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63
Chapter 2 Equations, Problem Solving, and Inequalities
2.1 The Addition and Subtraction Properties of Equality
Problems 2.1
1. a. If x is 7, then x − 5 = 3 becomes
7 − 5 = 3, which is a false statement.
Hence, 7 is not a solution of the equation.
b. If y is 4, then 1 = 5 − y becomes
1 = 5 − 4, which is a true statement.
Thus, 4 is a solution of the equation.
c. If z is 6, 1
2 03
z − = becomes
1(6) 2 0,
3− = which is a true statement.
Thus, 6 is a solution of the equation.
2. a. 5 7
5 5 7 5
12
x
x
x
− =− + = +
=
The solution is 12.
CHECK x − 5 0 7
12 − 5 7
7
b. 1 3
5 51 1 3 1
5 5 5 54
5
x
x
x
− =
− + = +
=
The solution is 4
.5
CHECK 315 5
x − 0
4 15 5
− 35
35
3. a. 5 2 4 3 17
5 17
5 5 17 5
12
y y
y
y
y
+ − + =+ =
+ − = −=
Thus 12 is the solution of the equation.
CHECK 5y + 2 − 4y + 3 0 17
5(12) + 2 − 4(12) + 3 17
60 + 2 − 48 + 3
17
b. 3 1 5
5 68 8 8
2 5
8 82 2 5 2
8 8 8 83
8
z z
z
z
z
− + + − =
+ =
+ − = −
=
Thus, 3
8 is the solution of the equation.
CHECK 3 18 8
5 6z z− + + − 0 58
( ) ( )3 3 3 18 8 8 8
5 6− + + − 58
15 3 18 18 8 8 8
− + + −
58
4. a. 5 7
5 7 7 7
2
2
x
x
x
x
= +− = − +− =
= −
The solution is −2.
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CHECK 5 0 7 + x
5 7 + (−2)
5
b. 5 2 4 3
5 2 2 4 3 2
5 4 5
5 4 4 4 5
5
x x
x x
x x
x x x x
x
− = +− + = + +
= +− = − +
=
The solution is 5.
CHECK 5x − 2 0 4x + 3
5(5) − 2 4(5) + 3
25 − 2 20 + 3
23 23
c. 0 3( 3) 7 2
0 3 9 7 2
0 2
0 2 2 2
2
y y
y y
y
y
y
= − + −= − + −= −
+ = − +=
The solution is 2.
CHECK 0 0 3(y − 3) + 7 − 2y
0 3(2 − 3) + 7 − 2(2)
3(−1) + 7 − 4
−3 + 7 − 4
0
d. 3( 1) 4 8
3 3 4 8
3 3 3 4 8 3
3 4 5
3 4 4 4 5
5
5
z z
z z
z z
z z
z z z z
z
z
+ = ++ = +
+ − = + −= +
− = − +− =
= −
The solution is −5.
CHECK 3(z + 1) 0 4z + 8
3(−5 + 1) 4(−5) + 8
3(−4) −20 + 8
−12 −12
5. 6 5 7 2
6 5 5 7 2 5
6 7 3
6 7 7 7 3
3
3
y y
y y
y y
y y y y
y
y
+ = ++ − = + −
= −− = − −
− = −=
The solution is 3.
CHECK 6y + 5 0 7y + 2
6(3) + 5 7(3) + 2
18 + 5 21 + 2
23 23
6. 5 3( 1) 4 3
5 3 3 4 3
2 3 4 3
2 4 3 4 4 3
2 3 3
2 3 3 3 3
2 0
z z
z z
z z
z z
z z
z z z z
+ − = ++ − = +
+ = +− + = − +− + =
− + − = −− =
Since this statement is false, the equation has no solution.
7. 3 4( 2) 11 4
3 4 8 11 4
11 4 11 4
y y
y y
y y
+ + = ++ + = +
+ = +
Since both sides are identical, this equation is an identity. The solution is all real numbers.
8. a. Let x = 2003 − 1998 = 5
400 7400 400(5) 7400
2000 7400
9400
x + = += +=
The estimated tuition and fees in 2003 are $9400.
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Chapter 2: Equations, Problem Solving, and Inequalities Bello, Introductory Algebra, 3e ISM
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b. 400 7400 9000
400 9000 7400
400 1600
4
x
x
x
x
+ == −==
This means 4 years after 1998 or in 1998 + 4 = 2002, tuition and fees are estimated to be $9000.
Exercises 2.1
1. If x is 3, then x − 1 = 2 becomes 3 − 1 = 2,
which is a true statement. Thus, 3 is a solution of the equation.
2. If x is 4, then 6 = x − 10 becomes
6 = 4 − 10, which is a false statement.
Hence, 4 is not a solution of the equation.
3. If y is −2, then 3y + 6 = 0 becomes
3(−2) + 6 = 0, which is a true statement.
Thus, −2 is a solution of the equation.
4. If z is −3, then −3z + 9 = 0 becomes
−3(−3) + 9 = 0, which is a false statement.
Hence, −3 is not a solution of the equation.
5. If n is 2, then 12 − 3n = 6 becomes
12 − 3(2) = 6, which is a true statement.
Thus, 2 is a solution of the equation.
6. If m is 1
3 ,2
then 1
3 72
m+ = becomes
1 13 3 7,
2 2+ = which is a true statement.
Thus 1
32
is a solution of the equation.
7. If d is 10, then 2
1 35
d + = becomes
2(10) 1 3,
5+ = which is a false statement.
Hence, 10 is not a solution of the equation.
8. If c is 2.3, then 3.4 = 2c − 1.4 becomes
3.4 = 2(2.3) − 1.4, which is a false
statement. Hence 2.3 is not a solution of the equation.
9. If a is 2.1, then 4.6 = 11.9 − 3a becomes
4.6 = 11.9 − 3(2.1), which is a false
statement. Hence 2.1 is not a solution of the equation.
10. If x is 1
,10
then 7
0.2 510
x= − becomes
7 10.2 5 ,
10 10
= −
which is a true
statement. Thus 1
10 is a solution of the
equation.
11. 5 9
5 5 9 5
14
x
x
x
− =− + = +
=
The solution is 14.
CHECK x − 5 0 9
14 − 5 9
9
12. 3 6
3 3 6 3
9
y
y
y
− =− + = +
=
The solution is 9.
CHECK y − 3 0 6
9 − 3 6
6
13. 11 8
11 8 8 8
19
m
m
m
= −+ = − +
=
The solution is 19.
CHECK 11 0 m − 8
11 19 − 8
11
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Bello, Introductory Algebra, 3e ISM Chapter 2: Equations, Problem Solving, and Inequalities
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14. 6 2
6 2 2 2
8
n
n
n
= −+ = − +
=
The solution is 8.
CHECK 6 0 n − 2
6 8 − 2
6
15. 2 8
3 32 2 8 2
3 3 3 310
3
y
y
y
− =
− + = +
=
The solution is 10
.3
CHECK x − 23
0 83
10 23 3
− 83
83
16. 4 35
3 34 4 35 4
3 3 3 339
313
R
R
R
R
− =
− + = +
=
=
The solution is 13.
CHECK y − 43
0 353
43
13 − 353
353
17. 2 6 10 5
16 5
16 16 5 16
21
k k
k
k
k
− − − =− =
− + = +=
The solution is 21.
CHECK 2k − 6 − k − 10 0 5
2(21) − 6 − 21 − 10 5
42 − 6 − 21 − 10
5
18. 3 4 2 6 7
2 7
2 2 7 2
9
n n
n
n
n
+ − − =− =
− + = +=
The solution is 9.
CHECK 3n + 4 − 2n − 6 0 7
3(9) + 4 − 2(9) − 6 7
27 + 4 − 18 − 6
7
19. 1 2
24 31 2
4 31 2 2 2
4 3 3 33 8
12 1211
12
z z
z
z
z
z
= − −
= −
+ = − +
+ =
=
The solution is 11
.12
CHECK 14
0 23
2z z− −
14
( )11 2 1112 3 12
2 − −
822 1112 12 12
− −
312
14
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Chapter 2: Equations, Problem Solving, and Inequalities Bello, Introductory Algebra, 3e ISM
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20. 7 1
3 22 57 1
2 57 1 1 1
2 5 5 535 2
10 1037
10
v v
v
v
v
v
= − −
= −
+ = − +
+ =
=
The solution is 37
.10
CHECK 72
0 15
3 2v v− −
72
( ) ( )37 37110 5 10
3 2− −
74111 210 10 10
− −
3510
72
21. 3
0 2 22
3 40
2 27
02
7 7 70
2 2 27
2
x x
x
x
x
x
= − − −
= − −
= −
+ = − +
=
The solution is 7
.2
CHECK 0 0 32
2 2x x− − −
0 ( )7 3 72 2 2
2 2− − −
3 72 2
7 2− − −
102
5 −
5 − 5
0
22. 1 1
0 3 24 2
1 20
4 43
04
3 3 30
4 4 43
4
y y
y
y
y
y
= − − −
= − −
= −
+ = − +
=
The solution is 3
.4
CHECK 0 0 1 14 2
3 2y y− − −
0 ( ) ( )3 31 14 4 4 2
3 2− − −
9 61 24 4 4 4
− − −
04
0
23. 1 1
4 35 51 1
5 51 1 1 1
5 5 5 50
c c
c
c
c
= + −
= +
− = + −
=
The solution is 0.
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Bello, Introductory Algebra, 3e ISM Chapter 2: Equations, Problem Solving, and Inequalities
68
CHECK 15
0 15
4 3c c+ −
15
15
4(0) 3(0)+ −
15
0 0+ −
15
24. 19 1
0 6 52 2
180
20 9
0 9 9 9
9
b b
b
b
b
b
= + − −
= +
= +− = + −− =
The solution is −9.
CHECK 0 0 19 12 2
6 5b b+ − −
0 19 12 2
6( 9) 5( 9)− + − − −
19 12 2
54 45− + + −
182
9− +
−9 + 9
0
25. 3 3 4 0
3 0
3 3 0 3
3
x x
x
x
x
− + + =+ =
− + = −= −
The solution is −3.
CHECK −3x + 3 + 4x 0 0
−3(−3) + 3 + 4(−3) 0
9 + 3 − 12
0
26. 5 4 6 0
4 0
4 4 0 4
4
y y
y
y
y
− + + =+ =
− + = −= −
The solution is −4.
CHECK −5y + 4 + 6y 0 0
−5(−4) + 4 + 6(−4) 0
20 + 4 − 24
0
27. 3 1 1 1
4 4 4 41 1
4 41 1 1 1
4 4 4 40
y y
y
y
y
+ + =
+ =
+ − = −
=
The solution is 0.
CHECK 3 1 14 4 4
y y+ + 0 14
3 1 14 4 4
(0) (0)+ + 14
14
0 0+ +
14
28. 1 2 1 1
2 3 2 32 1
3 32 2 1 2
3 3 3 31
3
y y
y
y
y
+ + =
+ =
+ − = −
= −
The solution is 1
.3
−
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Chapter 2: Equations, Problem Solving, and Inequalities Bello, Introductory Algebra, 3e ISM
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CHECK 1 2 12 3 2
y y+ + 0 13
( ) ( )1 1 2 1 12 3 3 2 3
− + + − 13
1 2 16 3 6
− + −
2 23 6
−
2 13 3
−
13
29. 3.4 3 0.8 2 0.1
3.4 0.9
3.4 0.9 0.9 0.9
2.5
2.5
c c
c
c
c
c
= − + + += − +
− = − + −= −
− =
The solution is −2.5.
CHECK
3.4 0 −3c + 0.8 + 2c + 0.1
3.4 −3(−2.5) + 0.8 + 2(−2.5) + 0.1
7.5 + 0.8 − 5 + 0.1
3.4
30. 1.7 3 0.3 4 0.4
1.7 0.7
1.7 0.7 0.7 0.7
1
c c
c
c
c
= − + + += +
− = + −=
The solution is 1.
CHECK 1.7 0 −3c + 0.3 + 4c + 0.4
1.7 −3(1) + 0.3 + 4(1) + 0.4
−3 + 0.3 + 4 + 0.4
1.7
31. 6 9 5
6 6 9 5 6
9
9
p p
p p p p
p
p
+ =− + = −
= −− =
The solution is −9.
CHECK 6p + 9 0 5p
6(−9) + 9 5(−9)
−54 + 9 −45
−45
32. 7 4 6
7 7 4 6 7
4
4
q q
q q q q
q
q
+ =− + = −
= −− =
The solution is −4.
CHECK 7q + 4 0 6q
7(−4) + 4 6(−4)
−28 + 4 −24
−24
33. 3 3 2 4
5 3 4
5 5 3 4 5
3
3
x x x
x x
x x x x
x
x
+ + =+ =
− + = −= −
− =
The solution is −3.
CHECK 3x + 3 + 2x 0 4x
3(−3) + 3 + 2(−3) 4(−3)
−9 + 3 − 6 −12
−12
34. 2 4 6 7
8 4 7
8 8 4 7 8
4
4
y y y
y y
y y y y
y
y
+ + =+ =
− + = −= −
− =
The solution is −4.
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Bello, Introductory Algebra, 3e ISM Chapter 2: Equations, Problem Solving, and Inequalities
70
CHECK 2y + 4 + 6y 0 7y
2(−4) + 4 + 6(−4) 7(−4)
−8 + 4 + (−24) −28
−28
35. 4( 2) 2 3 0
4 8 2 3 0
6 0
6 6 0 6
6
m m
m m
m
m
m
− + − =− + − =
− =− + = +
=
The solution is 6.
CHECK 4(m − 2) + 2 − 3m 0 0
4(6 − 2) + 2 − 3(6) 0
4(4) + 2 − 18
16 + 2 − 18
0
36. 3( 4) 2 2
3 12 2 2
3 14 2
3 3 14 2 3
14
14
n n
n n
n n
n n n n
n
n
+ + =+ + =
+ =− + = −
= −− =
The solution is −14.
CHECK 3(n + 4) + 2 0 2n
3(−14 + 4) + 2 2(−14)
3(−10) + 2 −28
−30 + 2
−28
37. 5( 2) 4 8
5 10 4 8
5 10 10 4 8 10
5 4 18
5 4 4 4 18
18
y y
y y
y y
y y
y y y y
y
− = +− = +
− + = + += +
− = − +=
The solution is 18.
CHECK 5(y − 2) 0 4y + 8
5(18 − 2) 4(18) + 8
5(16) 72 + 8
80 80
38. 3( 1) 4 1
3 3 4 1
3 3 3 4 1 3
3 4 4
3 4 4 4 4
4
4
z z
z z
z z
z z
z z z z
z
z
− = +− = +
− + = + += +
− = − +− =
= −
The solution is −4.
CHECK 3(z − 1) 0 4z + 1
3(−4 − 1) 4(−4) + 1
3(−5) −16 + 1
−15 −15
39. 3 1 2( 4)
3 1 2 8
3 1 1 2 8 1
3 2 7
3 2 2 2 7
7
a a
a a
a a
a a
a a a a
a
− = −− = −
− + = − += −
− = − −= −
The solution is −7.
CHECK 3a − 1 0 2(a − 4)
3(−7) − 1 2(−7 − 4)
−21 − 1 2(−11)
−22 −22
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Chapter 2: Equations, Problem Solving, and Inequalities Bello, Introductory Algebra, 3e ISM
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40. 4( 1) 5 3
4 4 5 3
4 4 4 5 3 4
4 5 7
4 5 5 7 5
7
7
b b
b b
b b
b b
b b b b
b
b
+ = −+ = −
+ − = − −= −
− = − −− = −
=
The solution is 7.
CHECK 4(b + 1) 0 5b − 3
4(7 + 1) 5(7) − 3
4(8) 35 − 3
32 32
41. 5( 2) 6 2
5 10 6 2
5 10 10 6 2 10
5 6 8
5 6 6 6 8
8
8
c c
c c
c c
c c
c c c c
c
c
− = −− = −
− + = − += +
− = − +− =
= −
The solution is −8.
CHECK 5(c − 2) 0 6c − 2
5(−8 − 2) 6(−8) − 2
5(−10) −48 − 2
−50 −50
42. 4 6 5 3 8
4 6 3 13
4 6 13 3 13 13
4 7 3
4 4 7 3 4
7
R R
R R
R R
R R
R R R R
R
− + = − +− + = − +
− + − = − + −− − = −
− + − = − +− =
The solution is −7.
CHECK −4R + 6 0 5 − 3R + 8
−4(−7) + 6 5 − 3(−7) + 8
28 + 6 5 + 21 + 8
34 34
43. 3 5 2 1 6 4 6
6 4
6 6 4 6
2
x x x x
x
x
x
+ − + = + −+ =
+ − = −= −
The solution is −2.
CHECK
3x + 5 − 2x + 1 0 6x + 4 − 6x
3(−2) + 5 − 2(−2) + 1 6(−2) + 4 − 6(−2)
−6 + 5 + 4 + 1 −12 + 4 + 12
4 4
44. 6 2 4 2 5 3
2 2 5
2 2 2 5 2
2 7
2 7
7
f f f f
f f
f f
f f
f f f f
f
− − = − + +− = +
− + = + += +
− = − +=
The solution is 7. CHECK
6f − 2 − 4f 0 −2f + 5 + 3f
6(7) − 2 − 4(7) −2(7) + 5 + 3(7)
42 − 2 − 28 −14 + 5 + 21
12 12
45. 2 4 5 6 1 14
7 4 8 1
7 4 4 8 1 4
7 8 3
7 8 8 8 3
3
g g g g
g g
g g
g g
g g g g
g
− + − = + −− + = − +
− + − = − + −− = − −
− + = − + −= −
The solution is −3.
CHECK
−2g + 4 − 5g 0 6g + 1 − 14g
−2(−3) + 4 − 5(−3) 6(−3) + 1 − 14(−3)
6 + 4 + 15 −18 + 1 + 42
25 25
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72
46. 2 3 9 6 1
7 3 6 1
7 3 3 6 1 3
7 6 4
7 6 6 6 4
4
x x x
x x
x x
x x
x x x x
x
− + + = −+ = −
+ − = − −= −
− = − −= −
The solution is −4.
CHECK
−2x + 3 + 9x 0 6x − 1
−2(−4) + 3 + 9(−4) 6(−4) − 1
8 + 3 − 36 −24 − 1
−25 −25
47. 6( 4) 4 2 4
6 24 4 2 4
4 28 4
4 4 28 4 4
28 0
x x x
x x x
x x
x x x x
+ + − =+ + − =
+ =− + = −
=
Since this statement is false, the equation has no solution.
48. 6( 1) 2 2 8 4
6 6 2 2 8 4
8 8 8 4
8 8 8 8 8 4
8 4
y y y
y y y
y y
y y y y
− − + = +− − + = +
− = +− − = − +
− =
Since this statement is false, the equation has no solution.
49. 10( 2) 10 2 8( 1) 18
10 20 10 2 8 8 18
8 10 8 10
z z z
z z z
z z
− + − = + −− + − = + −
− = −
Since both sides are identical, this equation is an identity. The solution is all real numbers.
50. 7( 1) 1 6( 1)
7 7 1 6 6
6 6 6 6
a a a
a a a
a a
+ − − = ++ − − = +
+ = +
Since both sides are identical, this equation is an identity. The solution is all real numbers.
51. 3 6 2 2( 2) 4
6 2 4 4
6 2
6 2
6
b b b
b b
b b
b b b b
b
+ − = − ++ = − ++ =
− + = −=
The solution is 6.
CHECK 3b + 6 − 2b 0 2(b − 2) + 4
3(6) + 6 − 2(6) 2(6 − 2) + 4
18 + 6 − 12 2(4) + 4
12 8 + 4
12
52. 3 2 3( 2) 5
2 2 3 6 5
2 2 3 1
2 3 3
3
b b b
b b
b b
b b
b
+ − = − +
+ = − ++ = −
+ ==
The solution is 3.
CHECK 3b + 2 − b 0 3(b − 2) + 5
3(3) + 2 − 3 3(3 − 2) + 5
9 + 2 − 3 3(1) + 5
8 8
53.
2 12 5 4 7
3 32 1
3 4 73 3
2 2 1 23 4 7
3 3 3 322 2
3 43 3
203 4
320
3 4 4 43
20
3
p p p
p p
p p
p p
p p
p p p p
p
+ − = − +
− = − +
− − = − + −
− = − + −
− = − +
− + = − + +
=
The solution is 20
.3
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Chapter 2: Equations, Problem Solving, and Inequalities Bello, Introductory Algebra, 3e ISM
73
CHECK
23
2 5p p+ − 0 13
4 7p− +
( ) ( )20 2023 3 3
2 5+ − ( )20 13 3
4 7− +
40 10023 3 3
+ − 80 223 3
− +
583
− 583
−
54. 2 2
4 6 3 27 7
2 162 3
7 72 3 2
2
q q q
q q
q q
q
+ − =− +
− + =− +
− =− +=
The solution is 2.
55.
3 15 9 5 1
8 23 3
4 58 2
3 3 3 34 5
8 8 2 89
4 58
94 5 5 5
89
8
r r r
r r
r r
r r
r r r r
r
+ − = − +
− + = − +
− + − = − + −
− = − +
− + = − + +
=
The solution is 9
.8
CHECK
38
5 9r r+ − 0 12
5 1r− +
( ) ( )9 3 98 8 8
5 9+ − ( )9 18 2
5 1− +
45 3 818 8 8
+ − 45 128 8
− +
338
− 338
−
56. Let p = old price. 7 23
16
p
p
+ =
=
The old price was $16.
57. Let h = average hourly earnings the previous year.
9.81 0.40
9.81 0.40 0.40 0.40
9.41
h
h
h
= +− = + −
=
The average hourly earnings the previous year were $9.41.
58. Let x = Consumer Price Index for previous year. 169.6 5.7
163.9
x
x
= +
=
The Consumer Price Index for housing was 163.9 the previous year.
59. Let y = cost 6 years ago. 142.2 326.9
142.2 142.2 326.9 142.2
184.7
y
y
y
+ =+ − = −
=
The cost of medical care was 184.7 points 6 years ago.
60. Let s = score 10 years ago. 476 16
492
s
s
= −
=
Ten years ago, the mathematics score was 492.
61. Let w = waste generated in 1960. 236.2 148.1
236.2 148.1 148.1 148.1
88.1
w
w
w
= +
− = + −=
In 1960 there was 88.1 million tons of waste generated. Let i = increase from 1960.
88.1 234
88.1 88.1 234 88.1
145.9
i
i
i
+ =
− + = −=
The increase from 1960 was 145.9 million tons.
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Bello, Introductory Algebra, 3e ISM Chapter 2: Equations, Problem Solving, and Inequalities
74
62. Let a = waste recovered in 1960. 66.7 72.3
5.6
a
a
+ =
=
In 1960 there was 5.9 million tons of waste recovered. Let b = waste recovered in 2000.
3.4 72.3
68.9
b
b
+ ==
In 2000 there was 68.9 million tons of waste recovered.
63. Let m = percent of males. 38 17
38 17 17 17
21
m
m
m
= +− = − +
=
21% of males engage in exercise-walking.
64. Let f = percent of females 26 1
25
f
f
= +=
25% of females participate in swimming.
65. Let p = percent that attended art museums. 35 8
35 8 8 8
27
p
p
p
= +
− = + −=
27% attended art museums.
66. Let x = consumption in 1970. 16 113.5
129.5
x
x
− =
=
The per capita consumption in 1970 was 129.5 pounds. Let y = consumption in 1999.
113.5 1.1
112.4
y
y
= +
=
The per capita consumption in 1999 was 112.4 pounds.
67. a. 100 5( 60)
120 100 5(62 60)
120 100 5(2)
120 110 is false
W h= + −
= + −
= +
=
No, her weight does not satisfy the equation.
b. 100 5( 60)
100 5(62 60)
100 5(2)
110
W h= + −
= + −
= +
=
Her weight should be 110 lb.
