Chapter 11 Additional Topics

Chapter 11
Additional Topics
Homework 11.1
1. x = 7
x = 7 or x = −7
3. x = − 3
Since x is always nonnegative, the solution set
for x = − 3 is the empty set, ∅ .
5. 5 x − 3 = 15
5 x = 18
18
x =
5
18 18
x = − or x =
5 5
7. −4 x − 11 = 2
− 4 x = 13
13
x = −
4
Since x is always nonnegative, the solution set
for
9. x + 2 = 5
13
x = − is the empty set, ∅ .
4
x + 2 = − 5 or x + 2 = 5
11. x − 5 = 0
x = − 7 or x = 3
x − 5 = 0
x = 5
13. 3x − 1 = 11
3x
− 1 = −11 or 3x
− 1 = 11
3x
= − 10 or 3x
= 12
10
x = − or x = 4
3
15. 5 − 2x
= 9
5 − 2x
= −9 or 5 − 2x
= 9
− 2x
= −14 or − 2x
= 4
17. 2x + 9 = − 6
x = 7 or x = −2
Since 2x + 9 is always nonnegative, the
solution set for 2x + 9 = − 6 is the empty set, ∅ .
19. 2 x + 5 = 8
x + 5 = 4
x + 5 = − 4 or x + 5 = 4
21. 3x
− 9 = 0
x = − 9 or x = −1
3x
− 9 = 0
3x
= 9
x = 3
23. −6 2x
− 7 + 4 = −2
−6 2x
− 7 = −6
2x
− 7 = 1
2x
− 7 = −1 or 2x
− 7 = 1
2x
= 6 or 2x
= 8
x = 3 or x = 4
25. 1 x − 5 =
7
2 3 6
1 5 7 1 5 7
x − = − or x − =
2 3 6 2 3 6
1 7 5 1 7 5
x = − + or x = +
2 6 3 2 6 3
1 1 1 17
x = or x =
2 2 2 6
17
x = 1 or x =
3
336

SSM: Intermediate Algebra Homework 11.1
27. 4.7 x − 3.9 = 8.8
4.7 x = 12.7
x ≈ 2.70
x ≈ −2.70 or x ≈ 2.70
29. 3.7 2.1x
+ 5.8 − 9.7 = 10.2
3.7 2.1x
+ 5.8 = 19.9
2.1x
+ 5.8 ≈ 5.38
2.1x
+ 5.8 = 5.38 or 2.1x
+ 5.8 = −5.38
31. x < 4
2.1x
= − 0.42 or 2.1x
= −11.18
− 4 < x < 4
− 4, 4
Interval: ( )
33. x ≥ 3
x = − 0.2 or x = −5.32
−4 0 4
x ≤ −3 or x ≥ 3
−∞, −3 ∪ 3, ∞
Interval: ( ] [ )
35. x < − 3
−3 0 3
Since x is nonnegative, the inequality x < − 3
has an empty set solution, ∅ .
37. 2 x − 5 > 3
2 x > 8
x > 4
x < − 4 or x > 4
−∞, −4 ∪ 4, ∞
Interval: ( ) ( )
39. 2 − 5 x ≤ −8
−5 x ≤ −10
x ≥ 2
−4 0 4
x ≤ −2 or x ≥ 2
−∞, −2 ∪ 2, ∞
Interval: ( ] [ )
41. x − 6 ≥ 7
x − 6 ≤ −7 or x − 6 ≥ 7
x ≤ −1 or x ≥ 13
−∞, −1 ∪ 13, ∞
Interval: ( ] [ )
43. 2x + 5 ≤ 15
−1 0 13
−15 ≤ 2x
+ 5 ≤ 15
−20 ≤ 2x
≤ 10
−10 ≤ x ≤ 5
− 10,5
Interval: [ ]
−10
45. 7x + 15 > − 4
0 5
Since 7x + 15 is always nonnegative, the
solution set for the inequality 7x + 15 > − 4 is
the set of all real numbers.
−∞,
∞
Interval: ( )
47. 2 x − 3 + 1 ≥ 17
2 x − 3 ≥ 16
x − 3 ≥ 8
0
x − 3 ≤ −8 or x − 3 ≥ 8
x ≤ −5 or x ≥ 11
−∞, −5 ∪ 11, ∞
Interval: ( ] [ )
−5 0
11
49. −3 5 − 2x
+ 1≤ −8
−3 5 − 2x
≤ −9
5 − 2x
≥ 3
5 − 2x
≤ −3 or 5 − 2x
≥ 3
−2x
≤ −8 or − 2x
≥ −2
x ≥ 4 or x ≤ 1
−∞,1 ∪ 4, ∞
Interval: ( ] [ )
0 1 4
−2
0
2
337

