6.3%, what was the principal? SOLUTION: Let p represent the principal on which the interest is earned at annual rate of 6.3% 0.063= . The formula for the simple interest is I prt= . The algebraic equation is: 47.25 0.063 1p= ⋅ ⋅ Solve the equation to obtain p:
47.25 0.063 147.25 0.063 1
0.063 1 0.063 1750
750
pp
pp
= ⋅ ⋅⋅ ⋅
=⋅ ⋅==
ANSWER: Hence, the principal on which a simple interest of $47.25 is earned at 6.3% annual rate is $750.
58. PROBLEM: If the simple interest earned for 2 years was $369.60 and the annual rate was 5¼%, what was the principal? SOLUTION: Let p represent the principal on which the interest is earned at
annual rate of 15 % 5.25% 0.05254
= = .
The formula for the simple interest is I prt= . The algebraic equation is: 369.60 0.0525 2p= ⋅ ⋅ Solve the equation to obtain p: 369.60 0.0525 2369.60 0.105369.60 0.1050.105 0.105
3,520
pp
p
p
= ⋅ ⋅= ⋅
=
=
ANSWER: Hence, the principal on which a simple interest of $369.60 is earned at 5¼% annual rate is $3,520.00.
59. PROBLEM: Joe invested last year’s $2,500 tax return in two different accounts. He put most of the money in a money market account earning 5% simple interest. He invested the rest in a CD earning 8% simple interest. How much did he put in each account if the total interest for the year was $138.50? SOLUTION: Let x represent the amount invested in the money market at 5% 0.05= . Let 2,500 x− represent the amount invested in the CD at 8% 0.08= . The formula for the simple interest is I prt= . Also, money market interest + CD interest = total interest The algebraic equation is: ( )0.05 1 2,500 0.08 1 138.50x x⋅ ⋅ + − ⋅ ⋅ = Solve the equation to obtain x:
( )0.05 1 2,500 0.08 1 138.500.05 2,500 0.08 0.08 138.50
0.05 200 0.08 138.500.03 200 138.50
0.03 200 200 138.50 2000.03 61.50.03 61.50.03 0.03
2,050
x xx x
x xx
xxx
x
⋅ ⋅ + − ⋅ ⋅ =+ ⋅ − ⋅ =
+ − =− + =
− + − = −− = −− −
=− −
=
Back substitute to obtain the amount invested in the CD: 2,500 2,500 2,050 450x− = − = ANSWER: Hence, Joe invested $2,050 at 5% in a money market and $450 at 8% in a CD.
60. PROBLEM: James invested $1,600 in two accounts. One account earns 4.25% simple interest and the other earns 8.5%. If the interest after one year was $85, how much did he invest in each account? SOLUTION: Let x represent the amount invested in account 1 at 4.25% 0.0425= . Let 1,600 x− represent the amount invested in account 2 at 8.5% 0.085= . The formula for the simple interest is I prt= . Also, interest from account 1 + interest from account 2 = total interest The algebraic equation is: ( )0.0425 1 1,600 0.085 1 85x x⋅ ⋅ + − ⋅ ⋅ = Solve the equation to obtain x:
( )0.0425 1 1,600 0.085 1 850.0425 1,600 0.085 0.085 85
0.0425 136 0.085 850.0425 136 85
0.0425 136 136 85 1360.0425 510.0425 510.0425 0.0425
1, 200
x xx x
x xx
xxx
x
⋅ ⋅ + − ⋅ ⋅ =+ ⋅ − ⋅ =
+ − =− + =
− + − = −− = −− −
=− −
=
Back substitute to obtain the amount invested in account 2: 1,600 1,600 1,200 400x− = − = ANSWER: Hence, James invested $1,200 at 4.25% in one account and $400 at 8.5% in the other account.
61. PROBLEM: Jane has her $5,400 savings invested in two accounts. She has part of it in a CD at 3% annual interest and the rest in a savings account that earns 2% annual interest. If the simple interest earned from both accounts is $140 for the year, then how much does she have in each account? SOLUTION: Let x represent the amount invested in the CD at 3% 0.03= . Let 5,400 x− represent the amount invested in the savings account at 2% 0.02= . The formula for the simple interest is I prt= . Also, CD interest + savings account interest = total interest The algebraic equation is: ( )0.03 1 5,400 0.02 1 140x x⋅ ⋅ + − ⋅ ⋅ = Solve the equation to obtain x:
( )0.03 1 5, 400 0.02 1 1400.03 5, 400 0.02 0.02 140
0.03 108 0.02 1400.01 108 140
0.01 108 108 140 1080.01 320.01 320.01 0.01
3, 200
x xx x
x xx
xxx
x
⋅ ⋅ + − ⋅ ⋅ =+ ⋅ − ⋅ =
+ − =+ =
+ − = −=
=
=
Back substitute to obtain the amount invested in the savings account: 5,400 5,400 3,200 2,200x− = − = ANSWER: Hence, Jane invested $3,200 at 3% in a CD and $2,200 at 2% in a savings account.
62. PROBLEM: Marty put last year’s bonus of $2,400 into two accounts. He invested part in a CD with 2.5% annual interest and the rest in a money market fund with 1.3% annual interest. His total interest for the year was $42.00. How much did he invest in each account? SOLUTION: Let x represent the amount invested in the CD at 2.5% 0.025= . Let 2,400 x− represent the amount invested in the a money market fund at 1.3% 0.013= . The formula for the simple interest is I prt= . Also, CD interest + money market interest = total interest The algebraic equation is: ( )0.025 1 2,400 0.013 1 42.00x x⋅ ⋅ + − ⋅ ⋅ = Solve the equation to obtain x:
( )0.025 1 2,400 0.013 1 42.000.025 2,400 0.013 0.013 42.00
0.025 31.2 0.013 42.000.012 31.2 42.00
0.012 31.2 31.2 42.00 31.20.012 10.80.012 10.80.012 0.012
900
x xx x
x xx
xxx
x
⋅ ⋅ + − ⋅ ⋅ =+ ⋅ − ⋅ =
+ − =+ =
+ − = −=
=
=
Back substitute to obtain the amount invested in the money market fund: 2,400 2,400 900 1,500x− = − = ANSWER: Hence, Marty invested $900 at 2.5% in a CD and $1,500 at 1.3% in a money market fund.
63. PROBLEM: Alice puts money into two accounts, one with 2% annual interest and another with 3% annual interest. She invests three times as much in the higher yielding account as she does in the lower yielding account. If her total interest for the year is $27.50, how much did she invest in each account? SOLUTION: Let x represent the amount invested in the lower yielding account at 2% 0.02= . Let 3x represent the amount invested in the higher yielding account at 3% 0.03= . The formula for the simple interest is I prt= . Also, lower yielding account interest + higher yielding account interest = total interest The algebraic equation is: 0.02 1 3 0.03 1 27.50x x⋅ ⋅ + ⋅ ⋅ = Solve the equation to obtain x:
0.02 1 3 0.03 1 27.500.02 0.09 27.50
0.11 27.500.11 27.500.11 0.11
250
x xx x
xx
x
⋅ ⋅ + ⋅ ⋅ =+ =
=
=
=
Back substitute to obtain the amount invested in the higher yielding account: 3 3 250 750x = ⋅ = ANSWER: Hence, Alice invested $250 at 2% in the lower yielding account and $750 at 3% the higher yielding account.
64. PROBLEM: Jim invested an inheritance in two separate banks. One bank offered 5.5% annual interest rate and the other 6¼%. He invested twice as much in the higher yielding bank account than he did in the other. If his total simple interest for one year was $4,860, then what was the amount of his inheritance? SOLUTION: Let x represent the amount invested in the lower yielding bank account at 5.5% 0.055= . Let 2x represent the amount invested in the higher yielding bank account at 6¼% 6.25% 0.0625= = . The formula for the simple interest is I prt= . Also, lower yielding account interest + higher yielding account interest = total interest The algebraic equation is: 0.055 1 2 0.0625 1 4,860x x⋅ ⋅ + ⋅ ⋅ = Solve the equation to obtain x:
0.055 1 2 0.0625 1 4,8600.055 0.125 4,860
0.18 4,8600.18 4,8600.18 0.18
27,000
x xx x
xx
x
⋅ ⋅ + ⋅ ⋅ =+ =
=
=
=
Back substitute to obtain the amount invested in the higher yielding account: 2 2 27,000 54,000x = ⋅ = Amount of Jim’s inheritance 27,000 54,000 81,000= + = ANSWER: Hence, amount of Jim’s inheritance was $81,000.
65. PROBLEM: If an item is advertised to cost $29.99 plus 9.25% tax, what is the total cost? SOLUTION: Let x represent the total cost.
Tax 9.259.25% 29.99 29.99 0.0925 29.99100
= ⋅ = ⋅ = ⋅
The algebraic equation is: 29.99 + 0.0925 29.99x = ⋅ Solve the equation to obtain x:
29.99 + 0.0925 29.9929.99 2.7732.76
xxx
= ⋅= +=
ANSWER: Hence, the total cost of the item $32.76.
66. PROBLEM: If an item is advertised to cost $32.98 plus 8¾% tax, what is the total cost? SOLUTION: Let x represent the total cost of the item.
Tax 8.758¾% 32.98 8.75% 32.98 32.98 0.0875 32.98100
= ⋅ = ⋅ = ⋅ = ⋅
The algebraic equation is: 32.98 0.0875 32.98x = + ⋅ Solve the equation to obtain x:
32.98 0.0875 32.9832.98 2.885...35.865...35.87
xxxx
= + ⋅= +=≈
ANSWER: Hence, the total cost of the item is $35.87.
67. PROBLEM: An item, including an 8.75% tax, cost $46.49. What is the original pretax cost of the item? SOLUTION: Let x represent the original pretax cost of the item.
Tax 8.758.75% 0.0875100
x x x= ⋅ = ⋅ = ⋅
The algebraic equation is: 0.0875 46.49x x⋅ + = Solve the equation to obtain x: 0.0875 46.49
1.0875 46.491.0875 46.491.0875 1.0875
42.75
x xxx
x
⋅ + ==
=
≈
ANSWER: Hence, the original pretax cost of the item is $42.75.
68. PROBLEM: An item, including a 5.48% tax, cost $17.82. What is the original pretax cost of the item? SOLUTION: Let x represent the original pretax cost of the item.
Tax 5.485.48% 0.0548100
x x x= ⋅ = ⋅ = ⋅
The algebraic equation is: 0.0548 17.82x x⋅ + = Solve the equation to obtain x: 0.0548 17.82
1.0548 17.821.0548 17.821.0548 1.0548
16.89
x xxx
x
⋅ + ==
=
≈
ANSWER: Hence, the original pretax cost of the item is $16.89.
69. PROBLEM: If a meal costs $32.75, what is the total after adding a 15% tip? SOLUTION: Let x represent the total cost of the meal.
Tip 1515% 32.75 32.75 0.15 32.75100
= ⋅ = ⋅ = ⋅
The algebraic equation is: 32.75 0.15 32.75x = + ⋅ Solve the equation to obtain x:
32.75 0.15 32.75 = 32.75 4.91
37.66
xxx
= + ⋅+
=
ANSWER: Hence, the total cost of the meal is $37.66.
70. PROBLEM: How much is a 15% tip on a restaurant bill that totals $33.33? SOLUTION: Let x represent the amount of the tip.
1515% 0.15100
= =
The algebraic equation is: 0.15 33.33x = ⋅ Solve the equation to obtain x:
0.15 33.334.99955
xxx
= ⋅=≈
ANSWER: Hence, the amount of the tip is $5.
71. PROBLEM: Ray has a handful of dimes and nickels valuing $3.05. He has 5 more dimes than he does nickels. How many of each coin does he have? SOLUTION: Let x represent the number of nickels. Let 5x + represent the number of dimes. The algebraic equation is: ( )0.05 5 0.10 3.05x x⋅ + + = Solve the equation to obtain x:
( )0.05 5 0.10 3.050.05 0.10 0.10 5 3.05
0.05 0.10 0.50 3.050.15 0.50 3.05
0.15 0.50 0.50 3.05 0.500.15 2.550.15 2.550.15 0.15
17
x xx x
x xx
xxx
x
⋅ + + =+ ⋅ + ⋅ =
+ + =+ =
+ − = −=
=
=
Back substitute to obtain the number of dimes: 5 17 5 22x + = + = ANSWER: Hence, Ray has 17 nickels and 22 dimes.
72. PROBLEM: Jill has 3 fewer half-dollars than she has quarters. The value of all 27 of her coins adds to $9.75. How many of each coin does Jill have? SOLUTION: Let x represent the number of quarters. Let 3x − represent the number of half-dollars. Total number of coins = 27 The algebraic equation is: ( )3 27x x+ − = Solve the equation to obtain x:
( )3 273 27
2 3 272 3 3 27 3
2 302 302 2
15
x xx x
xx
xx
x
+ − =+ − =
− =− + = +
=
=
=
Back substitute to obtain the number of half-dollars: 3 15 3 12x − = − = ANSWER: Hence, Jill has 15 quarters and 12 half-dollars. Alternatively, Let x represent the number of quarters. Let 3x − represent the number of half-dollars. Value of coins = $9.75 The algebraic equation is: ( )0.25 0.50 3 9.75x x⋅ + − = Solve the equation to obtain x:
( )0.25 0.50 3 9.750.25 0.50 0.50 3 9.75
0.25 0.50 1.5 9.750.75 1.5 9.75
0.75 1.5 1.5 9.75 1.50.75 11.250.75 11.250.75 0.75
15
x xx x
x xx
xxx
x
⋅ + − =+ ⋅ − ⋅ =
+ ⋅ − =− =
− + = +=
=
=
Back substitute to obtain the number of half-dollars: 3 15 3 12x − = − = ANSWER: Hence, Jill has 15 quarters and 12 half-dollars.
73. PROBLEM: Cathy has to deposit $410 worth of five- and ten-dollar bills. She has 1 fewer than three times as many tens as she does five-dollar bills. How many of each bill does she have to deposit? SOLUTION: Let x represent the number of five-dollar bills. Let 3 1x − represent the number of ten-dollar bills. The algebraic equation is: ( )5 10 3 1 410x x⋅ + ⋅ − = Solve the equation to obtain x:
( )5 10 3 1 4105 10 3 10 1 410
5 30 10 41035 10 410
35 10 10 410 1035 42035 42035 35
12
x xx x
x xx
xxx
x
⋅ + ⋅ − =+ ⋅ − ⋅ =
+ − =− =
− + = +=
=
=
3 1 3 12 1 36 1 35x − = ⋅ − = − = ANSWER: Hence, Cathy has 12 five-dollar bills and 35 ten-dollar bills.
74. PROBLEM: Billy has a pile of quarters, dimes, and nickels which values $4.05. He has 3 more dimes than quarters and 5 more nickels than quarters. How many of each coin does Billy have? SOLUTION: Let x represent the number of quarters. Let 3x + represent the number of dimes. Let 5x + represent the number of nickels. Value of 1 quarter is $0.25, total value of quarters present is 0.25x. Value of 1 dime is $0.10, total value of dimes present is 0.1(x + 3). Value of 1 nickel is $0.05, total value of nickels present is 0.05(x + 5). The algebraic equation is:
( ) ( )0.25 0.1 3 0.05 5 4.050.25 0.1 0.3 0.05 0.25 4.050.40 0.55 4.050.40 0.55 0.55 4.05 0.550.40 3.50.40 3.50.40 0.40
8.75 ~ 9
x x xx x xxxx
x
x
+ + + + =
+ + + + =+ =+ − = −=
=
=
Solve the equation to obtain x: 9 Back substitute to obtain the number of dimes: 3 9 3 12x + = + = Back substitute to obtain the number of nickels: 5 9 5 14x + = + = ANSWER: Hence, Billy has 9 quarters, 11 dimes, and 13 nickels.
75. PROBLEM: Mary has a jar with one-dollar bills, half-dollar coins, and quarters valuing $7.75. She has twice as many quarters than she does half-dollar coins and the same amount of half-dollar coins as one-dollar bills. How many of each does she have? SOLUTION: Let x represent the number of half-dollar coins and the number of one-dollar bills. Let 2x represent the number of quarters. The algebraic equation is: ( )1.00 0.50 0.25 2 7.75x x x⋅ + ⋅ + ⋅ = Solve the equation to obtain x:
( )1.00 0.50 0.25 2 7.750.50 0.5 7.75
2 7.752 7.752 2
3.875
x x xx x x
xx
x
⋅ + ⋅ + ⋅ =+ + =
=
=
=
Back substitute to obtain the number of quarters: 2x =7.75 Therefore we have 1.0 3.875 + 0.5 3.875 + 0.25 (2 3.875) = 7.75⋅ ⋅ ⋅ ⋅ ANSWER: Hence, Mary has 4 one-dollar bills, 4 half-dollar bills, and 8 quarters.
76. PROBLEM:
Chad has a bill-fold of one-, five-, and ten-dollar bills totaling $118. He has 2 more than 3 times as many ones as he does five-dollar bills and 1 fewer ten- than five-dollar bills. How many of each bill does Chad have? SOLUTION: Let x represent the number of five-dollar bills. Let 3 2x + represent the number of one-dollar bills. Let 1x − represent the number of ten-dollar bills. The algebraic equation is: ( ) ( )1 3 2 5 10 1 118x x x+ + ⋅ + − = Solve the equation to obtain x: ( ) ( )1 3 2 5 10 1 118
3 2 5 10 10 11818 8 118
18 8 8 118 818 12618 12618 18
7
x x xx x x
xx
xx
x
+ + ⋅ + − =+ + + − =
− =− + = +
=
=
=
Back substitute to obtain the number of one-dollar bills: 3 2 3 7 2 21 2 23x + = ⋅ + = + = Back substitute to obtain the number of ten-dollar bills: 1 7 1 6x − = − = ANSWER: Hence, Chad has 23 one-dollar bills, 7 five-dollar bills and 6 ten-dollar bills.
Part D: Uniform Motion (Distance Problems) Set up an algebraic equation then solve. 77. PROBLEM:
Two cars leave a location traveling in opposite directions. If one car averages 55 mph and the other averages 65 mph, then how long will it take for them to separate a distance of 300 miles? SOLUTION: Let t represent the time of travel of each car. Distance traveled by the Car 1: 55D r t t= ⋅ = ⋅ Distance traveled by the Car 2: 65D r t t= ⋅ = ⋅ The algebraic equation is: 55 65 300t t+ = Solve the equation to obtain t: 55 65 300
120 300120 300120 120
2.5
t ttt
t
+ ==
=
=
ANSWER: Hence, it will take 2.5 hours for the two cars to separate a distance of 300 miles.
Distance = Rate × Time Car 1 55t 55 mph t
Car 2 65t 65 mph t
Total 300 mi
78. PROBLEM: Two planes leave the airport at the same time traveling in opposite directions. The average speeds for the planes are 450 mph and 395 mph. How long will it take the planes to be a distance of 1,478.75 miles apart? SOLUTION: Let t represent the time of travel of each plane. Distance traveled by the Plane 1: 450D r t t= ⋅ = ⋅ Distance traveled by the Plane 2: 395D r t t= ⋅ = ⋅ The algebraic equation is: 450 395 1,478.75t t+ = Solve the equation to obtain t: 450 395 1, 478.75
845 1, 478.75845 1,478.75845 845
1.75
t ttt
t
+ ==
=
=
ANSWER: Hence, it will take 1.75 hours for the two planes to be a distance of 1,478.75 miles apart.
Distance = Rate × Time
Plane 1 450t 450 mph t
Plane 2 395t 395 mph t
Total 1,478.75 mi
79. PROBLEM: Bill and Ted are racing across the country. Bill leaves one hour earlier than Ted and travels at an average rate of 60 mph. If Ted intends to catch up at a rate of 70 mph, then how long will it take? SOLUTION: Let t represent the time of travel of Ted. Let 1t + represent the time of travel of Bill. Distance traveled by Bill: ( )60 1D r t t= ⋅ = ⋅ + Distance traveled by Ted: 70D r t t= ⋅ = ⋅ The distance traveled by both Bill and Ted is the same.
The algebraic equation is: ( )60 1 70t t+ = Solve the equation to obtain t:
( )60 1 7060 60 1 70
60 60 60 70 6060 1060 1010 10
66
t tt t
t t t ttt
tt
+ =⋅ + ⋅ =
+ − = −=
=
==
ANSWER: Hence, Ted will take 6 hours to catch up with Bill.
Distance = Rate × Time Bill 60(t + 1) 60 mph t + 1 Ted 70t 70 mph t
80. PROBLEM: Two brothers leave from the same location, one in a car and the other on a bicycle, to meet up at their grandmother’s house for dinner. If one brother averages 30 mph in the car and the other averages 12 mph on the bicycle, then it takes the brother on the bicycle 1 hour less than 3 times as long as the other in the car. How long does it take each of them to make the trip? SOLUTION: Let t represent the time of travel of the brother driving the car. Let 3 1t − represent the time of travel of the brother riding the bicycle. Distance traveled by car: 30D r t t= ⋅ = ⋅ Distance traveled by bicycle: ( )12 3 1D r t t= ⋅ = ⋅ − The distance traveled by both the brothers is the same. The algebraic equation is: ( )30 12 3 1t t= − Solve the equation to obtain t:
( )30 12 3 130 12 3 12 130 36 12
30 36 36 12 366 126 126 6
2
t tt tt t
t t t ttt
t
= −= ⋅ − ⋅= −
− = − −− = −− −
=− −
=
Back substitute to obtain the time of travel of the brother riding the bicycle: 3 1 3 2 1 6 1 5t − = ⋅ − = − = ANSWER: Hence, it takes 2 hours in the car and 5 hours on the bicycle.
Distance = Rate × Time Car 30t 30 mph t
Bicycle 12(3t − 1) 12 mph 3t − 1
81. PROBLEM: A commercial airline pilot flew at an average speed of 350 mph before being informed that his destination airfield may be closed due to poor weather conditions. In an attempt to arrive before the storm, he increased his speed 400 mph and flew for another 3 hours. If the total distance flown was 2, 950 miles, then how long did the trip take? SOLUTION: Let t represent the time for which the pilot flew at a speed of 350 miles. Distance traveled before increasing the speed: 350D r t t= ⋅ = ⋅ Distance traveled after increasing the speed: 400 3D r t= ⋅ = ⋅ The algebraic equation is: 350 400 3 2,950t + ⋅ = Solve the equation to obtain t:
350 400 3 2,950350 1,200 2,950
350 1,200 1,200 2,950 1,200350 1,750350 1,750350 350
5
tt
ttt
t
+ ⋅ =+ =
+ − = −=
=
=
Total time of trip 5 3 8= + = ANSWER: Hence, the trip took 8 hours.