68. a. 110 5( 60)
160 110 5(70 60)
160 160 is true
W h= + −
= + −
=
Yes, his weight does satisfy the equation.
b. 110 5(70 60)
110 5(10)
160 lb
W = + −
= +
=
69. a. 50 2.3( 60)
75 50 2.3(70 60)
75 50 2.3(10)
75 73 is false
W h= + −
= + −
= +
=
No, his weight doesn’t satisfy the equation.
b. He is 75 – 73 = 2 kg overweight.
70. a. 45.5 2.3( 60)
68.5 45.5 2.3(70 60)
68.5 68.5 is true
W h= + −
= + −
=
Yes, her weight does satisfy the equation.
b. 45.5 2.3( 60)
45.5 2.3(70 60) 68.5 kg
W h= + −
= + − =
71. a. 52 1.9( 60)
70 52 1.9(70 60)
70 52 1.9(10)
70 71 is false
W h= + −
= + −
= +
=
No, his weight does not satisfy the equation.
b. He is 71 – 70 = 1 kg underweight.
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Chapter 2: Equations, Problem Solving, and Inequalities Bello, Introductory Algebra, 3e ISM
75
72. a. 49 1.7( 60)
65 49 1.7(70 60)
65 66 is false
W h= + −
= + −
=
No, her weight doesn’t satisfy the equation.
b. She is 66 – 65 = 1 kg underweight.
73. a. 56.2 1.41( 60)
70.3 56.2 1.41(70 60)
70.3 56.2 14.1
70.3 70.3 is true
W h= + −
= + −
= +
=
Yes, his weight does satisfy the equation.
b. 56.2 1.41(70 60)
56.2 1.41(10)
56.2 14.1
70.3 lb
W = + −
= +
= +
=
74. a. 12 500
940 12(120) 500
940 940 is true
C W= −
= −
=
Yes, this caloric intake does satisfy the equation.
b. 12(120) 500 1440 500 940C = − = − =
75. a. 14 500
1700 12(150) 500
1700 1600 is false
C W= −
= −
=
No, this caloric intake does satisfy the equation.
b. He is 1700 – 1600 = 100 calories over.
76. a. 17 500
2900 17(200) 500
2900 2900 is true
C W= −
= −
=
Yes, this caloric intake does satisfy the equation.
b. 17(200) 500
3400 500
2900
C = −
= −
=
77. If f = 40 and H = 120, then H = 1.95f + 72.85 becomes 120 = 1.95(40) + 72.85 which is a false statement. Therefore, the bone cannot belong to the missing female.
78. If f = 40 and H = 150.85, then H = 1.95f + 72.85 becomes 150.85 = 1.95(40) + 72.85 which is a true statement. Therefore, the bone can belong to the missing female.
79. If t 36,L = a t144, 30L V= = and a 50,V =
then 2
2 a ta
t
L VV
L= becomes
22 144(30 )
50 ,36
=
which is a false statement. Therefore, you cannot believe him.
80. If t a t36, 144, 30L L V= = = and a 90,V =
then 2
2 a ta
t
L VV
L= becomes
22 144(30 )
90 ,36
=
which is a false statement. Therefore, you cannot believe him.
81. Answers may vary.
82. Answers may vary.
83. Answers may vary.
84. If −x = −5, then x = 5. Answers may vary.
85. a c b c+ = +
86. a c b c− = −
87. 5 4( 1) 3 4
5 4 4 3 4
9 4 3 4
9 4 4 3 4 4
9 3
x x
x x
x x
x x x x
+ + = ++ + = +
+ = ++ − = + −
=
Since this statement is false, there is no solution.
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Bello, Introductory Algebra, 3e ISM Chapter 2: Equations, Problem Solving, and Inequalities
76
88. 2 4( 1) 5 3
2 4 4 5 3
4 6 5 3
4 6 6 5 3 6
4 5 3
4 5 5 5 3
3
3
x x
x x
x x
x x
x x
x x x x
x
x
+ + = ++ + = +
+ = ++ − = + −
= −− = − −
− = −=
The solution is 3.
89. 5 4
5 5 4 5
9
x
x
x
− =− + = +
=
The solution is 9.
90. 1 3
5 51 1 3 1
5 5 5 54
5
x
x
x
− =
− + = +
=
The solution is 4
.5
91. 2.3 3.4
2.3 2.3 3.4 2.3
5.7
x
x
x
− =− + = +
=
The solution is 5.7.
92. 1 1
7 41 1 1 1
7 7 4 77 4
28 2811
28
x
x
x
x
− =
− + = +
= +
=
The solution is 11
.28
93. 2 6 2 12
8 12
8 8 12 8
4
x x
x
x
x
+ − + =+ =
+ − = −=
The solution is 4.
94. 3 5 2 3 7
8 7
8 8 7 8
1
x x
x
x
x
+ − + =+ =
+ − = −= −
The solution is −1.
95. 2 4 5
5 69 9 9
2 5
9 92 2 5 2
9 9 9 97
9
x x
x
x
x
− + + − =
− =
− + = +
=
The solution is 7
.9
96. 5 2 4 1
5 2 2 4 1 2
5 4 3
5 4 4 4 3
3
y y
y y
y y
y y y y
y
− = +− + = + +
= +− = − +
=
The solution is 3.
97. 0 4( 3) 5 3
0 4 12 5 3
0 7
0 7 7 7
7
z z
z x
z
z
z
= − + −= − + −= −
+ = − +=
The solution is 7.
98. 2 (4 1) 1 4
2 4 1 1 4
1 4 4
x x
x x
x x
− + = −− − = −
− = −
Since both sides are identical, this is an identity. The solution is all real numbers.
99. 3( 2) 3 2 (1 3 )
3 6 3 2 1 3
3 9 1 3
3 3 9 1 3 3
9 1
x x
x x
x x
x x x x
+ + = − −+ + = − +
+ = +− + = + −
=
This statement is false, so there is no solution.
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Chapter 2: Equations, Problem Solving, and Inequalities Bello, Introductory Algebra, 3e ISM
77
100. 3( 1) 3 2 5
3 3 3 2 5
3 2 5
3 2 2 2 5
5
x x
x x
x x
x x x x
x
+ − = −+ − = −
= −− = − −
= −
The solution is −5.
101. If z = −7, then 1
1 07
z − = becomes
1( 7) 1 0,
7− − = which is a false statement.
Hence, −7 is not a solution of the equation.
102. If 3
,5
x = − then 5
1 03
x + = becomes
5 31 0,
3 5
− + =
which is a true statement.
Thus, 3
5− is a solution of the equation.
103. 4(−5) = −20
104. 6(−3) = −18
105. 2 3 6 1
3 4 12 2
− = − = −
106. 5 7 35 1
7 10 70 2
− = − = −
107. The reciprocal of 3
2 is
2.
3
108. The reciprocal of 2
5 is
5.
2
109. 6 = 2 ⋅ 3
16 = 2 ⋅ 2 ⋅ 2 ⋅ 2
LCM = 2 ⋅ 2 ⋅ 2 ⋅ 2 ⋅ 3 = 48
110. 9 = 3 ⋅ 3
12 = 2 ⋅ 2 ⋅ 3
LCM = 2 ⋅ 2 ⋅ 3 ⋅ 3 = 36
111. 10 = 2 ⋅ 5
8 = 2 ⋅ 2 ⋅ 2
LCM = 2 ⋅ 2 ⋅ 2 ⋅ 5 = 40
112. 30 = 2 ⋅ 3 ⋅ 5
18 = 2 ⋅ 3 ⋅ 3
LCM = 2 ⋅ 3 ⋅ 3 ⋅ 5 = 90
2.2 The Multiplication and Division Properties of Equality
Problems 2.2
1. a. 35
5 5 35
15
x
x
x
=
⋅ = ⋅
=
The solution is 15.
CHECK 5x 0 3
155
3
3
b. 54
4 4( 5)4
20
y
y
y
=−
⋅ = −
=−
The solution is −20.
CHECK
4
y 0 −5
204
− −5
−5
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Bello, Introductory Algebra, 3e ISM Chapter 2: Equations, Problem Solving, and Inequalities
78
2. a. 3 12
3 12
3 34
x
x
x
=
=
=
The solution is 4.
CHECK 3x 0 12
3 ⋅ 4 12
12
b. 7 21
7 21
7 73
x
x
x
=−
−=
=−
The solution is −3.
CHECK 7x 0 −21
7(−3) −21
−21
c. 5 20
5
x− =
−
5
x
−
20
5
4x
=−=−
The solution is −4.
CHECK −5x 0 20
−5(−4) 20
20
3. a. 3
125
5 3 5(12)
3 5 3
5 121
3
x
x
x
=
=
⋅ = ⋅5 4
1 120x
⋅=
=
The solution is 20.
CHECK 35
x 0 12
35
(20) 12
12
b. 2
65
5 2 5(6)
2 5 2
5 6 5 31
2 1 115
x
x
x
x
− =
− − =−
⋅⋅ =− ⋅ =−
=−
The solution is −15.
CHECK 25
x− 0 6
25
( 15)− − 6
6
c. 4
85
5 4 5( 8)
4 5 4
5 ( 8)1
x
x
x
− =−
− − =− −
− −⋅ =
5( 2)
4 110x
− −=
=
The solution is 10.
CHECK 45
x− 0 −8
45
(10)− −8
−8
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Chapter 2: Equations, Problem Solving, and Inequalities Bello, Introductory Algebra, 3e ISM
79
4. a. 2 10 6
5 3
LCM of 10 and 6 is 2 ⋅ 5 ⋅ 3 = 30.
810 6
30 30 30 810 6
3 5 30 8
8 30 8
30
x x
x x
x x
x
x
+ =
⋅ + ⋅ = ⋅
+ = ⋅= ⋅=
The solution is 30.
CHECK 10 6x x+ 0 8
30 3010 6
+ 8
3 + 5
8
b. The LCM of 4 and 5 is 4 ⋅ 5 = 20.
14 5
20 20 20 14 5
5 4 20
20
x x
x x
x x
x
− =
⋅ − ⋅ = ⋅
− ==
The solution is 20.
CHECK 4 5x x− 0 1
20 204 5
− 1
5 − 4
1
5. a. The LCM of 3 and 4 is 3 ⋅ 4 = 12.
2 26
3 4
4 3
12
x x+ −+ =
212
3
x +⋅ +
212 6
44( 2) 3( 2) 72
4 8 3 6 72
7 2 72
7 70
10
x
x x
x x
x
x
x
−⋅ = ⋅
+ + − =+ + − =
+ ===
The solution is 10.
CHECK 2 23 4
x x+ −+ 0 6
10 2 10 23 4+ −+ 6
8123 4
+
4 + 2
6
b. The LCM of 5 and 3 is 5 ⋅ 3 = 15.
2 20
5 32 2
15 15 15(0)5 3
3( 2) 5( 2) 0
3 6 5 10 0
2 16 0
2 16
8
x x
x x
x x
x x
x
x
x
+ −− =
+ − − =
+ − − =+ − + =
− + =− = −
=
The solution is 8.
6. Forty percent of 30 is what number?
4030
1002
305
60
512
n
n
n
n
⋅ =
⋅ =
=
=
Thus, 40% of 30 is 12.
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Bello, Introductory Algebra, 3e ISM Chapter 2: Equations, Problem Solving, and Inequalities
80
7. What percent of 30 is 6? 30 6
30 6
30 301 20
5 100
x
x
x
⋅ =⋅
=
= =
Thus, 6 is 20% of 30.
8. 20 is 40% of what number?
10
4020
10040
20100
220
5
5 2 520
2 5 2
n
n
n
n
= ⋅
⋅ =
⋅ =
⋅ ⋅ = ⋅
1
50n =
Thus, 20 is 40% of 50.
9. a. 5.3% of 225 means 0.053 ⋅ 225 = 11.925
or 12 when rounded to the nearest whole number.
b. 10.6% of 226 means
0.106 ⋅ 226 = 23.956 or 24 when
rounded to the nearest whole number.
Exercises 2.2
Checks are left to the student.
1. 57
7 7 57
35
x
x
x
=
⋅ = ⋅
=
The solution is 35.
2. 92
2 2 92
18
y
y
y
=
⋅ = ⋅
=
The solution is 18.
3. 42
2 ( 4) 22
8
x
x
x
− =
⋅ − = ⋅
− =
The solution is −8.
4. 65
5 5 65
30
a
a
a
=
⋅ = ⋅
=
The solution is 30.
5. 53
3 3 53
15
b
b
b
=−
− ⋅ = − ⋅−
= −
The solution is −15.
6. 74
4 7 44
28
c
c
c
=−
− ⋅ = − − − =
The solution is −28.
7. 32
2( 3) 22
6
f
f
f
− =−
− − = − − =
The solution is 6.
8. 64
4 4( 6)4
24
g
g
g
= −−
− = − − − =
The solution is 24.
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Chapter 2: Equations, Problem Solving, and Inequalities Bello, Introductory Algebra, 3e ISM
81
9. 1
4 31
4 44 3
4
3
v
v
v
=
⋅ = ⋅
=
The solution is 4
.3
10. 2
3 72
3 33 7
6
7
w
w
w
=
⋅ = ⋅
=
The solution is 6
.7
11. 3
5 43
5 55 4
15
4
x
x
x
−=
− ⋅ = ⋅
= −
The solution is 15
.4
−
12. 8
9 28
2 29 2
16
9
y
y
y
−=
⋅ − = ⋅
− =
The solution is 16
.9
−
13. 3 33
3 33
3 311
z
z
z
=
=
=
The solution is 11.
14. 4 32
4 32
4 48
y
y
y
=
=
=
The solution is 8.
15. 42 6
42 6
6 67
x
x
x
− =−
=
− =
The solution is −7.
16. 7 49
7 49
7 77
b
b
b
= −−
=
= −
The solution is −7.
17. 8 56
8 56
8 87
c
c
c
− =−
=− −
= −
The solution is −7.
18. 5 45
5 45
5 59
d
d
d
− =−
=− −
= −
The solution is −9.
19. 5 35
5 35
5 57
x
x
x
− = −− −
=− −
=
The solution is 7.
20. 12 3
12 3
3 34
x
x
x
− = −− −
=− −
=
The solution is 4.
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Bello, Introductory Algebra, 3e ISM Chapter 2: Equations, Problem Solving, and Inequalities
82
21. 3 11
3 11
3 311
3
y
y
y
− =−
=− −
= −
The solution is 11
.3
−
22. 5 17
5 17
5 517
5
z
z
z
− =−
=− −
= −
The solution is 17
.5
−
23. 2 1.2
2 1.2
2 20.6
a
a
a
− =−
=− −
= −
The solution is −0.6.
24. 3 1.5
3 1.5
3 30.5
b
b
b
− =−
=− −
= −
The solution is −0.5.
25.
3
1
13 4
29
32
1 1 93
3 3 2
3
2
t
t
t
t
=
=
⋅ = ⋅
=
The solution is 3
.2
26. 2
4 63
204
35
3
r
r
r
=
=
=
The solution is 5
.3
27. 1
0.7531
3 3( 0.75)3
2.25
x
x
x
= −
⋅ = −
= −
The solution is −2.25.
28. 1
0.2541
4 4 0.254
1.00
y
y
y
=
⋅ = ⋅
=
The solution is 1.00.
29.
2
36
4
4( 6)
3
C
−
− =
⋅ −
1
4 3
3 4
8
C
C
= ⋅
− =
The solution is −8.
30. 2
29
9
F
F
− =
− =
The solution is −9.
31. 5
106
6 5 610
5 6 5
a
a
=
⋅ = ⋅
1
12a=
The solution is 12.
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Chapter 2: Equations, Problem Solving, and Inequalities Bello, Introductory Algebra, 3e ISM
83
32. 2
247
84
z
z
=
=
The solution is 84.
33. 4
0.45
5 4 5
4 5 4
y
y
− =
− ⋅ − =− 1
0.4⋅0.1
0.5y=−
The solution is −0.5.
34. 1
0.54
1 1
2 41
2
x
x
x
−=
−=
=−
The solution is 1
.2
−
35. 2
011
11 2 110
2 11 2
0
p
p
p
−=
− ⋅ = ⋅ − − =
The solution is 0.
36. 4
09
9 4 90
4 9 4
0
q
q
q
−=
− ⋅ = ⋅ − − =
The solution is 0.
37. 3
185
5
3
t− =
1
( 18)⋅ −6
5 3
3 5
30
t
t
−
= ⋅
− =
The solution is −30.
38. 2
87
28
R
R
− =
− =
The solution is −28.
39. 7
70.02
0.02 7 0.02( 7)
7 0.02 7
0.02
x
x
x
= −
= ⋅ −
= −
The solution is −0.02.
40. 6
60.03
0.03 0.03 6( 6)
6 6 0.030.03
y
y
y
− =
⋅ − = ⋅
− =
The solution is −0.03.
41. The LCM of 2 and 3 is 2 ⋅ 3 = 6.
102 3
6 6 6 102 33 2 60
5 60
12
y y
y y
y y
y
y
+ =
⋅ + ⋅ = ⋅
+ ===
The solution is 12.
42. The LCM of 4 and 3 is 4 ⋅ 3 = 12.
144 3
12 12 12 144 3
3 4 168
7 168
24
a a
a a
a a
a
a
+ =
⋅ + ⋅ = ⋅
+ ===
The solution is 24.
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Bello, Introductory Algebra, 3e ISM Chapter 2: Equations, Problem Solving, and Inequalities
84
43. The LCM of 7 and 3 is 7 ⋅ 3 = 21.
107 3
21 21 21 107 3
3 7 210
10 210
21
x x
x x
x x
x
x
+ =
⋅ + ⋅ = ⋅
+ ===
The solution is 21.
44. 2 6 4
3 2
The LCM of 6 and 4 is 2 ⋅ 3 ⋅ 2 = 12.
206 4
12 12 12 206 4
2 3 240
5 240
48
z z
z z
z z
z
z
+ =
⋅ + ⋅ = ⋅
+ ===
The solution is 48.
45. The LCM of 5 and 10 is 10.
65 10
10 10 10 65 10
2 60
3 60
20
x x
x x
x x
x
x
+ =
⋅ + ⋅ = ⋅
+ ===
The solution is 20.
46. The LCM of 2 and 6 is 6.
82 6
6 6 6 82 6
3 48
4 48
12
r r
r r
r r
r
r
+ =
⋅ + ⋅ = ⋅
+ ===
The solution is 12.
47. 2 6 8
3 4
The LCM of 6 and 8 is 2 ⋅ 3 ⋅ 4 = 24.
76 8
24 24 24 76 8
4 3 168
7 168
24
t t
t t
t t
t
t
+ =
⋅ + ⋅ = ⋅
+ ===
The solution is 24.
48. 3 9 12
3 4
The LCM of 9 and 12 is 3 ⋅ 3 ⋅ 4 = 36.
149 12
36 36 36 149 12
4 3 504
7 504
72
f f
f f
f f
f
f
+ =
⋅ + ⋅ = ⋅
+ ===
The solution is 72.
49. The LCM of 2 and 5 is 2 ⋅ 5 = 10.
7
2 5 107
10 10 102 5 10
5 2 7
7 7
1
x x
x x
x x
x
x
+ =
⋅ + ⋅ = ⋅
+ ===
The solution is 1.
50. The LCM of 3 and 7 is 3 ⋅ 7 = 21.
20
3 7 2120
21 21 213 7 21
7 3 20
10 20
2
a a
a a
a a
a
a
+ =
⋅ + ⋅ = ⋅
+ ===
The solution is 2.
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Chapter 2: Equations, Problem Solving, and Inequalities Bello, Introductory Algebra, 3e ISM
85
51. The LCM of 3 and 5 is 3 ⋅ 5 = 15.
23 5
15 15 15 23 5
5 3 30
2 30
15
c c
c c
c c
c
c
− =
⋅ − ⋅ = ⋅
− ===
The solution is 15.
52. The LCM of 4 and 7 is 4 ⋅ 7 = 28.
34 7
28 28 28 34 7
7 4 84
3 84
28
F F
F F
F F
F
F
− =
⋅ − ⋅ = ⋅
− ===
The solution is 28.
53. 6 = 2 ⋅ 3
8 = 2 ⋅2 ⋅ 2
12 = 2 ⋅ 2 ⋅ 3
LCM = 2 ⋅ 2 ⋅ 2 ⋅ 3 = 24
5
6 8 125
24 24 246 8 12
4 3 10
10
W W
W W
W W
W
− =
⋅ − ⋅ = ⋅
− ==
The solution is 10.
54. 6 = 2 ⋅ 3
10 = 2 ⋅ 5
3 = 3
LCM = 2 ⋅ 3 ⋅ 5 = 30
4
6 10 34
30 30 306 10 3
5 3 40
2 40
20
m m
m m
m m
m
m
− =
⋅ − ⋅ = ⋅
− ===
The solution is 20.
55. The LCM of 5, 10, and 2 is 10.
3 1
5 10 23 1
10 10 105 10 2
2 3 5
2 8
4
x
x
x
x
x
− =
⋅ − ⋅ = ⋅
− ===
The solution is 4.
56. The LCM of 7 and 14 is 14.
3 1 1
7 14 143 1 1
14 14 147 14 14
6 1 1
6 2
2 1
6 3
y
y
y
y
y
− =
⋅ − ⋅ = ⋅
− ==
= =
The solution is 1
.3
57. The LCM of 4, 3, and 2 is 4 ⋅ 3 = 12.
4 2 1
4 3 24 2 1
12 12 124 3 2
3( 4) 4( 2) 6
3 12 4 8 6
4 6
10
10
x x
x x
x x
x x
x
x
x
+ +− = −
+ + ⋅ − ⋅ = ⋅ −
+ − + = −+ − − = −
− + = −− = −
=
The solution is 10.
58. The LCM of 2, 8, and 16 is 16.
1 7 1
2 8 161 7 1
16 16 162 8 16
8( 1) 2 7 1
8 8 2 7 1
10 8 7 1
10 7 9
3 9
3
w w w
w w w
w w w
w w w
w w
w w
w
w
− ++ =
− +⋅ + ⋅ = ⋅
− + = +− + = +
− = += +==
The solution is 3.
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Bello, Introductory Algebra, 3e ISM Chapter 2: Equations, Problem Solving, and Inequalities
86
59. 2 6 4
3 2
The LCM of 6 and 4 is 2 ⋅ 3 ⋅ 2 = 12.
3 7
6 4 43 7
12 12 12 126 4 4
2 9 12 21
2 12 30
10 30
3
xx
xx
x x
x x
x
x
+ = −
⋅ + ⋅ = ⋅ − ⋅
+ = −= −
− = −=
The solution is 3.
60. The LCM of 6 and 3 is 6.
4 1
6 3 34 1
6 6 6 66 3 3
8 6 2
6 10
5 10
2
xx
xx
x x
x x
x
x
+ = −
⋅ + ⋅ = ⋅ − ⋅
+ = −= −
− = −=
The solution is 2.
61. 30% of 40 is what number?
3040
1003
4010
12
n
n
n
⋅ =
⋅ =
=
Thus, 30% of 40 is 12.
62. 17% of 80 is what number?
1780
10068
513.6
n
n
n
⋅ =
=
=
Thus, 17% of 80 is 13.6.
63. 40% of 70 is what number?
4070
100
2
5
n⋅ =
70⋅14
28
n
n
=
=
Thus, 40% of 70 is 28.
64. What percent of 40 is 8? 40 8
40 8
40 401 20
5 100
x
x
x
⋅ =⋅
=
= =
Thus, 8 is 20% of 40.
65. What percent of 30 is 15? 30 15
30 15
30 301 50
2 10
x
x
x
⋅ =⋅
=
= =
Thus, 15 is 50% of 30.
66. What percent of 40 is 4? 40 4
40 4
40 401 10
10 100
x
x
x
⋅ =⋅
=
= =
Thus, 4 is 10% of 40.
67. 30 is 20% of what number?
2030
1001
305
15 30 5
5150
n
n
n
n
= ⋅
= ⋅
⋅ = ⋅ ⋅
=
Thus, 30 is 20% of 150.