Homework 11.1
SSM: Intermediate Algebra
51. 5 2x
+ 3 + 7<
2
5 2x
+ 3 < −5
2x
+ 3 < −1
Since 2x + 3 is always nonnegative, the
solution set for the inequality 2x + 3 < − 1 is the
empty set, ∅ .
53. 2 x 3 9 + ≤
5 2 20
9 2x
3 9
− ≤ + ≤
20 5 2 20
39 2x
21
− ≤ ≤ −
20 5 20
39 21
− ≤ 2x
≤ −
4 4
39 21
− ≤ x ≤ −
8 8
⎡ 39 21⎤
Interval: ⎢−
, −
8 8 ⎥
⎣ ⎦
3 9
−
8
55. 2.9 x − 5.6 > 13.9
2.9 x > 19.5
x > 6.72
2 1
−
8
x < − 6.72 or x > 6.72
−∞, −6.72 ∪ 6.72, ∞
Interval: ( ) ( )
−6.72
57. −6.2 3.5x
−1.3 ≥ −14.5
0
3.5x
−1.3 ≤ 2.34
−2.34 ≤ 3.5x
−1.3 ≤ 2.34
−1.04 ≤ 3.5x
≤ 3.64
−0.30 ≤ x ≤ 1.04
− 0.30,1.04
[ ]
−0.30
0
0
6.72
1.04
59. False.
Fore example, let a = 5 and b = − 7 .
2 + ( − 7)
= 2 − 7 = − 5 = 5
2 + − 7 = 2 + 7 = 9
Since 5 ≠ 9 , the statement
a + b = a + b is not true for all real numbers a
and b.
61. a. 2x
+ 1 = 11
2x
+ 1 = − 11 or 2x
+ 1 = 11
b. 2x
+ 1 < 11
2x
= − 12 or 2x
= 10
x = − 6 or x = 5
− 11 < 2x
+ 1 < 11
− 12 < 2x
< 10
− 6 < x < 5
c. 2x
+ 1 > 11
2x
+ 1 < − 11 or 2x
+ 1 > 11
2x
< − 12 or 2x
> 10
x < − 6 or x > 5
d. Written responses will vary.
−6 0 5
When graphed together, this becomes
−6 0 5
which covers all real numbers.
63. a bx + c + d = k
a bx + c = k − d
k − d
bx + c =
a
⎛ k − d ⎞
k − d
bx + c = − ⎜ ⎟ or bx + c =
⎝ a ⎠
a
k − d k − d
bx = −c − or bx = − c +
a
a
−ac − k − d − ac + k − d
bx = or bx =
a
a
−ac − k − d − ac + k − d
x = or x =
ab
ab
− ac ± ( k − d )
x =
ab
( ) ( )
( ) ( )
338

SSM: Intermediate Algebra
Section 11.1 Quiz
65. No.
x + 3 < 10
− 10 < x + 3 < 10
− 13 < x < 7
Section 11.1 Quiz
1. 3 x − 4 = 11
3 x = 15
x = 5
x = 5 or x = − 5
2. 2x − 3 = 7
2x
− 3 = −7 or 2x
− 3 = 7
2x
= − 4 or 2x
= 10
x = − 2 or x = 5
3. − 2 3x
+ 5 + 9 = 1
− 2 3x
+ 5 = −8
3x
+ 5 = 4
3x
+ 5 = − 4 or 3x
+ 5 = 4
3x
= − 9 or 3x
= −1
1
x = − 3 or x = −
3
4. 5 6x
− 5 = 15
6x
− 5 = 3
6x
− 5 = −3 or 6x
− 5 = 3
6x
= 2 or 6x
= 8
5. 2x
− 5 = 0
1 4
x = or x =
3 3
2x
− 5 = 0
2x
= 5
x =
5
2
7. x ≤ 6
−6 ≤ x ≤ 6
− 6,6
Interval: [ ]
8. x ≥ 4
−6
x ≤ −4 or x ≥ 4
−∞, −4 ∪ 4, ∞
Interval: ( ] [ )
9. 4x − 8 > 12
0
−4 0 4
4x
− 8 < −12 or 4x
− 8 > 12
4x
< − 4 or 4x
> 20
x < − 1 or x > 5
−∞, −1 ∪ 5, ∞
Interval: ( ) ( )
10. 3x + 1 < 5
− 5 < 3x
+ 1 < 5
− 6 < 3x
< 4
−1 0 5
4
− 2 < x <
3
⎛ 4 ⎞
Interval: ⎜ −2, ⎟
⎝ 3 ⎠
−2 0 4
3
11. − 2 x + 5 + 10 ≥ 4
− 2 x + 5 ≥ −6
x + 5 ≤ 3
−3 ≤ x + 5 ≤ 3
−8 ≤ x ≤ −2
−8, − 2
Interval: [ ]
−8 −2 0
6
6. 7x + 1 = − 3
Since 7x + 1 is always nonnegative, the solution
set for 7x + 1 = − 3 is the empty set, ∅ .
339

Homework 11.2
SSM: Intermediate Algebra
12. 7 3x
− 2 ≤ 42
3x
− 2 ≤ 6
−6 ≤ 3x
− 2 ≤ 6
−4 ≤3x
≤8
4 8
− ≤ x ≤
3 3
⎡ 4 8⎤
Interval: ⎢−
,
3 3⎥
⎣ ⎦
4
−
3
0
8
3
3x
− 6 = 0
3x
− 6 = 0
3x
= 6
x = 2
Homework 11.2
1. y ≥ 2x
− 4
Graph the line y = 2x
− 4 with a solid line and
shade the region above it.
y
13. x − 5 < − 7
Since x − 5 is always nonnegative, the solution
set for x − 5 < − 7 is the empty set, ∅ .
14. 3 2x
− 9 > −1
1
2x
− 9 > −
3
Since 2x − 9 is always nonnegative, the
solution set for
numbers.
Interval: ( −∞,
∞ )
1
2x − 9 > − is the set of all real
3
15. False.
For example, let a = 4 and b = 9 .
a − b = 4 − 9 = − 5 = 5
a − b = 4 − 9 = 4 − 9 = − 5
0
Since 5 ≠ − 5 , the statement a − b = a − b is
not true for all real numbers a and b.
16. 3x
− 6 ≤ 0
0 ≤ 3x
− 6 ≤ 0
6 ≤3x
≤6
2 ≤ x ≤ 2
x = 2
As an alternative, remember that 3x − 6 is
always nonnegative, which means it is never less
than 0. Thus, the inequality can be written as
3.
5.
4
x
−4
4
−4
1
y < − x + 3
2
1
Graph the line y = − x + 3 with a dashed line
2
and shade the region below it.
y
4
−4
x
4
−4
2
y ≤ x − 5
3
2
Graph the line y = x − 5 with a solid line and
3
shade the region below it.
y
8
−8
x
8
−8
340