Distance = Rate × Time Before
increasing the speed 350t 350 mph t
After increasing the speed 400 × 3 400 mph 3
Total 2,950 mi
82. PROBLEM: Two brothers drove the 2,793 miles from Los Angeles to New York. One of the brothers, driving during the day, was able to average 70 mph, and the other, driving at night, was able to average 53 mph. If the brother driving at night drove 3 hours less than the brother driving in the day, then how many hours did they each drive? SOLUTION: Let t represent the time of travel of the brother driving during the day. Let 3t − represent the time of travel of the brother driving at night. Distance traveled during the day: 70D r t t= ⋅ = ⋅ Distance traveled at night: ( )53 3D r t t= ⋅ = − The algebraic equation is: ( )70 53 3 2,793t t+ − = Solve the equation to obtain t:
( )70 53 3 2,79370 53 53 3 2,793
70 53 159 2,793123 159 2,793
123 159 159 2,793 159123 2,952123 2,952123 123
24
t tt t
t tt
ttt
t
+ − =+ ⋅ − ⋅ =+ − =
− =− + = +
=
=
=
Back substitute to obtain the time of travel of the brother driving at night: 3 24 3 21t − = − =
ANSWER: Hence, the brother driving during the day drove for 24 hours and the other drove 21 hours at night.
Distance = Rate × Time During
day 70t 70 mph t At night 53(t − 3) 53 mph t − 3
Total 2,793 mi
83. PROBLEM: Joe and Ellen live 21 miles apart. Departing at the same time they cycle toward each other. If Joe averages 8 mph and Ellen averages 6 mph, how long will it take them to meet? SOLUTION: Let t represent the time of travel of Joe and Ellen. Distance traveled by Joe: 8D r t t= ⋅ = ⋅ Distance traveled by Ellen: 6D r t t= ⋅ = ⋅ The algebraic equation is: 8 6 21t t+ = Solve the equation to obtain t: 8 6 21
14 2114 2114 14
112
t ttt
t
+ ==
=
=
ANSWER: Hence, it will take Joe and Ellen 1½ hours to meet.
Distance = Rate × Time
Joe 8t 8 mph t
Ellen 6t 6 mph t Total 21 mi
84. PROBLEM: If it takes 6 minutes to drive to the automobile repair shop at an average speed of 30 mph, then how long will it take to walk back at an average rate of 4 mph? SOLUTION: Let t represent the time taken to walk back from the automobile repair shop. Distance traveled by the automobile: 30 6D r t= ⋅ = ⋅ Distance covered by walking: 4D r t t= ⋅ = ⋅ The distance covered both ways is the same. The algebraic equation is: 30 6 4 t⋅ = ⋅ Solve the equation to obtain t: 30 6 4180 4180 4
4 445
45
ttt
tt
⋅ = ⋅=
=
==
ANSWER: Hence, it will take 45 minutes to walk back.
Distance = Rate × Time
Automobile 30 × 6 30 mph 6 min
Walking 4t 4 mph t
85. PROBLEM: Jaime and Alex leave the same location and travel in opposite directions. Traffic conditions enabled Alex to average 14 mph faster than Jaime. After 1½ hours they are 159 miles apart. Find the speed at which each were able to travel. SOLUTION: Let r represent the speed at which Jaime was able to travel. Let 14r + represent the speed at which Alex was able to travel. Distance traveled by Jaime: 1.5D r t r= ⋅ = ⋅ Distance traveled by Alex: ( )14 1.5D r t r= ⋅ = + ⋅ The algebraic equation is: ( )1.5 1.5 14 159r r+ + = Solve the equation to obtain r:
( )1.5 1.5 14 1591.5 1.5 1.5 14 159
3 21 1593 21 21 159 21
3 1383 1383 3
46
r rr r
rr
rr
r
+ + =+ ⋅ + ⋅ =
+ =+ − = −
=
=
=
Speed at which Alex was able to travel 14 46 14 60t= + = + = ANSWER: Hence, Jaime was able to travel at a speed of 46 mph and Alex was able to travel at a speed of 60 mph.
Distance = Rate × Time Jaime r × 1.5 r 1.5 hrs Alex (r + 14) × 1.5 r + 14 1.5 hrs Total 159 mi
86. PROBLEM: Jane and Holly live 51 miles apart and leave at the same time toward each other to meet for lunch. Jane traveled on the freeway at twice the average speed as Holly. They were able to meet in a half hour. At what rate did each travel? SOLUTION: Let r represent the speed at which Holly was able to travel. Let 2r represent the speed at which Jane was able to travel. Distance traveled by Holly: 0.5D r t r= ⋅ = ⋅ Distance traveled by Jane: 2 0.5D r t r= ⋅ = ⋅ The algebraic equation is: 0.5 2 0.5 51r r⋅ + ⋅ = Solve the equation to obtain r:
0.5 2 0.5 510.5 1 51
1.5 511.5 511.5 1.5
34
r rr r
rr
r
⋅ + ⋅ =+ ⋅ =
=
=
=
Back substitute to obtain the speed at which Jane was able to travel: 2 2 34 68r = ⋅ = ANSWER: Hence, Holly traveled at a rate of 34 mph and Jane at 68 mph.
Part F: Discussion Board Topics 87. PROBLEM:
Discuss ideas for calculating taxes and tips mentally. ANSWER: a) Know what your local sales tax is. If you live an area that has both state and other local sales tax, add them together to get the appropriate amount. b) Take the rate and first mentally multiply it times the number of dollars that you have purchased. If the rate is 7% and your purchase is $1 the amount that you will have to pay is 7 cents. For every dollar that you purchase with a sales tax of 7% you need to add this amount. If your purchase is $10 then add 70 cents. If you're living in a state with a 5% sales tax, either multiply by .05 or 5 cents on the dollar or multiply by 10 cents and divide by 2 ($10 X.10=1.00/2=.50). c) Adjust for the dollars in between. Most taxes will have an equal divisor. If the amount is 5%, then every 20 cents another penny is added. For that 7% rate the amount is 100/7=14 or every 14 cents another penny is added. (Source: http://www.ehow.com/how_2093952_mentally-figure-sales-tax.html)
Distance = Rate × Time Holly r × 0.5 r 0.5 hr Jane 2r × 0.5 2r 0.5 hr Total 51 mi
88. PROBLEM: Research historical methods for representing unknowns. ANSWER: (Students answers may vary.) Take an equation with more than one unknown, and given certain information about all but one unknown, manipulate the equation so that it is easily solvable. The intuition of mathematical answers is developed by examining if an answer is reasonable or not according to logic. (Source:http://www.lessonplanet.com/search?keywords=unknown+quantity&media=lesson)
89. PROBLEM: Research and compare simple interest and compound interest. What is the difference? ANSWER: Simple interest is calculated on the original principal only. Accumulated interest from prior periods is not used in calculations for the following periods. Simple interest is normally used for a single period of less than a year, such as 30 or 60 days. Compound interest is calculated each period on the original principal and all interest accumulated during past periods. Although the interest may be stated as a yearly rate, the compounding periods can be yearly, semiannually, quarterly, or even continuously. (Source: http://www.getobjects.com/Components/Finance/TVM/iy.html)
90. PROBLEM:
Discuss why algebra is a required subject. ANSWER: (Students answers may vary.) You wouldn't be able to use your computer if it weren't for algebra. Algebra underlies all the technology we enjoy and is needed in all the hard sciences. Algebra was needed in designing your house and in creating medications you take when you are sick. If you are wondering why you need to take algebra, it's because, like Shakespeare, it is one of the great things our culture has to offer. And if you work to understand it, you will improve your ability to think and problem solve in other areas of your life. (Source: http://wiki.answers.com/Q/Why_is_algebra_important_to_society)
91. PROBLEM: Research ways to show that a repeating decimal is rational. Share your findings on the discussion board. ANSWER:
Fraction: 19
Ellipsis: 0.111… Vinculum: 0.1 Dots: 0.1 Brackets: 0. (1)
2.6RatioandProportionApplications Part A: Ratios and RatesExpress the ratio in reduced form. 1. PROBLEM:
100 inches : 250 inches
SOLUTION: 100 ÷ 50 : 250 ÷ 50 2:5
ANSWER: 2:5
2. PROBLEM:
480 pixels : 320 pixels
SOLUTION: 480 ÷ 160 : 320 ÷ 160 3:2 ANSWER: 3:2
3. PROBLEM:
96 feet : 72 feet
SOLUTION: 96 ÷ 24 : 72 ÷ 24 4:3
ANSWER: 4:3
4. PROBLEM: 2404
mileshours
SOLUTION: 240 240 604 4 1
mileshours
= =
ANSWER: 60 miles/hour
5. PROBLEM:
96 feet3 seconds
SOLUTION:
96 feet 96 323 seconds 3 1
= =
ANSWER: 32 feet/second
6. PROBLEM:
6000 revolutions4 minutes
SOLUTION: 6000 revolutions 6000 1500
4 minutes 4 1= =
ANSWER: 1500 revolutions/minute
7. PROBLEM:
Google’s average 2008 stock price and earnings per share was $465.66 and $14.89 respectively. What was Google’s average price to earnings ratio in 2008? (Source: WolframAlpha)
SOLUTION: Google’s average price to earnings ratio in 2008, $ 465.66 : $14.89 46566 : 1489 46566 ÷1489 : 1489 ÷1489 31.27
ANSWER: 31.27
8. PROBLEM: The F22 Raptor has two engines that can produce 35,000 lbs of thrust each. If the takeoff weight of this fighter jet is 50,000 lbs, calculate the thrust to weight ratio of the plane? (Source: USAF)
SOLUTION: Ratio of the thrust to weight of the plane is, Thrust for 2 engines = 2 ⋅ 35,000 lbs = 70,000 lbs 70,000 lbs : 50,000 lbs 70,000 ÷ 50000 : 50000 ÷ 50000 1.4 ANSWER: 1.4
9. PROBLEM:
A discount warehouse offers a box of 55 individual instant oatmeal servings for $11.10. The supermarket offers smaller boxes of the same product containing 12 individual servings for $3.60. Which store offers the better value?
SOLUTION: Value offered by the discount warehouse, 55 oat meal servings
$11.10
55 5500 4.9511.10 1110
= =
4.95 oat meal servings/$ Value offered by the supermarket, 12 oat meal servings
$3.60
12 1200 3.333.60 360
= =
3.33 oat meal servings/$
ANSWER: The discount warehouse offers the better value.
10. PROBLEM: Joe and Mary wish to take a road trip together and need to decide whose car they will take. Joe calculated that his car was able to travel 210 miles on 12 gallons of gasoline. Mary calculated that her car traveled 300 miles on 19 gallons. Which one of their cars gets more miles to the gallon?
SOLUTION: Joe’s car
Miles to the gallon: 210 12
milesgallon
210 17.5 /12
miles miles gallongallon
=
Mary’s car
Miles to the gallon: 300 19
milesgallon
300 15.8 /19
miles miles gallongallon
=
ANSWER: Joe’s car gets more miles per gallon.
Part B: Solving ProportionsSolve. 11.
PROBLEM:
23 150
n=
SOLUTION: 23 1502 150
32 50100
n
n
nn
=
⋅=
⋅ ==
ANSWER: 100 n=
12. PROBLEM:
7 21
5n=
SOLUTION: 7 21
57 521
53
n
n
n
=
⋅=
=
ANSWER: 53
n=
13. PROBLEM:
1 53 n=
SOLUTION: 1 53
3 51
15
n
n
n
=
⋅=
=
ANSWER: 15n =
14. PROBLEM:
12 65 n=
SOLUTION:
12 65
6 51252
n
n
n
=
⋅=
=
ANSWER: 52
n =
15. PROBLEM: 3
8 2n= −
SOLUTION:
38 2
8 32
12
n
n
n
= −
⋅= −
= −
ANSWER: 12n = −
16. PROBLEM:
53 7n= −
SOLUTION: 5
3 75 37
157
n
n
n
= −
⋅= −
= −
ANSWER: 157
n = −
17. PROBLEM:
283n
=
SOLUTION:
283
8 32
12
n
n
n
=
⋅=
=
ANSWER: 12 n=
18. PROBLEM: 5 30n= −
SOLUTION:
5 30
53016
n
n
n
= −
− =
− =
ANSWER: 16
n− =
19. PROBLEM:
111n
=−
SOLUTION:
111111
1 12
n
n
nn
=−
− =
= +=
ANSWER: 2n =
20. PROBLEM:
111n
− = −+
SOLUTION:
111
111
1 10
n
n
nn
− = −+
+ =
= −=
ANSWER: 0n =
21. PROBLEM: 40 5
3n− = −
SOLUTION:
40 53
40 35
24
n
n
n
− = −
⋅=
=
ANSWER: 24 n=
22. PROBLEM:
2 1 33 5
n += −
SOLUTION:
( )
2 1 33 5
5 2 1 3 310 5 910 1410 1410 10
75
n
nnn
n
n
+= −
+ = − ⋅
+ = −= −
= −
= −
ANSWER: 75
n = −
23. PROBLEM: 5 2
3 3 3n=
+
SOLUTION:
( )
5 23 3 35 3 2 3 315 6 615 6 69 69 66 632
nn
nn
n
n
n
=+
⋅ = +
= +− ==
=
=
ANSWER: 32
n=
24. PROBLEM:
1 12 1 3nn+
=−
SOLUTION:
( ) ( )
1 12 1 33 1 1 2 13 3 2 13 2 1 3
4
nnn n
n nn n
n
+=
−+ = −
+ = −− = − −= −
ANSWER: 4n = −
25. PROBLEM: 5 7 1
5 2n n+ −
=
SOLUTION:
( ) ( )
5 7 15 2
2 5 7 5 110 14 5 510 5 5 145 195 195 5
195
n n
n nn nn n
n
n
n
+ −=
+ = −
+ = −− = − −= −
= −
= −
ANSWER: 195
n = −
26. PROBLEM:
72 36
nn +− + =
SOLUTION:
( )
72 36
6 2 3 712 18 712 7 1813 1113 13
1113
nn
n nn nn n
n
n
+− + =
− + = +
− + = +− − = −− −
=− −
=
ANSWER: 1113
n =
27. PROBLEM: Find two numbers in the ratio of 3 to 5 whose sum is 160. (Hint: Use n and 160 − n)
SOLUTION:
3160 5
nn=
−
( )5 3 1605 480 35 3 4808 4808 4808 8
60
n nn nn nn
n
n
= −
= −+ ==
=
=
160 – n = 160 – 60 = 100
ANSWER: The two numbers are 60 and 100.
28. PROBLEM:
Find two numbers in the ratio of 2 to 7 whose sum is 90.
SOLUTION:
( )
290 77 2 907 180 27 2 1809 1809 1809 9
2090 90 20 70
nn
n nn nn nn
n
nn
=−= −
= −+ ==
=
=− = − =
ANSWER: The two numbers are 20 and 70.
29. PROBLEM: Find two numbers in the ratio of −3 to 7 whose sum is 80.
SOLUTION:
( )
( )
380 77 3 807 240 37 3 2404 2404 4
6080 80 60 140
nn
n nn nn n
n
nn
= −−= − −
= − +− = −
= −
= −
− = − − =
ANSWER: The two numbers are –60 and 140.
30. PROBLEM:
Find two numbers in the ratio of −1 to 3 whose sum is 90.
SOLUTION:
( )
( )
190 33 1 903 903 902 902 902 2
4590 90 45 135
nn
n nn nn nn
n
nn
= −−= − −
= − +− = −= −
= −
= −
− = − − =
ANSWER: The two numbers are –45 and 135.
31. PROBLEM: A larger integer is 5 more than a smaller integer. If the two integers have a ratio of 6 to 5 find the integers.
SOLUTION: The smaller integer is represented by n. The larger integer is, n + 5.
( )
55 6
6 5 56 5 256 5 25
255 25 5 30
nnn nn nn n
nn
=+= +
= +− ==+ = + =
ANSWER: The two integers are 25 and 30.
32. PROBLEM:
A larger integer is 7 less than twice a smaller integer. If the two integers have a ratio of 2 to 3 find the integers.
SOLUTION: The smaller integer is represented by n. The larger integer is, 2n – 7.
( )
( )
22 7 33 2 2 73 4 1414 4 3142 7 2 14 7 28 7 21
nnn nn n
n nn
n
=−= −
= −= −=
− = − = − =
ANSWER: The two integers are 14 and 21.
Given the following proportions determine :x y . 33. PROBLEM:
3 4x y=
SOLUTION:
3 434
x y
xy
=
=
ANSWER: 34
xy=
34.
PROBLEM:
2 33 5
x y y−= −
SOLUTION:
( )
2 33 5
5 2 3 35 10 95 9 105
15
x y y
x y yx y yx y yx y
xy
−= −
− = − ⋅
− = −= − +=
=
ANSWER: 15
xy=
35. PROBLEM: 2 4 32 4 2
x yx y+
=−
SOLUTION:
( ) ( )
2 4 32 4 22 2 4 3 2 44 8 6 124 6 12 8
2 20202
101
x yx y
x y x yx y x yx x y y
x yxyxy
+=
−
+ = −
+ = −− = − −
− = −
=
=
ANSWER: 101
xy=
36. PROBLEM:
35
x yx y+
=−
SOLUTION:
( ) ( )
35
5 35 5 3 35 3 3 52 8
8241
x yx y
x y x yx y x yx x y yx y
xyxy
+=
−
+ = −
+ = −− = − −= −
= −
= −
ANSWER: 41
xy= −
Part C: Applications Set up a proportion then solve. 37. PROBLEM:
If 4 out of every 5 voters support the Governor, then how many of the 1200 people surveyed support the Governor?
SOLUTION: People supporting the Governor are represented by x. 4 out of 5 voters support the Governor. We will set up our ratios with the number of voters who support the Governor in the numerator to the total number of voters in the denominator.
41200 55 4 12005 4 12005 5
960
x
x
x
x
=
= ⋅⋅
=
=
ANSWER: Out of 1200 people, 960 support the Governor.
38. PROBLEM:
If 1 out of every 3 voters surveyed said they voted yes on proposition 23, then how many of the 600 people surveyed voted yes?
SOLUTION: People voted yes on proposition 23 are represented by x. 1 out of 3 voters voted yes for proposition 23. We will set up our ratios with the number of voters who voted yes for proposition 23 in the numerator to the total number of voters in the denominator.
1600 33 1 6003 6003 6003 3
200
x
xx
x
x
=
= ⋅=
=
=
ANSWER: Out of 600 people, 200 people voted yes for proposition 23.
39. PROBLEM: Out of 460 students surveyed, the ratio to support the student union remodel project was 3 to 5. How many students were in favor of the remodel?
SOLUTION: Students in favor of student union remodel project are represented by x. 3 out of 5 students support the student union remodel project. We will set up our ratios with the number of students support the student union remodel project in the numerator to the total number of students in the denominator.
3460 55 3 4605 3 4605 5
276
x
x
x
x
=
= ⋅⋅
=
=
ANSWER: Out of 460 students, 276 students favored the student union remodel project.
40. PROBLEM:
The ratio of students who carry credit card debt is estimated to be 5 to 7. Estimate the number of students that carry credit card debt out of a total of 14,000 students.
SOLUTION: Students who carry credit card debt are represented by x. 5 out of 7 students carry credit card debt. We will set up our ratios with the number of students carry credit card debt in the numerator to the total number of students.
514,000 77 5 140007 5 140007 7
10,000
x
x
x
x
=
= ⋅⋅
=
=
ANSWER: Out of 14,000 students, 10,000 students carry credit card debt.
41. PROBLEM: If the ratio of female to male students at the college is 6 to 5, then determine the number of male students out of 11,000 total students.
SOLUTION: Number of female students is represented by x. The ratio of female to male students is 6 to 5. The total number of students is 11,000. Number of female students are, 11,000 – x. We will set up our ratios with the number of female students to the number of male students.
( )
11,000 65
5 11000 655000 5 655000 6 555000 1155000 11
11 115000
xx
x xx xx xx
x
x
−=
− =
− == +=
=
=
ANSWER: Number of male students are 5000.
42. PROBLEM:
In the year 2009 it was estimated that there will be 838 deaths in the United States for every 100,000 people. If the total US population was estimated to be 307,212,123 people, then how many deaths in the US may we expect in 2009? (Source: CIA World Factbook)
SOLUTION: Number of deaths in the US in 2009 is represented by x. 838 out of 100,000 people are expected to die in the year 2009. We will set up our ratios with the number deaths in the numerator to the total population.
838100,000 307, 212,123838 307, 212,123
100,0002,574, 437.59
2,574, 438
x
x
xx
=
⋅=
==
ANSWER: In the year 2009 out of 307,212,123 people, 2,574,438 deaths are expected in the US.
43. PROBLEM: In the year 2009 it was estimated that there will be 1,382 births in the United States for every 100,000 people. If the total US population was estimated to be 307,212,123 people, then how many births in the US may we expect in 2009?
SOLUTION: Number of births in the US in 2009 is represented by x. 1,382 births out of 100,000 people are expected in the year 2009. We will set up our ratios with the number of births in the numerator to the total population.
1,382307, 212,123 100,000
1,382 307, 212,123100,000
4245671.544, 245,672
x
x
xx
=
⋅=
==
ANSWER: In the year 2009 out of 307,212,123 people, 4,245,672 births are expected in the US.
44. PROBLEM:
If 2 out of every 7 voters approve of a sales tax increase then determine the number of voters out of the 588 surveyed who do not support the increase.
SOLUTION: Number of voters who approved the sales tax increase is represented by x. The ratio of voters who approved the sales tax increase to the total number of voters is 2 to 7. The total number of students is 588. Number of voters who did not approve an increase in the sales tax is 588 – x. We will set up our ratios with the number of voters who approved an increase in the sales tax to the total number of voters.
2588 7
2 5887
168588 420
x
x
xx
=
⋅=
=− =
ANSWER: Out of 588, 420 voters approved the increase in the sales tax.
45. PROBLEM: A recipe calls for 1 cup of lemon juice to make 4 cups of lemonade. How much lemon juice is needed to make 2 gallons of lemonade?(Hint:1 Gallon = 16cups)
SOLUTION: Amount of lemon juice needed to prepare lemonade is represented by x. 1 cup of lemon juice is required to make 4 cups of lemonade. We will set up our ratios with the amounts of lemon juice required to the total amount of lemonade.
1 2 4
1 24
0.5
x cup of lemon juicegallons cups of lemonade
x
x
=
⋅=
=
0.5 or 12
gallon of lemon juice is needed to make 2 gallons of lemonade.
Using the conversion,1 gallon = 16 cups Therefore ½ gallon = x cups 1 1
16 21 1
16 2162
8
x
x
x
x
=
=
=
=
ANSWER: x = 8 cups of lemonade
46. PROBLEM: The classic “Shirley Temple” cocktail requires 1 part cherry syrup to 4 parts lemon-lime soda. How much cherry syrup then is needed to mix the cocktail given a 12 oz can of lemon-lime soda?