68. 20 is 40% of what number?
4020
1002
205
5 5 220
2 2 550
n
n
n
n
= ⋅
= ⋅
⋅ = ⋅ ⋅
=
Thus, 20 is 40% of 50.
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Chapter 2: Equations, Problem Solving, and Inequalities Bello, Introductory Algebra, 3e ISM
87
69. 12 is 60% of what number?
6012
1003
125
5 5 312
3 3 520
n
n
n
n
= ⋅
= ⋅
⋅ = ⋅ ⋅
=
Thus, 12 is 60% of 20.
70. 24 is 50% of what number?
5024
1001
242
12 24 2
248
n
n
n
n
= ⋅
= ⋅
⋅ = ⋅ ⋅
=
Thus, 24 is 50% of 48.
71. 16 is what percent of 211? 16 211
16 211
211 2110.076
n
n
n
= ⋅⋅
=
≈
Thus, 7.6% of 211 is 16.
72. What percent of 171 is 7? 171 7
171 7
171 1710.041
n
n
n
⋅ =⋅
=
≈
Thus, 4.1% of the 171 patients died.
73. What percent of 211 is 20? 211 20
2 211 20
211 2110.095
n
n
⋅ =⋅
=
≈
Thus, 9.5% of the 211 patients had heart attacks.
74. What percent of 171 is 7? 171 7
171 7
171 1710.041
n
n
n
⋅ =⋅
=
≈
Thus, 4.1% of the 171 patients had heart attacks.
75. What percent of 211 is 8? 211 8
211 8
211 2110.038
n
n
n
⋅ =⋅
=
≈
Thus, 3.8% of the 211 patients had a stroke.
76. What percent of 171 is 2? 171 2
171 2
171 1710.012
n
n
n
⋅ =⋅
=
≈
Thus, 1.2% of the 171 patients had a stroke.
77. What percent of 171 is 9? 171 9
171 9
171 1710.053
n
n
n
⋅ =⋅
=
≈
Thus, 5.3% of the 171 patients died.
78. What percent of 171 is 8? 171 8
171 8
171 1710.047
n
n
n
⋅ =⋅
=
≈
Thus, 4.7% of the 171 patients had a heart attack.
79. What percent of 211 is 8? 211 8
211 8
211 2110.038
n
n
n
⋅ =⋅
=
=
Thus, 3.8% of the 211 patients had a stroke.
80. What percent of 171 is 3? 171 3
171 3
171 1710.018
n
n
n
⋅ =⋅
=
≈
Thus, 1.8% of the 171 patients had a stroke.
81. a. 85 3500h=
b. 85 3500
3500 341
85 17
h
h
=
= =
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Bello, Introductory Algebra, 3e ISM Chapter 2: Equations, Problem Solving, and Inequalities
88
82. a. 160 3500h=
b. 160 3500
3500 721
160 8
h
h
=
= =
83. a. 260 3500h=
b. 260 3500
3500 613
260 13
h
h
=
= =
84. a. 280 3500h=
b. 280 3500
3500 112
280 2
h
h
=
= =
85. a. 450 3500h=
b. 450 3500
3500 77
450 9
h
h
=
= =
86. 8 × 250 × 20 = 40,000
Her yearly salary is $40,000.
87. 8 × 100 × 20 = 16,000
Her lost wages are $16,000.
88. Let x = multiple of the old wages needed to make up the loss within 1 year. 40,000 56,000
40,000 56,000
40,000 40,000
1.40 or 140%
x
x
x
⋅ =⋅
=
=
Thus, the employee needs a 40% increase to recuperate her lost wages within 1 year.
89. a. Answers may vary.
b. Answers may vary.
90. Answers may vary. Sample answer: It
would be easier to divide by −3.
3 18
3 18
3 36
x
x
x
− =−
=− −
= −
91. Answers may vary. Sample answer: It would be easier to multiply by the
reciprocal of 3
.4
−
315
4
4 3 4
3 4 3
x
x
− =
− ⋅ − =− 1
15⋅5
20x=−
92. ac bc=
93. a b
c c=
94. LCM
95. LCM
96. 15 is 30% of what number?
5
3015
1003
1510
1015
3
n
n
= ⋅
= ⋅
⋅10 3
3 1050
n
n
= ⋅ ⋅
=
Thus, 15 is 30% of 50.
97. What percent of 45 is 9? 45 9
45 9
45 451 20
5 100
x
x
x
⋅ =⋅
=
= =
Thus, 9 is 20% of 45.
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Chapter 2: Equations, Problem Solving, and Inequalities Bello, Introductory Algebra, 3e ISM
89
98. 40% of 60 is what number?
12
4060
100
260
5
x⋅ =
⋅
24
x
x
=
=
Thus, 40% of 60 is 24.
99. The LCM of 4 and 5 is 4 ⋅ 5 = 20.
2 13
4 52 1
20 20 20 34 5
5( 2) 4( 1) 60
5 10 4 4 60
9 6 60
9 54
6
x x
x x
x x
x x
x
x
x
+ −+ =
+ −⋅ + ⋅ = ⋅
+ + − =+ + − =
+ ===
The solution is 6.
100. The LCM of 2 and 3 is 2 ⋅ 3 = 6.
3 25
2 33 2
6 6 6 52 3
3( 3) 2( 2) 30
3 9 2 4 30
13 30
17
x x
x x
x x
x x
x
x
+ −− =
+ −⋅ − ⋅ = ⋅
+ − − =+ − + =
+ ==
The solution is 17.
101. The LCM of 6 and 10 is 2 ⋅ 3 ⋅ 5 = 30.
86 10
30 30 30 86 10
5 3 240
8 240
30
y y
y y
y y
y
y
+ =
⋅ + ⋅ = ⋅
+ ===
102. The LCM of 5 and 8 is 5 ⋅ 8 = 40.
35 8
40 40 40 35 8
8 5 120
3 120
40
y y
y y
y y
y
y
− =
⋅ − ⋅ = ⋅
− ===
The solution is 40.
103.
1
48
55 4 5
84 5 4
10
y
y
y
=
⋅ = ⋅
=
The solution is 10.
104.
1
36
44 3 4
63 4 3
8
y
y
y
− =
− ⋅ − =− ⋅
=−
The solution is −8.
105. 2
47
7 2 7
2 7 2
y
y
− =−
− ⋅ − = −( 4)⋅ −
2
14y=
The solution is 14.
106. 7 16
16
7
y
y
− =
=−
The solution is 16
.7
−
107. 72
14
x
x
=−
=−
108. 34
12
y
y
− =−
=
The solution is 12.
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Bello, Introductory Algebra, 3e ISM Chapter 2: Equations, Problem Solving, and Inequalities
90
109. 4( 6) 4 4(6) 4 24x x x− = − = −
110. 3(6 ) 3 6 3 18 3x x x− = ⋅ − ⋅ = −
111. 5(8 ) 5 8 5 40 5y y y− = ⋅ − ⋅ = −
112. 6(8 2 ) 6 8 6 2 48 12y y y− = ⋅ − ⋅ = −
113. 9(6 3 ) 9 6 9 3 54 27y y y− = ⋅ − ⋅ = −
114. −3(4x − 2) = −3 ⋅ 4x − 3 ⋅ (−2) = −12x + 6
115. −5(3x − 4) = −5 ⋅ 3x − 5 ⋅ (−4) = −15x + 20
116.
5
3 2020
4⋅ =
3 1515
1 4 1⋅ = =
117.
41 24
246⋅ =
1 44
1 6 1⋅ = =
118.
1
4 55
5
−− ⋅ − =
4
1 5⋅−
44
1
= =
119.
1
7 77
3
−− ⋅ − =
3
1 7⋅−
33
1
= =
2.3 Linear Equations
Problems 2.3 1. a. 4 5 17
4 5 5 17 5
4 12
4 12
4 43
x
x
x
x
x
+ =+ − = −
=
=
=
The solution is 3.
CHECK 4x + 5 0 17
4(3) + 5 17
12 + 5
17
b. 3 5 2
3 5 5 2 5
3 7
3 7
3 37
3
x
x
x
x
x
− − =− − + = +
− =−
=− −
= −
The solution is 7
.3
−
CHECK −3x − 5 0 2
( )73
3 5− − − 2
7 − 5
2
2. 7( 1) 4( 2) 5
7 7 4 8 5
7 7 4 13
7 7 7 4 13 7
7 4 6
7 4 4 4 6
3 6
3 6
3 32
x x
x x
x x
x x
x x
x x x x
x
x
x
+ = + ++ = + ++ = +
+ − = + −= +
− = − +=
=
=
The solution is 2.
CHECK 7(x + 1) 0 4(x + 2) + 5
7(2 + 1) 4(2 + 2) + 5
7(3) 4(4) + 5
21 16 + 5
21
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Chapter 2: Equations, Problem Solving, and Inequalities Bello, Introductory Algebra, 3e ISM
91
3. a. The LCM of 3, 7, and 21 is 21.
3 7
20
21 7 3
21
x x= +
20
21⋅ 21= 21
7
x⋅ +
3
20 3 7
20 10
20 10
10 102
x
x x
x
x
x
⋅
= +=
=
=
The solution is 2.
CHECK 2021
0 7 3x x+
2021
2 27 3
+
6 1421 21
+
2021
b. The LCM of 4, 5, and 20 is 20.
5 4
1 17( 4)
4 5 20
20
x x +− =
120
4⋅ − 20
5
x⋅ =
17( 4)
20
x +⋅
5 4 17( 4)
5 4 17 68
5 5 4 17 68 5
4 17 63
4 17 17 17 63
21 63
21 63
21 213
x x
x x
x x
x x
x x x x
x
x
x
− = +− = +
− − = + −− = +
− − = − +− =−
=− −
= −
The solution is −3.
CHECK 1
4 5x− 0
17( 4)
20
x+
314 5
−− 17( 3 4)
20
− +
5 1220 20
+ 17(1)
20
1720
1720
4. ( )( )
14
61
6 6 46
6 4
6 4 4 4
6 4
6 4
S C
S C
S C
S C
S C
C S
= −
⋅ = ⋅ −
= −
+ = − +
+ == +
5. a. Let S = 12. 3 24
12 3 24
12 24 3 24 24
36 3
36 3
3 312
S L
L
L
L
L
L
= −= −
+ = − +=
=
=
The length of their foot is 12 inches.
b.
3 24
24 3 24 24
24 3
324
3 324
324
3
S L
S L
S L
LS
SL
SL
= −
+ = − +
+ =
+=
+=
+=
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Bello, Introductory Algebra, 3e ISM Chapter 2: Equations, Problem Solving, and Inequalities
92
6. 3 4 7
3 4 4 7 4
3 7 4
3 7 4
3 37 4
3
x y
x y y y
x y
x y
yx
+ =+ − = −
= −−
=
−=
7. a. 2001 2000 1;
992(1) 16,145 17,137
x
C
= − =
= + =
The tuition cost in 2001 was about $17,137.
b. 2005 2000 5;
992(5) 16,145 21,105
x
C
= − =
= + =
The tuition cost in 2001 was about $21,105.
c. 992 16,145
20,000 992 16,145
992 16,14520,000
16,145
C x
x
x
= +
= +
+=
− 16,145−
3855 992
3855
9924
x
x
x
=
=
≈
The cost will be $20,000 4 years after 2000, or in 2004.
Exercises 2.3
Checks are left to the student.
1. 3 12 0
3 12 12 0 12
3 12
3 12
3 34
x
x
x
x
x
− =− + = +
=
=
=
The solution is 4.
2. 5 10 0
5 10 10 0 10
5 10
2
a
a
a
a
+ =
+ − = −
=−=−
The solution is −2.
3. 2 6 8
2 6 6 8 6
2 2
2 2
2 21
y
y
y
y
y
+ =+ − = −
=
=
=
The solution is 1.
4. 4 5 3
4 5 5 3 5
4 8
4 8
4 42
b
b
b
b
b
− =− + = +
=
=
=
The solution is 2.
5. 3 4 10
3 4 4 10 4
3 6
3 6
3 32
z
z
z
z
z
− − = −− − + = − +
− = −− −
=− −
=
The solution is 2.
6. 4 2 6
4 2 2 6 2
4 8
4 8
4 42
r
r
r
r
r
− − =− − + = +
− =−
=− −
= −
The solution is −2.
7. 5 1 13
5 1 1 13 1
5 14
5 14
5 514
5
y
y
y
y
y
− + = −− + − = − −
− = −− −
=− −
=
The solution is 14
.5
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Chapter 2: Equations, Problem Solving, and Inequalities Bello, Introductory Algebra, 3e ISM
93
8. 3 1 9
3 1 1 9 1
3 10
3 10
3 310
3
x
x
x
x
x
− + = −− + − = − −
− = −− −
=− −
=
The solution is 10
.3
9. 3 4 10
3 4 4 10 4
3 6
3 6
2 6
2 6
2 23
x x
x x
x x
x x x x
x
x
x
+ = ++ − = + −
= +− = − +
=
=
=
The solution is 3.
10. 4 4 7
4 4 4 7 4
4 3
4 3
3 3
3 3
3 31
x x
x x
x x
x x x x
x
x
x
+ = ++ − = + −
= +− = − +
=
=
=
The solution is 1.
11. 5 12 6 8
5 12 12 6 8 12
5 6 4
5 6 6 6 4
4
4
x x
x x
x x
x x x x
x
x
− = −− + = − +
= +− = − +
− == −
The solution is −4.
12. 5 7 7 19
5 7 7 7 19 7
5 7 12
5 7 7 7 12
2 12
2 12
2 26
x x
x x
x x
x x x x
x
x
x
+ = ++ − = + −
= +− = − +− =−
=− −
= −
The solution is −6.
13. 4 7 6 9
4 7 7 6 9 7
4 6 16
4 6 6 6 16
2 16
2 16
2 28
v v
v v
v v
v v v v
v
v
v
− = +− + = + +
= +− = − +− =−
=− −
= −
The solution is −8.
14. 8 4 15 10
8 4 4 15 10 4
8 15 14
8 15 15 15 14
7 14
7 14
7 72
t t
t t
t t
t t t t
t
t
t
+ = −+ − = − −
= −− = − −
− = −− −
=− −
=
The solution is 2.
15. 6 3 12 0
3 12 0
3 12 12 0 12
3 12
3 12
3 34
m m
m
m
m
m
m
− + =+ =
+ − = −= −
−=
= −
The solution is −4.
16. 10 15 5 25
5 15 25
5 15 15 25 15
5 10
2
k k
k
k
k
k
+ − =
+ =+ − = −
==
The solution is 2.
17. 10 3 8 6
10 10 3 8 10 6
3 2 6
3 6 2 6 6
3 2
3 2
3 32
3
z z
z z
z z
z z z z
z
z
z
− = −− − = − −
− = − −− + = − − +
= −−
=
= −
The solution is 2
.3
−
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Bello, Introductory Algebra, 3e ISM Chapter 2: Equations, Problem Solving, and Inequalities
94
18. 8 4 10 6
8 8 4 10 8 6
4 2 6
4 6 2 6 6
10 2
10 2
10 101
5
y y
y y
y y
y y y y
y
y
y
− = +− − = − +
− = +− − = + −
− =−
=− −
= −
The solution is 1
.5
−
19. 5( 2) 3( 3) 1
5 10 3 9 1
5 10 3 10
5 10 10 3 10 10
5 3
5 3 3 3
2 0
2 0
2 20
x x
x x
x x
x x
x x
x x x x
x
x
x
+ = + ++ = + ++ = +
+ − = + −=
− = −=
=
=
The solution is 0.
20.
(4 2 ) 7( 1)
4 2 7 7
3 4 7 7
3 4 4 7 7 4
3 7 3
3 7 7 7 3
4 3
4 3
4 43
4
y y y
y y y
y y
y y
y y
y y y y
y
y
y
− − = −− + = −
− = −− + = − +
= −− = − −− = −− −
=− −
=
The solution is 3
.4
21. 5(4 3 ) 7(3 4 )a a− = −
20 15 21 28
20 20 15 21 20 28
15 1 28
15 28 1 28 28
13 1
13 1
13 131
13
a a
a a
a a
a a a a
a
a
a
− = −− − = − −
− = −− + = − +
=
=
=
The solution is 1
.13
22.
3 14.5 1.3
4 43 1
4.5 4.5 1.3 4.54 4
3 15.8
4 43 1 1 1
5.84 4 4 4
25.8
41
5.821
2 2 5.82
11.6
y y
y y
y y
y y y y
y
y
y
y
− = +
− + = + +
= +
− = − +
=
=
⋅ = ⋅
=
The solution is 11.6.
23.
7 55.6 3.3
8 87 5
5.6 5.6 3.3 5.68 8
7 58.9
8 87 5 5 5
8.98 8 8 8
28.9
81
8.94
14 4 ( 8.9)
4
35.6
c c
c c
c c
c c c c
c
c
c
c
− + = − −
− + − = − − −
− = − −
− + = − + −
− = −
− = −
− ⋅ − = − ⋅ −
=
The solution is 35.6.
Full file at http://TestbankCollege.eu/Solution-Manual-Introductory-Algebra-3rd-Edition-Ignacio-Bello
Chapter 2: Equations, Problem Solving, and Inequalities Bello, Introductory Algebra, 3e ISM
95
24.
2
210
33 2
103 3
510
3
3 5 310
5 3 5
x x
x x
x
x
+ =
+ =
=
⋅ = ⋅
1
6x =
The solution is 6.
25.
1 42 2
4 51 1 4 1
2 24 4 5 4
16 52 2
20 2011
2 220
112 2 2 2
2011
420
1 1 11( 4 )
4 4 2011
80
x x
x x
x x
x x
x x x x
x
x
x
− + = +
− + − = + −
− = + −
− = +
− − = − +
− =
− ⋅ − = − ⋅
= −
The solution is 11
.80
−
26.
1 26 2
7 71 1 2 1
6 27 7 7 7
36 2
73
6 2 2 27
34
71 1 3
44 4 7
3
28
x x
x x
x x
x x x x
x
x
x
+ = −
+ − = − −
= −
− = − −
= −
⋅ = ⋅ −
= −
The solution is 3
.28
−
27.
1 23
2 21 2
2 2 2 32 2
1 2 6
2 3 6
2 3 3 6 3
2 9
2 9
2 29
2
x x
x x
x x
x
x
x
x
x
− −+ =
− −⋅ + ⋅ = ⋅
− + − =− =
− + = +=
=
=
The solution is 9
.2
28.
3 5 312
3 33 5 3
3 3 3 123 33 5 3 36
4 8 36
4 8 8 36 8
4 28
4 28
4 47
x x
x x
x x
x
x
x
x
x
+ ++ =
+ +⋅ + ⋅ = ⋅
+ + + =+ =
+ − = −=
=
= The solution is 7.
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Bello, Introductory Algebra, 3e ISM Chapter 2: Equations, Problem Solving, and Inequalities
96
29. 15 4
20 20 20 15 4
4 5 20
20
20
x x
x x
x x
x
x
− =
⋅ − ⋅ = ⋅
− =− =
= −
The solution is −20.
30. 13 2
6 6 6 13 22 3 6
6
6
x x
x x
x x
x
x
− =
⋅ − ⋅ = ⋅
− =− =
= −
The solution is −6.
31.
1 2 23
4 31 2 2
12 12 12 34 3
3( 1) 4(2 2) 36
3 3 8 8 36
5 11 36
5 11 11 36 11
5 25
5 25
5 55
x x
x x
x x
x x
x
x
x
x
x
+ −− =
+ −⋅ − ⋅ = ⋅
+ − − =+ − + =
− + =− + − = −
− =−
=− −
= −
The solution is −5.
32. 4 6
3 44 6
12 123 4
4( 4) 3( 6)
4 16 3 18
4 16 16 3 18 16
4 3 2
4 3 3 3 2
2
z z
z z
z z
z z
z z
z z
z z z z
z
+ +=
+ +⋅ = ⋅
+ = ++ = +
+ − = + −= +
− = − +=
The solution is 2.
33.
2 1 4
3 122 1 4
12 123 12
4(2 1) 4
8 4 4
8 4 4 4 4
8
8
7 0
7 0
7 70
h h
h h
h h
h h
h h
h h
h h h h
h
h
h
− −=
− −⋅ = ⋅
− = −− = −
− + = − +=
− = −=
=
=
The solution is 0.
34.
5 6 7 42
7 35 6 7 4
21 21 21 27 3
3(5 6 ) 7( 7 4 ) 42
15 18 49 28 42
64 10 42
64 64 10 42 64
10 22
10 22
10 1011
5
y y
y y
y y
y y
y
y
y
y
y
− − −− =
− − −⋅ − ⋅ = ⋅
− − − − =− + + =
+ =− + = −
= −−
=
= −
The solution is 11
.5
−
35. 2 3 3 1
12 4
2 3 3 14 4 4 1
2 42(2 3) (3 1) 4
4 6 3 1 4
5 4
5 5 4 5
1
w w
w w
w w
w w
w
w
w
+ +− =
+ +⋅ − ⋅ = ⋅
+ − + =+ − − =
+ =+ − = −
= −
The solution is −1.
Full file at http://TestbankCollege.eu/Solution-Manual-Introductory-Algebra-3rd-Edition-Ignacio-Bello
Chapter 2: Equations, Problem Solving, and Inequalities Bello, Introductory Algebra, 3e ISM
97
36.
7 2 1
6 2 47 2 1
12 12 126 2 4
2(7 2) 6 3
14 4 6 3
14 10 3
14 14 10 3 14
10 11
10 11
11 1110
11
r r
r r
r r
r r
r r
r r r r
r
r
r
++ =
+⋅ + ⋅ = ⋅
+ + =+ + =
+ =− + = −
= −−
=− −
− =
The solution is 10
.11
−
37.
8 23 1 5
6 3 28 23 1 5
6 6 66 3 28 23 2 15
8 21 15
8 8 21 15 8
21 7
21 7
7 73
xx
xx
x x
x x
x x x x
x
x
x
−+ =
−⋅ + ⋅ = ⋅
− + =− =
− − = −− =−
=
− =
The solution is −3.
38.
1 2 48
2 3 41 2 4
12 12 12 12 ( 8)2 3 4
6( 1) 4( 2) 3( 4) 96
6 6 4 8 3 12 96
13 26 96
13 26 26 96 26
13 122
122
13
x x x
x x x
x x x
x x x
x
x
x
x
+ + ++ + =−
+ + +⋅ + ⋅ + ⋅ = ⋅ −
+ + + + + =−
+ + + + + =−+ =−
+ − =− −
=−
=−
The solution is 122
.13
−
39.
5 4 3( 2)
2 3 25 4 3
6 6 6 6 ( 2)2 3 2
3( 5) 2( 4) 3( 3) 6 12
3 15 2 8 3 9 6 12
7 3 3
7 7 3 3 7
3 10
3 3 3 10
4 10
4 10
4 45
2
x x xx
x x xx
x x x x
x x x x
x x
x x
x x
x x x x
x
x
x
− − −− = − −
− − −⋅ − ⋅ = ⋅ − ⋅ −
− − − = − − +− − + = − − +
− = − +− + = − + +
= − ++ = − + +
=
=
=
The solution is 5
.2
40.
1 2 316
2 3 41 2 3
12 12 12 12 162 3 4
6( 1) 4( 2) 3( 3) 192
6 6 4 8 3 9 192
13 23 192
13 23 23 192 23
13 169
13 169
13 1313
x x x
x x x
x x x
x x x
x
x
x
x
x
+ + ++ + =
+ + +⋅ + ⋅ + ⋅ = ⋅
+ + + + + =+ + + + + =
+ =+ − = −
=
=
=
The solution is 13.
41. 1 1
4 42 8
1 14 4
2 2
x x
x x
− + = −
− + = −
This is an identity. The solution is all real numbers.
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Bello, Introductory Algebra, 3e ISM Chapter 2: Equations, Problem Solving, and Inequalities
98
42.