SSM: Intermediate Algebra Homework 11.2
7. y > x
Graph the line y = x with a dashed line and
shade the region above it.
y
shade below it.
y
4
4
−4
4
x
−4
4
x
−4
−4
9. 2x
+ 5y
< 10
5y
< − 2x
+ 10
2
y < − x + 2
5
2
Graph the line y = − x + 2 with a dashed line
5
and shade the region below it.
y
15. y ≤ 2
Graph the line y = 2 with a solid line and shade
the region below it.
−4
4
−4
y
4
x
−4
4
4
x
17. y > − 5
Graph the line y = − 5 with a dashed line and
shade the region above it.
y
−4
4
11. 4x
− 6y
− 6 ≥ 0
−6y
≥ − 4x
+ 6
2
y ≤ x −1
3
2
Graph the line y = x − 1 with a solid line and
3
shade below it.
y
4
−4
x
4
−4
19.
x
−4
4
−4
1
y ≥ x − 2
3
y > − x + 3
1
Graph y = x − 2 with a solid line and the line
3
y = − x + 3 with a dashed line. The solution
region of the system is the intersection of the
1
solution regions of y ≥ x − 2 and y > − x + 3 .
3
y
3 − 2 + y ≤ −2
13. ( x )
3x
− 6 + y ≤ −2
y ≤ − 3x
+ 4
Graph the line y = − 3x
+ 4 with a solid line and
−4
2
−4
2
x
341

Homework 11.2
SSM: Intermediate Algebra
21. y ≤ x − 4
y ≥ −3x
Graph the line y = x − 4 with a solid line and the
line y = − 3x
with a solid line. The solution
region of the system is the intersection of the
solution regions of y ≤ x − 4 and y ≥ − 3x
.
y
4
y < − x + 5 , y ≤ x + 5 , and
y
4
−8
2
x
−4
1
y > x + 1.
2
−4
−4
23. y ≤ − 3x
+ 9
y ≥ 2x
− 3
x ≥ 0
4
x
y ≥ 0
Graph the lines y = − 3x
+ 9 , y = 2x
− 3 , x = 0 ,
and y = 0 with solid lines. The solution region
of the system is the intersection of the solution
regions of y ≤ − 3x
+ 9 , y ≥ 2x
− 3 , x ≥ 0 , and
y ≥ 0 .
−8
y
8
−8
25. y < − x + 5
y ≤ x + 5
8
x
1
y > x + 1
2
1
Graph the lines y = − x + 5 and y = x + 1 with
2
dashed lines, and the line y = x + 5 with a solid
line. The solution region of the system is the
intersection of the solution regions of
27. y ≤ −3
y ≥ −5
Graph the lines y = − 3 and y = − 5 with solid
lines. The solution region of the system is the
intersection of the solution regions of y ≤ − 3
and y ≥ − 5 .
−4
4
−4
29. 2x
− 4y
≤ 8
y
4
−4y
≤ − 2x
+ 8
x
3x
+ 5y
≤ 10
5y
≤ − 3x
+ 10
1
3
y ≥ x − 2 y ≤ − x + 2
2
5
1
3
Graph the lines y = x − 2 and y = − x + 2
2
5
with solid lines. The solution region of the
system is the intersection of the solution regions
1
3
of y ≥ x − 2 and y ≤ − x + 2 .
2
5
−4
4
−4
y
4
x
342

SSM: Intermediate Algebra
Section 11.2 Quiz
1
≥ − 4 −1
2
1
y ≥ x − 2 −1
2
1
31. y ( x )
( x )
y < −2 − 1 + 3
y < − 2x
+ 2 + 3
y < − 2x
+ 5
y ≥ x − 3
2
1
Graph the line y = x − 3 with a solid line and
2
the line y = − 2x
+ 5 with a dashed line. The
solution region of the system is the intersection
1
of the solution regions of y ≥ ( x − 4)
− 1 and
2
y < −2 x − 1 + 3 .
( )
y
4
Section 11.2 Quiz
1. y ≤ 2x
− 6
Graph the line y = 2x
− 6 with a solid line and
shade the region below it.
y
x
−4
4
−4
−8
2. y ≥ − 3x
+ 9
Graph the line y = − 3x
+ 9 with a solid line and
shade the region above it.
y
−4
2
x
8
−4
4
33. 2x
− 3y
< 6
− 3y
< − 2x
+ 6
2
y > x − 2
3
The graph of 2x
− 3y
< 6 is the region above
2
y = x − 2 or rewritten as 2x
− 3y
= 6 .
3
35. Answers may vary. One possible answer:
y ≥ x
( 3,4 ) is a solution but ( )
4,3 is not a solution.
37. The intersection of the solution regions of
y ≥ 2x
+ 1 and y ≤ 2x
+ 1 is the solid line
y = 2x
+ 1.
y
x
4 8
3. 4x
− 2y
> 8
− 2y
> − 4x
+ 8
y < 2x
− 4
Graph the line y = 2x
− 4 with a dashed line and
shade the region below it.
y
x
−4
4
−4
−8
4
−4
4
x
−4
39. Answers may vary. See Key Points.
343

Section 11.2 Quiz
SSM: Intermediate Algebra
4. 3y
− 5x
< 15
3y
< 5x
+ 15
5
y < x + 5
3
5
Graph the line y = x + 5 with a dashed line and
3
shade the region below it.
4
y
7. y ≤ − x + 4
y > 2x
− 3
Graph the line y = − x + 4 with a solid line and
the line y = 2x
− 3 with a dashed line. The
solution region of the system is the intersection
of the solution regions of y ≤ − x + 4 and
y > 2x
− 3 .
4
y
−4
4
x
−4
4
x
−4
−4
− 2 + 3 + 4x
≥ −8
5. ( y )
−2y
− 6 + 4x
≥ −8
−2y
≥ −4x
− 2
y ≤ 2x
+ 1
Graph the line y = 2x
+ 1 with a solid line and
shade the region below it.
y
4
−4
x
4
−4
8.
2
y ≤ x + 1
5
1
y < − x + 2
4
2
Graph the line y = x + 1 with a solid line and
5
1
the line y = − x + 2 with a dashed line. The
4
solution region of the system is the intersection
2
of the solution regions for y ≤ x + 1 and
5
1
y < − x + 2 .
4
y
6. y < − 2
Graph the line y = − 2 and shade the region
below it.
y
−4
4
4
x
4
−4
−4
4
x
−4
344