SOLUTION: Amount of cherry syrup needed to mix the cocktail is represented by x. 1 part of cheery syrup is mixed with 4 parts of lemon-lime soda. We will set up our ratios with the amount of cherry syrup required to the amount of lemon-lime soda.
1 12 4 -
1 124
3
x part cherry syrupoz parts lemon lime soda
x
x
=
⋅=
=
ANSWER: 3 oz of cherry syrup is required to mix the cocktail given a 12 oz can of lemon-lime soda.
47. PROBLEM: A printer can print 30 pages in one minute. How long will it take to print a 720 pages booklet?
SOLUTION: Time taken by the printer to print a 720 page booklet is represented by x. The printer can print 30 pages in one minute. We will set up our ratios with the time taken by the printer to the total number of pages.
1 720 30
720 1 2430
x minutepages pages
x
=
⋅= =
ANSWER: The printer can print a 720 pages booklet in 24 minutes.
48. PROBLEM:
A typist can type 75 words per minute. How long will it take to type 72 pages if there are approximately 300 words per page?
SOLUTION: Number of words per page is 300. Total number of pages to be typed is 72 pages. Total number of words = 300 ⋅ 72 = 21,600 Time taken by the typist to type 21,600 is represented by x. The typist can type 75 words in one minute. The total number of words to We will set up our ratios with the time taken by the typist to the total number of words.
1 21,600 75
1 21,600 28875
x minutewords words
x
=
⋅= =
ANSWER: To type 72 pages the typist takes 288 minutes.
49. PROBLEM: On a map every 1
16 inch represents 1 mile. How many miles does 123 inches
represent?
SOLUTION: The miles represented by 1
23 inches on the map is represented by x. Every 1
16 inch on the map represents 1 mile. We will set up our ratios with the miles to the inches on the graph.
17 12 16
71 72 16 561 216
x
x
=
⋅= = ⋅ =
ANSWER: 1
23 inches on the map represent 56 miles. 50. PROBLEM:
On a graph every 1 centimeter represents 100 feet. What measurement on the map will represent one mile?
SOLUTION: 1 mile is made up of 5,280 feet. The measurement on the map which represents 5,280 feet mile is represented by x. Every 1 centimeter on the map represents 100 feet. We will set up our ratios with the feet to the centimeters on the graph.
1 5,280 100
1 5,280100
52.8
x centimeterfeet feet
x
x
=
⋅=
=
ANSWER: 52.8 centimeter on the graph represents 1 mile.
51. PROBLEM: A candy store offers mixed candy at $3.75 for every half-pound. How much will 2.6 pounds of candy cost?
SOLUTION: Cost of 2.6 pounds of candy is represented by x. ½ pound of candy costs $3.75. We will set up our ratios with the cost of the candy to the weight of the candy.
$3.7512.6 2
2 3.75 2.6 $19.5
xpounds pound
x
=
= ⋅ ⋅ =
ANSWER: 2.6 pounds of candy costs $19.50.
52. PROBLEM:
Mixed nuts are priced at $6.45 per pound. How many pounds of mixed nuts can be purchased with $20.00?
SOLUTION: Weight of mixed nuts that can be purchased with $20.00 is represented by x. One pound of mixed nuts cost $6.45. We will set up our ratios with the weight of mixed nuts to the price of the mixed nuts.
1 $20.00 $6.45
1 20 3.106.45
x pound
x
=
⋅= =
ANSWER: 3.10 pounds can be purchased with $20.00.
53. PROBLEM:
Corn at the farmers market is bundled and priced at $1.33 for 6 ears. How many ears can be purchased with $15.00?
SOLUTION: Ears of corn purchased with $15.00 are represented by x. 6 ears of corn costs $1.33. We will set up our ratios with the number of ears of corn to the cost.
6 $15.00 $1.33
6 151.3367.6 ~ 68
x ears of corn
x
x
=
⋅=
=
ANSWER: 68 ears of corn can be purchased with $15.00.
54. PROBLEM: If 4 pizzas cost $21.00, then how much will 16 pizzas cost?
SOLUTION: Cost of 16 pizzas is represented by x. 4 pizzas cost $21.00. We will set up our ratios with the number of pizzas and the cost.
16 4 $21.00
16 214
$84
pizzas pizzasx
x
x
=
⋅=
=
ANSWER: 16 pizzas will cost $84.
55. PROBLEM:
Honey Nut Cheerios contains 110 calories in one 34 cup serving. How many
calories are in a 781 cup serving?
SOLUTION: Calories in a 7
81 cup serving are represented by x. Honey Nut Cheerios contains 110 calories in one 3
4 cup serving. We can set up our ratios with the calories to the cup servings.
110 calories31 4
110 15 4 2753 8
58
x
cups of serving cup of serving
x
=
⋅ ⋅= =
⋅
ANSWER: A 7
81 cup of serving contains 275 calories.
56. PROBLEM: Chicken flavored rice contains 300 calories in each 2.5 ounce serving. How many calories are in a 4 ounce scoop of chicken flavored rice?
SOLUTION: Calories in a 4 ounce scoop of chicken flavored rice are represented by x. Chicken flavored rice contains 300 calories in each 2.5 ounce serving. We will set up our ratios with the calories to the weight of the serving.
300 4 2.5
300 4 4802.5
x caloriesounce serving ounce serving
x
=
⋅= =
ANSWER: 4 ounce scoop of chicken flavored rice contains 480 calories.
57. PROBLEM:
A 200 lb man would weigh about 33.2 lbs on the moon. How much will a 150 lb man weigh on the moon?
SOLUTION: The weight of a 150 lb man on moon is represented by x. 200 lb man weighs about 33.2 lbs on the moon. We will set up our ratios as weight of the man on moon to the actual weight of the man.
33.2 150 200
33.2 150 24.9200
x lbs on moonlb lb
x
=
⋅= =
ANSWER: A 150 lb man weighs 24.9 lbs on the moon.
58. PROBLEM:
A 200 lb man would weigh about 75.4 lbs on Mars. How much will a 150 lb man weigh on Mars?
SOLUTION: The weight of a 150 lb man on Mars is represented by x. 200 lb man weighs about 75.4 lbs on Mars. We will set up our ratios as weight of the man on Mars to the actual weight of the man.
75.4 150 200
75.4 150 56.55 ~ 56.6200
x lbs on Marslb lb
x
=
⋅= =
ANSWER: A 150 lb man weighs 56.6 lbs on Mars.
59. PROBLEM:
There is a one out of six chance of rolling a one on a six sided dice. How many times can we expect a one to come up in 360 rolls of the dice?
SOLUTION: Number of times we can expect a one to come up in 360 rolls of the dice. There is a one out of six chance of rolling a one on a six sided die. We will set up our ratios as times we can expect a one to come up to the number of rolls of the dice.
1360 6
360 1 606
xrolls rolls
x
=
⋅= =
ANSWER: We can expect a one to come up 60 times in 360 rolls of the dice.
60. PROBLEM:
There is a one out of six chance of rolling a seven with two six sided dice. How many times can we expect a seven to come up in 300 rolls?
SOLUTION: Number of times we can expect a seven to come up in 300 rolls of the two six sided dice is represented by x. There is a one out of six chance of rolling a seven with two six sided dice. We will set up our ratios as times we can expect a seven to come up to the number of rolls of the two dices.
1300 6
300 1 506
xrolls rolls
x
=
⋅= =
ANSWER: We can expect a seven to come up 50 times in 300 rolls of the two dice.
61. PROBLEM: The ratio of peanuts to all nuts in a certain brand of packaged mixed nuts is 3 to 5. If the package contains 475 nuts, then how many peanuts can we expect?
SOLUTION: Number of peanuts in the package is represented by x. The ratio of peanuts to all nuts is 3 to 5. We will set up our ratios as the number of peanuts to the number of nuts.
3 475 5
475 3 2855
x peanutsnuts nuts
x
=
⋅= =
ANSWER: If the package contains 475 nuts, we can expect 285 peanuts.
62. PROBLEM:
A mixed bag of marbles is packaged with a ratio of 6 orange marbles for every 5 red marbles. If the package contains 216 orange marbles, then how many red marbles can we expect?
SOLUTION: Number of red marbles in the bag is represented by x. The bag contains a ratio of 6 orange marbles for every 5 red marbles. We will set up our ratios as the number of orange marbles to the red marbles.
216 6 5
216 5 1806
orange marbles orange marblesx red marbles
x
=
⋅= =
ANSWER: If the bag contains 216 orange marbles and 180 red marbles.
63. PROBLEM: A graphics designer wishes to create a 720 pixel wide screen capture. If the width to height ratio is to be 3:2, then how many pixels should he set the height?
SOLUTION: Number of pixels to be set as height is represented by x. The width to height ratio is 3:2. We can set up our ratios as the number of pixels to be set as width to the number of pixels to be set as the height.
720 32
720 23
480
pixelsx
x
x
=
⋅=
=
ANSWER: If the width to height ratio is to be 3:2, the height should be set as 480 pixels.
64. PROBLEM:
If a video monitor is produced in the width to height ratio of 16:9 and the width of the monitor is 40 inches, then what will be the height?
SOLUTION: Height of the monitor in inches is represented by x. The width to height ratio is 16:9. We can set up our ratios as the width of the monitor in inches to the height of the monitor in inches.
ANSWER: If the width to height ratio is to be 16:9, the height of the monitor is 22.5 inches.
40 169
40 916
22.5
inchesx
x
x
=
⋅=
=
Part D: Similar Triangles If triangle ABC is similar to triangle RST, find the remaining two sides given the information. 65. PROBLEM:
6a = , 8b = , 10c = , and 16s =
SOLUTION:
6 8 1016
6 816
6 168
12
8 1016
10 168
20
a b cr s t
r t
r
r
r
t
t
t
= =
= =
=
⋅=
=
=
⋅=
=
ANSWER: 12 r= , 20t =
66. PROBLEM: 36b = , 48c = , 20r = , and 32t =
SOLUTION:
36 4820 32
4820 32
20 4832
30
36 4832
36 3248
24
a b cr s ta
sa
a
a
s
s
s
= =
= =
=
⋅=
=
=
⋅=
=
ANSWER: 24s =
67. PROBLEM: 2b = , 4c = , 6r = , and 4s =
SOLUTION:
2 46 4
26 4
6 24
3
2 44
4 42
8
a b cr s ta
ta
a
a
t
t
t
= =
= =
=
⋅=
=
=
⋅=
=
ANSWER: 3a = , 8t =
68. PROBLEM: 3b = , 2c = , 10r = , and 12t =
SOLUTION:
3 210 12
210 12
2 1012
53
3 212
3 122
18
a b cr s ta
sa
a
a
s
s
s
= =
= =
=
⋅=
=
=
⋅=
=
ANSWER: 18s =
69. PROBLEM: 40a = , 50c = , 3s = , and 10t =
SOLUTION:
40 503 10
40 501
40 1050
8
50 310
15
0
503 10
a b cr s t
br
r
r
r
b
b
b
= =
= =
=
⋅=
=
=
⋅=
=
ANSWER: 8 r= , 15b =
70. PROBLEM: 2c = , 7r = , 9s = , and 4t =
SOLUTION:
27 9 4
27 4
2 74
72
29 4
2 94
92
a b cr s ta b
a
a
a
b
b
b
= =
= =
=
⋅=
=
=
⋅=
=
ANSWER: 92
b =
71. PROBLEM:
At the same time of day a tree casts a 12 ft shadow while a 6 ft man cast a 3 ft shadow. Estimate the height of the tree.
SOLUTION:
Height of the man Height of the man's shadowHeight of the tree Height of the tree's shadow6 3
126 12
324
ft ftx ft
x
x
=
=
⋅=
=
ANSWER: The height of the tree is 24 ft.
72. PROBLEM: At the same time of day a father and son, standing side by side, cast a 4 ft and 2 ft shadow respectively. If the father is 6 ft tall, then how tall is his son?
SOLUTION:
Height of the father Height of the father's shadowHeight of the son Height of the son's shadow
6 42
6 24
3
ft ftx ft
x
x
=
=
⋅=
=
ANSWER: The height of the son is 3 ft.
73. PROBLEM:
If the 6-8-10 right triangle ABC is similar to RST with a scale factor of 2/3, then find the perimeter of triangle RST.
SOLUTION: Perimeter of 6-8-10 right triangle ABC is, 6 + 8 + 10 = 24 units
Perimeter of triangle ABC 2Perimeter of triangle RST 324 2
324 3
236
x
x
x
=
=
⋅=
=
ANSWER: Perimeter of triangle RST is 36 units.
74. PROBLEM: If the 3-4-5 right triangle ABC is similar to RST with a scale factor of 5, then find the perimeter of triangle RST.
SOLUTION: Perimeter of 3-4-5 right triangle ABC is, 3 + 4 + 5 = 12 units
Perimeter of triangle ABC 5Perimeter of triangle RST12 5
125
x
x
=
=
=
ANSWER: Perimeter of triangle RST is125
units.
75. PROBLEM:
An equilateral triangle with sides measuring 6 units is similar to another with scale factor 3:1. Find the length of each side of the unknown triangle.
SOLUTION:
Length of each side of a known triangle 3Length of each side of unknown triangle 16 3
1632
unitsx
x
x
=
=
=
=
ANSWER: The length of each side of the unknown triangle is 2 units.
76. PROBLEM: The perimeter of an equilateral triangle ABC measures 45 units. If triangle ~ ABC RST and 20r = , then what is the scale factor?
SOLUTION: Perimeter of the equilateral triangle ABC is 45 units. Perimeter of the equilateral triangle RST is 20⋅3 = 60 units.
Perimeter of the equilateral triangle Scale factor = Perimeter of the equilateral triangle
45 3Scale factor =60 4
ABCRST
=
ANSWER: 3Scale factor4
=
77. PROBLEM:
The perimeter of an isosceles triangle ABC, where the two equal sides each measure twice that of the base, is 60 units. If the base of a similar triangle measures 6 units find its perimeter.
SOLUTION: Base of the triangle ABC is represented by x. The other two sides measure 2x each. Perimeter of the triangle ABC = x + 2x + 2x 60 units = 5x 60 5 = 5 5
12
x
x=
Base = 12 units The other two sides of ABC measure 2x = 2⋅12 = 24 units each. The base of the similar triangle is represented by y.
Side of Side of the similar triangleBase of triangle Base of the similar triangle24 units12 units 6 units24 612
12
ABCABC
y
y
y
=
=
⋅=
=
The other two sides of the similar triangle measure 12 units each.
ANSWER: Perimeter of the similar triangle = 6 + 12 + 12 = 30 units
78. PROBLEM: The perimeter of an isosceles triangle ABC measures 11 units and its two equal sides measure 4 units. If triangle ABC is similar to triangle RST and triangle RST has a perimeter of 22 units, then find all the sides of triangle RST.
SOLUTION: Base of the triangle ABC is represented by x. The other two sides measure 4 units each. Perimeter of the triangle ABC = x + 4 + 4 11 units = x + 8 3 units = x Base = 3 units Base of the similar triangle RST is represented by s.
The base of the similar triangle RST measures 6 units. The other two sides of the similar triangle are represented by r and t. Perimeter of the similar triangle RST = 6 + r + t r = t 22 units = 6 + 2r 22 – 6 = 2r 16 = 2r
ANSWER: The other two sides of the similar triangle RST measures 8 units each.
Perimeter of Perimeter of the similar triangle Base of triangle Base of the similar triangle 11 units 22 units3 units
3 2211
6
ABC RSTABC RST
s
s
s
=
=
⋅=
=
16 2 = 2 2
8
r
r=
79. PROBLEM: A 6-8-10 right triangle ABC is similar to a triangle RST with perimeter 72 units. Find the length of each leg of triangle RST.
SOLUTION: Perimeter of the right triangle ABC = a + b + c = 6 + 8 + 10 = 24 units
Perimeter of the right triangle 24 1Scale factor = Perimeter of the right triangle 72 3
ABCRST
= =
6 8
Scale factor
13
13
6 31
1
1
8
0
6
a b cr s t
r s t
r
r
r
= = =
= = =
=
⋅=
=
13
8 31
24
8s
s
s
=
⋅=
=
13
10 3
3
0
0
1
1
t
t
t
=
⋅=
=
ANSWER: The length of each leg of the triangle RST is r = 18 units, s = 24 units, and t = 30 units.
80. PROBLEM: The perimeter of triangle ABC is 60 units and 20b = units. If ~ ABC RST and
10s = units, then find the perimeter of triangle RST.
SOLUTION: 20Scale factor 210
bs
= = =
Perimeter Scale factor = Perimeter
60 units2 Perimeter
60Perimeter 30 units2
ABCRST
RST
RST
=
= =
ANSWER: Perimeter 30 unitsRST =
Part E: Discussion Board Topics 81. PROBLEM:
What is the golden ratio and where does it appear?
ANSWER: In mathematics and the arts, two quantities are in the golden ratio if the ratio of the sum of the quantities to the larger quantity is equal to the ratio of the larger quantity to the smaller one. The golden ratio is an irrational mathematical constant, approximately 1.6180339887. Other names frequently used for the golden ratio are the golden section and golden mean. a b a
a b+
=
(Source: http://en.wikipedia.org/wiki/Golden_ratio)
82. PROBLEM: Research and discuss the properties of similar triangles?
ANSWER: Similar triangles have the same shape, but not necessarily the same size. The measures of the corresponding angles of similar triangles are equal and the corresponding sides are proportional. Given similar triangles ABC and RST,
we may write ~ ABC RST and conclude that all of the corresponding angles are equal. The notation indicates that angle A corresponds to angle R and that the measures of these angles are equal, A = R.
In addition, we have the measures of angle B = S and C = T.
Typically, we use uppercase letters for angles and the side opposite of the given angle is denoted with the same letter in lower case. Hence, we may denote the proportionality of the sides as follows:
a b cr s t= =
83. PROBLEM: Discuss the mathematics of perspective.
ANSWER: Every time you look at a building from a different location, you see it from a different perspective. In addition to revealing different details of the building, each perspective provides important visual clues about the relative distance to objects in our field of view. These clues arise in the form of apparent distortions. For instance, features that are known to be of equal length, area, or angle appear to be different when seen in perspective. Far from misleading, these distortions are essential to our depth perception and understanding of our surroundings. (Source: http://www.math-ed.com/Resources/cr/sum.htm)
84. PROBLEM:
Research and discuss the various aspect ratios that are available in modern media devices.
ANSWER: The aspect ratio of an image is the ratio of the width of the image to its height, expressed as two numbers separated by a colon. That is, for an x : y aspect ratio, no matter how big or small the image is, if the width is divided into x units of equal length and the height is measured using this same length unit, the height will be measured to be y units. For example, consider a group of images, all with an aspect ratio of 16:9. One image is 16 inches wide and 9 inches high. Another image is 16 centimeters wide and 9 centimeters high. A third is 8 yards wide and 4.5 yards high. The most common aspect ratios used today in the presentation of films in movie theaters are 1.85:1 and 2.39:1. Two common video graphic aspect ratios are 4:3 (1.33:1), universal for standard-definition video formats, and 16:9 (1.78:1), universal for high-definition television and European digital television. Other cinema and video aspect ratios exist, but are used infrequently. In still camera photography, the most common aspect ratios are 4:3, 3:2, and more recently being found in consumer cameras 16:9. With television, DVD and Blu-ray, converting formats of unequal ratios is achieved by either, enlarging the original image (by the same factor in both directions) to fill the receiving format’s display area and cutting off any excess picture information (zooming and cropping), by adding horizontal mattes (letterboxing) or vertical mattes (pillarboxing) to retain the original format’s aspect ratio, or (for TV and DVD) by stretching (hence distorting) the image to fill the receiving format’s ratio, by scaling different factors in both directions, possibly scaling by a different factor in the center and at the edges (as in Wide Zoom mode). (Source: http://en.wikipedia.org/wiki/Aspect_ratio_%28image%29)
2.7IntroductiontoInequalitiesandIntervalNotationPart A: Simple Inequalities Graph all solutions on a number line and provide the corresponding interval notation. 1. PROBLEM:
10x ≤
SOLUTION: Use a closed dot at 10 and shade all real numbers strictly less than 10. Use negative infinity (−∞) to indicate that the solution set is unbounded to the left on a number line.
ANSWER: Interval notation: ( ,10]−∞ 2. PROBLEM:
5x > −
SOLUTION: Use an open dot at –5 and shade all real numbers strictly greater than –5. Use positive infinity (∞) to indicate that the solution set is unbounded to the right on a number line.
ANSWER: Interval notation: ( 5, )− ∞
3. PROBLEM:
0x >
SOLUTION: Use an open dot at 0 and shade all real numbers strictly greater than 0. Use positive infinity (∞) to indicate that the solution set is unbounded to the right on a number line.
ANSWER: Interval notation: (0, )∞ 4. PROBLEM:
0x ≤
SOLUTION: Use a closed dot at 0 and shade all real numbers strictly less than 0. Use negative infinity (–∞) to indicate that the solution set is unbounded to the left on a number line.
ANSWER: Interval notation: ( ,0]−∞
5. PROBLEM: 3x ≤ −
SOLUTION: Use a closed dot at –3 and shade all real numbers strictly less than –3. Use negative infinity (–∞) to indicate that the solution set is unbounded to the left on a number line.
ANSWER: Interval notation: ( , 3]−∞ − 6. PROBLEM:
1x ≥ −
SOLUTION: Use a closed dot at –1 and shade all real numbers strictly greater than –1. Use positive infinity (∞) to indicate that the solution set is unbounded to the right on a number line.
ANSWER: Interval notation:[ 1, )− ∞
7. PROBLEM:
4 x− <
SOLUTION: Use an open dot at –4 and shade all real numbers strictly greater than –4 . Use positive infinity (∞) to indicate that the solution set is unbounded to the right on a number line.
ANSWER: Interval notation: ( 4, )− ∞ 8. PROBLEM:
1 x≥
SOLUTION: Use a closed dot at 1 and shade all real numbers strictly lesser than 1. Use negative infinity (–∞) to indicate that the solution set is unbounded to the left on a number line.
ANSWER: Interval notation: ( ,1]−∞
9. PROBLEM: 12
x < −
SOLUTION: Use an open dot at 12
− and shade all real numbers strictly lesser
than 12
− . Use negative infinity (–∞) to indicate that the solution set is unbounded
to the left on a number line.
ANSWER: Interval notation: ( )12,−∞ −
10. PROBLEM:
32
x ≥ −
SOLUTION: Use a closed dot at 32
− and shade all real numbers strictly greater
than 32
− . Use positive infinity (∞) to indicate that the solution set is unbounded
to the right on a number line.