2 16 4
3 5
2 46 4
3 52 4
15( 6 ) 15 15 15 43 5
90 10 12 60
90 10 10 12 10 60
90 2 60
90 60 2 60 60
30 2
1
15
x x
x x
x x
x x
x x
x x
x x x x
x
x
− + = −
− + = −
− + ⋅ = ⋅ − ⋅
− + = −
− + − = − −
− = −− + = − +
− =
=−
The solution is 1
.15
−
43.
1 1(8 4) 5 (4 8) 1
2 44 2 5 2 1
4 3 3
4 3 3 3 3
4 6
4 6
3 6
3 6
3 32
x x
x x
x x
x x
x x
x x x x
x
x
x
+ − = + +
+ − = + +− = +
− + = + += +
− = − +=
=
=
The solution is 2.
44. 1 1
(3 9) 2 (9 18) 33 9
3 2 2 3
5 5
x x
x x
x x
+ + = + +
+ + = + ++ = +
This is an identity. The solution is all real numbers.
45. 3
92 5
310 10 10 10 9
2 510 5 6 90
9 90
9 90
9 910
x xx
x xx
x x x
x
x
x
+ − =
⋅ + ⋅ − ⋅ = ⋅
+ − ==
=
=
The solution is 10.
46.
5 3 110
3 4 65 3 11
12 12 12 12 03 4 6
20 9 22 0
11 22 0
11 22 22 0 22
11 22
11 22
11 112
x x
x x
x x
x
x
x
x
x
− + =
⋅ − ⋅ + ⋅ = ⋅
− + =+ =
+ − = −= −
−=
= −
The solution is −2.
47.
4 3 5 3
9 2 6 24 3 5 3
18 18 18 189 2 6 2
8 27 15 27
8 27 12
8 8 27 12 8
27 20
27 20
20 2027
20
x x x
x x x
x x x
x x
x x x x
x
x
x
− = −
⋅ − ⋅ = ⋅ − ⋅
− = −− = −
− − = − −− = −− −
=− −
=
The solution is 27
.20
48. 7 4 2 11
2 3 5 67 4 2 11
30 30 30 302 3 5 6
105 40 12 55
77 55
5
7
x x x
x x x
x x x
x
x
− + =−
⋅ − ⋅ + ⋅ = ⋅ −
− + =−=−
=−
The solution is 5
.7
−
Full file at http://TestbankCollege.eu/Solution-Manual-Introductory-Algebra-3rd-Edition-Ignacio-Bello
Chapter 2: Equations, Problem Solving, and Inequalities Bello, Introductory Algebra, 3e ISM
99
49.
3 4 1 7 18(19 3) 1
2 8 123 4 1 7 18
24 24 (19 3) 24 1 242 8 12
12(3 4) 3(19 3) 24 2(7 18)
36 48 57 9 24 14 36
21 57 14 12
21 57 57 14 12 57
21 14 69
21 14 14 14 69
7 69
69
7
x xx
x xx
x x x
x x x
x x
x x
x x
x x x x
x
x
+ +− − = −
+ +⋅ − ⋅ − = ⋅ − ⋅
+ − − = − +
+ − + = − −− + =− −
− + − =− − −− =− −
− + =− + −
− =−
=
The solution is 69
.7
50.
11 2 1 17 7 2(3 1) (7 2)
3 2 6 911 2 1 17 7 2
18 18 (3 1) 18 18 (7 2)3 2 6 9
6(11 2) 9(3 1) 3(17 7) 4(7 2)
66 12 27 9 51 21 28 8
39 3 23 29
39 3 3 23 29 3
39 23 32
39 23 23 23 32
16 32
16
16
x xx x
x xx x
x x x x
x x x x
x x
x x
x x
x x x x
x
x
− +− − = − −
− +⋅ − ⋅ − = ⋅ − ⋅ −
− − − = + − −− − + = + − +
− = +− + = + +
= +− = − +
=
=32
162x =
The solution is 2.
51. 2
2
2 2
2
2
C r
rC
Cr
Cr
= π
π=
π π
=π
=π
52. A b h
b hA
b bA
hb
Ah
b
=
=
=
=
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Bello, Introductory Algebra, 3e ISM Chapter 2: Equations, Problem Solving, and Inequalities
100
53. 3 2 6
3 3 2 6 3
2 6 3
2 6 3
2 26 3
2
x y
x x y x
y x
y x
xy
+ =− + = −
= −−
=
−=
54. 5 6 30
5 5 6 30 5
6 30 5
6 30 5
6 65 30
6
x y
x x y x
y x
y x
xy
− =− − = −
− = −− −
=− −
−=
55.
2
2
2 2 2
2
2
2
2
)A r rs
A r rs
A r r r rs
A r rs
A r rs
r r
A rs
r
A rs
r
= π( +
= π + π
− π = π − π + π
− π = π
− π π=
π π− π
=π
− π=
π
56.
2
2
2 2 2
2
2
2
2
2 ( )
2 2
2 2 2 2
2 2
2 2
2 2
2
2
2
2
T r rh
T r rh
T r r r rh
T r rh
T r rh
r r
T rh
r
T rh
r
= π +
= π + π
− π = π − π + π
− π = π
− π π=
π π− π
=π
− π=
π
57. 2 1
1 2
2 11 1
1 2
1 12
2
V P
V P
V PV V
V P
PVV
P
=
⋅ = ⋅
=
58. a c
b da c
b bb d
bca
d
=
⋅ = ⋅
=
59.
( ) ( )
fS
H hf
H h S H hH h
HS hS f
HS hS hS f hS
HS f hS
HS f hS
S Sf Sh
HS
=−
− ⋅ = − ⋅−
− =− + = +
= ++
=
+=
60.
( ) ( )
EI
R nrE
I R nr R nrR nr
IR Inr E
IR Inr Inr E Inr
IR E Inr
IR E Inr
I IE Inr
RI
=+
⋅ + = ⋅ ++
+ =+ − = −
= −−
=
−=
Full file at http://TestbankCollege.eu/Solution-Manual-Introductory-Algebra-3rd-Edition-Ignacio-Bello
Chapter 2: Equations, Problem Solving, and Inequalities Bello, Introductory Algebra, 3e ISM
101
61. a. 3 22
22 3 22 22
22 3
22 3
3 322
322
3
S L
S L
S L
S L
SL
SL
= −+ = − ++ =+
=
+=
+=
b. 3 21
21 3 21 21
21 3
21 3
3 321
321
3
S L
S L
S L
S L
SL
SL
= −+ = − ++ =+
=
+=
+=
62. 5 190
190 5 190 190
190 5
190 5
5 5190
5190
5
W H
W H
W H
W H
WH
WH
= −+ = − ++ =+
=
+=
+=
63. 5 200
200 5 200 200
200 5
200 5
5 5200
5200
5
W H
W H
W H
W H
WH
WH
= −+ = − ++ =+
=
+=
+=
64. 172
17 17 172
172
2 ( 17) 22
2 34
34 2
AH
AH
AH
AH
H A
A H
= −
− = − −
− = −
− ⋅ − = − ⋅ −
− + == −
65. S = 3L − 22
or 22
311 22 33
113 3
SL
L
+=
+= = =
Tyrone’s foot is 11 inches long.
66. S = 3L − 21
or 21 7 21 28 1
93 3 3 3
SL
+ += = = =
Maria’s foot is 28 1
93 3
= inches long.
67. First, find the length of a man’s foot for size 7.
22
37 22 29
3 3
SL
L
+=
+= =
Then, solve S = 3L − 21 for women with
29:
3L=
293 21 29 21 8
3S
= − = − =
Sue wears a size 8.
68. Let P = 7.27. 0.37 0.23( 1)
7.27 0.37 0.23( 1)
7.27 0.37 0.23 0.23
7.27 0.14 0.23
7.27 0.14 0.14 0.14 0.23
7.13 0.23
31
P x
x
x
x
x
x
x
= + −
= + −
= + −= +
− = − +=
=
The prices are the same when the weight is 31 ounces.
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Bello, Introductory Algebra, 3e ISM Chapter 2: Equations, Problem Solving, and Inequalities
102
69. Let C = 11.50. 1 0.75( 1)
11.50 1 0.75( 1)
11.50 1 0.75 0.75
11.50 0.25 0.75
11.50 0.25 0.25 0.25 0.75
11.25 0.75
11.25 0.75
0.75 0.7515
C h
h
h
h
h
h
h
h
= + −= + −= + −= +
− = − +=
=
=
After 15 hours, the cost is $11.50.
70. Let N = A and solve for x. N = 0.165x + 4.68
A = −0.185x + 5.38
0.165 4.68 0.185 5.38
0.165 4.68 4.68 0.185 5.38 4.68
0.165 0.185 0.7
0.165 0.185 0.185 0.185 0.7
0.35 0.7
2
x x
x x
x x
x x x x
x
x
+ =− +
+ − =− + −
=− ++ =− + +
==
The run production will be the same when
x = 2, or in 1996 + 2 = 1998.
71. a. 0.423(5) 1.392 0.723 hourH = − =
b. 0.68 hour
72. a. 0.423(6) 1.392 1.146 hoursH = − =
b. 1.19 hours
73. a. 0.423(7) 1.392 0.569 hourH = − =
b. 1.61 hours
74. 1980–1985
75. a. 0.423(8) 1.392 1.992 hourH = − =
b. 1.95 hours
c. younger employees
76. Cost is $30 plus $0.15 per mile. C = 30 + 0.15m
77. Let C = 40 and solve for m. 30 0.15
40 30 0.15
40 30 30 30 0.15
10 0.15
10 0.15
0.15 0.152
663
C m
m
m
m
m
m
= += +
− = − +=
=
=
The mileage rate and flat rate are the same
at 2
663
miles.
78. Flat rate: $40 ⋅ 7 = $280
Mileage rate: 30 7 0.15(300)
210 45
$255
C = ⋅ += +=
The mileage rate is cheaper.
79. a.
0.2
1.2
S C M
S C C
S C
= += +=
b. Let C = 8. S = 1.2C S = 1.2(8) S = 9.6 The selling price is $9.60.
80. a.
0.25
0.25 0.25 0.25
0.75
0.75
S C M
S C S
S S C S S
S C
C S
= += +
− = + −==
b. Let S = 80. C = 0.75(80) C = 60 The article cost $60.
81. Answers may vary.
82. Answers may vary.
83. The solution to the equation is x = 0. If x = 0, then dividing by x (or 0) is undefined.
84. ax b c+ =
85. literal
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Chapter 2: Equations, Problem Solving, and Inequalities Bello, Introductory Algebra, 3e ISM
103
86.
1 2
1 2
1 2
2 1 2 2
2 1
21
( )2
2 2( )
22
2
2
2
hA b b
hA b b
h hA
b bh
Ab b b b
hA
b bh
A b hb
h
= +
⋅ = ⋅ +
= +
− = + −
− =
−=
87. 50 40 0.2( 100)
50 40 0.2 20
50 20 0.2
50 20 20 20 0.2
30 0.2
30 0.2
0.2 0.2150
150
m
m
m
m
m
m
m
m
= + −= + −= +
− = − +=
=
==
88. 7
12 4 37
12 12 1212 4 3
7 3 4
7 7
1
x x
x x
x x
x
x
= +
⋅ = ⋅ + ⋅
= +
==
The solution is 1.
89.
1 8( 2)
3 5 151 8( 2)
15 15 153 5 15
5 3 8( 2)
5 3 8 16
5 5 3 8 16 5
3 8 11
3 8 8 8 11
11 11
11 11
11 111
x x
x x
x x
x x
x x
x x
x x x x
x
x
x
+− =
+⋅ − ⋅ = ⋅
− = +− = +
− − = + −− = +
− − = − +− =−
=− −
= −
The solution is −1.
90. 10( 2) 6( 1) 18
10 20 6 6 18
10 20 6 24
10 20 20 6 24 20
10 6 4
10 6 6 6 4
4 4
4 4
4 41
x x
x x
x x
x x
x x
x x x x
x
x
x
+ = + ++ = + ++ = +
+ − = + −= +
− = − +=
=
=
The solution is 1.
91. 5( 2) 3( 1) 9
5 10 3 3 9
5 10 3 12
5 10 10 3 12 10
5 3 2
5 3 3 3 2
2 2
2 2
2 21
x x
x x
x x
x x
x x
x x x x
x
x
x
− + = − + −− − = − − −− − = − −
− − + = − − +− = − −
− + = − + −− = −− −
=− −
=
The solution is 1.
92. 4 5 2
4 5 5 2 5
4 7
4 7
4 47
4
x
x
x
x
x
− − =− − + = +
− =−
=− −
= −
The solution is 7
.4
−
93. 3 8 11
3 8 8 11 8
3 3
3 3
3 31
x
x
x
x
x
+ =+ − = −
=
=
=
The solution is 1.
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Bello, Introductory Algebra, 3e ISM Chapter 2: Equations, Problem Solving, and Inequalities
104
94. 2n n+
95. Quotient means divide: a b
c
+
96. a
b c−
97. Product means multiply and sum means add: a(b + c)
98. (a − b)c
99. Difference means subtract and product means multiply: a bc−
2.4 Problem Solving: Integer, General, and Geometry Problems
Problems 2.4 1. (1) Read the problem.
We are asked to find three consecutive odd integers.
(2) Select the unknown. Let n = first odd integer. Then n + 2 = 2nd odd integer. and n + 4 = 3rd odd integer.
(3) Think of a plan. Translate the sentence into an equation. n + (n + 2) + (n + 4) = 129
(4) Use algebra to solve the resulting
equation.
( 2) ( 4) 129
2 4 129
3 6 129
3 6 6 129 6
3 123
41
n n n
n n n
n
n
n
n
+ + + + =+ + + + =
+ =+ − = −
==
Thus, the three consecutive odd integers are 41, 41 + 2 = 43, and 41 + 4 = 45.
(5) Verify the solution. Since 41 + 43 + 45 = 129, our result is correct.
2. (1) Read the problem. We are asked to find a certain number.
(2) Select the unknown. Let n represent the number.
(3) Think of a plan. Translate the sentence into an equation.
3n − 14 = n + 2
(4) Use algebra to solve the resulting
equation.
3 14 2
3 14 14 2 14
3 16
2 16
8
n n
n n
n n
n
n
− = +
− + = + += +
==
The number is 8.
(5) Verify the solution.
Is 3(8) − 14 = 8 + 2? Yes, since
24 − 14 = 10 is true.
3. (1) Read the problem. We are asked to find the number of calories in the cheeseburger and in the fries.
(2) Select the unknown. Let f = number of calories in the fries. Then f + 120 = number of calories in the cheeseburger.
(3) Think of a plan. Translate the problem. f + (f + 120) = 540
(4) Use algebra to solve the equation. ( 120) 540
120 540
2 120 540
2 420
210
f f
f f
f
f
f
+ + =
+ + =
+ ==
=
Thus, the fries have 210 calories and the cheeseburger has 210 + 120 = 330 calories.
(5) Verify the solution. Does 210 + 330 = 540? Yes, the equation is true.
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Chapter 2: Equations, Problem Solving, and Inequalities Bello, Introductory Algebra, 3e ISM
105
4. (1) Read the problem. We are asked to find the measure of an angle.
(2) Select the unknown. Let m = measure of the angle.
90 − m is its complement.
180 − m is its supplement.
(3) Think of a plan. Translate the problem statement into an equation.
180 − m = 4(90 − m) − 45
(4) Use algebra to solve the equation. 180 4(90 ) 45
180 360 4 45
180 315 4
180 180 315 180 4
135 4
4 135 4 4
3 135
45
m m
m m
m m
m m
m m
m m m m
m
m
− = − −− = − −
− = −− − = − −
− = −− + = − +
==
The measure of the angle is 45°.
(5) Verify the solution.
Is the supplement (180° − 45° = 135°)
45° less than 4 times its complement
(4 ⋅ (90° − 45°) = 180°)? Yes, since
135° = 180° − 45°.
Exercises 2.4
Students should use the RSTUV method to solve.
An outline of each solution is given.
1. Let n = first even integer. Then n + 2 = 2nd even integer, and n + 4 = 3rd even integer.
( 2) ( 4) 138
2 4 138
3 6 138
3 6 6 138 6
3 132
3 132
3 344
n n n
n n n
n
n
n
n
n
+ + + + =+ + + + =
+ =+ − = −
=
=
=
Thus, the three integers are 44, 44 + 2 = 46, and 44 + 4 = 48.
2. Let n = first odd integer. Then n + 2 = 2nd odd integer, and n + 4 = 3rd odd integer.
( 2) ( 4) 135
2 4 135
3 6 135
3 6 6 135 6
3 129
3 129
3 343
n n n
n n n
n
n
n
n
n
+ + + + =+ + + + =
+ =+ − = −
=
=
=
Thus, the three integers are 43, 43 + 2 = 45, and 43 + 4 = 47.
3. Let n = first even integer. Then n + 2 = 2nd even integer, and n + 4 = 3rd even integer.
( 2) ( 4) 24
2 4 24
3 6 24
3 6 6 24 6
3 30
3 30
3 310
n n n
n n n
n
n
n
n
n
+ + + + = −+ + + + = −
+ = −+ − = − −
= −−
=
= −
Thus, the integers are −10, −10 + 2 = −8,
and −10 + 4 = −6.
4. Let n = first odd integer. Then n + 2 = 2nd odd integer, and n + 4 = 3rd odd integer.
( 2) ( 4) 27
2 4 27
3 6 27
3 6 6 27 6
3 33
3 33
3 311
n n n
n n n
n
n
n
n
n
+ + + + = −+ + + + = −
+ = −+ − = − −
= −−
=
= −
Thus, the integers are −11, −11 + 2 = −9,
and −11 + 4 = −7.
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Bello, Introductory Algebra, 3e ISM Chapter 2: Equations, Problem Solving, and Inequalities
106
5. Let n = first integer. Then n + 1 = next consecutive integer.
( 1) 25
1 25
2 1 25
2 1 1 25 1
2 26
2 26
2 213
n n
n n
n
n
n
n
n
+ + = −+ + = −
+ = −+ − = − −
= −−
=
= −
Thus, the integers are −13 and
−13 + 1 = −12.
6. Let n = first integer. Then n + 1 = next consecutive integer.
( 1) 9
1 9
2 1 9
2 10
5
n n
n n
n
n
n
+ + =−
+ + =−+ =−
=−=−
Thus, the integers are −5 and −5 + 1 = −4.
7. Let n = first integer, n + 1 = second integer, n + 2 = last integer.
2 2 23
3 2 23
3 2 2 23 2
3 21
3 21
3 37
n n
n
n
n
n
n
+ + =+ =
+ − = −=
=
=
Thus, the integers are 7, 7 + 1 = 8, and 7 + 2 = 9.
8. Let n = Maria’s number, n + 1 = Latasha’s number n + 2 = Kim’s number.
2 2 47
3 2 47
3 2 2 47 2
3 45
15
n n
n
n
n
n
+ + =
+ =+ − = −
==
Thus, Maria’s apartment number is 15, Latasha’s is 15 + 1 = 16, and Kim’s is 15 + 2 = 17.
9. Let n = first locker number, n + 1 = middle locker number, n + 2 = last locker number.
( 2) 2( 1)
2 2 2
2 2 2 2
n n n
n n n
n n
+ + = ++ + = +
+ = +
This is an identity, so the solution is any consecutive integers.
10. Let e = price of English book. Then e + 27 = price of math book.
( 27) 141
27 141
2 27 141
2 27 27 141 27
2 114
57
e e
e e
e
e
e
e
+ + =+ + =
+ =+ − = −
==
Thus, the English book was $57 and the math book was $57 + $27 = $84.
11. Let p = price of cheaper book. Then p + 24 = price of more expensive book.
( 24) 64
24 64
2 24 64
2 24 24 64 24
2 40
2 40
2 220
p p
p p
p
p
p
p
p
+ + =+ + =
+ =+ − = −
=
=
=
Thus, the books cost $20 and $20 + $24 = $44.
12. Let x = points scored by losing team. Then x + 5 = points scored by winning team.
( 5) 179
5 179
2 5 179
2 5 5 179 5
2 174
2 174
2 287
x x
x x
x
x
x
x
x
+ + =+ + =
+ =+ − = −
=
=
=
Thus, the losing team scored 87 points and the winning team scored 87 + 5 = 92 points.
Full file at http://TestbankCollege.eu/Solution-Manual-Introductory-Algebra-3rd-Edition-Ignacio-Bello
Chapter 2: Equations, Problem Solving, and Inequalities Bello, Introductory Algebra, 3e ISM
107
13. Let x = points scored by losing team. Then x + 55 = points scored by winning team.
( 55) 133
55 133
2 55 133
2 55 55 133 55
2 78
39
x x
x x
x
x
x
x
+ + =+ + =
+ =+ − = −
==
Thus, the losing team scored 39 points and the winning team scored 39 + 55 = 94 pts.
14. Let m = amount Mida spent.
Then 2
amount Sandra spent.5
m =
2210
55 2
2105 5
7210
52
150; 150 605
m m
m m
m
m
+ =
+ =
=
= ⋅ =
Mida spent $150 and Sandra spent $60.
15. Let x charges on other card.
Then 3
4 charges on one card.8
x + =
34 147
8
34 147
88 3
4 1478 8
114 147
811
1438
8 11 8143
11 8 118 13
104
x x
x x
x x
x
x
x
x
x
+ + =
+ + =
+ + =
+ =
=
⋅ = ⋅
= ⋅=
Thus, the charges on the cards are $104 and
3(104) 4 3 13 4 39 4 $43.
8+ = ⋅ + = + =
16. Let c = amount of Marcus’s bill. Then c + 2 = amount of Liang’s bill, and (c + 2) + 2 = amount of Mourad’s bill.
( 2) [( 2) 2] 261
2 2 2 261
3 6 261
3 255
85
c c c
c c c
c
c
c
+ + + + + =
+ + + + + =+ =
==
Thus, Marcus’s bill was $85, Liang’s bill was $85 + $2 = $87, and Mourad’s bill was $87 + $2 = $89.
17. Let n = first number. Then 3n = second number,
and 3n − 5 = third number. 3 (3 5) 254
3 3 5 254
7 5 254
7 5 5 254 5
7 259
7 259
7 737
n n n
n n n
n
n
n
n
n
+ + − =+ + − =
− =− + = +
=
=
=
Thus, the three numbers are 37, 3(37) = 111,
and 111 − 5 = 106.
18. Let n = first integer. Then n + 1 = 2nd integer and n + 2 = 3rd integer.
( 1) ( 2) 4 17
1 2 4 17
3 3 4 17
3 20 4
20
n n n n
n n n n
n n
n n
n
+ + + + = −
+ + + + = −+ = −
+ ==
Thus, the integers are 20, 20 + 1 = 21, and 20 + 2 = 22.
19. Let n = smaller number. Then 6n = larger number.
6 147
7 147
7 147
7 721
n n
n
n
n
+ ==
=
=
Thus, the integers are 21 and 6(21) = 126.
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20. Let x = fraction. 5 3( 1)
5 3 3
5 3 3 3 3
2 3
3
2
x x
x x
x x x x
x
x
= += +
− = − +=
=
The fraction is 3
.2
21. Let w = percent accessing from work. Then w + 25 = percent accessing from home.
( 25) 15 66
25 15 66
2 40 66
2 40 40 66 40
2 26
2 26
2 213
25 38
w w
w w
w
w
w
w
w
w
+ + + =+ + + =
+ =+ − = −
=
=
=+ =
Thus 13% access the Internet from work and 38% from home.
22. Let a = Internet market audience in Los Angeles. Then a + 727 = Internet market audience in New York.
( 727) 7955
727 7955
2 727 7955
2 7228
3614
727 4341
a a
a a
a
a
a
a
+ + =+ + =
+ =
==
+ =
Thus, the Internet market audience in Los Angeles is 3614 (thousand) and in New York is 4341 (thousand).
23. Let x = salary per week for men 16−24
years old. Then x + 330 = salary per week for men over 25 years old.
330 722
330 330 722 330
392
x
x
x
+ =+ − = −
=
Men 16−24 years old make $392 per week.