SSM: Intermediate Algebra Homework 11.3
9. 3x
− 4y
≥ 12
−4y
≥ − 3x
+ 12
6y
− 2x
≤ 12
6y
≤ 2x
+ 12
3
1
y ≤ x − 3
y ≤ x + 2
4
3
3
1
Graph the lines y = x − 3 and y = x + 2 with
4
3
solid lines. The solution region of the system is
the intersection of the solution regions for
3
1
y ≤ x − 3 and y ≤ x + 2 .
4
3
16
8
y
8 16
10. x − y > 3 → y < x − 3
x + y < 5 → y < − x + 5
x > 0
x
y > 0
Graph the lines y = x − 3 , y = − x + 5 , x = 0 ,
and y = 0 with dashed lines. The solution region
of the system is the intersection of the solution
regions for y < x − 3 , y < − x + 5 , x > 0 , and
y > 0 .
−4
4
y
−2
4
x
11. 2x
− 5y
> 10
− 5y
> − 2x
+ 10
2
y < x − 2
5
Answers may vary.
Three possible solutions:
−5, −5 , 0, −4 , and 5, − 2
( ) ( ) ( )
2
− 5 < ( −5)
− 2
5
− 5 < −2 − 2
− 5 < −4
2
( )
− 4 < 0 − 2
5
− 4 < 0 − 2
− 4 < −2
true
true
Three possible non-solutions:
− 5,0 , 0,1 , and 5,2
( ) ( ) ( )
2
0 < ( −5)
− 2
5
0 < −2 − 2
0 < −4
false
2
1 < ( 0)
− 2
5
1 < 0 − 2
1 < −2
false
2
( )
− 2 < 5 − 2
5
− 2 < 2 − 2
− 2 < 0
true
2
2 < ( 5)
− 2
5
2 < 2 − 2
2 < 0
false
12. Answers may vary. One possible inequality
y ≤ − x + 5
( −2, −5 ),( −2,5 ), and ( 2, − 5)
are solutions but
( 2,5 ) is not a solution.
Homework 11.3
1.
3.
3 3 3
x + 27 = x + 3
( x
2
3)( x 3x
2
3 )
( x
2
3)( x 3x
9)
= + − +
= + − +
3 3 3
x − 64 = x − 4
( x
2
4)( x 4x
2
4 )
( x
2
4)( x 4x
16)
= − + +
= − + +
5.
3 3 3
x − 1000 = x −10
( x
2
10)( x 10x
2
10 )
( x
2
10)( x 10x
100)
= − + +
= − + +
345

Homework 11.3
SSM: Intermediate Algebra
7.
9.
3 3 3
x + 125 = x + 5
3 3 3
x + 1 = x + 1
( x
2
5)( x 5x
2
5 )
( x
2
5)( x 5x
25)
= + − +
= + − +
( x
2
1)( x x
2
1 )
( x
2
1)( x x 1)
= + − +
= + − +
3 3 3
11. x ( x)
64 − 125 = 4 − 5
13. x ( x)
2 2
( )
( x ) ( x) ( x)( )
= 4 − 5 4 + 4 5 + 5
2
( 4x 5)( 16x 20x
25)
= − + +
3 3 3
27 + 64 = 3 + 4
15. x − = ( x)
2 2
( )
( x ) ( x) ( x)( )
= 3 + 4 3 − 3 4 + 4
2
( 3x 4)( 9x 12x
16)
= + − +
3 3 3
1000 1 10 −1
17. x ( x)
2 2
( )
( x ) ( x) ( x)( )
= 10 − 1 10 + 10 1 + 1
2
( 10x 1)( 100x 10x
1)
= − + +
3 3 3
8 − 27 = 2 − 3
19. x ( x)
2 2
( )
( x ) ( x) ( x)( )
= 2 − 3 2 + 2 3 + 3
2
( 2x 3)( 4x 6x
9)
= − + +
3 3 3
125 + 8 = 5 + 2
2 2
( )
( x ) ( x) ( x)( )
= 5 + 2 5 − 5 2 + 2
2
( 5x 2)( 25x 10x
4)
= + − +
3 3
21. 4x
− 32 = 4( x − 8)
3 3
= 4( x − 2 )
2 2
= 4( x − 2)( x + 2x
+ 2 )
2
= 4( x − 2)( x + 2x
+ 4)
3 3
23. 7x
− 7000 = 7( x −1000)
3 3
= 7( x −10
)
2 2
= 7( x − 10)( x + 10x
+ 10 )
2
= 7( x − 10)( x + 10x
+ 100)
3 3
25. 10x
+ 640 = 10( x + 64)
3 3
= 10( x + 4 )
2 2
= 10( x + 4)( x − 4x
+ 4 )
2
= 10( x + 4)( x − 4x
+ 16)
4 3
27. x + 27x = x( x + 27)
3 3
= x( x + 3 )
2 2
= x( x + 3)( x − 3x
+ 3 )
2
= x( x + 3)( x − 3x
+ 9)
4 3
29. 3x − 24x = 3x( x − 8)
3 3
= 3x( x − 2 )
2 2
= 3x( x − 2)( x + 2x
+ 2 )
2
= 3x( x − 2)( x + 2x
+ 4)
4 3
31. 32x − 4x = 4x( 8x
−1)
3 3
(( x)
)
= 4x
2 −1
2 2
( )
( ) ( ) ( )( )
= 4x 2x − 1 2x + 2x
1 + 1
2
( )( )
= 4x 2x − 1 4x + 2x
+ 1
3 2
33. 2x − 50x = 2x( x − 25)
( )( x )
= 2x x − 5 + 5
3 3
35. 2x
− 16 = 2( x − 8)
3 3
= 2( x − 2 )
2 2
= 2( x − 2)( x + 2x
+ 2 )
2
= 2( x − 2)( x + 2x
+ 4)
346