ANSWER: Interval notation: )3
2 ,− ∞⎡⎣ 11. PROBLEM:
314
x ≥ −
SOLUTION: Use a closed dot at 74
− and shade all real numbers strictly greater
than 74
− . Use positive infinity (∞) to indicate that the solution set is unbounded to
the right on a number line.
ANSWER: Interval notation: )3
41 ,− ∞⎡⎣
12. PROBLEM: 34
x <
SOLUTION: Use an open dot at 34
and shade all real numbers strictly lesser
than 34
. Use negative infinity (–∞) to indicate that the solution set is unbounded to
the left on a number line.
ANSWER: Interval notation: ( )3
4,−∞ Part B: Compound Inequalities Graph all solutions on a number line and give the corresponding interval notation. 13. PROBLEM:
2 5x− < <
SOLUTION: The lower bound, –2 is included in the solution set using an open dot. Shade all real numbers between −2 and 5, and indicate that the upper bound, 5, is included in the solutions set by using an open dot.
ANSWER: Interval Notation: ( )2, 5− 14. PROBLEM:
5 1x− ≤ ≤ −
SOLUTION: The lower bound, –5 is included in the solution set using a closed dot. Shade all real numbers between −5 and −1, and indicate that the upper bound, −1, is included in the solutions set by using a closed dot.
ANSWER: Interval Notation: [ ]5, 1− −
15. PROBLEM: 5 20x− < ≤
SOLUTION: The lower bound, –5 is included in the solution set using an open dot. Shade all real numbers between −5 and 20, and indicate that the upper bound, 20, is included in the solutions set by using a closed dot.
ANSWER: Interval Notation: ( ]5, 20− 16. PROBLEM:
0 15x≤ <
SOLUTION: The lower bound, 0 is included in the solution set using a closed dot. Shade all real numbers between 0 and 15, and indicate that the upper bound, 15, is included in the solutions set by using an open dot.
ANSWER: Interval Notation: [ )0, 15
17. PROBLEM:
10 40x< ≤
SOLUTION: The lower bound, 10 is included in the solution set using an open dot. Shade all real numbers between 10 and 40, and indicate that the upper bound, 40, is included in the solutions set by using a closed dot.
ANSWER: Interval Notation: ( ]10, 40 18. PROBLEM:
40 10x− ≤ < −
SOLUTION: The lower bound, –40 is included in the solution set using a closed dot. Shade all real numbers between −40 and −10, and indicate that the upper bound, −10, is included in the solutions set by using an open dot.
ANSWER: Interval Notation: [ 40, 10)− −
19. PROBLEM: 0 50x< ≤
SOLUTION: The lower bound, 0 is included in the solution set using an open dot. Shade all real numbers between 0 and 50, and indicate that the upper bound, 50, is included in the solutions set by using a closed dot.
ANSWER: Interval Notation: (0, 50] 20. PROBLEM:
30 0x− < <
SOLUTION: The lower bound, –30 is included in the solution set using an open dot. Shade all real numbers between −30 and 0, and indicate that the upper bound, 0, is included in the solutions set by using an open dot.
ANSWER: Interval Notation: ( 30, 0)−
21. PROBLEM:
5 18 8
x− < <
SOLUTION: The lower bound, 58
− is included in the solution set using an open
dot. Shade all real numbers between 58
− and 18
, and indicate that the upper
bound, 18
, is included in the solutions set by using an open dot.
ANSWER: Interval Notation: ( )5 1
8 8,−
22. PROBLEM:
3 14 2
x− ≤ ≤
SOLUTION: The lower bound, 34
− is included in the solution set using a closed
dot. Shade all real numbers between 34
− and 12
, and indicate that the upper
bound, 12
, is included in the solutions set by using a closed dot.
ANSWER: Interval Notation:
3 14 2,−⎡ ⎤⎣ ⎦
23. PROBLEM:
11 12
x− ≤ <
SOLUTION: The lower bound, 1− is included in the solution set using a closed
dot. Shade all real numbers between 1− and 32
, and indicate that the upper
bound, 32
, is included in the solutions set by using an open dot.
ANSWER: Interval Notation: [ )1
21,1− 24. PROBLEM:
1 112 2
x− < < −
SOLUTION: The lower bound, 32
− is included in the solution set using an open
dot. Shade all real numbers between 32
− and 12
− , and indicate that the upper
bound, 12
− , is included in the solutions set by using an open dot.
ANSWER: Interval Notation: ( )1 1
2 21 , − −
25. PROBLEM: 3 3x or x< − >
SOLUTION:
3x < − : Use an open dot at 3− and shade all real numbers strictly lesser than 3− . Use negative infinity (–∞) to indicate that the solution set is unbounded to the left on a number line.
3x > : Use an open dot at 3 and shade all real numbers strictly greater than 3 . Use positive infinity (∞) to indicate that the solution set is unbounded to the right on a number line.
The symbol ∪ is used to denote “or.”
ANSWER: Interval notation: ( ) ( ), 3 3,−∞ − ∪ ∞
26. PROBLEM:
2 4x or x< − ≥
SOLUTION: 2x < − : Use an open dot at 2− and shade all real numbers strictly lesser than 2− .
Use negative infinity (–∞) to indicate that the solution set is unbounded to the left on a number line.
4x ≥ : Use a closed dot at 4 and shade all real numbers strictly greater than 4 . Use positive infinity (∞) to indicate that the solution set is unbounded to the right on a number line.
The symbol ∪ is used to denote “or.”
ANSWER: Interval notation: ( ) [ ), 2 4,−∞ − ∪ ∞ 27. PROBLEM:
0 10x or x≤ >
SOLUTION: 0x ≤ : Use a closed dot at 0 and shade all real numbers strictly lesser than 0 . Use
negative infinity (–∞) to indicate that the solution set is unbounded to the left on a number line.
10x > : Use an open dot at 10 and shade all real numbers strictly greater than 10 . Use positive infinity (∞) to indicate that the solution set is unbounded to the right on a number line.
The symbol ∪ is used to denote “or.”
ANSWER: Interval notation: ( ,0] (10, )−∞ ∪ ∞
28. PROBLEM: 20 10x or x≤ − ≥ −
SOLUTION:
20x ≤ − : Use a closed dot at 20− and shade all real numbers strictly lesser than20− . Use negative infinity (–∞) to indicate that the solution set is unbounded to
the left on a number line. 10x ≥ − : Use a closed dot at 10 and shade all real numbers strictly greater than10
. Use positive infinity (∞) to indicate that the solution set is unbounded to the right on a number line.
The symbol ∪ is used to denote “or.” ANSWER: Interval notation: ( ] [ ), 20 10,−∞ − ∪ − ∞
29. PROBLEM:
2 1 3 3
x or x< − >
SOLUTION:
2 3
x < − : Use an open dot at 2 3
− and shade all real numbers strictly lesser than
2 3
− . Use negative infinity (–∞) to indicate that the solution set is unbounded to
the left on a number line. 1 3
x > : Use an open dot at 13
and shade all real numbers strictly greater than 13
.
Use positive infinity (∞) to indicate that the solution set is unbounded to the right on a number line.
The symbol ∪ is used to denote “or.”
ANSWER: Interval notation: ( ) ( )2 1
3 3, , −∞ − ∪ ∞
30. PROBLEM: 4 13 3
x or x≤ − > −
SOLUTION:
43
x ≤ − : Use a closed dot at 43
− and shade all real numbers strictly lesser than
43
− . Use negative infinity (–∞) to indicate that the solution set is unbounded to
the left on a number line. 13
x > − : Use a closed dot at 13
− and shade all real numbers strictly greater than
13
− . Use positive infinity (∞) to indicate that the solution set is unbounded to the
right on a number line.
The symbol ∪ is used to denote “or.” ANSWER: Interval notation: ( ( )4 1
3 3, ,−∞ − ∪ − ∞⎤⎦ 31. PROBLEM:
5 5x or x> − <
SOLUTION: When we combine both solutions sets and form the union, we can see that all real numbers will solve the original compound inequality.
ANSWER: Interval notation: ( ),= −∞ ∞ 32. PROBLEM:
12 6x or x< > −
SOLUTION: When we combine both solutions sets and form the union, we can see that all real numbers will solve the original compound inequality.
ANSWER: Interval notation: ( ),= −∞ ∞
33. PROBLEM: 3 3x or x< ≥
SOLUTION: When we combine both solutions sets and form the union, we can see that all real numbers will solve the original compound inequality.
ANSWER: Interval notation: ( ),= −∞ ∞
34. PROBLEM:
0 0x or x≤ >
SOLUTION: When we combine both solutions sets and form the union, we can see that all real numbers will solve the original compound inequality.
ANSWER: Interval notation: ( ),= −∞ ∞
35. PROBLEM:
7 2x or x< − <
SOLUTION: Use an open dot at 2 and shade all real numbers strictly lesser than2 . Use negative infinity (–∞) to indicate that the solution set is unbounded to the left on a number line.
ANSWER: Interval notation: ( , 2)−∞ 36. PROBLEM:
3 0x or x≥ − >
SOLUTION: Use a closed dot at 3− and shade all real numbers strictly greater than 3− . Use positive infinity (∞) to indicate that the solution set is unbounded to the right on a number line. The other inequality 0x > is overlapped by the first inequality.
ANSWER: Interval notation: [ 3, )− ∞
37. PROBLEM: 5 0x or x≥ >
SOLUTION: Use an open dot at 0 and shade all real numbers strictly greater than 0 . Use positive infinity (∞) to indicate that the solution set is unbounded to the right on a number line. The other inequality 5x ≥ is overlapped by the second inequality.
ANSWER: Interval notation: (0, )∞
38. PROBLEM:
15 10x or x< ≤
SOLUTION: Use an open dot at 15 and shade all real numbers strictly lesser than15 . Use negative infinity (–∞) to indicate that the solution set is unbounded to the left on a number line. The other inequality 10x ≤ is overlapped by the first inequality.
ANSWER: Interval notation: ( ,15)−∞
39. PROBLEM:
2 3x and x> − <
SOLUTION: Use an open dot at 2− and shade all real numbers strictly greater than 2− but lesser than3 . Use an open dot at 3 and shade all real numbers strictly lesser than3 but greater than 2− .
ANSWER: Interval notation: ( 2, 3)− 40. PROBLEM:
0 5x and x≥ <
SOLUTION: Use a closed dot at 0 and shade all real numbers strictly greater than 0 but lesser than5 . Use an open dot at 5 and shade all real numbers strictly lesser than5 but greater than 0 .
ANSWER: Interval notation: [0, 5)
41. PROBLEM: 5 1x and x≥ − ≤ −
SOLUTION: Use a closed dot at 5− and shade all real numbers strictly greater than 5− but lesser than 1− . Use a closed dot at 1− and shade all real numbers strictly lesser than 1− but greater than 5− .
ANSWER: Interval notation:[ 5, 1]− − 42. PROBLEM:
4 2x and x< − >
SOLUTION:
ANSWER: Interval notation: ∅
43. PROBLEM:
3 3x and x≤ >
SOLUTION:
ANSWER: Interval notation: ∅
44. PROBLEM:
5 5x and x≤ ≥
SOLUTION: Only an open dot at 5 is graphed on the number line because the numbers cannot be greater than 5 or lesser than 5 only equal to 5.
ANSWER: Interval notation:{ }5
45. PROBLEM:
0 0x and x≤ ≥
SOLUTION: Only an open dot at 0 is graphed on the number line because the numbers cannot be greater than 0 or lesser than 0 only equal to 0.
ANSWER: Interval notation:{ }0
46. PROBLEM: 2 1x and x< ≤ −
SOLUTION: Use a closed dot at 1− and shade all real numbers strictly lesser than 1− . Use negative infinity (–∞) to indicate that the solution set is unbounded to the left on a number line. Here 2x < is not a solution because it solves only one of the inequalities.
ANSWER: Interval notation: ( , 1]−∞ −
47. PROBLEM:
0 1x and x> ≥ −
SOLUTION: Use an open dot at 0 and shade all real numbers strictly lesser than0 . Use positive infinity (∞) to indicate that the solution set is unbounded to the right on a number line. Here 1x ≥ − is not a solution because it solves only one of the inequalities.
ANSWER: Interval notation: (0, )∞ 48. PROBLEM:
5 2x and x< <
SOLUTION: Use an open dot at 2 and shade all real numbers strictly lesser than2 . Use negative infinity (–∞) to indicate that the solution set is unbounded to the left on a number line. Here 5x < is not a solution because it solves only one of the inequalities.
ANSWER: Interval notation: ( , 2)−∞
Part C: Interval Notation Determine the inequality given the answers expressed in interval notation. 49. PROBLEM:
( ],7−∞
SOLUTION: The negative infinity (–∞) indicates that the solution is unbounded to the left on a number line. The 7 enclosed within the square bracket indicates that the variable x is lesser than or equal to 7.
ANSWER: Inequality: 7x ≤
50. PROBLEM:
( )4,− ∞
SOLUTION: The positive infinity (∞) indicates that the solution is unbounded to the right on a number line. The –4 enclosed within the bracket indicates that the variable x is greater than –4. ANSWER: Inequality: 4x > −
51. PROBLEM:
[ )12 ,− ∞
SOLUTION: The positive infinity (∞) indicates that the solution is unbounded to
the right on a number line. The 12
− enclosed within the square bracket indicates
that the variable x is greater than or equal to 12
− .
ANSWER: Inequality:
12
x ≥ −
52. PROBLEM:
( ), 3−∞ −
SOLUTION: The negative infinity (–∞) indicates that the solution is unbounded to the left on a number line. The – 3 enclosed within the bracket indicates that the variable x is lesser than – 3. ANSWER: Inequality: 3x < −
53. PROBLEM: ( ]8,1 0−
SOLUTION: The –8 enclosed within the bracket indicates that the variable x is greater than –8. Inequality: 8x > − The 10 enclosed within the square bracket indicates that the variable x is lesser than or equal to 10. Inequality: 10x ≤
ANSWER: On combining the two inequalities: 8 10x− < ≤
54. PROBLEM:
( ]20, 0−
SOLUTION: The –20 enclosed within the bracket indicates that the variable x is greater than –20. Inequality: 20x > − The 0 enclosed within the square bracket indicates that the variable x is lesser than or equal to 0. Inequality: 0x ≤ ANSWER: On combining the two inequalities: 20 0x− < ≤
55. PROBLEM:
( )14, 2− −
SOLUTION: The –14 enclosed within the bracket indicates that the variable x is greater than –14. Inequality: 14x > − The –2 enclosed within the bracket indicates that the variable x is lesser than –2. Inequality: 2x < −
ANSWER: On combining the two inequalities: 14 2x− < < −
56. PROBLEM: 2 4, 3 3⎡ ⎤⎢ ⎥⎣ ⎦
SOLUTION: The 23
enclosed within the square bracket indicates that the
variable x is greater than or equal to 23
.
Inequality: 23
x ≥
The 43
enclosed within the square bracket indicates that the variable x is lesser
than or equal to 43
.
Inequality: 43
x ≤
ANSWER: On combining the two inequalities: 2 43 3
x≤ ≤
57. PROBLEM:
3 1,4 2
⎛ ⎞−⎜ ⎟⎝ ⎠
SOLUTION: The 34
− enclosed within the bracket indicates that the variable x is
greater than 34
− .
Inequality: 34
x > −
The 12
enclosed within the bracket indicates that the variable x is lesser than 12
.
Inequality: 12
x <
ANSWER: On combining the two inequalities:
3 14 2
x− < <
58. PROBLEM: ( ), 8−∞ −
SOLUTION: The negative infinity (–∞) indicates that the solution is unbounded to the left on a number line. The –8 enclosed within the bracket indicates that the variable x is lesser than –8. ANSWER: Inequality: 8x < −
59. PROBLEM:
( )8, ∞
SOLUTION: The positive infinity (∞) indicates that the solution is unbounded to the right on a number line. The 8 enclosed within the square bracket indicates that the variable x is greater 8.
ANSWER: Inequality: 8x >
60. PROBLEM:
( ) [ ), 4 8,−∞ ∪ ∞
SOLUTION: ( ), 4−∞ : The negative infinity (–∞) indicates that the solution is unbounded to the left on a number line. The 4 enclosed within the bracket indicates that the variable x is lesser than 4. Inequality: 4x < ∪ is used to represent “or.” [ )8,∞ : The positive infinity (∞) indicates that the solution is unbounded to the right on a number line. The 8 enclosed within the square bracket indicates that the variable x is greater than or equal to 8. Inequality: 8x ≥ ANSWER: Combining the two inequalities: 4 8x or x< ≥
61. PROBLEM: ( ] [ ), 2 0,−∞ − ∪ ∞
SOLUTION: ( ], 2−∞ − : The negative infinity (–∞) indicates that the solution is unbounded to the left on a number line. The –2 enclosed within the square bracket indicates that the variable x is lesser than or equal to –2. Inequality: 2x ≤ − ∪ is used to represent “or.” [ )0,∞ : The positive infinity (∞) indicates that the solution is unbounded to the right on a number line. The 0 enclosed within the square bracket indicates that the variable x is greater than or equal to 0. Inequality: 0x ≥
ANSWER: Combining the two inequalities: 2 0x or x≤ − ≥
62. PROBLEM:
( ] ( ), 5 5,−∞ − ∪ ∞
SOLUTION: ( ], 5−∞ − : The negative infinity (–∞) indicates that the solution is unbounded to the left on a number line. The –5 enclosed within the square bracket indicates that the variable x is lesser than or equal to –5. Inequality: 5x ≤ − ∪ is used to represent “or.” ( )5,∞ : The positive infinity (∞) indicates that the solution is unbounded to the right on a number line. The 5 enclosed within the bracket indicates that the variable x is greater than 5. Inequality: 5x > ANSWER: Combining the two inequalities: 5 5x or x≤ − >
63. PROBLEM: ( ) ( ),0 2,−∞ ∪ ∞
SOLUTION: ( ),0−∞ : The negative infinity (–∞) indicates that the solution is unbounded to the left on a number line. The 0 enclosed within the bracket indicates that the variable x is lesser than 0. Inequality: 0x < ∪ is used to represent “or.” ( )2,∞ : The positive infinity (∞) indicates that the solution is unbounded to the right on a number line. The 2 enclosed within the bracket indicates that the variable x is greater than 2. Inequality: 2x >
ANSWER: Combining the two inequalities: 0 2x or x< >
64. PROBLEM:
( ) ( ), 15 5,−∞ − ∪ − ∞
SOLUTION: ( ), 15−∞ − : The negative infinity (–∞) indicates that the solution is unbounded to the left on a number line. The –15 enclosed within the bracket indicates that the variable x is lesser than –15. Inequality: 15x < − ∪ is used to represent “or.” ( )5,− ∞ : The positive infinity (∞) indicates that the solution is unbounded to the right on a number line. The –5 enclosed within the bracket indicates that the variable x is greater than –5. Inequality: –5x > ANSWER: Combining the two inequalities: 15 5x or x< − > −
Write an equivalent inequality. 65. PROBLEM:
All real numbers less than 27.
SOLUTION: The variable x is less than 27. ANSWER: Inequality: 27x <
66. PROBLEM: All real numbers less than or equal to zero.
SOLUTION: The variable x is lesser than or equal to zero. ANSWER: Inequality: 0x ≤
67. PROBLEM:
All real numbers greater than 5.
SOLUTION: The variable x is greater than 5. ANSWER: Inequality: 5x >
68. PROBLEM:
All real numbers greater than or equal to −8.
SOLUTION: The variable x is greater than or equal to −8. ANSWER: Inequality: 8x ≥ −
69. PROBLEM:
All real numbers strictly between −6 and 6.
SOLUTION: The variable x is lesser than 6 but greater than –6. ANSWER: Inequality: 6 6x− < <
70. PROBLEM:
All real numbers strictly between −80 and 0.
SOLUTION: The variable x is lesser than 10 but greater than −80. ANSWER: Inequality: 80 0x− < <
Part D: Discussion Board Topics 71. PROBLEM:
Compare interval notation with set-builder notation. Share an example of a set described using both systems.
ANSWER: Set-builder notation is a system for describing sets using familiar mathematical notation. Interval notation is a textual system of expressing solutions to an algebraic inequality. For e.g.: The real numbers greater than or equal to 2.
Since the set is too large to list, set-builder notation allows us to describe it using familiar mathematical notation. We can do this as follows:
{ }| 2x x∈ ≥ The vertical bar ( | ) is read, “such that.” It is assumed that all variables represent real numbers. For this reason, we can omit the “∈ ” and write { }| 2x x ≥ , which is read, “the set of all real numbers x such that x is greater than or equal to 2.”
Interval notation for the above example is as follows. Positive infinity (∞) indicates that the solution is unbounded to the right on a number line. 2 is enclosed within a square bracket which will indicate that the variable x is greater than or equal to 2. Interval notation: [ )2, ∞
72. PROBLEM:
Infinity as an endpoint, explain why we do not use a bracket in interval notation. ANSWER: If brackets are used in the interval notation the inclusiveness of the particular inequality is lost.
73. PROBLEM: Research and discuss the different compound inequalities, in particular, unions and intersections.
ANSWER: A compound inequality is actually two or more inequalities in one statement joined by the word “and” or by the word “or”. Compound inequalities with the logical “or” require that either condition must be satisfied. Therefore, the solution set of this type of compound inequality consists of all the elements of the solution sets of each in equality. When we join these individual solution sets it is called the union, denoted ∪ . For example, the solutions to the compound inequality 3 6x or x< ≥ can be graphed as follows:
Sometimes we will encounter compound inequalities where the separate solution sets overlap. In the case where the compound inequality contains the word “or” we combine all the elements of both sets to create one set containing all the elements of each.
An inequality such as, 1 3x− ≤ < reads, “−1 one is less than or equal to x and x is less than three.” This is actually a compound inequality because it can be decomposed as: 1 3x and x− ≤ < The logical “and” requires that both conditions must be true. Both inequalities will be satisfied by all the elements in the intersection, denoted ∩ , of the solution sets of each.