24. Let y = salary per week for women 16−24
years old. Then y + 188 = salary per week for women over 25 years old.
188 542
188 188 542 188
354
y
y
y
+ =+ − = −
=
Women 16−24 years old make $354 per
week.
25. Let a = height of antenna. 1250 1472
1250 1250 1472 1250
222
a
a
a
+ =− + = −
=
The antenna is 222 feet high.
26. 0.10 10
17.50 0.10 10
17.50 10 0.10 10 10
7.50 0.10
7.50 0.10
0.10 0.1075
C m
m
m
m
m
m
= += +
− = + −=
=
=
Thus 75 miles were traveled.
27. 96 32
0 96 32
0 96 96 96 32
96 32
96 32
32 323
v t
t
t
t
t
t
= −= −
− = − −− = −− −
=− −
=
The rocket will reach its highest point at 3 seconds.
28. 96 32
16 96 32
16 96 96 96 32
80 32
80 32
32 322.5
v t
t
t
t
t
t
= −= −
− = − −− = −− −
=− −
=
The rocket’s velocity decreases to 16 feet per second at 2.5 seconds.
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Chapter 2: Equations, Problem Solving, and Inequalities Bello, Introductory Algebra, 3e ISM
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29. Let r votes received by loser. Then r + 372 = votes received by winner.
( 372) 980
372 980
2 372 980
2 372 372 980 372
2 608
2 608
2 2304
372 676
r r
r r
r
r
r
r
r
r
+ + =+ + =
+ =+ − = −
=
=
=+ =
Thus, the candidates received 304 votes and 676 votes.
30. Let r = cost of Russian program.
Then r − 19 = cost of U.S. program. ( 19) 71
19 71
2 19 71
2 19 19 71 19
2 90
45
19 26
r r
r r
r
r
r
r
r
+ − =
+ − =
− =− + = +
==
− =
The cost of the Russian program is $45 billion and the cost of the U.S. program is $26 billion.
31. Let m = number of minutes after the first 3 minutes.
3.05 0.7 7.95
3.05 3.05 0.7 7.95 3.05
0.7 4.90
0.7 4.90
0.7 0.77
m
m
m
m
m
+ =− + = −
=
=
=
He talked for 7 minutes after the first 3 minutes. The call was 10 minutes long.
32. a. Let h = number of hours after first hour. 1.25 0.75 7.25
1.25 1.25 0.75 7.25 1.25
0.75 6
0.75 6
0.75 0.758
h
h
h
h
h
+ =− + = −
=
=
=
Since she parked 8 hours at 75¢ per hour, she parked 9 hours total.
b. 1.25 0.75 10
1.25 1.25 0.75 10 1.25
0.75 8.75
0.75 8.75
0.75 0.7511.6 or 11
h
h
h
h
h
+ =− + = −
=
=
=
(round down to the nearest whole number) She can park 11 hours at 75¢ per hour plus the 1st hour at $1.25. So, she can park 12 hours for $10 or less.
33. a. Let m = number of miles. 0.95 1.25 28.45
0.95 0.95 1.25 28.45 0.95
1.25 27.50
1.25 27.50
1.25 1.2522
m
m
m
m
m
+ =− + = −
=
=
=
The ride was 22 miles long.
b. Find the cost of a 12-mile ride using the equation from a. 0.95 1.25 0.95 1.25(12)
0.95 15
15.95
m+ = += +=
The limo is cheaper, since $15 < $15.95.
34. Let x = number in the 18−25 category.
Then x + 844,000 = # in the 35-and-older category.
( 844,000) 7,072,000
844,000 7,072,000
2 844,000 7,072,000
2 6,228,000
2 6,228,000
2 23,144,000
844,000 3,958,000
x x
x x
x
x
x
x
x
+ + =
+ + =
+ ==
=
=
+ =
Thus, 3,114,000 are in the 18−25 group and
3,958,000 are in the 35-and-older group.
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110
35. Let m = measure of the angle.
90 − m = complement and
180 − m = supplement. 180 2(90 ) 20
180 180 2 20
180 200 2
180 180 200 180 2
20 2
2 20 2 2
20
m m
m m
m m
m m
m m
m m m m
m
− = − +− = − +− = −
− − = − −− = −
− + = − +=
The measure of the angle is 20°.
36. Let m = measure of the angle.
90 − m = complement and
180 − m = supplement.
180 3(90 ) 10
180 270 3 10
180 280 3
100 3
2 100
50
m m
m m
m m
m m
m
m
− = − +− = − +
− = −− = −
==
The measure of the angle is 50°.
37. Let m = measure of the angle.
90 − m = complement and
180 − m = supplement.
180 3(90 )
180 270 3
180 180 270 180 3
90 3
3 90 3 3
2 90
2 90
2 245
m m
m m
m m
m m
m m m m
m
m
m
− = −− = −
− − = − −− = −
− + = − +=
=
=
The measure of the angle is 45°.
38. Let m = measure of the angle.
90 − m = complement and
180 − m = supplement.
180 4(90 )
180 360 4
180 4
3 180
60
m m
m m
m m
m
m
− = −
− = −− = −
==
The measure of the angle is 60°.
39. Let m = measure of the angle.
90 − m = complement and
180 − m = supplement.
180 4(90 ) 54
180 360 4 54
180 306 4
180 180 306 180 4
126 4
4 126 4 4
3 126
3 126
3 342
m m
m m
m m
m m
m m
m m m m
m
m
m
− = − −− = − −− = −
− − = − −− = −
− + = − +=
=
=
The measure of the angle is 42°.
40. Let m = measure of the angle.
90 − m = complement and
180 − m = supplement.
180 3(90 ) 41
180 270 3 41
180 229 3
49 3
2 49
24.5
m m
m m
m m
m m
m
m
− = − −− = − −
− = −− = −
==
The measure of the angle is 24.5°.
41. Let A = amount of garbage.
0.5 236
1.5 236
1.5 236
1.5 1.5157.333
A A
A
A
A
+ =
=
=
=
The amount, A, of garbage is about 157 million tons.
42. Let n = number of landfills.
6157 1767
7924
n
n
− =
=
The number n of landfills is 7924.
43. Let g = amount of garbage sent to landfills.
9 131
9 9 131 9
140
g
g
g
− =
− + = +
=
The amount, g, of garbage sent to landfills is 140 million tons.
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Chapter 2: Equations, Problem Solving, and Inequalities Bello, Introductory Algebra, 3e ISM
111
44. Let p = pounds of garbage discarded in 2002.
3.09 0.05
3.14
p
p
= −
=
The amount, p, of garbage discarded in 2002 was 3.14 lb per person per day.
45. Let R = amount of recycled materials.
66.7 72.3
66.7 66.7 72.3 66.7
5.6
R
R
R
+ =
+ − = −
=
The total amount, R, of recycled materials is 5.6 million tons.
46. Let M = the amount of material recovered for recycling.
49.8 55.4
5.6
M
M
+ =
=
The amount, M, of material recovered for recycling is 5.6 million tons.
47. Let g = amount of garbage burned.
33 0.14
33 0.14
0.14 0.14235.714
g
g
g
=
=
≈
The amount, g, of garbage burned was about 236 million tons.
48. Let g = amount of garbage gone to landfills.
131 0.55
238.182
g
g
=
=
The amount, g, of garbage gone to landfills is about 2387 million tons.
49. Let t = amount of yard trimmings.
35.2% 23.1%
35.2% 23.1% 23.1% 23.1%
12.1%
t
t
t
= +
− = + −
=
The portion, p, of yard trimmings in 2003 was 12.1%.
50. Let b = amount of lead.
0.93 2
2.1505
b
b
=
≈
The amount, b, of lead in recycled batteries is about 2.15 billion pounds.
51. Let x = the number of years Diophantus lived.
1youth years
6x =
1grew beard
12x =
1married
7x =
1one-half life span of father's age.
2x =
1 1 1 15 4
6 12 7 214 7 12 42
984 84 84 84
759
8475 75 84 75
984 84 84 84
99
8484 84 9
99 9 84
84
x x x x x
x x x x x
x x
x x x x
x
x
x
+ + + + + =
+ + + + =
+ =
− + = −
=
⋅ = ⋅
=
Diophantus lived 84 years.
52. Answers may vary. Sample answer: Determine what the problem is asking for.
53. Answers may vary.
54. Answers may vary.
55. RSTUV
56. consecutive
57. complementary
58. supplementary
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59. Let n = first odd integer. Then n + 2 = 2nd odd integer, and n + 4 = 3rd odd integer.
( 2) ( 4) 249
2 4 249
3 6 249
3 6 6 249 6
3 243
3 243
3 381
2 83
4 85
n n n
n n n
n
n
n
n
n
n
n
+ + + + =+ + + + =
+ =+ − = −
=
=
=+ =+ =
The integers are 81, 83, and 85.
60. Let n = number. 4 3 9
4 12
3 12
4
n n
n n
n
n
− = +
= +=
=
The number is 4.
61. Let p = calories in the pizza. Then p + 70 = calories in the shake.
( 70) 530
70 530
2 70 530
2 70 70 530 70
2 460
2 460
2 2230
70 300
p p
p p
p
p
p
p
p
p
+ + =+ + =
+ =+ − = −
=
=
=+ =
The pizza had 230 calories and the shake had 300 calories.
62. Let m = the measure of the angle.
90 − m = complement and
180 − m = supplement.
180 3(90 ) 47
180 270 3 47
180 223 3
43 3
2 43
21.5
m m
m m
m m
m m
m
m
− = − −− = − −
− = −− = −
==
The measure of the angle is 21.5°.
63. 55 100
55 100
55 5520
11
T
T
T
=
=
=
64. 88 3240
88 3240
88 88405
11
R
R
R
=
=
=
65. 15 120
15 120
15 158
T
T
T
=
=
=
66. 81 3240
40
T
T
=
=
67. 75 600
75 600
75 758
x
x
x
− = −− −
=− −
=
68. 45 900
20
x
x
− =−
=
69. 0.02 70
0.02 70
0.02 0.023500
P
P
P
− = −− −
=− −
=
70. 0.05 100
2000
R
R
− =−
=
71. 0.04 40
0.04 40
0.04 0.041000
x
x
x
− = −− −
=− −
=
72. 0.03 30
1000
x
x
− =
=
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2.5 Problem Solving: Motion, Mixture, and Investment Problems
Problems 2.5
1. (1) Read the problem.
We are asked to find the average speed of the bus.
(2) Select the unknown. Let R = average speed.
(3) Think of a plan.
T = 214 − 12 − 24 = 178
178 6000
R T D
R
× =× =
(4) Use algebra to solve the equation. 178 6000
178 6000
178 17833.7
R
R
R
=
=
≈
The average speed of the bus is 33.7 miles per hour.
(5) Verify the solution.
33.7 178 5998.6
R T D× =× =
2. (1) Read the problem. We are asked to find the number of hours if the car is going 60 mph.
(2) Select the unknown. Let T = number of hours.
(3) Think of a plan.
R × T = D
car 60 T 60T
bus 40 T + 3 40(T + 3)
60T = 40(T + 3)
(4) Use algebra to solve the equation. 60 40( 3)
60 40 120
60 40 40 40 120
20 120
20 120
20 206
T T
T T
T T T T
T
T
T
= += +
− = − +=
=
=
It takes the car 6 hours to overtake the bus.
(5) Verify the solution.
car: 60 × 6 = 360
bus: 40 × 9 = 360
3. (1) Read the problem. We are asked to find how many hours to meet if the faster bus travels at 50 mph.
(2) Select the unknown. Let T = hours until they meet.
(3) Think of a plan.
R × T = D
Supercruiser 40 T 40T
bus 50 T 50T
40T + 50T = 3240
(4) Use algebra to solve the equation. 40 50 3240
90 3240
90 3240
90 9036
T T
T
T
T
+ ==
=
=
Each bus traveled 36 hours before they met.
(5) Verify the solution.
40 × 36 = 1440
50 × 36 = 1800
1440 + 1800 = 3240
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4. (1) Read the problem. We are asked to find the number of ounces of the 50% solution that should be added to make a 30% solution.
(2) Select the unknown. Let x = number of ounces of 50% solution to added.
(3) Think of a plan.
%
×
oz
=
Amt of Pure acid in Final Mix
50% sol. 0.50 x 0.50x
5% sol. 0.05 32 1.60
30% sol. 0.30 x + 32 0.30(x + 32)
0.50x + 1.60 = 0.30(x + 32)
(4) Use algebra to solve the equation. 10 0.50 10 1.60 10 [0.30( 32)]
5 16 3( 32)
5 16 3 96
5 16 16 3 96 16
5 3 80
5 3 3 3 80
2 80
2 80
2 240
x x
x x
x x
x x
x x
x x x x
x
x
x
⋅ + ⋅ = ⋅ ++ = ++ = +
+ − = + −= +
− = − +=
=
=
The photographer must add 40 ounces of the 30% solution.
(5) Verify the solution. 0.50(40) + 1.60 = 21.6 0.30(40 + 32) = 21.6
5. (1) Read the problem. We are asked to find how much is invested if her income is only $400.
(2) Select the unknown. Let s = amount invested in stocks.
Then 6000 − s = amount invested in
bonds.
(3) Think of a plan.
P × r = I
stocks s 0.05 0.05s
bonds 6000 − s 0.10 0.1(6000 − s)
0.05s + 0.10(6000 − s) = 400
(4) Use algebra to solve the equation. 0.05 0.10(6000 ) 400
0.05 600 0.10 400
600 0.05 400
0.05 200
0.05 200
0.05 0.054000
s s
s s
s
s
s
s
+ − =
+ − =− =
− =−− −
=− −
=
The woman invested $4000 in stocks and $2000 in bonds.
(5) Verify the solution. 5% of $4000 is $200 and 10% of $2000 is $200, so the total interest is $400.
Exercises 2.5
Students should use the RSTUV method to solve.
An outline of each solution is given.
1. Let D = 400 and T = 8.
8 400
50
R T D
R
R
× =
× ==
The bus’ average speed is 50 miles per hour.
2. Let T = 12 − 7 = 5 and D = 260.
5 260
52
R T D
R
R
× =
× ==
The car’s speed is 52 miles per hour.
3. Let D = 246 and T = 120.
120 246
120 246
120 1202.05 2
R T D
R
R
R
× =× =
=
= ≈
The rate of play of the tape is 2 meters per minute.
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Chapter 2: Equations, Problem Solving, and Inequalities Bello, Introductory Algebra, 3e ISM
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4. a. Let D = 12 and T = 6.
6 12
2
R T D
R
R
× =
× ==
or let D = 1 and T = 6.
6 1
1
6
R T D
R
R
× =× =
=
The rate of output is 2 inches per second
or 1
6 page per second.
b. Let 1
6R = and D = 60.
160
6360
R T D
T
T
× =
× =
=
It would take 360 seconds or
3606 minutes.
60=
5. Let D = 200 and R = 400.
400 200
400 200
400 4001
2
R T D
T
T
T
× =× =
=
=
It takes 1
2 hour to go from Tampa to
Miami.
6. R × T = D
freight 30 T + 1 30(T + 1)
passenger 60 T 60T
30( 1) 60
30 30 60
30 30
1
T T
T T
T
T
+ =+ =
==
It will take the passenger train 1 hour to overtake the freight train.
7. R × T = D
bus 60 T + 2 60(T + 2)
car 90 T 90T
60( 2) 90
60 120 90
60 60 120 90 60
120 30
120 30
30 304
T T
T T
T T T T
T
T
T
+ =+ =
− + = −=
=
=
It will take her 4 hours to catch the bus.
8. R × T = D
train 50 T 50T
car 60 T − 1 60(T − 1)
50 60( 1)
50 60 60
10 60
6
T T
T T
T
T
= −= −
− =−=
Find either the train’s distance or the car’s distance.
50 6 300
R T D× =× =
The towns are 300 miles apart.
9. R × T = D
bicycle 15 T 15T
car 60 12
T − ( )12
60 T +
115 60
2
15 60 30
45 30
2
3
T T
T T
T
T
= −
= −
− =−
=
Find either the bicycle’s distance or the car’s distance.
215 10
3
R T D× =
× =
Their house is 10 miles from the school.
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Bello, Introductory Algebra, 3e ISM Chapter 2: Equations, Problem Solving, and Inequalities
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10. R × T = D
jet 480 T − 1 480(T − 1)
plane 480 T 480T
480( 1) 480 1632
480 480 480 1632
960 480 1632
960 2112
2.2
T T
T T
T
T
T
− + =
− + =− =
==
They will pass each other in 2.2 hours on 2 hours 12 minutes.
11. R × T = D
car 50 T 50T
another car 55 T 55T
50 55 630
105 630
105 630
105 1056
T T
T
T
T
+ ==
=
=
The cars will meet in 6 hours.
12. R × T = D
crew 70 T 70T
her 65 T–2 55(T–2)
50 55 630
105 630
6
T T
T
T
+ =
==
The cars will meet in 6 hours. 70 65( 2) 275
70 65 130 275
135 130 275
135 130 130 275 130
135 405
135 405
135 1353
T T
T T
T
T
T
T
T
+ − =+ − =
− =− + = +
=
=
=
Find the distance the crew traveled.
70 3 210
R T D× =× =
The first crew traveled 210 kilometers.
13. R × T = D
out 480 T 480T
back 640 7 − T 640(7 − T)
480 640(7 )
480 4480 640
480 640 4480 640 640
1120 4480
1120 4480
1120 11204
T T
T T
T T T T
T
T
T
= −= −
+ = − +=
=
=
Find the distance out or the distance back.
480 4 1920
R T D× =× =
The base is 1920 miles from the target.
14. R × T = D
driving 40 T 40T
walking 5 14
2 T− ( )14
5 2 T−
140 5 2
4
940 5
4
4540 5
445
40 5 5 5445
454
1 45 145
45 4 451
4
T T
T T
T T
T T T T
T
T
T
= − = −
= −
+ = − +
=
⋅ = ⋅
=
Find the distance he drove.
140 10
4
R T D× =
⋅ =
His car broke down 10 miles from his house.
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15. R × T = D
first seconds 20R 180 20R(180)
last seconds R 60 R(60)
20 (180) (60) 366
3600 60 366
3660 366
3660 366
3660 36600.1
R R
R R
R
R
R
+ =+ =
=
=
=
During the first 180 seconds, the shuttle was traveling at 20(0.1) = 2 feet per second or 24 inches per second. During the last 60 seconds, the shuttle was traveling at 0.1 feet per second or 1.2 inches per second.
16. R × T = D
first seconds 13R 180 13R(180)
last seconds R 60 R(60)
13 (180) (60) 366
2340 60 366
2400 366
2400 366
2400 24000.1525
R R
R R
R
R
R
+ =+ =
=
=
=
The shuttle’s final rate of approach is 0.1525 feet per second or 12(0.1525) = 1.83 inches per second.
17. % × liters = Amount of glycerin in final mixture
40% solution 0.40 x 0.40x
80% solution 0.80 10 8
65% solution 0.65 x + 10 0.65(x + 10)
0.40 8 0.65( 10)
0.40 8 0.65 6.5
0.40 0.65 1.5
0.25 1.5
6
x x
x x
x x
x
x
+ = ++ = +
= −− = −
=
Mix 6 liters of the 40% glycerin solution.
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118
18. % × parts = Amount of acetic acid
99.5% acid 0.995 x 0.995x
10% acid 0.10 100 10
28% acid 0.28 x + 100 0.28(x + 100)
0.995 10 0.28( 100)
0.995 10 0.28 28
0.995 0.28 18
0.715 18
0.715 18
0.715 0.71525
x x
x x
x
x
x
x
+ = ++ = +
= +=
=
≈
Add 25 parts of glacial acetic acid.
19. price per pound × pounds = price
copper 0.65 x 0.65x
zinc 0.30 70 − x 0.30(70 − x)
brass 0.45 70 0.45(70)
0.65 0.30(70 ) 0.45(70)
100 0.65 100 0.30(70 ) 100 0.45(70)
65 30(70 ) 45(70)
65 2100 30 3150
35 2100 3150
35 1050
30
70 40
x x
x x
x x
x x
x
x
x
x
+ − =⋅ + ⋅ − = ⋅
+ − =+ − =
+ ===
− =
Mix 30 pounds of copper with 40 pounds of zinc.
20. price per pound × pounds = price
Oolong 19 x 19x
Another 4 50 − x 4(50 − x)
Mixture 7 50 7(50)
19 4(50 ) 7(50)
19 200 4 350
15 200 350
15 150
10
x x
x x
x
x
x
+ − =+ − =
+ ===
Mix 10 pounds of Oolong tea.
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Chapter 2: Equations, Problem Solving, and Inequalities Bello, Introductory Algebra, 3e ISM
119
21. price per pound × pounds = price
Blue Jamaican 5 x 5x
Regular 2 80 2(80)
Mixture 2.60 x + 80 2.6(x + 80)
5 2(80) 2.6( 80)
5 160 2.6 208
5 2.6 48
2.4 48
20
x x
x x
x x
x
x
+ = ++ = +
= +==
Mix 20 pounds of Blue Jamaican coffee.
22. % × ounces = Amount of alcohol in final mixture
vermouth 0.10 x 0.1x
gin 0.60 20 0.6(20)
martini 0.30 x + 20 0.3(x + 20)
0.1 0.6(20) 0.3( 20)
10 0.1 10 0.6(20) 10 0.3( 20)
6(20) 3( 20)
120 3 60
60 3
2 60
30
x x
x x
x x
x x
x x
x
x
+ = +⋅ + ⋅ = ⋅ +
+ = ++ = +
= − +− = −
=
Mix 30 ounces of vermouth.
23. % × ounces = amount of alcohol in final mixture
40% 0.4 x 0.4x
20% 0.2 64 − x 0.2(64 − x)
30% 0.3 64 0.3(64)
0.4 0.2(64 ) 0.3(64)
10 0.4 10 0.2(64 ) 10 0.3(64)
4 2(64 ) 3(64)
4 128 2 192
2 128 192
2 64
32
64 32
x x
x x
x x
x x
x
x
x
x
+ − =⋅ + ⋅ − = ⋅
+ − =+ − =
+ ===
− =
Mix 32 ounces of each.
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24. % × quarts = Amount of antifreeze in final solution
50% 0.5 30 − x 0.5(30 − x)
100% 1 x x
70% 0.7 30 0.7(30)
0.5(30 ) 0.7(30)
15 0.5 21
15 0.5 21
0.5 6
12
x x
x x
x
x
x
− + =− + =
+ ===
Drain 12 quarts of 50% solution and replace with 12 quarts of pure antifreeze.
25. % × quarts = Amount of antifreeze in final solution
50% 0.5 30 − x 0.5(30 − x)
0% 0 x 0
30% 0.3 30 0.3(30)
0.5(30 ) 0 0.3(30)
15 0.5 9
0.5 6
12
x
x
x
x
− + =− =− = −
=
Drain 12 quarts of 50% solution and replace with 12 quarts of water.
26. % × ounces = Amount of juice in mixture
Concentrate x 12 12x
Water 0 3(12) = 36 0 ⋅ 36 = 0
10% juice 0.1 12 + 3(12) = 48 0.1(48)
12 0 0.1(48)
12 4.8
0.4
x
x
x
+ ===
The concentrate is 40% juice.
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121
27. % × ounces = Amount of juice in mixture
Welch’s x 12 12x
Water 0 3(12) = 36 0 ⋅ 36 = 0
30% juice 0.3 12 + 3(12) = 48 0.3(48)
12 0 0.3(48)
12 14.4
1.2
x
x
x
+ ===
This is impossible, since it would need 120% of juice in the concentrate.
28. P × r = I
5% x 0.05 0.05x
7% 15,000 − x 0.07 0.07(15,000 − x)
0.05 0.07(15,000 ) 870
100 0.05 100 0.07(15,000 ) 100 870
5 7(15,000 ) 87,000
5 105,000 7 87,000
105,000 2 87,000
2 18,000
9000
15,000 6000
x x
x x
x x
x x
x
x
x
x
+ − =⋅ + ⋅ − = ⋅
+ − =+ − =
− =− = −
=− =
$9000 is invested at 5% and $6000 is invested at 7%.