SSM: Intermediate Algebra
Section 11.3 Quiz
37. x 2 12x 36 ( x 6) 2
− + = −
4 3 3
39. x + x + x + 1 = x ( x + 1) + 1( x + 1)
( x
3
1)( x 1)
( x
3
1)( x
3
1 )
( x 1)( x
2
1)( x x
2
1 )
( x 1)( x
2
1)( x x 1)
( x
2 2
1) ( x x 1)
= + +
= + +
= + + − +
= + + − +
= + − +
41. 6x 2 + 11x − 10 = ( 3x − 2)( 2x
+ 5)
3 2 2
43. 2x − 20x + 32x = 2x( x − 10x
+ 16)
45.
4 3
8x + 32x − x − 4
3
( ) ( )
= 8x x + 4 − 1 x + 4
3
( x 4)( 8x
1)
= + −
( x ) ( x)
( )( x )
= 2x x − 8 − 2
3 3
( )
2 2
( )
= + 4 2 −1
( x )( x ) ( x) ( x)( )
= + 4 2 − 1 2 + 2 1 + 1
2
( x 4)( 2x 1)( 4x 2x
1)
= + − + +
3 2 2
47. x − x − 30x = x( x − x − 30)
( 6)( x 5)
= x x − +
3 3
49. 125 − 64x
= 5 − ( 4x)
3
2
( )
2
( 5 4x) 5 ( 5)( 4x) ( 4x)
= − + +
2
( 5 4x)( 25 20x 16x
)
2
( 4x 5)( 16x 20x
25)
= − + +
= − − + +
3 3
51. No. When factoring a − b , the middle term of
the trinomial is only a ⋅ b .
( )( )
3 2
x − 8 = x − 2 x + 2x
+ 4
53. No. The middle term of the trinomial is
incorrect.
( )( )
3 3 2 2
x + a = x + a x − ax + a
55. Written response. Answers may vary.
( )( )
( )( )
3 3 2 2
a − b = a − b a + ab + b
3 3 2 2
a + b = a+ b a − ab+
b
Section 11.3 Quiz
1.
2.
3 3 3
x − 216 = x − 6
( x
2
6)( x 6x
2
6 )
( x
2
6)( x 6x
36)
= − + +
= − + +
3 3 3
x + 216 = x + 6
3. x ( x)
( x
2
6)( x 6x
2
6 )
( x
2
6)( x 6x
36)
= + − +
= + − +
3 3 3
64 + 1 = 4 + 1
4. x ( x)
2 2
( )
( x ) ( x) ( x)( )
= 4 + 1 4 − 4 1 + 1
2
( 4x 1)( 16x 4x
1)
= + − +
3 3 3
27 − 125 = 3 − 5
2 2
( )
( x ) ( x) ( x)( )
= 3 − 5 3 + 3 5 + 5
2
( 3x 5)( 9x 15x
25)
= − + +
4 3
5. 16x + 54x = 2x( 8x
+ 27)
3 3
(( x)
)
= 2x
2 + 3
2 2
( )
( ) ( ) ( )( )
= 2x 2x + 3 2x − 2x
3 + 3
2
( )( )
= 2x 2x + 3 4x − 6x
+ 9
4 3
6. 375x − 24x = 3x( 125x
− 8)
3 3
(( x)
)
= 3x
5 − 2
2 2
( )
( ) ( ) ( )( )
= 3x 5x − 2 5x + 5x
2 + 2
2
( )( )
= 3x 5x − 2 25x + 10x
+ 4
4 3
7. 5x − 5x = 5x( 1−
x )
2
= 5x( 1− x)( 1+ x + x )
2
= −5x( x − 1)( x + x + 1)
347

Homework 11.4
SSM: Intermediate Algebra
3 3
8. 56 − 7x
= 7( 8 − x )
3 3
= 7( 2 − x )
2 2
= 7( 2 − x)( 2 + 2x + x )
2
= −7( x − 2)( 4 + 2x + x )
2
= −7( x − 2)( x + 2x
+ 4)
4 3 3
9. x + 2x − 8x − 16 = x ( x + 2) − 8( x + 2)
( x
3
2)( x 8)
( x
3
2)( x
3
2 )
( x 2)( x
2
2)( x 2x
2
2 )
( x 2)( x
2
2)( x 2x
4)
= + −
= + −
= + − + +
= + − + +
4 3 3
10. x − 5x + x − 5 = x ( x − 5) + 1( x −5)
Homework 11.4
1. − 36 = i 36 = 6i
3. − − 25 = − i 25 = − 5i
5. − 13 = i 13
( x
3
5)( x 1)
( x 5)( x
2
1)( x x 1)
= − +
= − + − +
7. − 20 = i 20 = i 4 ⋅ 5 = 2i
5
11.
13.
15.
17.
20 + − 50 20 + i 50
=
10 10
20 + i 25⋅
2
=
10
20 + 5i
2
=
10
20 5i
2
= +
10 10
= 2 +
2
i
2
− 9 + − 9 = 2 −9
= 2i
9
( )
= 2i
3
= 6i
4 + 2 −25 −1− 8 −4
= 4 + 2i
25 −1−
8i
4
( ) i( )
= 4 + 2i
5 −1−
8 2
= 4 + 10i
−1−16i
= 4 − 1+ 10i
−16i
= 3 − 6i
−4 − 25 = i 4 ⋅i
25
= 2i
⋅5i
= 10i
2
( )
= 10 −1
= −10
9.
4 − −24 4 − i 24
=
8 8
4 − i 4⋅6
=
8
4 − 2i
6
=
8
4 2i
6
= −
8 8
1 6
= − i
2 4
19. −3 − 5 = i 3 ⋅i
5
2
= i 15
= − 15
21. ( ) ( )
4 − 7i + 3 + 10i = 4 − 7i + 3 + 10i
23. ( ) ( )
= 4 + 3 − 7i
+ 10i
= 7 + 3i
6 − 5i − 2 + 13i = 6 − 5i − 2 −13i
= 6 − 2 − 5i
−13i
= 4 −18i
348