74. PROBLEM:
Research and discuss the history of infinity. ANSWER: Ancient cultures had various ideas about the nature of infinity. The ancient Indians and Greeks, unable to codify infinity in terms of a formalized mathematical system approached infinity as a philosophical concept. Early Greek In accordance with the traditional view of Aristotle, the Hellenistic Greeks generally preferred to distinguish the potential infinity from the actual infinity; for example, instead of saying that there is infinity of primes; Euclid instead said that there are more prime numbers than contained in any given collection of prime numbers Early Indian The Isha Upanishad of the Yajurveda states that “if you remove a part from infinity or add a part to infinity, still what remains is infinity”. (Source: http://en.wikipedia.org/wiki/Infinity#History)
75. PROBLEM: Research and discuss the contributions of Georg Cantor.
ANSWER: Georg Ferdinand Ludwig Philipp Cantor as a German mathematician, best known as the inventor of set theory, which has become a fundamental theory in mathematics. Cantor established the importance of one-to-one correspondence between sets, defined infinite and well-ordered sets, and proved that the real numbers are “more numerous” than the natural numbers. In fact, Cantor’s theorem implies the existence of an “infinity of infinities.” He defined the cardinal and ordinal numbers and their arithmetic. Cantor’s work is of great philosophical interest, a fact of which he was well aware. (Source: http://en.wikipedia.org/wiki/Georg_Cantor)\
76. PROBLEM:
What is a Venn diagram? Explain and post an example. ANSWER: Venn diagrams or set diagrams are diagrams that show all hypothetically possible logical relations between a finite collection of sets (aggregation of things). A Venn diagram is constructed with a collection of simple closed curves drawn in the plane. Venn diagrams normally comprise overlapping circles. The following example involves two sets, A and B, represented here as colored circles. The orange circle set A, represents all living creatures that are two-legged. The blue circle, set B, represents the living creatures that can fly. Each separate type of creature can be imagined as a point somewhere in the diagram. Living creatures that both can fly and have two legs—for example, parrots—are then in both sets, so they correspond to points in the area where the blue and orange circles overlap. That area contains all such and only such living creatures.
(Source: http://en.wikipedia.org/wiki/File:Venn-diagram-AB.svg)
2.8LinearInequalities(onevariable) Part A: Checking for Solutions Determine whether or not the given number is a solution to the given inequality. 1. PROBLEM:
2 3 6; 1x x− < = −
SOLUTION:
( )2 3 62 1 3 6
2 3 55 6
x − <
− − <
− − < −− <
ANSWER: 1x = − is a solution.
2. PROBLEM:
3 1 0; 2x x− + ≤ = −
SOLUTION:
( )3 1 03 2 1 0
6 1 07 0
x− + ≤
− − + ≤
+ ≤≤
ANSWER: 2x = − is not a solution.
3. PROBLEM:
5 20 0; 3x x− > =
SOLUTION:
( )5 20 05 3 20 015 20 0
5 0
x − >
− >
− >− >
ANSWER: 3x = is not a solution.
4. PROBLEM: 1 3 11 ; 2 4 4
x x+ > − = −
SOLUTION:
1 312 41 1 312 4 4
1 318 4
7 38 4
x + > −
⎛ ⎞− + > −⎜ ⎟⎝ ⎠
− + > −
> −
ANSWER: 14
x = − is a solution.
5. PROBLEM:
5 7 1 9; 0x x− < + < =
SOLUTION:
( )5 7 1 95 7 0 1 95 1 9
x− < + <
− < + <
− < <
ANSWER: 0x = is a solution.
6. PROBLEM:
20 3 5 10; 5x x− ≤ − − ≤ − =
SOLUTION:
( )20 3 5 1020 3 5 5 1020 20 10
x− ≤ − − ≤ −
− ≤ − − ≤ −
− ≤ − ≤ −
ANSWER: 5x = is a solution.
7. PROBLEM: 3 3x or x< − > ; 10x = −
SOLUTION:
3 310 3 10 3
x or xor
< − >− < − − >
ANSWER: 10x = − is a solution.
8. PROBLEM:
10 1; 2
x or x x< ≥ =
SOLUTION:
0 11 10 12 2
x or x
or
< ≥
< ≥
ANSWER: 12
x = is not a solution.
9. PROBLEM:
2 1 3 2 1 5x or x+ < − + ≥ ; 2x =
SOLUTION:
( ) ( )2 1 3 2 1 52 2 1 3 2 2 1 55 3 5 5
x or xor
or
+ < − + ≥
+ < − + ≥
< − ≥
ANSWER: 2x = is a solution.
10. PROBLEM:
4 1 17 3 2 6x or x− < − + ≥ ; 1x =
SOLUTION:
( ) ( )4 1 17 3 2 64 1 1 17 3 1 2 63 17 5 6
x or xor
or
− < − + ≥
− < − + ≥
< − ≥
ANSWER: 1x = is not a solution.
Part B: Solving Linear Inequalities Solve and graph the solution set. In addition, present the solution set in interval notation. 11. PROBLEM:
5 1x + >
SOLUTION: 5 15 5 1 5
4
xxx
+ >+ − > −> −
ANSWER: Interval notation: ( 4, )− ∞ 12. PROBLEM:
3 4x − < −
SOLUTION: 3 43 3 4 3
1
xxx
− < −− + < − +< −
ANSWER: Interval notation: ( , 1)−∞ −
13. PROBLEM:
6 24x ≤
SOLUTION: 6 246 246 6
4
x
x
x
≤
≤
≤
ANSWER: Interval notation: ( , 4]−∞
14. PROBLEM: 4 8x > −
SOLUTION:
4 84 84 4
2
x
x
x
> −
> −
> −
ANSWER: Interval notation: ( 2, )− ∞
15. PROBLEM:
7 14x− ≤
SOLUTION: 7 147 147 7
22
x
x
xx
− ≤
− ≤
− ≤≥ −
ANSWER: Interval notation:[ 2, )− ∞ 16. PROBLEM:
2 5 9x− + >
SOLUTION: 2 5 92 5 5 9 52 42 42 2
22
xxx
x
xx
− + >− + − > −− >
− >
− >< −
ANSWER: Interval notation: ( , 2)−∞ −
17. PROBLEM: 7 3 25x − ≤
SOLUTION:
7 3 257 3 3 25 37 287 287 7
4
xxx
x
x
− ≤− + ≤ +≤
≤
≤
ANSWER: Interval notation: ( , 4]−∞ 18. PROBLEM:
12 7 53x + > −
SOLUTION: 12 7 5312 7 7 53 712 6012 6012 12
5
xxx
x
x
+ > −+ − > − −> −
> −
> −
ANSWER: Interval notation: ( 5, )− ∞
19. PROBLEM:
2 5 7x− + < −
SOLUTION: 2 5 72 5 5 7 52 122 2
66
xx
x
xx
− + < −− + − < − −
− < −
− < −>
ANSWER: Interval notation: (6, )∞
20. PROBLEM: 2 4 4x− + ≤
SOLUTION:
2 4 42 4 4 4 42 0
00
xxx
xx
− + ≤− + − ≤ −− ≤− ≤≥
ANSWER: Interval notation:[0, )∞
21. PROBLEM:
15 10 20x− + >
SOLUTION: 15 10 2015 10 10 20 1015 1015 1015 15
2323
xxx
x
x
x
− + >− + − > −− >
− >
− >
< −
ANSWER: Interval notation: ( )23, −∞ −
22. PROBLEM: 8 1 29x− + ≤
SOLUTION:
8 1 298 1 1 29 18 288 288 8
7272
xxx
x
x
x
− + ≤− + − ≤ −− ≤
− ≤
− ≤
≥ −
ANSWER: Interval notation: )7
2 , − ∞⎡⎣ 23. PROBLEM:
1 3 17
x − <
SOLUTION:
1 3 171 3 3 1 371 47
28
x
x
x
x
− <
− + < +
<
<
ANSWER: Interval notation: ( , 28)−∞
24. PROBLEM: 1 1 22 3 3
x − >
SOLUTION:
1 1 22 3 31 1 1 2 12 3 3 3 31 12
2
x
x
x
x
− >
− + > +
>
>
ANSWER: Interval notation: (2, )∞
25. PROBLEM:
5 1 13 2 3
x + ≤
SOLUTION:
5 1 13 2 35 1 1 1 13 2 2 3 25 13 6
1 36 51
10
x
x
x
x
x
+ ≤
+ − ≤ −
≤ −
⋅≤ −
⋅
≤ −
ANSWER: Interval notation: ( 110, −∞ − ⎤⎦
26. PROBLEM:
3 1 54 2 2
x− − ≥
SOLUTION:
3 1 54 2 23 1 1 5 14 2 2 2 23 34
3 43
44
x
x
x
x
xx
− − ≥
− − + ≥ +
− ≥
⋅− ≥
− ≥≤ −
ANSWER: Interval notation: ( , 4]−∞ −
27. PROBLEM:
1 3 15 4 5
x− + < −
SOLUTION:
1 3 15 4 51 3 3 1 35 4 4 5 41 195 20
19 520
194
194
x
x
x
x
x
x
− + < −
− + − < − −
− < −
⋅− < −
− < −
>
ANSWER: Interval notation: ( )194 , ∞
28. PROBLEM:
2 1 33
x− + < −
SOLUTION:
2 1 332 1 1 3 132 43
4 32
66
x
x
x
x
xx
− + < −
− + − < − −
− < −
⋅− < −
− < −>
ANSWER: Interval notation: (6, )∞
29. PROBLEM:
( )2 3 1 14x− + <
SOLUTION: ( )2 3 1 146 2 146 2 2 14 26 126 126 6
22
xxxx
x
xx
− + <
− + <− + − < −− <
− <
− <> −
ANSWER: Interval notation: ( 2, )− ∞
30. PROBLEM: ( )7 2 1 15x− − + <
SOLUTION:
( )7 2 1 157 14 1 157 15 157 15 15 15 157 0
00
xxxxx
xx
− − + <
− + + <− + <− + − < −− <− <>
ANSWER: Interval notation: (0, )∞
31. PROBLEM:
( )9 3 3 4 12x x− + > −
SOLUTION: ( )9 3 3 4 12
9 9 12 1212 1212 12
x xx x− + > −
− + > −> −> −
∅
ANSWER: Interval notation: ∅
32. PROBLEM:
( )12 4 3 5 2x x− + ≤ −
SOLUTION: ( )12 4 3 5 2
12 12 20 220 2
x xx x− + ≤ −
− − ≤ −− ≤ −
ANSWER: Interval notation:
33. PROBLEM: ( )5 3 2 6 1x− − ≥ −
SOLUTION:
( )5 3 2 6 15 6 18 1
6 23 16 23 23 1 236 246 246 6
44
xx
xxx
x
xx
− − ≥ −
− + ≥ −− + ≥ −− + − ≥ − −− ≥ −
− ≥ −
− ≥ −≤
ANSWER: Interval notation: ( , 4]−∞ 34. PROBLEM:
( )9 10 12 22x x− − <
SOLUTION: ( )9 10 12 22
9 10 12 2212 2212 12 22 121010
x xx xxxx
x
− − <
− + <− + <− + − < −− <> −
ANSWER: Interval notation: ( 10, )− ∞
35. PROBLEM: ( ) ( )2 7 3 3 3x x− − + ≤ −
SOLUTION:
( ) ( )2 7 3 3 32 14 3 9 3
23 323 23 3 232020
x xx xxxx
x
− − + ≤ −
− − − ≤ −− − ≤ −− − + ≤ − +− ≤≥ −
ANSWER: Interval notation:[ 20, )− ∞ 36. PROBLEM:
5 3 3 7x x− > +
SOLUTION: 5 3 3 75 3 3 3 3 72 3 72 3 3 7 32 102 102 2
5
x xx x x xxxx
x
x
− > +− − > − +− >− + > +>
>
>
ANSWER: Interval notation: (5, )∞
37. PROBLEM: ( ) ( )4 3 2 2 3 12x x− ≤ − + +
SOLUTION:
( ) ( )4 3 2 2 3 1212 8 2 6 1212 8 2 612 2 8 2 2 614 8 614 8 8 6 814 1414 1414 14
1
x xx xx xx x x xxxx
x
x
− ≤ − + +
− ≤ − − +− ≤ − ++ − ≤ − + +− ≤− + ≤ +≤
≤
≤
ANSWER: Interval notation: ( ,1]−∞ 38. PROBLEM:
( ) ( )5 3 15 10 4x x x− ≥ − +
SOLUTION: ( ) ( )5 3 15 10 4
5 15 15 10 45 15 5 4
15 4
x x xx x xx x
− ≥ − +
− ≥ − −− ≥ −
− ≥ −
∅
ANSWER: Interval notation: ∅
39. PROBLEM: ( )12 1 2 6 3 5x x+ > − −
SOLUTION:
( )12 1 2 6 3 512 1 12 6 51 11
x xx x+ > − −
+ > − −> −
ANSWER: Interval notation: 40. PROBLEM:
( ) ( )3 2 5 2 3 5 2x x− + > + +
SOLUTION: ( ) ( )3 2 5 2 3 5 2
3 6 5 6 10 23 1 6 123 6 1 6 6 12
3 1 123 1 1 12 13 133 133 3
133133
x xx xx xx x x x
xxx
x
x
x
− + > + +
− + > + +− > +− − > − +
− − >− − + > +− >
− >
− >
< −
ANSWER: Interval notation: 13( , ) 3
−∞ −
41. PROBLEM: ( ) ( )4 3 1 2 2 4 1 3x x x− − + ≤ − −
SOLUTION:
( ) ( )4 3 1 2 2 4 1 312 4 2 8 2 310 4 8 510 8 4 8 8 518 4 518 4 4 5 418 918 918 18
12
12
x x xx x xx xx x x xxxx
x
x
x
− − + ≤ − −
− + + ≤ − −− + ≤ −− − + ≤ − −− + ≤ −− + − ≤ − −− ≤ −
− ≤ −
− ≤ −
≥
ANSWER: Interval notation: [ )12 , ∞
42. PROBLEM: 2( 2) 14 7(2 1)x x x− − + < +
SOLUTION:
2( 2) 14 7(2 1)2 4 14 14 7
12 4 14 712 14 4 14 14 7
2 4 72 4 4 7 42 32 32 2
3232
x x xx x xx xx x x xxxx
x
x
x
− − + < +− + + < +
+ < +− + < − +
− + <− + − < −− <
− <
− <
> −
ANSWER: Interval notation: ( )3
2 ,− ∞ Set up an algebraic inequality and then solve it. 43. PROBLEM:
The sum of three times a number and 4 is greater than negative 8.
SOLUTION: The unknown number is represented by n. 3 4 83 4 4 8 43 123 123 3
4
nnn
n
n
+ > −+ − > − −> −
> −
> −
ANSWER: 4n > −
44. PROBLEM: The sum of 7 and three times a number is less than or equal to 1.
SOLUTION: The unknown number is represented by n.
3 7 13 7 7 1 73 63 63 3
2
nnn
n
n
+ ≤+ − ≤ −≤ −
≤ −
≤ −
ANSWER: 2n ≤ −
45. PROBLEM:
When a number is subtracted from 10 the result is at most 12.
SOLUTION: The unknown number is represented by n. 10 1210 1210 10 12 10
22
nn
nn
n
− ≤− ≤− − ≤ −
− ≤≥ −
ANSWER: 2n ≥ −
46. PROBLEM:
When 5 times a number is subtracted from 6 the result is at least 26.
SOLUTION: The unknown number is represented by n. 6 5 266 6 5 26 6
5 205 205 5
44
nn
n
n
nn
− ≥− − ≥ −
− ≥
− ≥
− ≥≤ −
ANSWER: 4n ≤ −
47. PROBLEM: If five is added to three times a number, then the result is less than twenty.
SOLUTION: The unknown number is represented by n.
5 3 203 5 5 20 53 153 153 3
5
nnn
n
n
+ <+ − < −<
<
<
ANSWER: 5n <
48. PROBLEM:
If three is subtracted from two times a number, then the result is greater than or equal to nine.
SOLUTION: The unknown number is represented by n.
2 3 92 3 3 9 32 122 122 2
6
nnn
n
n
− ≥− + ≥ +≥
≥
≥
ANSWER: 6n ≥
49. PROBLEM:
Bill earns $12.00 plus $0.25 for every person he gets to register to vote. How many people must he register to earn at least $50.00 for the day?
SOLUTION: Number of people registered by Bill is represented by x.
$12.00 + $0.25x ≥ $50.00 12 0.25 5012 12 0.25 50 120.25 380.25 380.25 0.25
152
xx
x
x
x
+ ≥− + ≥ −
≥
≥
≥
ANSWER: Bill must register at least 152 people.
50. PROBLEM: With a golf club membership, costing $100 per month, each round of golf only cost $25.00. How many rounds of golf can a member play if he wishes to keep his costs at most $250 per month?
SOLUTION: Number of rounds of golf members can play is represented by x. $100 + $25.00x ≤ $250
100 25 250100 100 25 250 10025 15025 15025 25
6
xx
x
x
x
+ ≤− + ≤ −≤
≤
≤
ANSWER: Members may play 6 rounds or less.
51. PROBLEM:
Joe earned a 72, 85, and 75 on his first three algebra exams. What must he score on the fourth exam to average at least 80?
SOLUTION: The score Joe needs to earn in the fourth exam is represented by x.
72 85 75 804
232 320232 232 320 232
88
x
xx
x
+ + +≥
+ ≥− + ≥ −
≥
ANSWER: Joe must earn at least an 88 on the fourth exam.
52. PROBLEM:
Maurice earned 4, 7, and 9 points out of ten on the first three quizzes. What must he score on the fourth quiz to average at least 7?
SOLUTION: Points Maurice needs to score in the fourth quiz is represented by x.
4 7 9 74
20 2820 20 28 20
8
x
xx
x
+ + +≥
+ ≥− + ≥ −≥
ANSWER: Maurice must earn at least 8 points on the fourth quiz.
53. PROBLEM: A computer is set to shut down if the temperature exceeds 40°C. Give an equivalent statement using degrees Fahrenheit. (Hint: 5
9 ( 32)C F= − )
SOLUTION:
ANSWER: The computer will shut down when it exceeds 104°F. 54. PROBLEM:
A certain brand of makeup is guaranteed not to run if the temperature is less than 35°C. Give an equivalent statement using degrees Fahrenheit.
SOLUTION:
59
0
0
( 32)535 ( 32)9
315 5( 32)315 5 160315 160 5 160 160475 5475 55 5
95
C F
C F
C FF
FF
F
F
≥ −
≥ −
≥ −≥ −+ ≥ − +≥
≥
≥
ANSWER: The product is guaranteed not run if the temperature is less than 95°F.
59
0
( 32)540 ( 32)9
360 5 160360 160 5 160 160520 5520 5
5 5104
C F
C F
FF
F
F
F
≥ −
≥ −
≥ −+ ≥ − +≥
≥
≥
Part C: Compound Inequalities Solve and graph the solution set. In addition, present the solution set in interval notation. 55. PROBLEM:
1 3 5x− < + <
SOLUTION: 1 3 51 3 3 3 5 34 2
xx
x
− < + <− − < + − < −− < <
ANSWER: Interval notation: ( 4, 2)− 56. PROBLEM:
10 5 20x− ≤ <
SOLUTION: 10 5 2010 5 205 5 5
2 4
x
x
x
− ≤ <
− ≤ <
− ≤ <
ANSWER: Interval notation:[ 2, 4)−
57. PROBLEM:
2 4 6 10x− ≤ + <
SOLUTION: 2 4 6 102 6 4 6 6 10 68 4 48 4 44 4 42 1
xx
x
x
x
− ≤ + <− − ≤ + − < −− ≤ <
− ≤ <
− ≤ <
ANSWER: Interval notation:[ )2,1 −
58. PROBLEM: 10 3 1 4x− ≤ − ≤ −
SOLUTION:
10 3 1 410 1 3 1 1 4 19 3 33 3 33 1
xx
x
x
− ≤ − ≤ −− + ≤ − + ≤ − +
− ≤ ≤ −
− ≤ ≤ −
ANSWER: Interval notation:[ 3, 1]− −
59. PROBLEM:
15 3 6 6x− < − ≤
SOLUTION: 15 3 6 615 6 3 6 6 6 69 3 129 3 123 3 33 4
xx
x
x
x
− < − ≤− + < − + ≤ +− < ≤
− < ≤
− < ≤
ANSWER: Interval notation: ( 3, 4]− 60. PROBLEM:
22 5 3 3x− < + ≤
SOLUTION: 22 5 3 322 3 5 3 3 3 325 5 025 5 05 5
5 0
xx
x
x
x
− < + ≤− − < + − ≤ −− < ≤
− < ≤
− < ≤
ANSWER: Interval notation: ( 5, 0]−
61. PROBLEM: 11 5 12
x− ≤ − ≤
SOLUTION:
11 5 12
11 5 5 5 1 52
14 62
12 4 2 2 62
8 12
x
x
x
x
x
− ≤ − ≤
− + ≤ − + ≤ +
≤ ≤
⋅ ≤ ⋅ ≤ ⋅
≤ ≤
ANSWER: Interval notation:[8,1 2] 62. PROBLEM:
1 8 5 5x< + <
SOLUTION: 1 8 5 51 5 8 5 5 5 5
4 8 04 8 08 81 02
xx
x
x
x
< + <− < + − < −− < <
− < <
− < <
ANSWER: Interval notation: ( )1
2 , 0−
63. PROBLEM: 1 2 1 45 3 5 5
x− ≤ − <
SOLUTION:
1 2 1 45 3 5 51 1 2 1 1 4 15 5 3 5 5 5 5
20 133 2 30 12 3 2
302
x
x
x
x
x
− ≤ − <
− + ≤ − + < +
≤ <
≤ ⋅ < ⋅
≤ <
ANSWER: Interval notation: )320,⎡⎣
64. PROBLEM:
1 3 2 12 4 3 2
x− < − ≤
SOLUTION:
1 3 2 12 4 3 21 2 3 2 2 1 22 3 4 3 3 2 3
1 3 76 4 64 1 4 3 4 73 6 3 4 3 62 149 9
x
x
x
x
x
− < − ≤
− + < − + ≤ +
< ≤
⋅ < ⋅ ≤ ⋅
< ≤
ANSWER: Interval notation: ( 2 14
9 9, ⎤⎦
65. PROBLEM: 3 3( 1) 3x− ≤ − ≤
SOLUTION:
3 3( 1) 33 3 3 33 3 3 3 3 3 3
0 3 63 603 3
0 2
xx
xx
x
x
− ≤ − ≤− ≤ − ≤− + ≤ − + ≤ +≤ ≤
≤ ≤
≤ ≤
ANSWER: Interval notation:[0, 2] 66. PROBLEM:
12 6( 3) 0x− < − ≤
SOLUTION: 12 6( 3) 012 6 18 012 18 6 18 18 0 18
6 6 186 6 186 6 61 3
xx
xx
x
x
− < − ≤− < − ≤− + < − + ≤ +< ≤
< ≤
< ≤
ANSWER: Interval notation: (1, 3]
67. PROBLEM: ( )4 2 3 6x< − + <
SOLUTION:
( )4 2 3 64 2 6 64 6 2 6 6 6 610 2 1210 2 122 2 2
5 65 66 5
xx
xx
x
xxx
< − + <
< − − <+ < − − + < +< − <
< − <
< − <− > > −− < < −
ANSWER: Interval notaion: ( 6, 5)− − 68. PROBLEM:
( )5 5 1 15x− ≤ − + <
SOLUTION: ( )5 5 1 15
5 5 5 155 5 5 5 5 15 510 5 1010 5 105 5 5
2 22 2
2 2
xx
xx
x
xx
x
− ≤ − + <
− ≤ − + <− − ≤ − + − < −− ≤ − <
− ≤ − <
− ≤ − <≥ > −
− < ≤
ANSWER: Interval notation: ( 2, 2]−
69.