29. P × r = I
6% x 0.06 0.06x
8% 20,000 − x 0.08 0.08(20,000 − x)
0.06 0.08(20,000 ) 1500
100 0.06 100 0.08(20,000 ) 100 1500
6 8(20,000 ) 150,000
6 160,000 8 150,000
160,000 2 150,000
2 10,000
5000
20,000 15,000
x x
x x
x x
x x
x
x
x
x
+ − =⋅ + ⋅ − = ⋅
+ − =+ − =
− =− = −
=− =
$5000 is invested at 6% and $15,000 is invested at 8%.
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Bello, Introductory Algebra, 3e ISM Chapter 2: Equations, Problem Solving, and Inequalities
122
30. P × r = I
7.5% x 0.075 0.075x
6% 25,000 − x 0.06 0.06(25,000 − x)
0.075 0.06(25,000 ) 1620
0.075 1500 0.06 1620
1500 0.015 1620
0.015 120
8000
25,000 17,000
x x
x x
x
x
x
x
+ − =+ − =
+ ===
− =
$8000 is invested at 7.5% and $17,000 is invested at 6%.
31. P × r = I
5% x 0.05 0.05x
7% 18,000 − x 0.07 0.07(18,000 − x)
0.05 0.07(18,000 ) 1100
0.05 1260 0.07 1100
1260 0.02 1100
0.02 160
8000
x x
x x
x
x
x
+ − =+ − =
− =− = −
=
$8000 is invested at 5% in the savings account.
32. P × r = I
8% 20,000 0.08 0.08(20,000)
6% x 0.06 0.06x
0.08(20,000) 0.06 2200
1600 0.06 2200
0.06 600
10,000
x
x
x
x
+ =+ =
==
She must invest $10,000 at 6%.
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Chapter 2: Equations, Problem Solving, and Inequalities Bello, Introductory Algebra, 3e ISM
123
33. P × r = I
5% x 0.05 0.05x
6% 10,000 − x 0.06 0.06(10,000 − x)
0.05 0.06(10,000 ) 60
0.05 600 0.06 60
0.05 660 0.06
0.11 660
6000
10,000 4000
x x
x x
x x
x
x
x
= − += − += −==
− =
$6000 is invested at 5% and $4000 is invested at 6%.
34. P × r = I
8% x + 500 0.08 0.08(x + 500)
6% x 0.06 0.06x
0.08( 500) 0.06 600
0.08 40 0.06 600
40 0.14 600
0.14 560
4000
500 4500
x x
x x
x
x
x
x
+ + =+ + =
+ ===
+ =
$4000 is invested at 6% and $4500 is invested at 8%.
35. P × r = I
6% x 0.06 0.06x
10% 40,000 − x 0.10 0.10(40,000 − x)
0.06 0.10(40,000 )
0.06 4000 0.1
0.16 4000
25,000
40,000 15,000
x x
x x
x
x
x
= −= −==
− =
$25,000 is invested at 6% and $15,000 is invested at 10%.
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124
36. P × r = I
lost 5% x 0.05 0.05x
gained 11% 10,000 − x 0.11 0.11(10,000 − x)
0.05 0.11(10,000 )
0.05 1100 0.11
0.16 1100
6875
10,000 3125
x x
x x
x
x
x
= −= −==
− =
$6875 was invested at 5% and $3125 was invested at 11%.
37. Answers may vary.
38. Answers may vary.
39. Answers may vary.
40. D RT=
41. I Pr=
42. P × r = I
5% 8000 − x 0.05 0.05(8000 − x)
10% x 0.10 0.10x
0.05(8000 ) 0.1 650
400 0.05 0.1 650
400 0.05 650
0.05 250
5000; 8000 3000
x x
x x
x
x
x x
− + =− + =
+ =
== − =
$3000 is invested at 5% and $5000 is invested at 10%.
43. % × gallons = Amount of salt in mixture
10% 0.1 x 0.1x
20% 0.2 15 0.2(15)
16% 0.16 x + 15 0.16(x + 15)
0.1 0.2(15) 0.16( 15)
0.1 3 0.16 2.4
0.1 0.16 0.6
0.06 0.6
10
x x
x x
x x
x
x
+ = ++ = +
= −− = −
=
Thus 10 gallons of 10% salt solution should be added.
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125
44. R × T = D
1st train 40 T 40T
2nd train 35 T 35T
40 35 300
75 300
4
T T
T
T
+ ===
They meet after 4 hours.
45. R × T = D
bus 50 T + 1 50(T + 1)
car 60 T 60T
50( 1) 60
50 50 60
50 10
5
T T
T T
T
T
+ =+ =
==
The car overtakes the bus in 5 hours.
46. Let D = 250 and T = 5.
5 250
5 250
5 550
R T D
R
R
R
× =× =
=
=
The average speed is 50 miles per hour.
47. 2.9(30) 71 87 71 158+ = + =
48. 5 5
(95 32) (63) 359 9
− = =
49. 1
(20)(10) 10(10) 1002
= =
50. 2(38) 10 76 10 66− = − =
51. 22 1 23 2
73 3 3
+= =
52. 21 1 22 1
73 3 3
+= =
2.6 Formulas and Geometry Applications
Problems 2.6
1. a. Let h = 15.
2.8 28.1
2.8(15) 28.1
42 28.1
70.1
H h= += += +=
Thus a woman with a 15-inch humerus should be about 70.1 inches tall.
b. 2.8 28.1
28.1 2.8
2.828.1
2.8 2.828.1
2.828.1
2.8
H h
H h
hH
Hh
Hh
= +
− =
−=
−=
−=
c. Let H = 61.7.
61.7 28.112
2.8h
−= =
Thus the length of the humerus of a 61.7-inch-tall woman is 12 inches.
2. a. Let C = 22.
1( 4)
61 1
(22 4) (18) 36 6
S C= −
= − = =
The ant’s speed is 3 cm/sec.
b. ( )
( )
14
61
6 6 46
6 4
6 4
6 4
S C
S C
S C
S C
C S
= −
⋅ = ⋅ −
= −
+ =
= +
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126
c. Let S = 2. 6(2) 4
12 4
16
C = += +=
Thus, the temperature is 16°C.
3. a. The final cost is obtained by adding the original cost C and the tax T. F = C + T
b. Let C = 10.
10
10
10
F C T
F T
F T
T F
= += +
− == −
4. a. 2(2.4 in.) (1.2 in.) 2.88 in.
A LW=
= ⋅ =
b. 2 2
2(2.4 in.) 2(1.2 in.)
4.8 in. 2.4 in.
7.2 in.
P L W= += += +=
c. Let P = 100 and L = 20. 2 2
100 2(20) 2
100 40 2
100 40 2
60 2
30
P L W
W
W
W
W
W
= += += +
− ===
The width is 30 inches.
5. a.
2
1
21
(30 in.)(15 in.)2
225 in.
A bh=
=
=
b. Let A = 300 and b = 20.
1
21
300 (20)2
300 10
30
A bh
h
h
h
=
=
==
Thus, the height of the sail is 30 feet.
6. a. Let C = 9π.
2
2
9
24.5
C r
r
r
r
= π9π = π
=
=
The radius is 4.5 inches.
b. Find the area of the entire CD. 2 2(4.5) 20.25A r= π = π = π
Find the area of the nonrecorded part. 2 2(0.75) 0.5625A r= π = π = π
The area of the recorded region is
20.25 π − 0.5625π = 19.6875π 2in.
7. a. (2 10) 8 180
10 10 180
10 190
19
x x
x
x
x
− + =− =
==
2(19) − 10 = 28 and 8(19) = 152
The angles measure 28° and 152°.
b. 8 15 4 3
8 4 12
4 12
3
x x
x x
x
x
− = −= +==
8(3) − 15 = 9 and 4(3) − 3 = 9
Both measure 9°.
c. (3 3) ( 17) 90
4 14 90
4 76
19
x x
x
x
x
− + + =+ =
==
3(19) − 3 = 54 and 19 + 17 = 36
The angles measure 54° and 36°.
8. a. Let S = 2.
22
322 2 24
83 3
SL
+=
+= = =
The foot is 8 inches long.
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Chapter 2: Equations, Problem Solving, and Inequalities Bello, Introductory Algebra, 3e ISM
127
b. Let S = 3.
21
321 3 24
83 3
SL
+=
+= = =
The foot is 8 inches long.
c. Let L = 17. 3 22
3(17) 22
51 22
29
S L= −= −= −=
His right foot needs a size 29.
Exercises 2.6
1. a. Let R = 30 and T = 4. D = RT = 30(4) = 120
b. Let R = 55 and T = 5. D = RT = 55(5) = 275 The car traveled 275 miles.
c.
or
D RT
D RT
T TD D
R RT T
=
=
= =
d. Let D = 180 and T = 3.
18060
3
DR
T= = =
The rate is 60 miles per hour.
2. a. Let D = 240 and T = 4.
24060
4
DR
T= = =
The rate is 60 miles per hour.
b. Let D = 140 and T = 4.
14035
4
DR
T= = =
The rate is 35 miles per hour.
c. D
RT
DT R T
TTR D
DT
R
=
⋅ = ⋅
=
=
d. Let R = 35 and D = 105.
1053
35
DT
R= = =
It would take the train 3 hours.
3. a. Let H = 60. 5 190
5(60) 190
300 190
110
W H= −= −= −=
He should weigh 110 pounds.
b. 5 190
190 5
190
5190
5
W H
W H
WH
WH
= −+ =+
=
+=
c. Let W = 200.
200 190 390 178 in. 6 ft
5 5 2H
+= = = =
He should be 78 inches or 6 feet 6 inches tall.
4. a. Let A = 6.
617 17 17 3 14
2 2
AH = − = − = − =
Thus, a 6-year-old should sleep 14 hours.
b. 172
172
2( 17) 22
2 34
34 2
AH
AH
AH
H A
A H
= −
− = −
− − = − ⋅ −
− + == −
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128
c. Let H = 11. 34 2
34 2(11) 34 22 12
A H= −
= − = − =
Thus, a child should sleep 11 hours at 12 years of age.
5. a. Let C = 15. 9
3259
(15) 32527 32
59
F C= +
= +
= +=
The corresponding temperature is 59°F.
b. 9
3259
325
5 5 9( 32)
9 9 55 5
( 32) or ( 32)9 9
F C
F C
F C
F C C F
= +
− =
⋅ − = ⋅
− = = −
c. Let F = 50.
5( 32)
95
(50 32)95
(18)910
C F= −
= −
=
=
The corresponding temperature is 10°C.
6. a. P = I − E
b. Let I = 2700 and E = 347.
P = I − E = 2700 − 347 = 2353
The profit is $2353.
c. P I E
P I I I E
P I E
E I P
= −− = − −− = −
= −
d. Let P = 750 and I = 1300.
E = I − P = 1300 − 750 = 550
The expenses are $550.
7. a. Btu
EERw
=
b. Let Btu = 9000 and w = 1000.
Btu 9000EER 9
1000w= = =
The EER is 9.
c. Btu
EER
Btu(EER)
(EER)( ) Btu
Btu (EER)( )
w
w ww
w
w
=
=
==
d. Let w = 2000 and EER = 10. Btu = (EER)(w) = 10(2000) = 20,000 It produces 20,000 Btu per hour.
8. a. C = A − L
b. Let A = 4800 and L = 2300.
C = A − L = 4800 − 2300 = 2500
The capital is $2500.
c. C A L
C A L
L A C
= −− = −
= −
d. Let C = 18,200 and A = 30,000.
L = 30,000 − 18,200 = 11,800
Its liabilities are $11,800.
9. a. S = C + M
b. Let M = 15 and C = 52. S = C + M = 52 + 15 = 67 The selling price should be $67.
c. S C M
S C M
M S C
= +− =
= −
d. Let S = 18.75 and C = 10.50.
M = S − C = 18.75 − 10.50 = 8.25
The markup is $8.25.
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Chapter 2: Equations, Problem Solving, and Inequalities Bello, Introductory Algebra, 3e ISM
129
10. a. TS dN= π
b. Let d = 2 and N = 100.
T (3.14)(2)(100) 628S dN= π = =
The tip speed is 628 meters per second.
c. T
T
T
T
S dN
S dN
d dS
Nd
SN
d
= ππ
=π π
=π
=π
d. Let T 275S = π and d = 2.
T 275137.5
SN
d
π= = =
π π(2)
The number of revolutions per second is 137.5.
11. a. Let W = 10 and L = 20. 2 2
2(20) 2(10)
40 20
60
P L W= += += +=
The perimeter is 60 cm.
b. Let P = 220 and W = 20. 2 2
220 2 2(20)
220 2 40
180 2
90
P L W
L
L
L
L
= += += +==
The length is 90 cm.
12. a. Let W = 15 and L = 30. 2 2
2(30) 2(15) 60 30 90
P L W= +
= + = + =
The perimeter is 90 cm.
b. Let P = 180 and L = 60. 2 2
180 2(60) 2
180 120 2
60 2
30
P L W
W
W
W
W
= += += +==
The width is 30 cm.
13. a. Let r = 10.
C = 2πr = 2(3.14)(10) = 62.8
The circumference is 62.8 inches.
b. 2
2
2 2
2
2
C r
C r
Cr
Cr
= ππ
=π π
=π
=π
c. Let C = 20π.
2010
2
Cr
π= = =
π 2π
The radius is 10 inches.
14. Let C = 26π. 2
26 2
26 2
213
C r
r
r
r
= ππ = ππ π
=2π π
=
The radius is 13 inches.
15. a. Let L = 4.2 and W = 3.1. A = LW = (4.2)(3.1) = 13.02
The area is 13.02 2m .
b.
or
A LW
A LW
L LA A
W WL L
=
=
= =
c. Let A = 60 and L = 10.
606
10
AW
L= = =
The width is 6 meters.
16. Vertical angles must be equal. 3 5 2 25
3 2 30
30
x x
x x
x
− = += +=
3(30) − 5 = 85 and 2(30) + 25 = 85
The angles each measure 85°.
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Bello, Introductory Algebra, 3e ISM Chapter 2: Equations, Problem Solving, and Inequalities
130
17. Vertical angles must be equal. 15 4 25 2
4 10 2
2 10
5
x x
x x
x
x
− = −− = −− =
= −
15 − 4(−5) = 35 and 25 − 2(−5) = 35
The angles each measure 35°.
18. Vertical angles must be equal. 80 3 40 5
3 40 5
2 40
20
x x
x x
x
x
− = −− = − −
= −= −
80 − 3(−20) = 140 and 40 − 5(−20) = 140
The angles each measure 140°.
19. Vertical angles must be equal. 80 3 40 5
3 40 5
2 40
20
x x
x x
x
x
+ = += − +
− = −=
80 + 3(20) = 140 and 40 + 5(20) = 140
The angles each measure 140°.
20. Vertical angles must be equal. 17 4 27 2
4 10 2
2 10
5
x x
x x
x
x
− = −− = −− =
= −
17 − 4(−5) = 37 and 27 − 2(−5) = 37
The angles each measure 37°.
21. Vertical angles must be equal. 6 5 5 25
6 5 30
30
x x
x x
x
− = += +=
6(30) − 5 = 175 and 5(30) + 25 = 175
The angles each measure 175°.
22. The angles are complementary. 2 (3 10) 90
5 10 90
5 100
20
x x
x
x
x
+ − =− =
==
2(20) = 40 and 3(20) − 10 = 50
The angles measure 40° and 50°.
23. The angles are complementary. 7 (5 30) 90
12 30 90
12 120
10
x x
x
x
x
+ − =− =
==
7(10) = 70 and 5(10) − 30 = 20
The angles measure 70° and 20°.
24. The angles are complementary. (2 15) 3 90
5 15 90
5 105
21
x x
x
x
x
− + =− =
==
2(21) − 15 = 27 and 3(21) = 63
The angles measure 27° and 63°.
25. The angles are complementary. (4 25) 90
5 25 90
5 115
23
x x
x
x
x
− + =− =
==
4(23) − 25 = 67 and 23
The angles measure 67° and 23°.
26. The angles are supplementary. (3 20) 7 180
10 20 180
10 160
16
x x
x
x
x
+ + =+ =
==
3(16) + 20 = 68 and 7(16) = 112
The angles measure 68° and 112°.
27. The angles are supplementary. (8 30) (2 10) 180
10 40 180
10 140
14
x x
x
x
x
+ + + =+ =
==
8(14) + 30 = 142 and 2(14) + 10 = 38
The angles measure 142° and 38°.
28. The angles are supplementary. (3 9) (5 11) 180
8 20 180
8 160
20
x x
x
x
x
+ + + =+ =
==
3(20) + 9 = 69 and 5(20) + 11 = 111
The angles measure 69° and 111°.
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Chapter 2: Equations, Problem Solving, and Inequalities Bello, Introductory Algebra, 3e ISM
131
29. The angles are supplementary. (3 6) (5 6) 180
8 12 180
8 168
21
x x
x
x
x
+ + + =+ =
==
3(21) + 6 = 69 and 5(21) + 6 = 111
The angles measure 69° and 111°.
30. Let P = 80 and L = 30. 2 2
80 2(30) 2
80 60 2
20 2
10
P L W
W
W
W
W
= += += +==
It was 10 feet wide.
31. Let P = 1110 and L = 480. 2 2
1110 2(480) 2
1110 960 2
150 2
75
P L W
W
W
W
W
= += += +==
The pool is 75 meters wide.
32. Let L = 120 and P = 346. 2 2
346 2(120) 2
346 240 2
106 2
53
P L W
W
W
W
W
= += += +==
The football field is 53 yards wide.
33. Let C = 14.13. 2
14.13 2(3.14)
2.25
C r
r
r
= π==
The diameter is 2(2.25) = 4.5 inches.
34. Let C = 251.2 2
251.2 2(3.14)
40
C r
r
r
= π==
The diameter is 2(40) = 80 feet.
35. a. 10
10
10
C A
C A
A C
= +− =
= −
b. Let C = 50.
A = C − 10 = 50 − 10 = 40
The American size is 40.
36. a. 30
30
30
C A
C A
A C
= +− =
= −
b. Let C = 42.
A = C − 30 = 42 − 30 = 12
The American size is 12.
37. a. Let t = 1985 − 1975 = 10.
9.74 0.40
9.74 0.40(10)
9.74 4
13.74
N t= += += +=
In 1985, there were 13.74 million recreational boats.
b. 9.74 0.40
9.74 0.40
9.74
0.409.74 5( 9.74)
or 0.40 2
N t
N t
Nt
N Nt t
= +− =−
=
− −= =
c. Let N = 17.74.
9.74 17.74 9.74 820
0.40 0.40 0.40
Nt
− −= = = =
1975 + 20 = 1995 In 1995, the expected number of recreational boats is 17.74 million.
38. a. Let t = 2000 − 1980 = 20.
720 5
720 5(20)
720 100
820
N t= += += +=
You would expect 820 teams in 2000.
b. 720 5
720 5
720
5720
5
N t
N t
Nt
Nt
= +− =−
=
−=
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132
c. Let N = 795.
720 795 720 7515
5 5 5
Nt
− −= = = =
1980 + 15 = 1995 In 1995, the expected number of teams is 795.
39. a. 2010 2000 10;
28 420
28(10) 420
280 420
700
x
A x
= − =
= += +
= +=
They would pay $700.
b. 28 420
420 28 420 420
420 28
420 28
28 28420
28
A x
A x
A x
A x
Ax
= +
− = + −
− =−
=
−=
c. 420
281000 420 580
20.71428 28
Ax
−=
−= = ≈
Vehicle insurance will be $1000 in about 21 years (2021).
40. a. 2010 2000 10;
40 786
40(10) 786
1186
x
A x
= − =
= +
= +=
They would pay $1186.
b. 40 786
786 40
786
40
A x
A x
Ax
= +
− =
−=
c. 786
401000 786
5.3540
Ax
−=
−= =
Vehicle insurance will be $1000 in about 5 years (2005).
41. a. 3.3 34
3.3(10) 34 33 34 67
H r= +
= + = + =
He would be 67 in. tall.
b. 3.3 34
34 3.3
34 3.3
3.3 3.334
3.3
H r
H r
H r
Hr
= +
− =
−=
−=
c. 34
3.370.3 34 36.3
113.3 3.3
Hr
−=
−= = =
His radius is 11 in. long.
42. a. 3.3 32 3.3(10) 32 65H r= + = + =
She would be 65 in. tall.
b. 3.3 32
32 3.3
32
3.3
H r
H r
Hr
= +
− =
−=
c. 32 61.7 32
93.3 3.3
Hr
− −= = =
Her radius is 9 in. long.
43. a. 2.9 62
2.9(30) 62
87 62
149
H t= +
= +
= +=
She would be 149 cm tall.
b. 2.9 62
62 2.9
62 2.9
2.9 2.962
2.9
H t
H t
H t
Ht
= +
− =
−=
−=
c. 62
2.9151.9 62 89.9
312.9 2.9
Ht
−=
−= = =
Her tibia is 31 cm long.
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Chapter 2: Equations, Problem Solving, and Inequalities Bello, Introductory Algebra, 3e ISM
133
44. The room is 320
408
= square yards or
40 ⋅ 9 = 360 square feet.
Let A = 360 and L = 20.
360 20
18
A LW
W
W
===
The room is 18 feet wide.
45. Let A = 5400 and W = 60.
5400 (60)
90
A LW
L
L
===
The sod area can be 90 feet long.
46. Let L = 70 and P = 240. 2 2
240 2 2(70)
240 2 140
100 2
50
P W L
W
W
W
W
= += += +==
The yard is 50 feet wide.
47. Answers may vary.
48. Answers may vary.
49. Answers may vary.
50. Answers may vary. Sample answer: Compare volumes since two hamburgers could have the same area or the same circumference and still have different volumes.
51. Answers may vary. Sample answer: C = 2πr
Let 4
22
r = = and 6
32
r = = and compare.
2 (3) 31.5
(2) 2
π= =
2π or 150%
Using this method, the Monster burger is 50% bigger than the Lite burger.
52. LW
53. 2L + 2W
54. 1
2bh
55. 2
rπ
56. 2 rπ
57. equal
58. The angles are complementary. (3 15) 2 90
5 15 90
5 105
21
x x
x
x
x
− + =− =
==
3(21) − 15 = 48 and 2(21) = 42
The angles measure 48° and 42°.
59. The angles are complementary. (3 25) 2 90
5 25 90
5 115
23
x x
x
x
x
− + =− =
==
3(23) − 25 = 44 and 2(23) = 46
The angles measure 44° and 46°.
60. The angles are supplementary. 6 (4 20) 180
10 20 180
10 160
16
x x
x
x
x
+ + =+ =
==
6(16) = 96 and 4(16) + 20 = 84
The angles measure 96° and 84°.
61. The angles are supplementary. (7 30) (3 10) 180
10 40 180
10 140
14
x x
x
x
x
+ + + =+ =
==
7(14) + 30 = 128 and 3(14) + 10 = 52
The angles measure 128° and 52°.
62. The angles are supplementary. (4 9) (4 11) 180
8 20 180
8 160
20
x x
x
x
x
+ + + =+ =
==
4(20) + 9 = 89 and 4(20) + 11 = 91
The angles measure 89° and 91°.
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134
63. The angles are supplementary. (2 6) (6 14) 180
8 20 180
8 160
20
x x
x
x
x
+ + + =+ =
==
2(20) + 6 = 46 and 6(20) + 14 = 134
The angles measure 46° and 134°.
64. a. Let r = 10. 2
2(10) 100 100(3.14) 314
A r=π
=π = π= =
The area is 100π 2in. or 314 2in. .
2
2 (10) 20
C r= π= π = π=20(3.14)= 62.8
The circumference is 20π in. or 62.8 in.
b. Let C = 40π. 2
40 2
40 2
220
C r
r
r
r
= ππ = ππ π
=2π π
=
The radius is 20 inches.