SSM: Intermediate Algebra Homework 11.4
25. ( ) ( )
27.
2 3 − 5i − 4 2 − 6i = 6 −10i − 8 + 24i
2i ⋅ 9i = 18i
2
( )
= 18 −1
= −18
29. ( )
−10i − 5i = 50i
31. ( )
2
( )
= 50 −1
= −50
5i 3 − 2i = 5i ⋅3 − 5i ⋅ 2i
33. ( )
= 15i
−10i
= 6 − 8 − 10i
+ 24i
= − 2 + 14i
2
( )
= 15i
−10 −1
= 15i
+ 10
= 10 + 15i
20 − 3i 2 − 7i = 20 − 3i ⋅ 2 + 3i ⋅7i
35. ( )( )
= 20 − 6i
+ 21i
2
( )
= 20 − 6i
+ 21 −1
= 20 − 6i
− 21
= −1−
6i
2 + 5i 3 + 4i = 2⋅ 3 + 2⋅ 4i + 5i ⋅ 3 + 5i ⋅ 4i
37. ( )( )
= 6 + 8i + 15i + 20i
( )
= 6 + 23i
+ 20 −1
= 6 + 23i
− 20
= − 14 + 23i
3 − 6i 5 + 2i = 3⋅ 5 + 3⋅ 2i − 6i ⋅5 − 6i ⋅ 2i
= 15 + 6i − 30i −12i
2
( )
= 15 − 24i
−12 −1
= 15 − 24i
+ 12
= 27 − 24i
2
39. ( 5 + 4i)( 5 − 4i) = 5 − ( 4i)
= 25 −16i
2
2
( )
= 25 −16 −1
= 25 + 16
= 41
2
41. ( )( )
2 2
1+ i 1− i = 1 − i
( )
= 1− −1
= 1+
1
= 2
2 2
2
43. ( 2 + 3i) = 2 + 2( 2)( 3i) + ( 3i)
= 4 + 12i
+ 9i
2
( )
= 4 + 12i
+ 9 −1
= 4 + 12i
− 9
= − 5 + 12i
2 2 2
45. ( 4 ) 4 2( 4)( )
= 16 − 8i
+ ( −1)
47.
49.
− i = − i + i
= 16 − 8i
−1
= 15 − 8i
3 3 2 − 5i
= ⋅
2 + 5i 2 + 5i 2 − 5i
6 −15i
=
2
4 − 25i
6 −15i
=
4 − 25 − 1
( )
6 −15i
=
4 + 25
6 −15i
=
29
6 15
= − i
29 29
7 7 6 − 3i
= ⋅
6 + 3i 6 + 3i 6 − 3i
42 − 21i
=
2
36 − 9i
42 − 21i
=
36 − 9 −1
( )
42 − 21i
=
36 + 9
42 − 21i
=
45
14 7
= − i
15 15
349

Homework 11.4
SSM: Intermediate Algebra
51.
53.
55.
2 − 3i 2 − 3i 7 − i
= ⋅
7 + i 7 + i 7 − i
14 − 2i − 21i + 3i
=
2
49 − i
14 − 23i
+ 3 −1
=
49 − −1
11−
23i
=
50
11 23
= − i
50 50
( )
( )
3 + 4i 3 + 4i 3 + 4i
= ⋅
3 − 4i 3 − 4i 3 + 4i
9 + 12i+ 12i+
16i
=
2
9 −16i
9 + 24i
+ 16( −1)
=
9 −16 −1
( )
− 7 + 24i
=
25
7 24
= − + i
25 25
3 + 5i 3 + 5i 2 − 9i
= ⋅
2 + 9i 2 + 9i 2 − 9i
6 − 27i + 10i − 45i
=
2
4 − 81i
6 −17i
− 45( −1)
=
4 − 81 −1
51−17i
=
85
51 17
= − i
85 85
3 1
= − i
5 5
( )
2
2
2
57. x
2 + 2x
+ 3 = 0
2
( ) − ( )( )
2( 1)
− 2 ± 2 4 1 3
x =
=
− 2 ± −8
2
− 2 ± i 8
=
2
− 2 ± 2i
2
=
2
= − 1±
i 2
59. x
2 − 2x
+ 5 = 0
2
( 2) ( 2) 4( 1)( 5)
2( 1)
− − ± − −
x =
2 ± −16
=
2
2 ± i 16
=
2
2 ± 4i
=
2
= 1±
2i
61. 2x
2 + x + 1 = 0
2
( ) − ( )( )
2( 2)
− 1±
1 4 2 1
x =
=
− 1±
−7
4
− 1±
i 7
=
4
1 7
= − ± i
4 4
350

SSM: Intermediate Algebra Homework 11.4
63.
2
2
5x
− 4x
= −1
5x
− 4x
+ 1 = 0
2
( 4) ( 4) 4( 5)( 1)
2( 5)
− − ± − −
x =
=
4 ± −4
10
4 ± i 4
=
10
4 ± 2i
=
10
2 1
= ± i
5 5
65. x( x − )
2 2 1 = −3
2
2
4x
− 2x
= −3
4x
− 2x
+ 3 = 0
2
( 2) ( 2) 4( 4)( 3)
2( 4)
− − ± − −
x =
=
2 ± −44
8
2 ± i 44
=
8
2 ± 2i
11
=
8
1 11
= ± i
4 4
67. ( 1)( 2)
x + x + = x
2
x + 2x + x + 2 = x
x
2
+ 2x
+ 2 = 0
2
( ) − ( )( )
2( 1)
− 2 ± 2 4 1 2
x =
=
− 2 ± −4
2
− 2 ± i 4
=
2
− 2 ± 2i
=
2
= − 1±
i
69. x( 3x − 2) = 2 + 2( x − 3)
2
2
3x − 2x = 2 + 2x
− 6
3x
− 4x
+ 4 = 0
2
( 4) ( 4) 4( 3)( 4)
2( 3)
− − ± − −
x =
=
4 ± −32
6
4 ± i 32
=
6
4 ± 4i
2
=
6
2 2 2
= ± i
3 3
71. x
2 = −9
73.
x = ± −9
x = ± i
x = ± 3i
2
− 3x
= 5
x
2
9
5
= −
3
5
x = ± −
3
x = ± i
5
3
5 3
x = ± i ⋅
3 3
x = ± i
x = ±
75. ( x ) 2
15
3
15
3
5 + 3 = −7
5x
+ 3 = ± −7
5x
+ 3 = ± i 7
5x
= − 3 ± i 7
i
3 7
x = − ± i
5 5
351