PROBLEM:
3 1 1 3 312 4 2 4 2
x⎛ ⎞− ≤ − + <⎜ ⎟⎝ ⎠
SOLUTION:
3 1 1 3 312 4 2 4 23 1 1 3 32 8 4 4 23 1 1 32 8 2 23 1 1 1 1 3 12 2 8 2 2 2 2
12 18
18 2 8 8 18
16 8
x
x
x
x
x
x
x
⎛ ⎞− ≤ − + <⎜ ⎟⎝ ⎠
− ≤ − + <
− ≤ + <
− − ≤ + − < −
− ≤ <
− ⋅ ≤ ⋅ < ⋅
− ≤ <
ANSWER: Interval notation:[ 16, 8)− 70. PROBLEM:
( )14 3 12 43
x− ≤ − + <
SOLUTION:
( )14 3 12 43
4 4 44 4 4 4 4 4
0 80 8
8 0
x
xx
xx
x
− ≤ − + <
− ≤ − − <− + ≤ − − + < +≤ − <≥ > −
− < ≤
ANSWER: Interval notation: ( 8, 0]−
71. PROBLEM: 2 12 2( 3) 20x− ≤ − − ≤
SOLUTION:
2 12 2( 3) 202 12 2 6 202 2 18 202 18 2 18 18 20 1820 2 220 2 22 2 2
10 110 1
1 10
xx
xx
x
x
xxx
− ≤ − − ≤− ≤ − + ≤− ≤ − + ≤− − ≤ − + − ≤ −− ≤ − ≤
− ≤ − ≤
− ≤ − ≤≥ ≥ −
− ≤ ≤
ANSWER: Interval notation:[ 1,1 0]− 72. PROBLEM:
( ) ( )5 2 1 3 2 5x x− < − − + <
SOLUTION: ( ) ( )5 2 1 3 2 5
5 2 2 3 6 55 8 55 8 8 8 5 8
3 133 1313 3
x xx xx
xxxx
− < − − + <
− < − − − <− < − − <− + < − − + < +< − <
− > > −− < < −
ANSWER: Interval notation: ( 13, 3)− −
73. PROBLEM: 3 15 2 6x or x≤ − >
SOLUTION:
3 15 2 63 15 2 6 3 3 2 2
5 3
x or x
x or x
x or x
≤ − >
≤ − >
≤ − >
ANSWER: Interval notation: ( ], 5 (3, )−∞ − ∪ ∞ 74. PROBLEM:
4 1 17 3 2 8x or x− < − + ≥
SOLUTION: 4 1 17 3 2 84 1 1 17 1 3 2 2 8 24 16 3 64 16 3 6 4 4 3 3
4 2
x or xx or xx or x
x or x
x or x
− < − + ≥− + < − + + − ≥ −< − ≥
< − ≥
< − ≥
ANSWER: Interval notation: ( , 4) [2, )−∞ − ∪ ∞
75. PROBLEM:
2 1 1 2 1 1x or x− + < − − + >
SOLUTION: 2 1 1 2 1 12 1 1 1 1 2 1 1 1 12 2 2 02 2 02 2
1 01 0
x or xx or xx or x
x or x
x or xx or x
− + < − − + >− + − < − − − + − > −− < − − >
− < − − >
− < − − >> <
ANSWER: Interval notation: ( ),0 (1, )−∞ ∪ ∞
76. PROBLEM: 7 4 4 6 5 1x or x+ ≤ − ≥
SOLUTION:
7 4 4 6 5 17 4 4 4 4 6 5 5 1 57 0 6 6
6 60 6 6
0 1
x or xx or xx or x
x or x
x or x
+ ≤ − ≥+ − ≤ − − + ≥ +≤ ≥
≤ ≥
≤ ≥
ANSWER: Interval notation: ( ,0] [1, )−∞ ∪ ∞
77. PROBLEM:
3 7 14 2 3 7x or x− < + >
SOLUTION: 3 7 14 2 3 73 7 7 14 7 2 3 3 7 33 21 2 43 21 2 4 3 3 2 2
7 2
x or xx or xx or x
x or x
x or x
− < + >− + < + + − > −< >
< >
< >
ANSWER: Interval notation:
78. PROBLEM: 3 1 5 4 3 23x or x− + < − − − > −
SOLUTION:
3 1 5 4 3 233 1 1 5 1 4 3 3 23 33 6 4 203 6 4 20 3 3 4 4
2 52 5
x or xx or xx or x
x or x
x or xx or x
− + < − − − > −− + − < − − − − + > − +− < − − > −
− < − − > −
− < − − > −> <
ANSWER: Interval notation:
79. PROBLEM:
1 12 1 2 12 2
x or x− < − − >
SOLUTION:
1 12 1 2 12 21 12 2 1 2 2 2 1 22 21 11 32 2
2 6
x or x
x or x
x or x
x or x
− < − − >
− + < − + − + > +
< >
< >
ANSWER: Interval notation: ( ), 2 (6, )−∞ ∪ ∞
80. PROBLEM:
1 13 2 3 23 3
x or x+ ≥ − + ≤
SOLUTION:
1 13 2 3 23 31 13 3 2 3 3 3 2 33 31 15 13 3
15 3
x or x
x or x
x or x
x or x
+ ≥ − + ≤
+ − ≥ − − + − ≤ −
≥ − ≤ −
≥ − ≤ −
ANSWER: Interval notation:
81. PROBLEM:
3 7 7 5 6 6x or x+ ≤ − + >
SOLUTION: 3 7 7 5 6 63 7 7 7 7 5 6 6 6 63 0 5 0
0 00 00
x or xx or xx or x
x or xx or xx
+ ≤ − + >+ − ≤ − − + − > −≤ − >≤ − >≤ <≤
ANSWER: ( ,0]−∞
82. PROBLEM: 10 3 17 20 6 26x or x− − ≤ − > −
SOLUTION:
10 3 17 20 6 2610 3 3 17 3 20 6 6 26 610 20 20 2010 20 20 20 10 10 20 20
2 12 12
x or xx or xx or x
x or x
x or xx or xx
− − ≤ − > −− − + ≤ + − + > − +− ≤ > −
− ≤ > −
− ≤ > −≥ − > −≥ −
ANSWER: Interval notation:[ 2, )− ∞
83. PROBLEM:
2 10 2 3 4 5x or x− < − − + > −
SOLUTION: 2 10 2 3 4 52 10 10 2 10 3 4 4 5 42 8 3 92 8 3 9 2 2 3 3
4 34 34
x or xx or xx or x
x or x
x or xx or xx
− < − − + > −− + < − + − + − > − −< − > −
< − > −
< − > −< <<
ANSWER: Interval notation: ( , 4)−∞
84. PROBLEM: 5 3 4 5 10 4x or x+ < − >
SOLUTION:
5 3 4 5 10 45 3 3 4 3 5 5 10 4 55 1 10 15 1 10 15 5 10 10
1 1 5 101 1 5 1015
x or xx or xx or x
x or x
x or x
x or x
x
+ < − >+ − < − − − > −< − > −
< − > −
< − > −
< <
<
ANSWER: Interval notation: ( )1
5,−∞ 85. PROBLEM:
3 18 5 20x and x< > −
SOLUTION: 3 18 5 203 18 5 20 3 3 5 5
6 44 6
x and x
x and x
x and xx
< > −
< > −
< > −− < <
ANSWER: Interval notation: ( 4, 6)−
86. PROBLEM: 7 5 3 10x and x+ ≤ − ≥ −
SOLUTION:
7 5 3 107 7 5 7 3 3 10 3
2 77 2
x and xx and xx and x
x
+ ≤ − ≥ −+ − ≤ − − + ≥ − +≤ − ≥ −
− ≤ ≤ −
ANSWER: Interval notation:[ 7, 2]− −
87. PROBLEM:
2 1 5 3 1 10x and x− < − <
SOLUTION: 2 1 5 3 1 102 1 1 5 1 3 1 1 10 12 6 3 112 6 3 11 2 2 3 3
113 3
3
x and xx and xx and x
x and x
x and x
x
− < − <− + < + − + < +< <
< <
< <
<
ANSWER: Interval notation: ( ,3)−∞ 88. PROBLEM:
5 2 13 3 4 13x and x+ < − + >
SOLUTION: 5 2 13 3 4 135 2 2 13 2 3 4 4 13 45 15 3 95 15 3 9 5 5 3 3
3 3
x and xx and xx and x
x and x
x and x
+ < − + >+ − < − − + − > −< − >
< − >
< − >
∅
ANSWER: Interval notation: ∅
Set up a compound inequality for the following then solve. 89. PROBLEM:
Five more than two times some number is to be between 15 and 25.
SOLUTION: The unknown number is represented by n. 15 < 2 5 2515 5< 2 5 5 25 510< 2 2010 2 20< 2 2 2
5 < < 10
nn
n
n
n
+ <− + − < −
<
<
ANSWER: 5 < < 10n
90. PROBLEM:
Four subtracted from three times some number is to be between −4 and 14.
SOLUTION: The unknown number is represented by n. 4 3 4 144 4 3 4 4 14 4
0 3 183 1803 3
0 6
nn
n
n
n
− < − <− + < − + < +< <
< <
< <
ANSWER: 0 6n< <
91. PROBLEM:
Clint wishes to earn a B which is at least 80 but less than 90. What range must he score on the fourth exam if the first three were 65, 75 and 90?
SOLUTION: The score of the fourth exam is represented by x.
65 75 9080 904
2304 80 4 4 904
320 230 360320 230 230 230 360 23090 130
x
x
xx
x
+ + +≤ <
+⋅ ≤ ⋅ < ⋅
≤ + <− ≤ − + < −≤ <
ANSWER: The maximum marks in an exam are 100. Hence, Clint must earn a score in the range from 90 to 100.
92. PROBLEM: A certain antifreeze is effective for a temperature range of −35°C to 120°C. Find the equivalent range in degrees Fahrenheit.
SOLUTION:
0 0535 ( 32) 1209
9 9 5 935 ( 32) 1205 5 9 563 32 21663 32 32 32 216 3231 248
C F C
F
FF
F
− < − <
− ⋅ < ⋅ − < ⋅
− < − <− + < − + < +− < <
ANSWER: The effective range is from −31°F to 248°F.
93. PROBLEM:
The average temperature in London ranges from 23°C in the summer to 14°C in the winter. Find the equivalent range in degrees Fahrenheit.
SOLUTION:
0 0514 ( 32) 239
9 9 5 914 ( 32) 235 5 9 525.2 32 41.425.2 32 32 32 41.4 3257.2 73.4
C F C
F
FF
F
< − <
⋅ < ⋅ − < ⋅
< − <+ < − + < +< <
ANSWER: The average temperature London ranges from 57.2°F to 73.4°F.
94. PROBLEM: If the base of a triangle measures 5 inches, then what range must the height be in for the area to be between 10 square inches and 20 square inches.
SOLUTION: The height of the triangle is represented by x.
1 1 5Area of the triangle = base height = 52 2 2
x x× × ⋅ ⋅ =
510 202
2 2 5 210 205 5 2 54 8
x
x
x
< <
⋅ < ⋅ < ⋅
< <
ANSWER: The height must be between 4 in and 8 in.
95. PROBLEM:
A rectangle has a length of 7 inches. Find all possible widths if the area is to be at least 14 square inches and at most 28 square inches.
SOLUTION: The width of the rectangle is represented by x. Area of the rectangle = Length ⋅ width = 7x
14 7 2814 7 287 7 7
2 4
x
x
x
≤ ≤
≤ ≤
≤ ≤
ANSWER: The width must be at least 2 in. and at most 4 in.
96. PROBLEM:
A rectangle has a width of 3 cm. Find all possible lengths if the perimeter must be at least 12 cm and at most 26 cm.
SOLUTION: Length of the rectangle is represented by x. Area of a rectangle = Length⋅width = 3x
23
12 3 2612 3 263 3 3
4 8
x
x
x
≤ ≤
≤ ≤
≤ ≤
ANSWER: The length must be at least 4 cm and at most 2
38 cm.
97. PROBLEM: The perimeter of a square must be between 40 ft and 200 ft. Find the length of all possible sides that satisfy this condition.
SOLUTION: Side length of the square is represented by x. Perimeter = 4⋅side length = 4x
40 4 20040 4 2004 4 4
10 50
x
x
x
< <
< <
< <
ANSWER: Sides must be between 10 ft and 50 ft.
98. PROBLEM:
If two times an angle is between 180 degrees and 270 degrees then what are the bounds of the angle?
SOLUTION: The unknown angle is represented by x.
180 2 270180 2 270
2 2 290 135
x
x
x
< <
< <
< <
ANSWER: The angle is between 90 degrees and 135 degrees.
99. PROBLEM:
If three times an angle is between 270 degrees and 360 degrees then what are the bounds of the angle?
SOLUTION: The unknown angle is represented by x.
270 3 360270 3 360
3 3 390 120
x
x
x
< <
< <
< <
ANSWER: The angle is between 90 degrees and 120 degrees.
Part D: Discussion Board Topics 100. PROBLEM:
Research and discuss the use of set builder notation with intersections and unions.
SOLUTION: Set-builder notation is a system for describing sets using familiar mathematical notation. Interval notation is a textual system of expressing solutions to an algebraic inequality. For e.g.: The real numbers greater than or equal to 2.
Since the set is too large to list, set-builder notation allows us to describe it using familiar mathematical notation. We can do this as follows:
{ }| 2x x∈ ≥ The vertical bar ( | ) is read, “such that.” It is assumed that all variables represent real numbers. For this reason, we can omit the “∈ ” and write { }| 2x x ≥ , which is read, “the set of all real numbers x such that x is greater than or equal to 2.” To describe compound inequalities such as 3 6x or x< ≥ we write { } 3 | 6x orx x< ≥ , which is read, “the set of all real numbers x such that x is less than 3 or x is greater than or equal to 6.”
Bounded intervals, such as 1 3x− ≤ < , can be written { } 1| 3xx − ≤ < , which is read, “the set of all real numbers x such that x is greater than or equal to −1 and less than 3.” ANSWER:
101. PROBLEM: Can we combine logical “or” into one statement like we do for logical “and”?
ANSWER: Logical “or” cannot be combined into one statement like logical “and.” If “or” statements are combined in the case of inequalities, both the inequalities need to be satisfied. If the statements are combined, the meaning of the word “or” is negated.
2.9ReviewExercisesandSampleExamChapter2ReviewExercises 2.1 Introduction to Algebra Evaluate. 1. PROBLEM:
2 7x + when 4x = −
SOLUTION: ( )2 7 2 4 7
8 7 1
x + = − +
= − += −
ANSWER: 1−
2. PROBLEM:
4 1x− + when 2x = −
SOLUTION: ( )4 1 4 2 1
= 8 1 = 9
x− + = − − +
+
ANSWER: 9
3. PROBLEM: 2 13 2
y − when 35
y =
SOLUTION:
2 1 2 3 13 2 3 5 2
2 1 5 2
1 10
y ⎛ ⎞− = −⎜ ⎟⎝ ⎠
= −
= −
ANSWER: 110
−
4. PROBLEM:
3 54 3
y− + when 23
y =
SOLUTION:
3 5 3 2 54 3 4 3 3
1 5 2 3
7 6
y ⎛ ⎞− + = − +⎜ ⎟⎝ ⎠
= − +
=
ANSWER: 76
5. PROBLEM:
2 4b ac− when 5a = , 2b = − , and 12
c =
SOLUTION:
( )22 14 2 4 52
4 10 6
b ac− = − − ⋅ ⋅
= −= −
ANSWER: 6−
6. PROBLEM:
2 4b ac− when 14
a = − , 1b = − , and 3c = −
SOLUTION:
( ) ( )22 14 1 4 34
1 3 2
b ac ⎛ ⎞− = − − ⋅ − ⋅ −⎜ ⎟⎝ ⎠
= −= −
ANSWER: 2−
7. PROBLEM:
22 3x x− + when 3x = −
SOLUTION: ( ) ( )222 3 2 3 3 3
18 3 3 24
x x− + = − − − +
= + +=
ANSWER: 24
8. PROBLEM:
25 2 4x x− + when 1x = −
SOLUTION: ( ) ( )225 2 4 5 1 2 1 4
5 2 4 11
x x− + = − − − +
= + +=
ANSWER: 11
9. PROBLEM:
Calculate the simple interest earned on a 3 year investment of $750 at an annual interest rate of 8%.
SOLUTION: I Prt=
$750 3 0.08I = ⋅ ⋅ = $180
ANSWER: $180
10. PROBLEM: A bus traveled for 2
31 hours at an average speed of 48 miles per hour. What distance did the bus travel?
SOLUTION: D rt=
548 80 3
milesD hours mileshour
= ⋅ =
ANSWER: 80 miles
11. PROBLEM:
Calculate the area of a rectangle with dimensions 4½ feet by 6 feet.
SOLUTION: Area of a rectangle = length ⋅ width = 4½ feet ⋅ 6 feet
= 9 feet 6 feet = 27 sq.ft2
⋅
ANSWER: 27 sq.ft
12. PROBLEM:
Calculate the volume of a rectangular box with dimensions 4½ feet by 6 feet by 1 foot.
SOLUTION: Volume = lwh
9Volume 6 1 27 cubic feet2
= ⋅ ⋅ =
ANSWER: 27 cubic feet
2.2 Simplifying Algebraic Expressions Multiply. 13. PROBLEM:
( )5 3 2x− −
SOLUTION: ( ) ( )5 3 2 5 3 5 2
15 10x x
x− − = − ⋅ − − ⋅
= − +
ANSWER: 15 10x− +
14. PROBLEM: ( )6 9 3x − ⋅
SOLUTION:
( )6 9 3 6 3 9 3 18 27x x x− ⋅ = ⋅ − ⋅ = −
ANSWER: 18 27x − 15. PROBLEM:
( )23 4 8 324
x x− +
SOLUTION:
( )2 2
2
3 3 3 34 8 32 4 8 324 4 4 4 3 6 24
x x x x
x x
− + = ⋅ − ⋅ + ⋅
= − +
ANSWER: 23 6 24x x− +
16. PROBLEM:
21 2 52010 5 4
x x⎛ ⎞− − −⎜ ⎟⎝ ⎠
SOLUTION:
( ) ( )2 2
2
1 2 5 1 2 520 20 20 2010 5 4 10 5 4
2 8 25
x x x x
x x
⎛ ⎞− − − = − ⋅ − − ⋅ − − ⋅⎜ ⎟⎝ ⎠
= − + +
ANSWER: 22 8 25x x− + +
17. PROBLEM:
( )3 2 5 1a b c− − + −
SOLUTION: ( ) ( )
( ) ( ) ( )3 2 5 1 1 3 2 5 1
1 3 1 2 1 5 1 1 3 2 5 1
a b c a b c
a b ca b c
− − + − = − − + −
= − ⋅ − − ⋅ + − ⋅ − − ⋅
= − + − +
ANSWER: 3 2 5 1a b c− + − +
18. PROBLEM: ( )3 26 3 7 5y y y− + − +
SOLUTION:
( ) ( ) ( ) ( )3 2 3 2
3 2
6 3 7 5 6 6 3 6 7 6 5
6 18 42 30
y y y y y y
y y y
− + − + = − ⋅ + − ⋅ − − ⋅ + − ⋅
= − − + −
ANSWER: 3 26 18 42 30y y y− − + −
Simplify. 19. PROBLEM:
5 7 3 5a b a b− − +
SOLUTION: 5 7 3 5 5 3 7 5 = 2 2a b a b a a b b
a b− − + = − − +
−
ANSWER: 2 2a b−
20. PROBLEM:
2 26 4 7 3x x x x− + −
SOLUTION: 2 2 2 2
2
6 4 7 3 6 7 4 3 13 7
x x x x x x x xx x
− + − = + − −
= −
ANSWER: 213 7x x−
21. PROBLEM: 3 1 1 15 2 10 4
xy xy+ − −
SOLUTION:
3 1 1 1 3 1 1 15 2 10 4 5 10 2 4
3 2 1 2 1 1 6 1 2 1 10 4 10 4
5 1 10 41 1 2 4
xy xy xy xy
xy xy
xy
xy
+ − − = − + −
⋅ − ⋅ − − −= + = +
= +
= +
ANSWER: 1 12 4
xy +
22. PROBLEM:
3 4 1 14 21 3 7
a b a b− − + −
SOLUTION:
3 4 1 1 3 1 4 14 21 3 7 4 3 21 7
9 4 4 3 12 215 7
12 215 1
12 3
a b a b a a b b
a b
a b
a b
− − + − = − + − −
− + − −= +
= − −
= − −
ANSWER: 5 112 3
a b− −
23. PROBLEM: 2 2 2 22 7 9a b ab a b ab+ − +
SOLUTION:
2 2 2 2 2 2 2 2
2 2
2 7 9 7 2 9 = 6 11a b ab a b ab a b a b ab ab
a b ab+ − + = − + +
− +
ANSWER: 2 26 11a b ab− +
24. PROBLEM:
2 23 5 9y y y− + − +
SOLUTION: 2 2 2 23 5 9 3 5 9
3 14y y y y y y
y− + − + = − − + +
= − +
ANSWER: 3 14y− +
25. PROBLEM:
( )8 8 3 7x− − −
SOLUTION: ( ) ( )8 8 3 7 8 8 8 3 7 64 24 7
64 17x x x
x− − − = − ⋅ − − ⋅ − = − + −
= − +
ANSWER: 64 17x− + 26. PROBLEM:
( )7 6 9x− −
SOLUTION: ( )7 6 9 7 6 9
6 16x x
x− − = − +
= − +
ANSWER: 6 16x− +
27. PROBLEM: ( ) ( )22 3 2 1 5 7x x x− + − −
SOLUTION:
( ) ( )2 2
2
2
2 3 2 1 5 7 2 3 2 2 2 1 5 7
6 4 5 2 7 6 9 9
x x x x x x
x x xx x
− + − − = ⋅ − ⋅ + ⋅ − +
= − − + +
= − +
ANSWER: 26 9 9x x− +
28. PROBLEM:
( ) ( )2 22 6 8 5 12 1y y y y+ − − − +
SOLUTION: ( ) ( )2 2 2 2
2
2 6 8 5 12 1 2 6 8 5 12 1
3 18 9
y y y y y y y y
y y
+ − − − + = + − − + −
= − + −
ANSWER: 23 18 9y y− + −
29. PROBLEM:
( ) ( )6 3 2 7 5 3a b a b− − + −
SOLUTION: ( ) ( )6 3 2 7 5 3 6 3 6 35 21
32 15 6a b a b a b a b
a b− − + − = − + + −
= − +
ANSWER: 32 15 6a b− + 30. PROBLEM:
( ) ( )2 210 5 1 3 5 1x x x x− − + − + −
SOLUTION:
( ) ( )2 2 2 2
2
10 5 1 3 5 1 10 5 5 5 3 5 1
8 6
x x x x x x x x
x
− − + − + − = − + − − − +
= − +
ANSWER: 28 6x− +
31. PROBLEM: Subtract 5 1x − from 2 3x − .
SOLUTION:
( )2 3 5 1 2 3 5 1 3 2
x x x xx
− − − = − − +
= − −
ANSWER: 3 2x− − 32. PROBLEM:
Subtract x-3 from twice the quantity x-1
SOLUTION: 2(x-1) –(x-3) =2x-2 –x+3 = x+1 ANSWER: x+1 2.3 Solving Linear Equations: Part I Is the given value a solution to the linear equation? 33. PROBLEM:
3 18x− + = − ; 15x = −
SOLUTION:
( )3 18
15 3 1815 3 1818 18
x− + = −
− − + = −
+ = −≠ −
ANSWER: 15x = − is not a solution.