65. a. Let h = 20. 2.75 71.48
2.75(20) 71.48
126.48
H h= += +=
The woman is 126.48 centimeters.
b. 2.75 71.48
71.48 2.75
71.48
2.7571.48
2.75
H h
H h
Hh
Hh
= +− =−
=
−=
c. Let H = 140.23.
71.48 140.23 71.4825
2.75 2.75
Hh
− −= = =
The woman’s humerus is 25 centimeters.
61. a. T = C + t
b. Let T = 8.48 and C = 8.
8.48 8
0.48
T C t
t
t
= += +=
The tax is $0.48.
67. a. Let b = 10 and h = 15.
1
21
(10)(15)275
A bh=
=
=
The area is 75 2in. .
b. 1
22
2 2 or
A bh
A bh
A Ab b
h h
=
=
= =
c. Let A = 18 and h = 9.
2 2(18)4
9
Ab
h= = =
The base is 4 inches.
68. a. Let t = 1985 − 1975 = 10.
15 0.60
15 0.60(10) 15 6 21
N t= +
= + = + =
There were 21 thousand theaters in 1985.
b. 15 0.60
15 0.60
15
0.6015 5( 15)
or 0.60 3
N t
N t
Nt
N Nt t
= +− =−
=
− −= =
c. Let N = 27.
27 15
0.6020
t
t
−=
=
1975 + 20 = 1995 In 1995, the number of theaters totaled 27,000.
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Chapter 2: Equations, Problem Solving, and Inequalities Bello, Introductory Algebra, 3e ISM
135
69. a. Let F = 41.
5 160
9 95 160
(41)9 9205 160
9 945
95
C F= −
= −
= −
=
=
The temperature is 5°C.
b. 5 160
9 99 9 5 9 160
5 5 9 5 99
325
9 932 or 32
5 5
C F
C F
C F
C F F C
= −
⋅ = ⋅ − ⋅
= −
+ = = +
c. Let C = 20. 9
3259
(20) 32536 32
68
F C= +
= +
= +=
The temperature is 68°F.
70. 3 2 2( 2)
3 2 2 4
3 2 2
2
x x
x x
x x
x
− = −
− = −
= −=−
71. 2 1 3
2 1 1 3 1
2 4
2 4
4
x x
x x
x x
x x x x
x
− = +− + = + +
= +− = − +
=
72. 3 2 2( 1)
3 2 2 2
3 2
0
x x
x x
x x
x
− = −
− = −
==
73. 4( 1) 3 7
4 4 3 7
4 4 4 3 7 4
4 3 3
4 3 3 3 3
3
x x
x x
x x
x x
x x x x
x
+ = ++ = +
+ − = + −= +
− = − +=
74.
3
4 6 63
12 12 124 6 6
3 2 2( 3)
2 6
2 2 2 6
3 6
3 6
3 32
x x x
x x x
x x x
x x
x x x x
x
x
x
− −+ =
− − ⋅ + ⋅ = ⋅
− + = −− = −
− − = − −− = −− −
=− −
=
75. 13 2
6 6 6 13 22 3 6
6
6
x x
x x
x x
x
x
− =
⋅ − ⋅ = ⋅
− =− =
= −
2.7 Properties of Inequalities
Problems 2.7 1. a. Since 5 is to the right of 3, 5 > 3.
b. Since −1 is to the right of −4, −1 > −4.
c. Since −5 is to the left of −4, −5 < −4.
2. a. For x ≤ −2, any number less than or
equal to −2 is a solution.
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Bello, Introductory Algebra, 3e ISM Chapter 2: Equations, Problem Solving, and Inequalities
136
b. For x > −3, any number greater than −3
is a solution.
3. a. 4 3 3( 2)
4 3 3 6
4 3 3 3 6 3
4 3 3
4 3 3 3 3
3
x x
x x
x x
x x
x x x x
x
− < −− < −
− + < − +< −
− < − −< −
Any number less than −3 is a solution.
b. 3( 2) 2 5
3 6 2 5
3 6 6 2 5 6
3 2 1
3 2 2 2 1
1
x x
x x
x x
x x
x x x x
x
+ ≥ ++ ≥ +
+ − ≥ + −≥ −
− ≥ − −≥ −
Any number greater than or equal to −1
is a solution.
4. a. 4 3 2 5
4 3 3 2 5 3
4 2 2
4 2 2 2 2
2 2
2 2
2 21
x x
x x
x x
x x x x
x
x
x
+ ≤ ++ − ≤ + −
≤ +− ≤ − +
≤
≤
≤
Any number less than or equal to 1 is a solution.
b. 5( 1) 3 1
5 5 3 1
5 5 5 3 1 5
5 3 6
5 3 3 3 6
2 6
2 6
2 23
x x
x x
x x
x x
x x x x
x
x
x
− > +− > +
− + > + +> +
− > − +>
>
>
Any number greater than 3 is a solution.
5. a. 4 20
4 20
4 45
x
x
x
− <−
>− −
> −
Any number greater than −5 is a
solution.
b. 23
3 3 23
6
x
x
x
−>
− − < − ⋅
< −
Any number less than −6 is a solution.
c. 2( 1) 4 1
2 2 4 1
2 2 2 4 1 2
2 4 3
2 4 4 4 3
2 3
2 3
2 23
2
x x
x x
x x
x x
x x x x
x
x
x
− ≤ +− ≤ +
− + ≤ + +≤ +
− ≤ − +− ≤−
≥− −
≥ −
Any number greater than or equal to
3
2− is a solution.
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Chapter 2: Equations, Problem Solving, and Inequalities Bello, Introductory Algebra, 3e ISM
137
6.
8
3 4 48
12 12 123 4 4
4 3 3( 8)
3 24
3 3 3 24
4 24
4 24
4 46
x x x
x x x
x x x
x x
x x x x
x
x
x
− −+ <
− − ⋅ + ⋅ < ⋅
− + < −− < −
− − < − −− < −− −
>− −
>
Any number greater than 6 is a solution.
7. a. 2 < x and x < 4 can be written as 2 < x < 4. The solution consists of all numbers between 2 and 4.
b. 3
1 3 1 ( )
3
x
x
x
≥ −− ⋅ ≤ − ⋅ −
− ≤
Now −3 ≤ x and x ≤ −1 can be written as
−3 ≤ x ≤ −1. The solution consists of all
numbers between −3 and −1, inclusive.
c. 2 6
2 2 6 2
4
x
x
x
+ ≤+ − ≤ −
≤
3 6
3 6
3 32
x
x
x
− <−
>− −
> −
Rearranging, we have −2 < x ≤ 4. The
solution consists of all numbers
between −2 and 4 and the number 4.
8. P = 20 − 2x < 10
20 2 10
20 20 2 10 20
2 10
2 10
2 25
x
x
x
x
x
− <− − < −
− < −− −
>− −
>
This means that more than 5 years after 1997, or in 2003, the percent of smokers in this group is less than 10%.
Exercises 2.7
1. Since 8 is to the left of 9, 8 < 9.
2. Since −8 is to the right of −9, −8 > −9.
3. Since −4 is to the right of −9, −4 > −9.
4. Since 7 is to the right of 3, 7 > 3.
5. Since 1
4 is to the left of
1,
3 1 1
.4 3
<
6. Since 1
5 is to the left of
1,
2 1 1
.5 2
<
7. Since 2
3− is to the right of −1,
21.
3− > −
8. Since 1
5− is to the right of −1,
11.
5− > −
9. Since 1
34
− is to the left of −3, 1
3 3.4
− < −
10. Since 1
45
− is to the left of −4, 1
4 4.5
− < −
11. 2 6 8
2 6 6 8 6
2 2
2 2
2 21
x
x
x
x
x
+ ≤+ − ≤ −
≤
≤
≤
Any number less than or equal to 1 is a solution.
12. 4 5 3
4 8
2
y
y
y
− ≤
≤
≤
Any number less than or equal to 2 is a solution.
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Bello, Introductory Algebra, 3e ISM Chapter 2: Equations, Problem Solving, and Inequalities
138
13. 3 4 10
3 4 4 10 4
3 6
3 6
3 32
y
y
y
y
y
− − ≥ −− − + ≥ − +
− ≥ −− −
≤− −
≤
Any number less than or equal to 2 is a solution.
14. 4 2 6
4 8
2
z
z
z
− − ≥
− ≥
≤−
Any number less than or equal to −2 is a
solution.
15. 5 1 14
5 1 1 14 1
5 15
5 15
5 53
x
x
x
x
x
− + < −− + − < − −
− < −−
>− −
>
Any number greater than 3 is a solution.
16. 3 1 8
3 9
3
x
x
x
− + <−
− <−>
Any number greater than 3 is a solution.
17. 3 4 10
3 4 4 10 4
3 6
3 6
2 6
2 6
2 23
a a
a a
a a
a a a a
a
a
a
+ ≤ ++ − ≤ + −
≤ +− ≤ − +
≤
≤
≤
Any number less than or equal to 3 is a solution.
18. 4 4 7
4 3
3 3
1
b b
b b
b
b
+ ≤ +
≤ +≤
≤
Any number less than or equal to 1 is a solution.
19. 5 12 6 8
5 12 12 6 8 12
5 6 4
5 6 6 6 4
4
4
z z
z z
z z
z z z z
z
z
− ≥ −− + ≥ − +
≥ +− ≥ − +
− ≥≤ −
Any number less than or equal to −4 is a
solution.
20. 5 7 7 19
5 7 12
2 12
6
z z
z z
z
z
+ ≥ +
≥ +− ≥
≤−
Any number less than or equal to −6 is a
solution.
21. 10 3 7 6
10 10 3 7 10 6
3 3 6
3 6 3 6 6
3 3
3 3
3 31
x x
x x
x x
x x x x
x
x
x
− ≤ −− − ≤ − −
− ≤ − −− + ≤ − − +
≤ −−
≤
≤ −
Any number less than or equal to −1 is a
solution.
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Chapter 2: Equations, Problem Solving, and Inequalities Bello, Introductory Algebra, 3e ISM
139
22. 8 4 12 6
4 20 6
10 20
2
y y
y y
y
y
− ≤− +
− ≤− +
− ≤−≥
Any number greater than or equal to 2 is a solution.
23. 5( 2) 3( 3) 1
5 10 3 9 1
5 3
5 3 3 3
2 0
2 0
2 20
x x
x x
x x
x x x x
x
x
x
+ < + +
+ < + +<
− < −<
<
<
Any number less than 0 is a solution.
24. 5(4 3 ) 7(3 4 ) 12
20 15 21 28 12
15 13 28
13 13
1
x x
x x
x x
x
x
− < − +
− < − +− < −
<<
Any number less than 1 is a solution.
25.
1 42 2
4 51 4
20( 2 ) 20 20(2 ) 204 5
40 5 40 16
40 40 11
80 11
80 11
80 8011
80
x x
x x
x x
x x
x
x
x
− + ≥ +
− + ≥ +
− + ≥ +
− ≥ +
− ≥−
≤− −
≤−
Any number less than or equal to 11
80− is a
solution.
26.
1 26 2
7 71 2
7 6 7 7 2 77 7
42 1 14 2
42 1 1 14 2 1
42 14 3
42 14 14 14 3
28 3
28 3
28 283
28
x x
x x
x x
x x
x x
x x x x
x
x
x
+ ≥ −
⋅ + ⋅ ≥ ⋅ − ⋅
+ ≥ −+ − ≥ − −
≥ −− ≥ − −
≥ −−
≥
≥ −
Any number greater than or equal to 3
28−
is a solution.
27. 15 4
20 20 20 15 4
4 5 20
20
20
x x
x x
x x
x
x
− ≤
⋅ − ⋅ ≤ ⋅
− ≤− ≤
≥ −
Any number greater than or equal to −20 is
a solution.
28. 13 2
6 6 6 13 22 3 6
6
6
x x
x x
x x
x
x
− ≤
⋅ − ⋅ ≤ ⋅
− ≤− ≤
≥ −
Any number greater than or equal to −6 is a
solution.
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Bello, Introductory Algebra, 3e ISM Chapter 2: Equations, Problem Solving, and Inequalities
140
29.
7 2 1 3
6 2 47 2 1 3
12 12 126 2 4
2(7 2) 6 3 3
14 4 6 9
14 14 10 9 14
10 5
10 5
5 52
2
xx
xx
x x
x x
x x x x
x
x
x
x
++ ≥
+⋅ + ⋅ ≥ ⋅
+ + ≥ ⋅+ + ≥
− + ≥ −≥ −
−≤
− −− ≤
≥ −
Any number greater than or equal to −2 is a
solution.
30.
8 23 1 5
6 3 28 23 1 5
6 6 66 3 28 23 2 15
8 8 21 15 8
21 7
21 7
7 73
3
xx
xx
x x
x x x x
x
x
x
x
−+ ≥
−⋅ + ⋅ ≥ ⋅
− + ≥− − ≥ −
− ≥−
≥
− ≥≤ −
Any number less than or equal to −3 is a
solution.
31. x < 3 and −x < −2 (or x > 2) can be rewritten
as 2 < x < 3. The solution consists of all numbers between 2 and 3.
32. −x < 5 (or x > −5) and x < 2 can be rewritten
as −5 < x < 2. The solution consists of all
numbers between −5 and 2.
33. 1 4
1 1 4 1
3
x
x
x
+ <+ − < −
<
and 1
1 ( ) 1 ( 1)
1
x
x
x
− < −− ⋅ − > − ⋅ −
>
Now x < 3 and x > 1 can be rewritten as 1 < x < 3. The solution consists of all numbers between 1 and 3.
34. 2 1
2 2 1 2
3
x
x
x
− <− + < +
<
and 2
1( ) 1(2)
2
x
x
x
− <− − > −
> −
Now x < 3 and x > −2 can be rewritten as
−2 < x < 3. The solution consists of all
numbers between −2 and 3.
35. 2 3
2 2 3 2
5
x
x
x
− <− + < +
<
and 2
1 2 1 ( )
2
x
x
x
> −− ⋅ < − ⋅ −
− <
Now x < 5 and −2 < x can be rewritten as
−2 < x < 5. The solution consists of all the
numbers between −2 and 5.
36. 3 1
3 3 1 3
4
x
x
x
− <− + < +
<
and 1
1 1 1 ( )
1
x
x
x
> −− ⋅ < − ⋅ −
− <
Now x < 4 and −1 < x can be rewritten as
−1 < x < 4. The solution consists of all the
numbers between −1 and 4.
37. 2 3
1
x
x
+ <<
and 4 1
5
x
x
− < +− <
Now x < 1 and −5 < x can be rewritten as
−5 < x < 1. The solution consists of all
numbers between −5 and 1.
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Chapter 2: Equations, Problem Solving, and Inequalities Bello, Introductory Algebra, 3e ISM
141
38. 4 5
1
x
x
+ <<
and 1 2
3
x
x
− < +− <
Now x < 1 and −3 < x can be rewritten as
−3 < x < 1. The solution consists of all
numbers between −3 and 1.
39. 1 2
3
x
x
− >>
and 7 12
5
x
x
+ <<
Now x > 3 and x < 5 can be rewritten as 3 < x < 5. The solution consists of all numbers between 3 and 5.
40. 2 1
3
x
x
− >>
and 5
5
1 15
x
x
x
− > −− −
<− −
<
Now x > 3 and x < 5 can be rewritten as 3 < x < 5. The solution consists of all numbers between 3 and 5.
41. 20°F < t < 40°F
42. h ≤ 29,029 ft
43. $12,000 < s < $13,000
44. 18 < m < 22
45. 2 ≤ e ≤ 7
46. a > $150 billion
47. $3.50 < c < $4.00
48. r < 19,000 mi
49. a < 41 ft
50. d ≤ 370
51. N = 720 + 5t > 800 720 5 800
720 720 5 800 720
5 80
5 80
5 516
t
t
t
t
t
+ >− + > −
>
>
>
1980 + 16 = 1996 After 1996, you would expect that the NCAA would have more than 800 teams.
52. D = −10t + 260 < 200
10 260 200
10 260 260 200 260
10 60
10 60
10 106
t
t
t
t
t
− + <− + − < −
− < −− −
>− −
>
1999 + 6 = 2005 After 2005, you would expect the number of deaths per 100,000 to be less than 200.
53. C = 127 + 17t > 300 127 17 300
127 127 17 300 127
17 173
17 173
17 1710.2 or 11
t
t
t
t
t
+ >− + > −
>
>
>
1980 + 11 = 1991 After 1991, you would expect the average hospital room rate to surpass the $300 per day mark.
54. C = 45 + 2t > 60 45 2 60
45 45 2 60 45
2 15
2 15
2 27.5 or 8
t
t
t
t
t
+ >− + > −
>
>
>
1985 + 8 = 1993 In 1993, you would expect the per capita consumption of poultry products to exceed 60 pounds per year.
55. J = 5 ft = 60 in.
56. B > F
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Bello, Introductory Algebra, 3e ISM Chapter 2: Equations, Problem Solving, and Inequalities
142
57. F = S − 3
58. F > J
59. S = 6 ft 5 in. = 77 in.
60. B > F and F > J can be rewritten as B > F > J.
61. F = S − 3 = 77 − 3 = 74
74 in.
or 6 ft 2 in.
B F
B
B
>>>
62. Answers may vary.
63. You have to change the direction of the inequality when multiplying or dividing by a negative number.
64. Answers may vary. Sample answer: A number cannot simultaneously be less than
−5 and greater than 2.
65. Answers may vary.
66. x a y a+ < +
67. x a y a− < −
68. ax ay<
69. ax ay>
70. x y
a a<
71. x y
a a>
72. 2 6
2 2 6 2
4
x
x
x
+ ≤+ − ≤ −
≤
and 3 9
3 9
3 33
x
x
x
− ≤−
≥− −
≥ −
Now x ≤ 4 and x ≥ −3 can be rewritten as
−3 ≤ x ≤ 4. The solution consists of all
numbers between −3 and 4, inclusive.
73. 3
1 3 1 ( )
3
x
x
x
≥ −− ⋅ ≤ − ⋅ −
− ≤
and x ≤ −1
Now −3 ≤ x and x ≤ −1 can be rewritten as
−3 ≤ x ≤ −1. The solution consists of all
numbers between −3 and −1, inclusive.
74. 2 < x and x < 4 can be rewritten as 2 < x < 4. The solution consists of all numbers between 2 and 4.
75.
4
3 4 44
12 12 123 4 4
4 3 3( 4)
3 12
3 3 3 12
4 12
4 12
4 43
x x x
x x x
x x x
x x
x x x x
x
x
x
− −+ <
− − ⋅ + ⋅ < ⋅
− + < −− < −
− − < − −− < −− −
>− −
>
Any number greater than 3 is a solution.
76. 4 5 11
4 5 5 11 5
4 6
4 6
3 6
3 6
3 32
x x
x x
x x
x x x x
x
x
x
+ < ++ − < + −
< +− < − +
<
<
<
Any number less than 2 is a solution.
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Chapter 2: Equations, Problem Solving, and Inequalities Bello, Introductory Algebra, 3e ISM
143
77. 3( 1) 3
3 3 3
3 3 3 3 3
3 6
3 6
2 6
2 6
2 23
x x
x x
x x
x x
x x x x
x
x
x
− > +− > +
− + > + +> +
− > − +>
>
>
Any number greater than 3 is a solution.
78. 4 7 3( 2)
4 7 3 6
4 7 7 3 6 7
4 3 1
4 3 3 3 1
1
x x
x x
x x
x x
x x x x
x
− < −− < −
− + < − +< +
− < − +<
Any number less than 1 is a solution.
79. 3( 1) 2 5
3 3 2 5
3 3 3 2 5 3
3 2 2
3 2 2 2 2
2
x x
x x
x x
x x
x x x x
x
+ ≥ ++ ≥ +
+ − ≥ + −≥ +
− ≥ − +≥
Any number greater than or equal to 2 is a solution.
80. x ≥ −2
Any number greater than or equal to −2 is a
solution.
81. x < 1 Any number less than 1 is a solution.
82. Since 5 is to the left of 7, 5 < 7.
83. Since −2 is to the left of −1, −2 < −1.
84. Since −4 is to the right of −5, −4 > −5.
85. Since 1
3 is to the right of −3,
13.
3> −
86. F = 181.5 + 0.8t > 189.5 181.5 0.8 189.5
0.8 8
10
t
t
t
+ >>
>
2000 + 10 = 2010 After 2010, you would expect the daily consumption of grams of fat to exceed 189.5.
87.
88.
89.
90.
91.
92.
93.
94.
95.
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Bello, Introductory Algebra, 3e ISM Chapter 2: Equations, Problem Solving, and Inequalities
144
Collaborative Learning
Answers may vary. Sample answers:
1. Let t = 2001 − 1985 = 16.
45 2
45 2(16)
45 32
77
C t= += += +=
The per capita consumption of poultry in 2001 is estimated to be 77 pounds per year. The table value for 2001 is approximately 83 pounds per year.
2. Answers may vary.
3. C is highest in 2001 and lowest in 1980.
4. Answers may vary.
5. Answers may vary.
Review Exercises
1. a. If x = 5, then 7 = 14 − x becomes
7 = 14 − 5, which is a false statement.
Hence, 5 is not a solution of the equation.
b. If x = 4, then 13 = 17 − x becomes
13 = 17 − 4 which is a true statement.
Thus 4 is a solution of the equation.
c. If x = −2, then 8 = 6 − x becomes
8 = 6 − (−2), which is a true statement.
Thus −2 is a solution of the equation.
2. a. 1 1
3 31 1 1 1
3 3 3 32
3
x
x
x
− =
− + = +
=
The solution is 2
.3
b. 5 2
7 75 5 2 5
7 7 7 77
71
x
x
x
x
− =
− + = +
=
=
The solution is 1.
c. 5 1
9 95 5 1 5
9 9 9 96
92
3
x
x
x
x
− =
− + = +
=
=
The solution is 2
.3
3. a. 5 2 5
3 49 9 9
3 5
9 93 3 5 3
9 9 9 92
9
x x
x
x
x
− + + − =
+ =
+ − = −
=
The solution is 2
.9
b. 4 2 6
2 37 7 7
2 6
7 72 2 6 2
7 7 7 74
7
x x
x
x
x
− + + − =
+ =
+ − = −
=
The solution is 4
.7
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Chapter 2: Equations, Problem Solving, and Inequalities Bello, Introductory Algebra, 3e ISM
145
c. 5 1 5
4 56 6 6
4 5
6 64 4 5 4
6 6 6 61
6
x x
x
x
x
− + + − =
+ =
+ − = −
=
The solution is 1
.6
4. a. 3 4( 1) 2 3
3 4 4 2 3
3 2
3 2 2 2
5 or 5
x x
x x
x
x
x x
= − + −= − + −= −
+ = − += =
The solution is 5.
b. 4 5( 1) 9 4
4 5 5 9 4
4 4
4 4 4 4
0 or 0
x x
x x
x
x
x x
= − + −= − + −= +
− = + −= =
The solution is 0.
c. 5 6( 1) 8 5
5 6 6 8 5
5 2
5 2 2 2
3 or 3
x x
x x
x
x
x x
= − + −= − + −= +
− = + −= =
The solution is 3.
5. a. 6 3( 1) 2 3
6 3 3 2 3
9 3 2 3
9 3 3 2 3 3
9 2
x x
x x
x x
x x x x
+ + = ++ + = +
+ = ++ − = + −
=
Since this statement is false, the equation has no solution.
b.
2 4( 1) 7 4
2 4 4 7 4
6 4 7 4
6 6 4 7 6 4
4 1 4
4 4 1 4 4
8 1
8 1
8 81
8
x x
x x
x x
x x
x x
x x x x
x
x
x
− + − = − −− + − = − −
− + = − −− + + = − + −
= − −+ = − − +
= −−
=
= −
The solution is 1
.8
−
c. 1 2( 1) 3 2
1 2 2 3 2
3 2 3 2
3 2 2 3 2 2
3 3
x x
x x
x x
x x x x
− − + = −− − − = −
− − = −− − + = − +
− =
Since this statement is false, the equation has no solution.