Section 11.4 Quiz
SSM: Intermediate Algebra
77. ( x ) 2
−2 3 − 1 + 4 = 12
( x )
2
−2 3 − 1 = 8
( x − )
2
3 1 = −4
3x
− 1 = ± −4
3x
− 1 = ± i 4
3x
− 1= ± 2i
3x
= 1±
2i
1 2
x = ± i
3 3
79. Student 2 did the work correctly. When working
with even roots of negative numbers, it is
necessary to first write the problem using
complex numbers of the form a + bi . Not all
rules that apply to the real numbers will apply to
complex numbers.
81. Answers may vary. Possible answers:
1− i + 1+ 2i = 1− i + 1+ 2i = 2 + i
a. ( ) ( )
1− i + 1+ i = 1− i + 1+ i = 2
b. ( ) ( )
1+ i + − 1+ i = 1+ i − 1+ i = 2i
c. ( ) ( )
83. If the discriminant is positive, there are two
distinct real solutions. If the discriminant is zero,
there is one real solution (a repeated root). If the
discriminant is negative, there are two distinct
imaginary solutions.
Section 11.4 Quiz
1.
2.
3 + 4 −36 − 2 − 8 − 36 = 3 + 4i
36 − 2 − 8i
36
−2 − 7 = i 2 ⋅i
7
= i
2
= −
14
14
3. ( ) ( )
( ) i( )
= 3 + 4i
6 − 2 − 8 6
= 3 + 24i
− 2 − 48i
= 3 − 2 + 24i
− 48i
= 1−
24i
6 − 2i + 3 − 4i = 6 − 2i + 3 − 4i
= 6 + 3 − 2i
− 4i
= 9 − 6i
4.
−4i ⋅ 3i = −12i
2
( )
= −12 −1
= 12
5. ( )( )
5 − 3i 7 + i = 5⋅ 7 + 5⋅i − 3i ⋅ 7 − 3i ⋅i
= 35 + 5i − 21i − 3i
( )
= 35 −16i
− 3 −1
= 35 − 16i
+ 3
= 38 −16i
2 2 2
6. ( 4 − 3i) = ( 4) − 2( 4)( 3i) + ( 3i)
= 16 − 24i
+ 9i
2
( )
= 16 − 24i
+ 9 −1
= 16 − 24i
− 9
= 7 − 24i
2 2
7. ( 8 + 5i)( 8 − 5i) = ( 8) − ( 5i)
8.
9.
2 2 4 + i
= ⋅
4 − i 4 − i 4 + i
8 + 2i
=
2
16 − i
8 + 2i
=
16 − −1
( )
= 64 − 25i
2
( )
= 64 − 25 −1
= 64 + 25
= 89
8 + 2i
=
17
8 2
= + i
17 17
3 + 2i 3 + 2i 5 + 4i
= ⋅
5 − 4i 5 − 4i 5 + 4i
15 + 12i + 10i + 8i
=
2
25 −16i
15 + 22i
+ 8 −1
=
25 −16 1
7 + 22i
=
41
7 22
= + i
41 41
( )
( − )
2
2
352

SSM: Intermediate Algebra
Section 11.4 Quiz
1− i 1− i 1−
i
10. = ⋅
1+ i 1+ i 1−
i
2 2
1 − i − i + i
=
2 2
1 − i
1−
2i
1
=
1− −1
−2i
=
2
= −i
11. x
2 − 2x
+ 3 = 0
12.
+ ( − )
( )
2
( 2) ( 2) 4( 1)( 3)
2( 1)
− − ± − −
x =
2 ± −8
=
2
2 ± i 8
=
2
2 ± 2i
2
=
2
= 1±
i 2
2
− 3x
+ x − 4 = 0
2
( ) ( )( )
2( −3)
− 1± 1 − 4 −3 −4
x =
− 1±
−47
=
−6
1±
i 47
=
6
1 47
= ± i
6 6
13. ( 4)( 5)
2
x + x + = x
x + 4x + 5x + 20=
x
x
2
+ 8x
+ 20 = 0
2
( ) − ( )( )
2( 1)
− 8 ± 8 4 1 20
x =
− 8 ± −16
=
2
− 8 ± i 16
=
2
− 8 ± 4i
=
2
= − 4 ± 2i
14. ( x )
−3 2 − 5 + 1 = 13
2
( x )
−3 2 − 5 = 12
2
( x − )
2
2 5 = −4
2x
− 5 = ± −4
2x
− 5 = ± i 4
2x
− 5 = ± 2i
2x
= 5 ± 2i
5
x = ± i
2
15. The discriminant of the quadratic equation
2
ax + bx − a = 0 is b 2 − 4( a)( − a)
= b 2 + 4a
2 .
Since a ≠ 0 for quadratic equations, this
discriminant will always equal a positive number
and each solution will be a real number.
16. False.
The number 3i (or 0 + 3i ) is a complex number
and i is a pure imaginary number. Then
( )( )
2
i 3i = 3i
= − 3 . Since − 3 is not an imaginary
number, the statement is false.
353