34. PROBLEM:
4 3 3x x− = − ; 2x = −
SOLUTION:
( ) ( )4 3 34 2 3 3 2
8 3 611 6
x x− = −
− − = − −
− − =− ≠
ANSWER: 2x = − is not a solution.
35. PROBLEM:
8 2 5 1x x+ = + ; 13
x = −
SOLUTION: 8 2 5 1
1 18 2 5 13 3
8 52 13 38 6 5 33 3
2 23 3
x x+ = +
⎛ ⎞ ⎛ ⎞− + = − +⎜ ⎟ ⎜ ⎟⎝ ⎠ ⎝ ⎠
− + = − +
− + − +=
− = −
ANSWER: 13
x = − is a solution.
36. PROBLEM:
2 4 3 2x x+ = − ; 1x = −
SOLUTION:
( ) ( )2 4 3 22 1 4 3 1 2
2 4 3 22 5
x x+ = −
− + = − −
− + = − −≠ −
ANSWER: 1x = − is not a solution.
Solve. 37. PROBLEM:
23 25y + =
SOLUTION: 23 2523 23 25 232
yyy
+ =+ − = −=
ANSWER: 2y =
38. PROBLEM: 3 54x− =
SOLUTION:
3 543 543 3
18
x
x
x
− =−
=− −= −
ANSWER: 18x = −
39. PROBLEM:
84x=
SOLUTION:
84
4 4 8432
x
x
x
=
⋅ = ⋅
=
ANSWER: 32x =
40. PROBLEM:
5 22 3
x =
SOLUTION:
5 22 32 5 2 25 2 5 3
415
x
x
x
=
⋅ = ⋅
=
ANSWER: 415
x =
41. PROBLEM: 7 5 54x − = −
SOLUTION:
7 5 547 5 5 54 57 497 497 7
7
xxx
x
x
− = −− + = − += −
= −
= −
ANSWER: 7x = −
42. PROBLEM:
2 7 43x− + =
SOLUTION: 2 7 432 7 7 43 72 362 362 2
18
xxx
x
x
− + =− + − = −− =−
=− −= −
ANSWER: 18x = −
43. PROBLEM:
7 3 0x + =
SOLUTION: 7 3 07 3 3 0 37 37 37 7
37
xxx
x
x
+ =+ − = −= −
= −
= −
ANSWER: 37
x = −
44. PROBLEM: 4 5 5x + =
SOLUTION:
4 5 54 5 5 5 54 04 04 4
0
xxx
x
x
+ =+ − = −=
=
=
ANSWER: 0x =
45. PROBLEM:
1 10 3x= −
SOLUTION: 1 10 31 10 10 10 3
9 39 33 3
33
xx
x
x
xx
= −− = − −− = −− −
=− −==
ANSWER: 3x =
46. PROBLEM:
10 5 15y− =
SOLUTION: 10 5 1510 10 5 15 10
5 55 55 5
1
yy
y
y
y
− =− − = −
− =−
=− −= −
ANSWER: 1y = −
47. PROBLEM: 7 28y− =
SOLUTION:
7 287 7 28 7
2121
yy
yy
− =− − = −
− == −
ANSWER: 21y = −
48. PROBLEM:
33 16x− =
SOLUTION: 33 1633 33 16 33
1717
xx
xx
− =− − = −
− = −=
ANSWER: 17x =
49. PROBLEM:
5 1 36 3 2
x + =
SOLUTION:
5 1 36 3 25 1 1 3 16 3 3 2 35 9 26 65 76 66 5 6 75 6 5 6
75
x
x
x
x
x
x
+ =
+ − = −
−=
=
⋅ = ⋅
=
ANSWER: 75
x =
50. PROBLEM: 2 1 13 5 3
y− + = −
SOLUTION:
2 1 13 5 32 1 1 1 13 5 5 3 52 5 33 152 83 153 2 3 82 3 2 1545
y
y
y
y
y
y
− + = −
− + − = − −
− −− =
− = −
⎛ ⎞ ⎛ ⎞ ⎛ ⎞ ⎛ ⎞− ⋅ − = − ⋅ −⎜ ⎟ ⎜ ⎟ ⎜ ⎟ ⎜ ⎟⎝ ⎠ ⎝ ⎠ ⎝ ⎠ ⎝ ⎠
=
ANSWER: 45
y =
51. PROBLEM:
The sum of 9x and 6 is 51.
SOLUTION: 9 6 519 6 6 51 69 459 459 9
5
xxx
x
x
+ =+ − = −=
=
=
ANSWER: 5x =
52. PROBLEM: The difference of 3x and 8 is 25.
SOLUTION:
3 8 253 8 8 25 83 333 333 3
11
xxx
x
x
− =− + = +=
=
=
ANSWER: 11x =
2.4 Solving Linear Equations: Part II Solve. 53. PROBLEM:
5 2 3 6x x− = + SOLUTION:
5 2 3 65 3 2 3 3 62 2 62 2 2 6 22 82 82 2
4
x xx x x xxxx
x
x
− = +− − = − +− =− + = +=
=
=
ANSWER: 4x =
54. PROBLEM: 7 1 2 29x x+ = −
SOLUTION:
7 1 2 297 2 1 2 2 295 1 295 1 1 29 15 305 305 5
6
x xx x x xxxx
x
x
+ = −− + = − −+ = −+ − = − −= −
= −
= −
ANSWER: 6x = −
55. PROBLEM:
14 1 15 11x x+ = −
SOLUTION: 14 1 15 1114 15 1 15 15 11
1 111 1 11 1
1212
x xx x x x
xxx
x
+ = −− + = − −
− + = −− + − = − −− = −=
ANSWER: 12x =
56. PROBLEM:
6 13 3 7y y− = +
SOLUTION: 6 7 13 3 7 7
13 313 13 3 131616
y y y yyyy
y
− − = + −− − =− − + = +− == −
ANSWER: 16y = −
57. PROBLEM: 8 6 3 22 3y y y+ − = −
SOLUTION:
8 6 3 22 35 6 22 35 3 6 22 3 38 6 228 6 6 22 68 168 168 8
2
y y yy yy y y yyyy
y
y
+ − = −+ = −+ + = − ++ =+ − = −=
=
=
ANSWER: 2y =
58. PROBLEM:
12 5 6 6y y− + = −
SOLUTION: 12 5 6 618 5 618 5 618 6 618 18 6 6 18
6 246 246 6
4
y yy yy y y yy
yy
y
y
− + = −− = −− − = − −− = −− − = − −
− = −− −
=− −=
ANSWER: 4y =
59. PROBLEM: ( )5 2 7 1 2 1x x− − = +
SOLUTION:
( )5 2 7 1 2 15 14 2 2 17 14 2 17 14 2 2 2 17 16 17 7 16 1 7
16 616 616 16
38
x xx xx xx x x xx
xx
x
x
− − = +
− + = +− = +− − = − +− =− − = −
− = −− −
=− −
=
ANSWER: 38
x =
60. PROBLEM:
( )10 5 1 5x x− − = −
SOLUTION: ( )10 5 1 5
10 5 5 515 5 515 5 515 4 515 15 4 5 15
4 104 104 4
52
x xx xx xx x x xx
xx
x
x
− − = −
− + = −− = −− + = − +− =− − = −
− = −− −
=− −
=
ANSWER: 52
x =
61. PROBLEM: ( )2 3 4 7x x x− − = −
SOLUTION: ( )2 3 4 7
2 3 4 74 7
4 74 7
x x xx x xx xx x x x
− − = −
− + = −− + = −− + + = − +≠
ANSWER: ∅
62. PROBLEM:
( )9 3 2 1 3 3x x x− + = −
SOLUTION: ( )9 3 2 1 3 3
9 6 3 3 33 3 3 33 3 3 3 3 3
3 3
x x xx x xx xx x x x
− + = −
− − = −− = −− − = − −
− = −
ANSWER:
63. PROBLEM:
( ) ( ) ( )2 5 2 3 2 1 5 3x x x− − + = −
SOLUTION: ( ) ( ) ( )2 5 2 3 2 1 5 3
10 4 6 3 5 154 7 5 154 5 7 5 5 15
7 157 7 15 7
88
x x xx x x
x xx x x xxxx
x
− − + = −
− − − = −− = −− − = − −
− − = −− − + = − +− = −=
ANSWER: 8x =
64. PROBLEM: ( ) ( ) ( )3 5 1 4 4 5 2 10x x x− − − = − +
SOLUTION:
( ) ( ) ( )3 5 1 4 4 5 2 1015 3 4 16 10 5011 13 10 5011 10 13 10 10 5021 13 5021 13 13 50 1321 6321 6321 21
3
x x xx x xx xx x x xxxx
x
x
− − − = − +
− − + = − −+ = − −+ + = − + −+ = −+ − = − −= −
= −
= −
ANSWER: 3x = −
65. PROBLEM:
( )3 14 3 12 4
x − + =
SOLUTION:
( )3 14 3 12 4
9 16 12 418 16 1
4176 14
17 17 176 14 4 44 176
42164
6 216 4 6
78
x
x
x
x
x
x
x
x
x
− + =
− + =
−⎛ ⎞− =⎜ ⎟⎝ ⎠
− =
− + = +
+=
=
=⋅
=
ANSWER: 78
x =
66. PROBLEM:
( )3 1 4 9 24 6
x− − =
SOLUTION:
( )3 1 4 9 24 63 2 3 24 3 2
2 3 6 23 42 9 23 42 9 9 923 4 4 42 8 93 42 13 43 2 3 12 3 2 438
x
x
x
x
x
x
x
x
x
− − =
− + =
+− + =
− + =
− + − = −
−− =
− = −
⎛ ⎞ ⎛ ⎞ ⎛ ⎞ ⎛ ⎞− ⋅ − = − ⋅ −⎜ ⎟ ⎜ ⎟ ⎜ ⎟ ⎜ ⎟⎝ ⎠ ⎝ ⎠ ⎝ ⎠ ⎝ ⎠
=
ANSWER: 38
x =
67. PROBLEM:
( )2 1 19 3 3 23 2 2
x x⎛ ⎞− + = −⎜ ⎟⎝ ⎠
SOLUTION:
( )2 1 19 3 3 23 2 2
1 36 2 62 2
4 1 36 62 2
3 36 62 2
3 36 6 6 62 2
3 32 2
x x
x x
x x
x x
x x x x
⎛ ⎞− + = −⎜ ⎟⎝ ⎠
− + = −
−⎛ ⎞− = −⎜ ⎟⎝ ⎠
− = −
− − = − −
− = −
ANSWER:
68. PROBLEM:
( )5 11 4 1 54 2
x x⎛ ⎞− − = −⎜ ⎟⎝ ⎠
SOLUTION:
( )5 11 4 1 54 2
5 51 5 54 2
4 5 55 54 2
9 55 54 29 55 5 5 54 29 55 54 29 504 29 54 2
x x
x x
x x
x x
x x x x
x x
⎛ ⎞− − = −⎜ ⎟⎝ ⎠
− + = −
+− = −
− = −
− + = − +
− + =
− =
≠
ANSWER: ∅
69. PROBLEM: The sum of 4x and 3 is equal to the difference of 7x and 8.
SOLUTION:
4 3 7 84 7 3 7 7 8
3 3 83 3 3 8 33 113 113 3
113
x xx x x x
xxx
x
x
+ = −− + = − −
− + = −− + − = − −− = −− −
=− −
=
ANSWER: 113
x =
70. PROBLEM:
The difference of 5x and 1 is equal to the sum of 12x and 1.
SOLUTION: 5 1 12 15 12 1 12 12 1
7 1 17 1 1 1 17 27 27 7
27
x xx x x x
xxx
x
x
− = +− − = − +
− − = +− − + = + +− =−
=− −
= −
ANSWER: 27
x = −
71. PROBLEM: Solve for x: 9 1y x= +
SOLUTION:
9 11 9 1 11 91 9
9 91
91
9
y xy xy xy x
y x
yx
= +− = + −− =−
=
−=
−=
ANSWER: 19
yx −=
72. PROBLEM:
Solve for y: 5 2 3x y+ =
SOLUTION: 5 2 35 5 2 3 52 3 52 3 52 2
5 32
x yx x y xy x
xy
xy
+ =− + = −= −
−=
− +=
ANSWER: 5 32xy − +
=
73. PROBLEM: Solve for l: 2 2P l w= +
SOLUTION:
2 22 2 2 22 22 2
2 22
2
P l wP w l w wP w lP w l
P wl
= +− = + −− =−
=
−=
ANSWER: 22
P wl −=
74. PROBLEM:
Solve for b: 12
A bh=
SOLUTION:
12
2 2 12
2
A bh
A bhh h
Abh
=
⋅ = ⋅
=
ANSWER: 2Abh
=
2.5 Applications of Linear Equations 75. PROBLEM:
A larger integer is 3 more than twice a smaller integer. If their sum is 39, then find the integers.
SOLUTION: The smaller integer is represented by x. Larger integer: 2x + 3
2 3 393 3 393 3 3 39 33 363 363 3
12
x xxxx
x
x
+ + =+ =+ − = −=
=
=
Larger integer: ( )2 3 = 2 12 + 3 = 24 + 3 = 27x +
ANSWER: The integers are 12 and 27. 76. PROBLEM:
A larger integer is 5 more than 3 times a smaller integer. If their sum is 49, then find the integers.
SOLUTION: The smaller integer is represented by x. Larger integer: 3x + 5
3 5 494 5 494 5 5 49 54 444 444 4
11
x xxxx
x
x
+ + =+ =+ − = −=
=
=
Larger integer: ( )3 5 3 11 5 33 5 38x + = + = + = ANSWER: The two integers are 11 and 38.
77. PROBLEM: The sum of three consecutive odd integers is 45. Find the integers.
SOLUTION:
2 4 453 6 453 6 6 45 63 393 393 3
132 13 2 154 13 4 17
n n nnnn
n
nnn
+ + + + =+ =+ − = −=
=
=+ = + =+ = + =
ANSWER: The three integers are 13, 15, and 17.
78. PROBLEM:
The sum of three consecutive even integers is 72. Find the integers.
SOLUTION: 2 4 72
3 6 723 6 6 72 63 663 663 3
222 22 2 244 22 4 26
n n nnnn
n
nnn
+ + + + =+ =+ − = −=
=
=+ = + =+ = + =
ANSWER: The even integers are 22, 24, and 26.
79. PROBLEM: The sum of three consecutive integers is 60. Find the integers.
SOLUTION:
1 2 603 3 603 3 3 60 33 573 573 3
191 19 1 202 19 2 21
n n nnnn
n
nnn
+ + + + =+ =+ − = −=
=
=+ = + =+ = + =
ANSWER: The three integers are 19, 20, and 21.
80. PROBLEM:
The length of a rectangle is 7 cm less than twice its width. If the perimeter measures 46 cm, then find the dimensions of the rectangle.
SOLUTION: The width of the rectangle is represented by x. Length: 2x – 7 Perimeter = 2 (l + w) 46 = 2 (x + 2x – 7)
( )
( )
( )
2 2 – 7 462 46 2 72 2
2 7 233 7 233 7 7 23 73 303 303 3
10Length: 2 7 2 10 7 20 7 13
x x
x x
x xxxx
x
xx
+ =
+ − =
+ − =− =− + = +=
=
=
− = − = − =
ANSWER: Width = 10 cm; length = 13 cm
81. PROBLEM: A triangle has sides whose measures are consecutive even integers. If the perimeter is 24 m, then find the measure of each side.
SOLUTION: The sides of the triangle are represented by n, n + 2, and n + 4.
2 4 Perimeter2 4 24
3 6 243 6 6 24 63 183 183 3
61 6 2 82 6 4 10
n n nn n nnnn
n
nnn
+ + + + =+ + + + =+ =+ − = −=
=
=+ = + =+ = + =
ANSWER: The sides of the triangle are 6 m, 8 m, and 10 m respectively.
82. PROBLEM:
The circumference of a circle measures 24π in. Find the radius of the circle.
SOLUTION: The radius of the circle is represented by x. Circumference = 224 224 22 2
12 in.
rr
r
r
ππ ππ ππ π
=
=
=
ANSWER: The radius of the circle is 12 in.
83. PROBLEM: Mary invested $1,800 in two different accounts. On account earned 3.5% simple interest and the other earned 4.8%. If the total interest after one year was $79.25, then how much did she invest in each account?
SOLUTION: Let x represent the amount invested in account 1 at 3.5% 0.035= . Let 1800 x− represent the amount invested in account 2 at 4.8% 0.048= . The formula for the simple interest is I prt= . Also, interest from account 1 + interest from account 2 = total interest The algebraic equation is: ( )0.035 1 1800 0.048 1 79.25x x⋅ ⋅ + − ⋅ ⋅ = Solve the equation to obtain x:
( )( )
0.035 1 1800 0.048 1 79.25
35 1800 48 7925035 48 86,400 79,25035 48 86,400 86,400 79, 250 86,400
13 715013 715013 13
550
x x
x xx xx x
x
x
x
⋅ ⋅ + − ⋅ ⋅ =
+ − =
− + =− + − = −
− = −− −
=− −=
Back substitute to obtain the amount invested in account 2: 1800 1800 550 1250x− = − =
ANSWER: Hence, Mary invested $550 at 3.5% in one account and $1250 at 4.8% in the other account.
84. PROBLEM: James has $6 in dimes and quarters. If he has 4 less quarters than he does dimes, then how many of each coin does he have?
SOLUTION: The number of dimes is represented by x. Number of quarters: x – 4 Value of 1 quarter is $0.25. Hence, the total value of quarters is(x-4)0.25. Value of 1 dime is $0.10. Hence, the total value of dimes is (x )0.1. Sum of the total value of dimes and quarters is $6 0.25(x-4) + (x)0.1 = 6 0.25x -1+ 0.1(x) = 6 0.35x –1 = 6 0.35x – 1 + 1 = 6 + 1 0.35x = 7 0.35 70.35 0.35
20
x
x
=
=
Number of quarters = x – 4 = 20 – 4 = 16 ANSWER: James has 20 dimes and 16 quarters.
85. PROBLEM: Two brothers leave the house at the same time traveling in opposite directions. One averages 40 mph and the other 36 mph. How long does it take for the distance between them to reach 114 miles?
SOLUTION: Time taken by the brothers is represented by t. Distance travelled by the 1st brother, D1= rt = 40mph⋅x = 40t Distance travelled by the 2nd brother, D2= rt = 36mph⋅x = 36t The sum of the distances travelled should be 114 miles. 40t + 36t = 114 76t =114 76 11476 76
1.5
t
t
=
=
ANSWER: It takes 1.5 hours for the distance between the brothers to reach 114 miles.
86. PROBLEM:
Driving to her grandmother’s house, Jill made several stops and was only able to average 40 mph. The return trip took 2 hours less time because she drove nonstop and was able to average 60 mph. How long did it take Jill to drive home from her grandmother’s house?
SOLUTION: Driving time Jill takes to reach her grandmother’s house is represented by t. Driving time Jill takes to drive back home: t – 2. Distance travelled by Jill to her grandmother’s house = D1 = rt = 40mph⋅t = 40t Distance travelled back by Jill = D2 = rt = 60mph⋅(t – 2) = 60(t – 2) Both the distances are equal. D1 = D2 40t = 60(t – 2) 40t = 60t –120 40t – 60t = 60t – 60t –120 –20t = –120
20 12020 20
6
t
t
− −=
− −=
ANSWER: Driving time Jill takes to drive back home: t – 2 = 6 – 2 = 4 hours.
2.6 Ratio and Proportion Applications Solve. 87. PROBLEM:
34 8
n=
SOLUTION: 34 83 84
66
n
n
nn
=
⋅=
==
ANSWER: 6n =
88. PROBLEM:
7 283 n=
SOLUTION:
7 283
28 37
12
n
n
n
=
⋅=
=
ANSWER:
12n =
89. PROBLEM: 6 30
11n=
SOLUTION:
6 3011
6 1130
115
n
n
n
=
⋅=
=
ANSWER: 115
n=
90. PROBLEM:
25 3n=
SOLUTION:
25 3
5 23
103
n
n
n
=
⋅=
=
ANSWER:
103
n =
91.
PROBLEM: 3 1 1
3 2n −
=
SOLUTION:
( )
3 1 13 2
2 3 1 3 16 2 36 2 2 3 26 56 56 6
56
n
nnnn
n
n
−=
− = ⋅
− =− + = +=
=
=
ANSWER: 56
n =
92. PROBLEM:
4 12 5 3n
= −+
SOLUTION:
( )
4 12 5 33 4 1 2 512 2 512 5 2 5 517 217 2
2 2172
nn
nn
n
n
n
= −+
⋅ = − ⋅ +
= − −+ = − − += −
−=
− −
= −
ANSWER:
172
n = −
93. PROBLEM: 13
1n− =
−
SOLUTION:
( )
131
3 1 13 3 13 3 3 1 33 23 23 3
23
nn
nnn
n
n
− =−
− − =
− + =− + − = −− = −− −
=− −
=
ANSWER: 23
n =
94. PROBLEM:
2 16 2 1n n=
− +
SOLUTION:
( ) ( )
2 16 2 1
2 2 1 1 64 2 64 2 63 2 63 2 2 6 23 83 83 3
83
n nn n
n nn n n nnnn
n
n
=− +
+ = ⋅ −
+ = −− + = − −+ = −+ − = − −= −
= −
= −
ANSWER:
83
n = −
95. PROBLEM: Find two numbers in the proportion 4 to 5 whose sum is 27.
SOLUTION:
( )
27 45
5 27 4135 5 4135 5 5 4 5135 9135 9
9 915
1527 27 15 12
nn
n nn nn n n nn
n
nn
n
−=
− =
− =− + = +=
=
==− = − =
ANSWER: The two numbers are 15 and 12.