6. a. 5 2( 1) 2 7
5 2 2 2 7
2 7 2 7
x x
x x
x x
+ + = ++ + = +
+ = +
Since both sides are identical, the equation is an identity. The solution is all real numbers.
b. 2 3( 1) 5 3
2 3 3 5 3
5 3 5 3
x x
x x
x x
− + − = − +− + − = − +
− + = − +
Since both sides are identical, the equation is an identity. The solution is all real numbers.
c. 3 4( 1) 1 4
3 4 4 1 4
1 4 1 4
x x
x x
x x
− − − = −− − + = −
− = −
Since both sides are identical, the equation is an identity. The solution is all real numbers.
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Bello, Introductory Algebra, 3e ISM Chapter 2: Equations, Problem Solving, and Inequalities
146
7. a. 1
351
5 5 ( 3)5
15
x
x
x
= −
⋅ = ⋅ −
= −
The solution is −15.
b. 1
271
7 7 ( 2)7
14
x
x
x
= −
⋅ = ⋅ −
= −
The solution is −14.
c. 5 10
5 10
5 52
x
x
x
= −−
=
= −
The solution is −2.
8. a. 3
94
4 3 4( 9)
3 4 3
12
x
x
x
− = −
− ⋅ − = − ⋅ −
=
The solution is 12.
b. 3
95
5 3 5( 9)
3 5 3
15
x
x
x
− = −
− ⋅ − = − ⋅ −
=
The solution is 15.
c. 2
63
3 2 3( 6)
2 3 2
9
x
x
x
− = −
− ⋅ − = − ⋅ −
=
The solution is 9.
9. a. 2
53 4
212 12 12 5
3 44 6 60
10 60
10 60
10 106
x x
x x
x x
x
x
x
+ =
⋅ + ⋅ = ⋅
+ ==
=
=
The solution is 6.
b. 3
64 2
34 4 4 6
4 26 24
7 24
7 24
7 7
x x
x x
x x
x
x
+ =
⋅ + ⋅ = ⋅
+ ==
=
The solution is 24
.7
c. 3
105 10
310 10 10 10
5 102 3 100
5 100
5 100
5 520
x x
x x
x x
x
x
x
+ =
⋅ + ⋅ = ⋅
+ ==
=
=
The solution is 20.
10. a. 13 4
12 12 12 13 4
4 3 12
12
x x
x x
x x
x
− =
⋅ − ⋅ = ⋅
− ==
The solution is 12.
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Chapter 2: Equations, Problem Solving, and Inequalities Bello, Introductory Algebra, 3e ISM
147
b. 102 7
14 14 14 102 7
7 2 140
5 140
5 140
5 528
x x
x x
x x
x
x
x
− =
⋅ − ⋅ = ⋅
− ==
=
=
The solution is 28.
c. 24 5
20 20 20 24 5
5 4 40
40
x x
x x
x x
x
− =
⋅ − ⋅ = ⋅
− ==
The solution is 40.
11. a. 1 1
14 6
1 112 12 12 1
4 63( 1) 2( 1) 12
3 3 2 2 12
5 12
5 5 12 5
17
x x
x x
x x
x x
x
x
x
− +− =
− +⋅ − ⋅ = ⋅
− − + =− − − =
− =− + = +
=
The solution is 17.
b. 1 1
06 8
1 124 24 24 0
6 84( 1) 3( 1) 0
4 4 3 3 0
7 0
7 7 0 7
7
x x
x x
x x
x x
x
x
x
− +− =
− +⋅ − ⋅ = ⋅
− − + =− − − =
− =− + = +
=
The solution is 7.
c. 1 1
08 10
1 140 40 40 0
8 105( 1) 4( 1) 0
5 5 4 4 0
9 0
9 9 0 9
9
x x
x x
x x
x x
x
x
x
− +− =
− +⋅ − ⋅ = ⋅
− − + =− − − =
− =− + = +
=
The solution is 9.
12. a. What percent of 30 is 6? 30 6
30 6
30 301 20
5 100
x
x
x
⋅ =⋅
=
= =
Thus, 6 is 20% of 30.
b. What percent of 40 is 4? 40 4
40 4
40 401 10
10 100
x
x
x
⋅ =⋅
=
= =
Thus, 4 is 10% of 40.
c. What percent of 50 is 10? 50 10
50 10
50 501 20
5 100
x
x
x
⋅ =⋅
=
= =
Thus, 10 is 20% of 50.
13. a. 20 is 40% of what number?
4020
1002
205
5 5 220
2 2 550
n
n
n
n
= ⋅
= ⋅
⋅ = ⋅ ⋅
=
Thus 20 is 40% of 50.
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Bello, Introductory Algebra, 3e ISM Chapter 2: Equations, Problem Solving, and Inequalities
148
b. 30 is 90% of what number?
9030
1009
3010
10 10 930
9 9 10100
31
333
n
n
n
n
n
= ⋅
= ⋅
⋅ = ⋅ ⋅
=
=
Thus, 30 is 90% of 1
33 .3
c. 25 is 75% of what number?
7525
1003
254
4 4 325
3 3 4100
31
333
n
n
n
n
n
= ⋅
= ⋅
⋅ = ⋅ ⋅
=
=
Thus, 25 is 75% of 1
33 .3
14. a.
1 19( 4)
5 4 201 19( 4)
20 20 205 4 20
4 5 19( 4)
4 5 19 76
4 4 5 19 76 4
5 19 72
5 19 19 19 72
24 72
24 72
24 243
x x
x x
x x
x x
x x
x x
x x x x
x
x
x
+− =
+⋅ − ⋅ = ⋅
− = +− = +
− − = + −− = +
− − = − +− =−
=− −
= −
The solution is −3.
b.
1 6( 5)
5 4 51 6( 5)
20 20 205 4 5
4 5 24( 5)
4 5 24 120
5 24 116
29 116
29 116
29 294
x x
x x
x x
x x
x x
x
x
x
+− =
+⋅ − ⋅ = ⋅
− = +
− = +− = +
− =
−=
− −=−
The solution is −4.
c.
1 29( 6)
5 4 201 29( 6)
20 20 205 4 20
4 5 29( 6)
4 5 29 174
5 29 170
34 170
34 170
34 345
x x
x x
x x
x x
x x
x
x
x
+− =
+⋅ − ⋅ = ⋅
− = +
− = +− = +
− =
−=
− −=−
The solution is −5.
15. a. 1
21
2 22
2
2
2 2 or
A bh
A bh
A bh
A bh
b bA A
h hb b
=
⋅ = ⋅
=
=
= =
b. 2
2
2 2
or 2 2
C r
C r
C Cr r
= ππ
=π π
= =π π
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Chapter 2: Equations, Problem Solving, and Inequalities Bello, Introductory Algebra, 3e ISM
149
c. 3
3 33
3
3
3 3 or
bhV
bhV
V bh
V bh
h hV V
b bh h
=
⋅ = ⋅
=
=
= =
16. a. Let n = first number. Then n + 20 = other number.
( 20) 84
20 84
2 20 84
2 64
32
20 52
n n
n n
n
n
n
n
+ + =
+ + =+ =
==
+ =
The numbers are 32 and 52.
b. Let n = first number. Then n + 19 = other number.
( 19) 47
19 47
2 19 47
2 28
14
19 33
n n
n n
n
n
n
n
+ + =+ + =
+ =
==
+ =
The numbers are 14 and 33.
c. Let n = first number. Then n + 23 = second number.
( 23) 81
23 81
2 23 81
2 58
29
23 52
n n
n n
n
n
n
n
+ + =+ + =
+ =
==
+ =
The numbers are 29 and 52.
17. a. Let c = calories in chicken breast. Then c + 22 = calories in pie.
( 22) 578
2 22 578
2 556
2 556
2 2278
22 300
c c
c
c
c
c
c
+ + =+ =
=
=
=
+ =
The chicken breast has 278 calories and the pie has 300 calories.
b. Let c = calories in chicken breast. Then c + 38 = calories in pie.
( 38) 620
2 38 620
2 582
2 582
2 2291
38 329
c c
c
c
c
c
c
+ + =+ =
=
=
=
+ =
The chicken breast has 291 calories and the pie has 329 calories.
c. Let c = calories in chicken breast. Then c + 42 = calories in pie.
( 42) 650
2 42 650
2 608
2 608
2 2304
42 346
c c
c
c
c
c
c
+ + =+ =
=
=
=
+ =
The chicken breast has 304 calories and the pie has 346 calories.
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Bello, Introductory Algebra, 3e ISM Chapter 2: Equations, Problem Solving, and Inequalities
150
18. Let m = measure of angle.
90 − m = complement and
180 − m = supplement.
a. 180 3(90 ) 20
180 270 3 20
180 250 3
180 180 250 180 3
70 3
3 70 3 3
2 70
2 70
2 235
m m
m m
m m
m m
m m
m m m m
m
m
m
− = − −− = − −− = −
− − = − −− = −
− + = − +=
=
=
The measure of the angle is 35°.
b. 180 3(90 ) 30
180 270 3 30
180 240 3
180 180 240 180 3
60 3
3 60 3 3
2 60
2 60
2 230
m m
m m
m m
m m
m m
m m m m
m
m
m
− = − −− = − −− = −
− − = − −− = −
− + = − +=
=
=
The measure of the angle is 30°.
c. 180 3(90 ) 40
180 270 3 40
180 230 3
180 180 230 180 3
50 3
3 50 3 3
2 50
2 50
2 225
m m
m m
m m
m m
m m
m m m m
m
m
m
− = − −− = − −− = −
− − = − −− = −
− + = − +=
=
=
The measure of the angle is 25°.
19. a. R × T = D
car 1 40 T + 1 40(T + 1)
car 2 50 T 50T
40( 1) 50
40 40 50
40 10
40 10
10 104
T T
T T
T
T
T
+ =+ =
=
=
=
It takes the second car 4 hours to overtake the first car.
b. R × T = D
car 1 30 T + 1 30(T + 1)
car 2 50 T 50T
30( 1) 50
30 30 50
30 20
30 20
20 203 1
or 12 2
T T
T T
T
T
T T
+ =+ =
=
=
= =
It takes the second car 1
12
hours to
overtake the first car.
c. R × T = D
car 1 40 T + 1 40(T + 1)
car 2 60 T 60T
40( 1) 60
40 40 60
40 20
40 20
20 202
T T
T T
T
T
T
+ =+ =
=
=
=
It takes the second car 2 hours to overtake the first car.
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Chapter 2: Equations, Problem Solving, and Inequalities Bello, Introductory Algebra, 3e ISM
151
20. a. price per pound × pounds = price
product 1.50 x 1.5x
another product 3.00 15 3(15)
mixture 2.40 15 + x 2.4(15 + x)
1.5 3(15) 2.4(15 )
1.5 45 36 2.4
1.5 9 2.4
9 0.9
10
x x
x x
x x
x
x
+ = ++ = +
+ ===
10 pounds should be mixed.
b. price per pound × pounds = price
product 2.00 x 2x
another product 3.00 15 3(15)
mixture 2.50 15 + x 2.5(15 + x)
2 3(15) 2.5(15 )
2 45 37.5 2.5
2 7.5 2.5
7.5 0.5
15
x x
x x
x x
x
x
+ = ++ = ++ =
==
15 pounds should be mixed.
c. price per pound × pounds = price
product 6.00 x 6x
another product 2.00 15 2(15)
mixture 4.50 15 + x 4.5(15 + x)
6 2(15) 4.5(15 )
6 30 67.5 4.5
6 37.5 4.5
1.5 37.5
25
x x
x x
x x
x
x
+ = ++ = +
= +==
25 pounds should be mixed.
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Bello, Introductory Algebra, 3e ISM Chapter 2: Equations, Problem Solving, and Inequalities
152
21. a. P × r = I
5% x 0.05 0.05x
6% 30,000 − x 0.06 0.06(30,000 − x)
0.05 0.06(30,000 ) 1600
0.05 1800 0.06 1600
1800 0.01 1600
0.01 200
20,000
30,000 10,000
x x
x x
x
x
x
x
+ − =+ − =
− =− = −
=− =
$20,000 is invested at 5% and $10,000 is invested at 6%.
b. P × r = I
7% x 0.07 0.07x
9% 30,000 − x 0.09 0.09(30,000 − x)
0.07 0.09(30,000 ) 2300
0.07 2700 0.09 2300
2700 0.02 2300
0.02 400
20,000
30,000 10,000
x x
x x
x
x
x
x
+ − =+ − =
− =− = −
=− =
$20,000 is invested at 7% and $10,000 is invested at 9%.
c. P × r = I
6% x 0.06 0.06x
10%
30,000 − x 0.10 0.10(30,000 − x)
0.06 0.10(30,000 ) 2000
100 0.06 100 0.10(30,000 ) 100 2000
6 10(30,000 ) 200,000
6 300,000 10 200,000
300,000 4 200,000
4 100,000
25,000
30,000 5000
x x
x x
x x
x x
x
x
x
x
+ − =⋅ + ⋅ − = ⋅
+ − =+ − =
− =− = −
=− =
$25,000 is invested at 6% and $5000 is invested at 10%.
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Cumulative Review Chapters 1−2
153
22. a. 3.05 3
3 3.05
3 3 or
3.05 3.05
C m
C m
C Cm m
= +− =− −
= =
Let C = 27.40.
27.40 38
3.05m
−= =
The call lasted 8 minutes.
b. 3.15 3
3 3.15
3 3 or
3.15 3.15
C m
C m
C Cm m
= +− =− −
= =
Let C = 34.50.
34.50 310
3.15m
−= =
The call lasted 10 minutes.
c. 3.25 2
2 3.25
2 2 or
3.25 3.25
C m
C m
C Cm m
= +− =− −
= =
Let C = 21.50.
21.50 26
3.25m
−= =
The call lasted 6 minutes.
23. a. The angles are supplementary. (3 20) 2 180
5 20 180
5 200
40
x x
x
x
x
− + =− =
==
3(40) − 20 = 100 and 2(40) = 80
The angles measure 100° and 80°.
b. Vertical angles must be equal. 7 10 3 30
7 3 40
4 40
10
x x
x x
x
x
− = += +==
7(10) − 10 = 60 and 3(10) + 30 = 60
The angles each measure 60°.
c. The angles are complementary. (5 15) (2 5) 90
7 20 90
7 70
10
x x
x
x
x
+ + + =+ =
==
5(10) + 15 = 65 and 2(10) + 5 = 25
The angles measure 65° and 25°.
24. a. Since −8 is to the left of −7, −8 < −7.
b. Since 1
2 is to the right of −3,
13.
2> −
c. Since 4 is to the left of 1
4 ,3
1
4 4 .3
<
25. a. 4 2 2( 2)
4 2 2 4
4 2 2 2 4 2
4 2 6
4 2 2 2 6
2 6
2 6
2 23
x x
x x
x x
x x
x x x x
x
x
x
− < +− < +
− + < + +< +
− < − +<
<
<
Any number less than 3 is a solution.
b. 5 4 2( 1)
5 4 2 2
5 4 4 2 2 4
5 2 6
5 2 2 2 6
3 6
3 6
3 32
x x
x x
x x
x x
x x x x
x
x
x
− < +− < +
− + < + +< +
− < − +<
<
<
Any number less than 2 is a solution.
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Bello, Introductory Algebra, 3e ISM Chapter 2: Equations, Problem Solving, and Inequalities
154
c. 7 1 3( 1)
7 1 3 3
7 1 1 3 3 1
7 3 4
7 3 3 3 4
4 4
4 4
4 41
x x
x x
x x
x x
x x x x
x
x
x
− < +− < +
− + < + +< +
− < − +<
<
<
Any number less than 1 is a solution.
26. a. 6( 1) 4 2
6 6 4 2
6 6 6 4 2 6
6 4 8
6 4 4 4 8
2 8
2 8
2 24
x x
x x
x x
x x
x x x x
x
x
x
− ≥ +− ≥ +
− + ≥ + +≥ +
− ≥ − +≥
≥
≥
Any number greater than or equal to 4 is a solution.
b. 5( 1) 2 1
5 5 2 1
5 5 5 2 1 5
5 2 6
5 2 2 2 6
3 6
3 6
3 32
x x
x x
x x
x x
x x x x
x
x
x
− ≥ +− ≥ +
− + ≥ + +≥ +
− ≥ − +≥
≥
≥
Any number greater than or equal to 2 is a solution.
c. 4( 2) 2 2
4 8 2 2
4 8 8 2 2 8
4 2 10
4 2 2 2 10
2 10
2 10
2 25
x x
x x
x x
x x
x x x x
x
x
x
− ≥ +− ≥ +
− + ≥ + +≥ +
− ≥ − +≥
≥
≥
Any number greater than or equal to 5 is a solution.
27. a.
1
3 6 61
6 6 63 6 6
2 1
1
1
2 1
2 1
2 21
2
x x x
x x x
x x x
x x
x x x x
x
x
x
−− + ≤
− ⋅ − + ⋅ ≤ ⋅
− + ≤ −− ≤ −
− − ≤ − −− ≤ −− −
≥− −
≥
Any number greater than or equal to 1
2
is a solution.
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Cumulative Review Chapters 1−2
155
b.
1
4 7 71
28 28 284 7 7
7 4 4( 1)
3 4 4
3 4 4 4 4
7 4
7 4
7 74
7
x x x
x x x
x x x
x x
x x x x
x
x
x
−− + ≤
− ⋅ − + ⋅ ≤ ⋅
− + ≤ −− ≤ −
− − ≤ − −− ≤ −− −
≥− −
≥
Any number greater than or equal to 4
7
is a solution.
c.
1
5 3 31
15 15 155 3 3
3 5 5( 1)
2 5 5
2 5 5 5 5
3 5
3 5
3 35
3
x x x
x x x
x x x
x x
x x x x
x
x
x
−− + ≤
− ⋅ − + ⋅ ≤ ⋅
− + ≤ −≤ −
− < − −− ≤ −− −
≥− −
≥
Any number greater than or equal to 5
3
is a solution.
28. a. 2 4
2
x
x
+ ≤≤
and 2 6
2 6
2 23
x
x
x
− <−
>− −
> −
Now x ≤ 2 and x > −3 can be rewritten
as −3 < x ≤ 2. The solution consists of
all numbers between −3 and 2,
including 2.
b. 3 5
2
x
x
+ ≤≤
and 3 9
3 9
3 33
x
x
x
− <−
>− −
> −
Now x ≤ 2 and x > −3 can be rewritten
as −3 < x ≤ 2. The solution consists of
all numbers between −3 and 2,
including 2.
c. 1 2
1
x
x
+ ≤≤
and 4 8
4 8
4 42
x
x
x
− <−
>− −
> −
Now x ≤ 1 and x > −2 can be rewritten
as −2 < x ≤ 1. The solution consists of
all numbers between −2 and 1,
including 1.
Cumulative Review Chapters 1−−−−2
1. The additive inverse of −7 is 7.
2. 9 9
9 910 10
− =
3. The LCD is 63.
2 18
7 63− = − and
2 14
9 63− = −
2 2 18 14 32
7 9 63 63 63
− + − = − + − = −
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Bello, Introductory Algebra, 3e ISM Chapter 2: Equations, Problem Solving, and Inequalities
156
4. −0.7 − (−8.9) = −0.7 + 8.9 = 8.2
5. (−2.4)(3.6) = −8.64
6. 4(2 ) (2 2 2 2) (16) 16− = − ⋅ ⋅ ⋅ = − = −
7. 7 5 7 24 168 21
8 24 8 5 40 5
− ÷ − = − ⋅ − = =
8. 5 60 5 6 3
12 6 3
72 3
69
y x z÷ ⋅ − = ÷ ⋅ −= ⋅ −= −=
9. In 9 ⋅ (8 ⋅ 5) = 9 ⋅ (5 ⋅ 8), we changed the
order of multiplication. The commutative property of multiplication was used.
10. 6(5 7) 6(5 ) 6(7)
30 42
x x
x
+ = += +
11. 5 ( 6 ) 5 6
( 5 6)
1
cd cd cd cd
cd
cd
cd
− − − = − += − +==
12. 2 2( 4) 3( 1)
2 2 8 3 3
(2 2 3 ) ( 8 3)
3 ( 11)
3 11
x x x
x x x
x x x
x
x
− + − += − − − −= − − + − −= − + −= − −
13. The quotient of (a − 4b) and c is written as
4.
a b
c
−
14. If x = 4, then 11 = 15 − x becomes
11 = 15 − 4 which is a true statement. Thus
4 is a solution of the equation.
15. 5 4( 3) 4 3
5 4 12 4 3
5 8
5 8 8 8
13 or 13
x x
x x
x
x
x x
= − + −= − + −= −
+ = − += =
The solution is 13.
16. 7
213
3 7 3( 21)
7 3 7
3 3
9
x
x
x
x
− = −
− ⋅ − = − ⋅ −
= ⋅=
The solution is 9.
17. 23 5
15 15 15 23 5
5 3 30
2 30
2 30
2 215
x x
x x
x x
x
x
x
− =
⋅ − ⋅ = ⋅
− ==
=
=
The solution is 15.
18.
2( 1)4
4 92( 1)
36 4 36 364 9
144 9 4 2( 1)
144 9 8( 1)
144 9 8 8
144 144 9 8 8 144
9 8 136
9 8 8 8 136
17 136
17 136
17 178
x x
x x
x x
x x
x x
x x
x x
x x x x
x
x
x
+− =
+⋅ − ⋅ = ⋅
− = ⋅ +− = +− = +
− − = + −− = −
− − = − −− = −− −
=− −
=
The solution is 8.
19. 2
2
2 2
2 2
6
6
6 6
or 6 6
S a b
S a b
a aS S
b ba a
=
=
= =
20. Let n = first number. Then n + 35 = second number.
Full file at http://TestbankCollege.eu/Solution-Manual-Introductory-Algebra-3rd-Edition-Ignacio-Bello
Cumulative Review Chapters 1−2
157
( 35) 155
2 35 155
2 35 35 155 35
2 120
2 120
2 260
35 95
n n
n
n
n
n
n
n
+ + =+ =
+ − = −=
=
=+ =
The numbers are 60 and 95.
21. Let b = annual return from bonds. Then b + 105 = annual return from stocks.
( 105) 595
2 105 595
2 105 105 595 105
2 490
2 490
2 2245
105 350
b b
b
b
b
b
b
b
+ + =+ =
+ − = −=
=
=+ =
The bonds return $245 and the stocks return $350.
22. R × T = D
Train A 40 T + 6 40(T + 6)
Train B 50 T 50T
40( 6) 50
40 240 50
240 10
24
T T
T T
T
T
+ =+ =
==
It takes 24 hours for train B to catch train A.
23. P × r = I
bonds x 0.12 0.12x
certificates
5000 − x 0.14 0.14(5000 − x)
0.12 0.14(5000 ) 660
0.12 700 0.14 660
700 0.02 660
0.02 40
2000
5000 3000
x x
x x
x
x
x
x
+ − =+ − =
− =− = −
=− =
Arlene invested $2000 in bonds and $3000 in certificates of deposit.
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Bello, Introductory Algebra, 3e ISM Chapter 2: Equations, Problem Solving, and Inequalities
158
24.
5
6 5 55
30 30 306 5 5
5 6 6( 5)
6 30
6 6 6 30
5 30
5 30
5 56
x x x
x x x
x x x
x x
x x x x
x
x
x
−− + ≤
− ⋅ − + ⋅ ≤ ⋅
− + ≤ −≤ −
− ≤ − −− ≤ −− −
≥− −
≥
Any number greater than or equal to 6 is a solution.
Full file at http://TestbankCollege.eu/Solution-Manual-Introductory-Algebra-3rd-Edition-Ignacio-Bello
Cumulative Review Chapters 1−2
159
Full file at http://TestbankCollege.eu/Solution-Manual-Introductory-Algebra-3rd-Edition-Ignacio-Bello