96. PROBLEM:
A larger number is 2 less than twice a smaller. If the two numbers are in the proportion 5 to 9, then find the numbers.
SOLUTION: The smaller number is represented by n. Larger number: 2n – 2
( )
( )
52 2 99 5 2 29 10 109 10 10 10 10
1010
2 2 2 10 2 20 2 18
nnn nn nn n n nn
nn
=−= −
= −− = − −
− = −=
− = − = − =
ANSWER: The two numbers are 10 and 18.
97. PROBLEM: A recipe calls for 1½ teaspoons of vanilla extract for every 3 cups of batter. How many teaspoons of vanilla extract should be used with 7 cups of batter?
SOLUTION:
For 3 cups of batter 32
teaspoons of vanilla extract is used.
For 7 cups of batter:
372
3
⋅
37 21 7 12 33 6 2 2
⋅= = =
ANSWER: 132
teaspoons of vanilla extract is used for 7 cups of batter.
98. PROBLEM:
The ratio of female to male employees at a certain bank is 4 to 5. If there are 80 female employees at the bank, then determine the total number of employees.
SOLUTION: Number of male employees is represented by x. 80 4
580 5
4100
x
x
x
=
⋅=
=
ANSWER: Total number of employees = 100 + 80 = 180.
If triangle ABC is similar to triangle RST, then find the remaining two sides given the following:
99. PROBLEM:
4a = , 9b = , 12c = , and 3s =
SOLUTION:
4 9 123
4 933 4 49 3
a b cr s t
r t
r
r
= =
= =
=
⋅= =
9 123
12 3 49
t
t
=
⋅= =
ANSWER: r =43
, t = 4
100. PROBLEM:
7b = , 10c = , 15t = , and 6r =
SOLUTION:
7 106 15
a b cr s ta
s
= =
= =
106 15
10 6 415
a
a
=
⋅= =
7 10157 15 2110 2
s
s
=
⋅= =
ANSWER: 4a = , 212
s =
101. PROBLEM:
At the same time of day, a pole casts a 27 ft shadow and 4 ft boy casts a 6 ft shadow. Calculate the height of the pole.
SOLUTION:
Height of the pole Height of the pole's shadowHeight of the boy Height of the boy's shadowHeight of the pole 27
4 627 4Height of the pole 18
6
=
=
⋅= =
ANSWER: Height of the pole is 18 ft.
102. PROBLEM:
An equilateral triangle with sides measuring 10 units is similar to another equilateral triangle with scale factor of 2:3. Find perimeter of the unknown triangle.
SOLUTION:
Side of the first triangle 2Side of the second triangle 3
10 2Side of the second triangle 3
3 10Side of the second triangle 152
=
=
⋅= =
ANSWER: Perimeter of the second triangle = 3⋅15 = 45 units
2.7 Introduction to Inequalities and Interval Notation Graph all solutions on a number line and provide the corresponding interval notation. 103. PROBLEM:
1x < −
SOLUTION: Use an open dot at –1 and shade all real numbers strictly lesser than –1. Use negative infinity (–∞) to indicate that the solution set is unbounded to the left on a number line.
ANSWER: Interval notation: ( ), 1−∞ −
104. PROBLEM: 10x ≤
SOLUTION: Use a closed dot at 10 and shade all real numbers strictly lesser than 10. Use negative infinity (–∞) to indicate that the solution set is unbounded to the left on a number line.
ANSWER: Interval notation: ( ],10−∞
105. PROBLEM:
0x ≥
SOLUTION: Use a closed dot at 0 and shade all real numbers strictly greater than 0. Use positive infinity (∞) to indicate that the solution set is unbounded to the right on a number line.
ANSWER: Interval notation:[ )0,∞ 106. PROBLEM:
2x > −
SOLUTION: Use an open dot at –2 and shade all real numbers strictly greater than –2. Use positive infinity (∞) to indicate that the solution set is unbounded to the right on a number line.
ANSWER: Interval notation: ( )2,− ∞
107. PROBLEM: 1 32 2
x− ≤ <
SOLUTION: The lower bound, 12
− is included in the solution set using a closed
dot. Shade all real numbers between 12
− and 32
, and indicate that the upper
bound, 32
, is included in the solutions set by using an open dot.
ANSWER: Interval Notation: )312 2,−⎡⎣
108. PROBLEM:
20 30x− < <
SOLUTION: The lower bound, 20− is included in the solution set using an open dot. Shade all real numbers between 20− and 30 , and indicate that the upper bound, 30 , is included in the solutions set by using an open dot.
ANSWER: Interval Notation: ( )20,30−
109. PROBLEM:
5 15x or x< ≥
SOLUTION: 5x < : Use an open dot at 5 and shade all real numbers strictly lesser than5 . Use
negative infinity (–∞) to indicate that the solution set is unbounded to the left on a number line.
15x ≥ : Use a closed dot at 15 and shade all real numbers strictly greater than15 . Use positive infinity (∞) to indicate that the solution set is unbounded to the right on a number line.
The symbol ∪ is used to denote “or.”
ANSWER: Interval notation: ( ) [ ),5 15,−∞ ∪ ∞
110. PROBLEM: 2 0x or x< >
SOLUTION: When we combine both solutions sets and form the union, we can see that all real numbers will solve the original compound inequality.
ANSWER: Interval notation: ( ),= −∞ ∞
Determine the inequality given the answers expressed in interval notation. 111. PROBLEM:
( ),3−∞
SOLUTION: The negative infinity (–∞) indicates that the solution is unbounded to the left on a number line. The 3 enclosed within the bracket indicates that the variable x is lesser than 3.
ANSWER: Inequality: 3x <
112. PROBLEM:
[ )4,− ∞
SOLUTION: The positive infinity (∞) indicates that the solution is unbounded to the right on a number line. The –4 enclosed within the square bracket indicates that the variable x is greater than or equal to –4. ANSWER: Inequality: 4x ≥ −
113. PROBLEM:
( )2,2−
SOLUTION: The –2 enclosed within the bracket indicates that the variable x is greater than –2. Inequality: 2x > − The 2 enclosed within the bracket indicates that the variable x is lesser than 2. Inequality: 2x <
ANSWER: On combining the two inequalities: 2 2x− < ≤
114. PROBLEM: ( ]3,8−
SOLUTION: The –3 enclosed within the bracket indicates that the variable x is greater than –3. Inequality: 3x > − The 8 enclosed within the square bracket indicates that the variable x is lesser than or equal to 8. Inequality: 8x ≤ ANSWER: On combining the two inequalities: 3 8x− < ≤
115. PROBLEM:
( ) [ ),1 3,−∞ ∪ ∞
SOLUTION: ( ),1−∞ : The negative infinity (–∞) indicates that the solution is unbounded to the left on a number line. The 1 enclosed within the bracket indicates that the variable x is lesser than 1. Inequality: 1x < ∪ is used to represent “or.” [ )3,∞ : The positive infinity (∞) indicates that the solution is unbounded to the right on a number line. The 3 enclosed within the square bracket indicates that the variable x is greater than or equal to 3. Inequality: 3x ≥
ANSWER: Combining the two inequalities: 1 3x or x< ≥
116. PROBLEM:
( ] [ ), 8 8,−∞ − ∪ ∞
SOLUTION: ( ], 8−∞ − : The negative infinity (–∞) indicates that the solution is unbounded to the left on a number line. The –8 enclosed within the square bracket indicates that the variable x is lesser than or equal to –8. Inequality: 8x ≤ − ∪ is used to represent “or.” [ )8,∞ : The positive infinity (∞) indicates that the solution is unbounded to the right on a number line. The 8 enclosed within the square bracket indicates that the variable x is greater than or equal to 8. Inequality: 8x ≥ ANSWER: Combining the two inequalities: 8 8x or x≤ − ≥
2.8 Linear Inequalities (one variable) Solve and Graph. In addition, present the solution set in interval notation. 117. PROBLEM:
2 1x + > −
SOLUTION:
2 12 2 1 2
3
xxx
+ > −+ − > − −> −
ANSWER: Interval notation: ( )3,− ∞ 118. PROBLEM:
4 16x− ≥
SOLUTION: 4 164 164 4
44
x
x
xx
− ≥
− ≥
− ≥≤ −
ANSWER: Interval notation: ( ], 4−∞ −
119. PROBLEM:
9 4 5x + ≤ −
SOLUTION: 9 4 59 4 4 5 49 99 99 9
1
xxx
x
x
+ ≤ −+ − ≤ − −≤ −
≤ −
≤ −
ANSWER: Interval notation: ( ], 1−∞ −
120. PROBLEM: 5 7 13x − <
SOLUTION:
5 7 135 7 7 13 75 205 205 5
4
xxx
x
x
− <− + < +<
<
<
ANSWER: Interval notation: ( ), 4−∞
121. PROBLEM:
7 5 8 15x x+ − ≥
SOLUTION: 7 5 8 15
5 155 5 15 51010
x xxxx
x
+ − ≥− + ≥− + − ≥ −− ≥≤ −
ANSWER: Interval notation: ( ], 10−∞ −
122. PROBLEM: 5 6 3 2 9 5x x x− + < + −
SOLUTION:
5 6 3 2 9 58 6 9 38 9 6 9 9 3
6 36 6 3 633
x x xx xx x x xxxx
x
− + < + −− < −− − < − −
− − < −− − + < − +− <> −
ANSWER: Interval notation: ( )3,− ∞
123. PROBLEM:
( )3 4 4x x x− − > +
SOLUTION: ( )3 4 4
3 4 42 4 42 4 4
4 44 4 4 40
x x xx x xx xx x x x
xxx
− − > +
− + > ++ > +− + > − ++ >+ − > −>
ANSWER: Interval notation: ( )0,∞
124. PROBLEM: ( ) ( ) ( )3 2 1 3 2 2 4x x x− − − ≤ +
SOLUTION:
( ) ( ) ( )3 2 1 3 2 2 46 3 3 6 2 83 3 2 83 2 3 2 2 8
3 83 3 8 35
x x xx x xx xx x x x
xxx
− − − ≤ +
− − + ≤ ++ ≤ +− + ≤ − ++ ≤+ − ≤ −≤
ANSWER: Interval notation: ( ],5−∞
125. PROBLEM:
( )2 5 4 12x− − >
SOLUTION: ( )2 5 4 12
2 5 20 125 22 125 22 22 12 225 105 105 5
22
xx
xxx
x
xx
− − >
− + >− + >− + − > −− > −
− > −
− > −<
ANSWER: Interval notation: ( ), 2−∞
126. PROBLEM:
( )3 5 2 11 5x x x− − ≥ −
SOLUTION: ( )3 5 2 11 5
3 5 10 11 52 10 11 52 5 10 11 5 5
3 10 113 10 10 11 103 13 13 3
13
x x xx x x
x xx x x x
xxx
x
x
− − ≥ −
− + ≥ −− + ≥ −− + + ≥ − +
+ ≥+ − ≥ −≥
≥
≥
ANSWER: Interval notation: )1
3 ,∞⎡⎣ 127. PROBLEM:
1 2 5 11x− < + ≤
SOLUTION: 1 2 5 111 5 2 5 5 11 56 2 66 2 62 2 23 3
xx
x
x
x
− < + ≤− − < + − ≤ −− < ≤
− < ≤
− < ≤
ANSWER: Interval notation: ( ]3,3−
128. PROBLEM: 1 72 24 2
x− ≤ − ≤
SOLUTION:
1 72 24 27 1 7 7 72 22 4 2 2 2
3 1 112 4 23 4 11 42 2
6 22
x
x
x
x
x
− ≤ − ≤
− + ≤ − + ≤ +
≤ ≤
⋅ ⋅≤ ≤
≤ ≤
ANSWER: Interval notation:[ ]6,22
129. PROBLEM:
5 3 2 6 5 7x or x+ < − − ≥
SOLUTION: 5 3 2 6 5 75 3 3 2 3 6 5 5 7 55 5 6 125 5 6 125 5 6 6
1 2
x or xx or xx or x
x or x
x or x
+ < − − ≥+ − < − − − + ≥ +< − ≥
< − ≥
< − ≥
ANSWER: Interval notation: ( ) [ ), 1 2,−∞ − ∪ ∞
130. PROBLEM: 20 3 5 5 2 25x or x− ≤ − ≥
SOLUTION:
20 3 5 5 2 2520 20 3 5 20 5 5 2 25 5
3 15 2 203 15 2 203 3 2 2
5 105 10
x or xx or x
x or x
x or x
x or xx or x
− ≤ − ≥− − ≤ − − − ≥ −
− ≤ − − ≥
− ≤ − − ≥
− ≤ − − ≥≥ ≤ −
ANSWER: Interval notation: ( ] [ ), 10 5,−∞ − ∪ ∞
Chapter2SampleExam1. PROBLEM:
Evaluate 2 4b ac− when 1a = − , 2b = − , and 12c = .
SOLUTION:
( ) ( )22 14 2 4 1 4 2 62
b ac ⎛ ⎞− = − − ⋅ − = + =⎜ ⎟⎝ ⎠
ANSWER: 6 2. PROBLEM:
Determine the area of a triangle given that the base measures 10 cm and the height measures 5 cm. ( 1
2A bh= )
SOLUTION: 12
1 10 52
A bh
A
=
= ⋅ ⋅
25 sq.cmA = ANSWER: 25 sq.cmA =
3. PROBLEM: ( )5 2 4 1x− −
SOLUTION:
( )5 2 4 1 5 8 2 8 7
x xx
− − = − +
= − +
ANSWER: 8 7x− + 4. PROBLEM:
1 2 1 34 3 2 5
x y x y− + −
SOLUTION:
1 2 1 3 1 1 2 34 3 2 5 4 2 3 5
1 2 10 9 4 15
3 19 4 15
x y x y x x y y
x y
x y
− + − = + − −
+ +⎛ ⎞= − ⎜ ⎟⎝ ⎠
= −
ANSWER: 3 194 15
x y−
5. PROBLEM:
( ) ( )5 4 2 3 2 3a ab b a ab b+ − − + −
SOLUTION: ( ) ( )5 4 2 3 2 3 5 4 2 3 2 3 5 3 4 2 2 3 2 2
a ab b a ab b a ab b a ab ba a ab ab b ba ab b
+ − − + − = + − − − +
= − + − − += + +
ANSWER: 2 2a ab b+ +
6. PROBLEM: ( ) ( )2 23 5 1 4x x x x x− + − + − +
SOLUTION:
( ) ( )2 2 2 2
2 2
3 5 1 4 3 5 1 4
5 3 4 1 3 5
x x x x x x x x x x
x x x x xx
− + − + − + = − − + + − +
= − + − + − + += − +
ANSWER: 3 5x− +
Solve: 7. PROBLEM:
2 5 27x− =
SOLUTION: 2 5 272 2 5 27 2
5 255 255 5
5
xx
x
x
x
− =− − = −
− =−
=− −= −
ANSWER: 5x = −
8. PROBLEM: 1 3 12 4 8
x − = −
SOLUTION:
1 3 12 4 81 3 3 1 32 4 4 8 41 1 62 81 52 8
1 52 22 8
54
x
x
x
x
x
x
− = −
− + = − +
− +=
=
⋅ = ⋅
=
ANSWER: 54
x =
9. PROBLEM:
5 7 3 5x x− = −
SOLUTION: 5 7 3 55 3 7 3 3 52 7 52 7 7 5 72 22 22 2
1
x xx x x xxxx
x
x
− = −− − = − −− = −− + = − +=
=
=
ANSWER: 1x =
10. PROBLEM:
( ) ( )3 3 4 2 1y y− − + =
SOLUTION: ( ) ( )3 3 4 2 1
3 9 4 2 111 111 11 1 111212
y yy yyyy
y
− − + =
− − − =− − =− − + = +− == −
ANSWER: 12y = −
11. PROBLEM:
( ) ( )5 2 3 2 2 3x x x− − + = −
SOLUTION: ( ) ( )5 2 3 2 2 3
5 10 3 6 2 32 16 2 32 2 16 2 2 3
16 3
x x xx x xx xx x x x
− − + = −
− − − = −− = −− − = − −
− ≠ −
ANSWER: ∅
12. PROBLEM:
58 32
n=
SOLUTION: 58 32
5 32 208
n
n
=
⋅= =
ANSWER: 20n =
13. PROBLEM: 3 6
1 4n= −
+
SOLUTION:
( )
3 61 4
3 4 6 112 6 612 6 6 6 618 6
6 63
3
nn
nn
n
nn
= −+⋅ = − +
= − −+ = − − +
−=
− −− == −
ANSWER: 3n = −
14. PROBLEM:
Solve for b: 2A a b= +
SOLUTION: 2
222
2 2
2
A a bA a a a bA a bA a b
A ab
= +− = − +− =−
=
−=
ANSWER: 2
A ab −=
Solve and graph the solution set. In addition, present the solution set in interval notation. 15. PROBLEM:
2 3 23x + >
SOLUTION: 2 3 232 3 3 23 32 202 202 2
10
xxx
x
x
+ >+ − > −>
>
>
ANSWER: Interval notation: ( )10,∞ 16. PROBLEM:
( )5 2 1 35x− + ≤
SOLUTION: ( )5 2 1 3510 5 3510 5 5 35 510 3010 3010 10
33
xxxx
x
xx
− + ≤
− + ≤− + − ≤ −− ≤
− ≤
− ≤≥ −
ANSWER: Interval notation:[ )3,− ∞
17. PROBLEM: ( ) ( )4 3 2 3 2 1 1x x− < + +
SOLUTION:
( ) ( )4 3 2 3 2 1 112 8 6 3 112 8 6 412 6 8 6 6 46 8 46 8 8 4 86 126 126 6
2
x xx xx xx x x x
xxx
x
x
− < + +
− < + +− < +− − < − +− <− + < +<
<
<
ANSWER: Interval notation: ( ), 2−∞ 18. PROBLEM:
( )0 6 3 1 30x≤ − − ≤
SOLUTION: ( )
( )( )
0 6 3 1 30
0 6 6 6 3 1 30 6
6 3 1 246 3 3 246 3 3 3 3 24 39 3 219 3 213 3 33 7
3 77 3
x
x
xx
xx
x
xx
x
≤ − − ≤
− ≤ − − − ≤ −
− ≤ − − ≤
− ≤ − + ≤− − ≤ − + − ≤ −− ≤ − ≤
− ≤ − ≤
− ≤ − ≤≥ ≥ −
− ≤ ≤
ANSWER: Interval notation:[ ]7,3−
19. PROBLEM:
( )1 16 2 10 33 5
x or x⎛ ⎞− < − + ≥⎜ ⎟⎝ ⎠
SOLUTION:
( )1 16 2 10 33 5
16 2 2 2 35
16 2 2 2 2 2 2 3 25
16 0 15
0 5
x or x
x or x
x or x
x or x
x or x
⎛ ⎞− < − + ≥⎜ ⎟⎝ ⎠
− < − + ≥
− + < − + + − ≥ −
< ≥
< ≥
ANSWER: Interval notation: ( ) [ ),0 5,−∞ ∪ ∞ 20. PROBLEM:
An algebra student earned 75, 79, and 89 points on the first 3 quizzes. What must she score on the fourth quiz to earn an average of at least 80?
SOLUTION: Score on the fourth quiz is represented by x.
75 79 89 804
75 79 89 320243 320243 243 320 243
77
x
xx
xx
+ + +≥
+ + + ≥+ ≥− + ≥ −
≥
ANSWER: She must earn at least 77 points on the fourth quiz.
21. PROBLEM: The sum of three consecutive odd integers is 117. Find the integers.
SOLUTION:
2 4 1173 6 1173 6 6 117 63 1113 1113 3
372 37 2 394 37 4 41
n n nnnn
n
nnn
+ + + + =+ =+ − = −=
=
=+ = + =+ = + =
ANSWER: 37, 39, 41
22. PROBLEM:
The length of a rectangle is 6 inches less than twice the width. If the perimeter measures 39 inches, then find the dimensions of the rectangle. SOLUTION: The width of the rectangle is represented by x. Length: 2x – 6 Perimeter = 2(l + w) 39 = 2(2x – 6 + x) 39 = 4x – 12 + 2x 39 = 6x – 12 39 + 12 = 6x – 12 + 12 51 = 6x 51 66 6
172
x
x
=
=
17Length: 2 6 2 6 112
x ⎛ ⎞− = − =⎜ ⎟⎝ ⎠
ANSWER: Width = 18 inches2
; Length = 11 inches
23. PROBLEM: Millie invested her $5,350 savings in two accounts. One account earns 5% annual interest and the other earns 6.2% in annual interest. If she earned $317.30 simple interest in one year, then how much was in each account? SOLUTION: Let x represent the amount invested in account 1 at 5% 0.05= . Let 5350 x− represent the amount invested in account 2 at 6.2% 0.062= . The formula for the simple interest is I prt= . Also, interest from account 1 + interest from account 2 = total interest The algebraic equation is: ( )0.05 1 5350 0.062 1 317.30x x⋅ ⋅ + − ⋅ ⋅ = Solve the equation to obtain x:
( )( )
( )
0.05 1 5350 0.062 1 317.30
0.05 5350 0.062 317.30
5 5350 6.2 31,7305 33,170 6.2 31,730
1.2 14401.2 14401.2 1.2
1200
x x
x x
x xx x
x
x
x
⋅ ⋅ + − ⋅ ⋅ =
+ − =
+ − =
+ − =− = −− −
=− −=
Back substitute to obtain the amount invested in account 2: 5350 5350 1200 4150x− = − =
ANSWER: Hence, Millie invested $1,200 at 5% in one account and $4150 at 6.2% in the other account.
24. PROBLEM: Because of traffic, Joe was only able to drive an average 42 mph on the trip to a conference. He was able to average 63 mph on the return trip and it took one hour less time. How long did it take Joe to drive home from the conference?
SOLUTION: Time taken by Joe to drive to the conference is represented by t. Time taken by Joe to return back home from the conference: t – 1. Distance travelled by Joe from home to the conference = D1 = rt = 42 mph⋅t = 42t Distance travelled by Joe back home = D2 = rt = 63 mph⋅ (t – 2) = 63(t – 1) Both the distances are equal. D1 = D2 42t = 63(t – 1) 42t = 63t – 63 42t – 63t = 63t – 63t – 63 –21t = –63
21 –6321 21
3
t
t
−=
− −=
ANSWER: Time taken by Joe to return back home from the conference: t – 1 = 3 – 1 = 2 hours.
25. PROBLEM:
A graphics designer wishes to crop an image in the width to height ratio of 3:2. If the height is required to be 400 pixels, then how many pixels should the width be set to?
SOLUTION: Number of pixels the width should be set to is represented by x. We should set up the ratios as the number of pixels set up as the width to the number of pixels set up as the height.
3400 2
3 400 6002
x
x
=
⋅= =
ANSWER: The width should be set to 600 pixels.
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