Homework 11.5
SSM: Intermediate Algebra
Homework 11.5
1. x = dollars invested in CD
y = dollars invested in mutual fund
a. x + y = 10000 or y = 10000 − x
I = .0287 x + .081y
f ( x) = .0287 x + .081(10000 − x)
= .0287 x + .081⋅10000 − .081x
f ( x) = − .0523x + 810 0≤ x ≤10000
b. f is decreasing. The more money invested
in the CD, the less interest will be earned.
c. We want f ( x) = 500
− .0695x
+ 850.5 = 500
− .0695x
= −350.5
−350.5
x = = $5043.17
−.0695
You should invest $5043.17 in the CD and
$3956.83 in the mutual fund in order to earn
$500 interest.
d. f (10000) = − .0695(10000) + 850.5
= $155.50
Although the formula for f ( x)
yields a
plausible result at x = 10000 , this value is
outside the domain of f, since only $9000 is
available for investment. Thus model
breakdown has occurred.
5. x = dollars invested in CD
c. We want f ( x) = 400
− .0523x
+ 810 = 400
− .0523x
= − 410
− 410
x = = $7839.39
−.0523
You should invest $7839.39 in the CD and
$2160.61 in the mutual fund in order to earn
$400 interest.
3. x = dollars invested in CD
y = dollars invested in mutual fund
a. x + y = 9000 or y = 9000 − x
I = .025 x + .0945y
f ( x) = .025 x + .0945(9000 − x)
= .025 x + .0945⋅9000 − .0945x
f ( x) = − .0695x + 850.5 0≤ x ≤9000
b. f (500) = − .0695(500) + 850.5
f (500) = $815.75
If $500 is invested in the CD and $8500 in
the mutual fund, $815.75 in interest will be
earned.
y = dollars invested in mutual fund
a. x + y = 8000 or y = 8000 - x
I = .015 x + .116y
f ( x) = .015 x + .116(8000 − x)
= .015 x + .116⋅8000 − .116x
f ( x) = − .101x
+ 928
2500≤
x ≤8000 ( see part b.)
b. Since f is a decreasing function and
2500 ≤ x ≤ 8000
the minimum interest possible will be
f (8000) = − .101(8000) + 928 = $120.00
and the maximum interest possible will be
f (2500) = − .101(2500) + 928 = $675.50
Thus the investment can earn from $120 to
$675.50 in interest.
c. We want f ( x) = 400
− .101x
+ 928 = 400
− .101x
= −528
−528
x = = $5227.72
−.101
You should invest $5227.72 in the CD and
$2772.28 in the mutual fund in order to earn
$400 interest.
354

SSM: Intermediate Algebra Homework 11.5
7. x = dollars invested in CD
y = dollars invested in mutual fund
a. x + y = 6000 or y = 6000 - x
I = .0285 x + .09y
f ( x) = .0285 x + .09(6000 − x)
= .0285 x + .09⋅6000 − .09x
f ( x) = − .0615x + 540 0≤ x ≤6000
b. We evaluate f (0) = $540. Thus the I-
intercept is (0, 540). This means that if the
entire $6000 is invested in the mutual fund,
$540 in interest will be earned. Since f is a
decreasing function, $540 is also the
maximum interest that can be earned.
c. The x-intercept occurs when f (x) = 0.
So
− .0615x
+ 540 = 0
− .0615x
= − 540
−540
x = = $8780.49
−.0615
Thus the x-intercept is (8780.49, 0).
Although the formula for f ( x)
yields a
plausible result for f (x) = 0 , this value is
outside the domain of f , since only $6000 is
available for investment. It is also illogical
to invest a large sum of money that returns
no interest. Thus model breakdown has
occurred.
d. The slope of f is − 0.0615 . It means that
for every additional dollar invested in the
CD, 6.15 cents less interest is received.
b. f is decreasing. The more $50 tickets sold,
the less revenue will be earned.
c. f (16000) = − 25(16000) + 1500000
= 1,100,000.
If 16000 of the $50 tickets are sold and 4000
of the $75 tickets are sold, the total revenue
will be $1,100,000.
d. We want R = cost + profit = $1,075,000.
− 25x
+ 1500000 = 1075000.
− 25x
= − 425000.
−425000
x = = 17000
−25
Thus we should sell 17000 of the $50 tickets
and 3000 of the $75 tickets to make
$600,000 profit.
11. x = number of $45 tickets sold
y = number of $70 tickets sold
a. x + y = 12000 or y = 12000 − x
b.
R = 45x + 70y
f ( x) = 45x + 70(12000 − x)
= 45x
+ 70⋅12000 − 70x
f ( x) = − 25x + 840000 0≤ x ≤12000
9. x = number of $50 tickets sold
y = number of $75 tickets sold
a. x + y = 20000 or y = 20000 − x
R = 50x + 75y
f ( x) = 50x + 75(20000 − x)
= 50x
+ 75⋅ 20000 − 75x
f ( x) = − 25x + 1500000 0≤ x ≤ 20000
c. Since f is a decreasing function and
0 ≤ x ≤ 12000 , the minimum revenue
possible will be
f (12000) = − 25(12000) + 840000 = 540,000
and the maximum revenue possible will be
f (0) = − 25(0) + 840000 = 840,000.
Thus the earned revenue will be between
$540,000 and $840,000.
355

Homework 11.5
SSM: Intermediate Algebra
d. We want R = $602,500.
− 25x
+ 840000 = 602500.
− 25x
= − 237500
−237500
x = = 9500
−25
Thus we should sell 9500 of the $45 tickets
and 2500 of the $70 tickets to earn $602,500
in revenue.
13. x = average price of a coach ticket
y = average price of a first-class ticket
a. y = x + 242
R = 126x + 8y
f ( x) = 126x + 8( x + 242)
= 126x
+ 8x
+ 8⋅
242
f ( x) = 134x + 1936 x ≥ 0
b. The slope of f is 134. This means that for
each dollar increase in the average price of a
coach ticket (and therefore a dollar increase
in the average price of a first-class ticket
also), the revenue from the flight will
increase by $134.
c. We want R = $14,130.
134x
+ 1936 = 14130
134x
= 12194
12194
x = = $91.
134
The average selling price for a coach ticket
should be $91 and the average price for a
first-class ticket should be $333 in order to
earn revenue of $14,130.
15. x = number of part-time students per semester
y = number of full-time students per semester
y
=
3x
R = 13(3x + 14 y)
f ( x) = 13⋅ 3x + 13⋅14(3 x )
= 39x
+ 546x
f ( x) = 585x x ≥ 0
For revenue of $900,000 per semester, we need
f ( x) = 900000
585x
= 900000
900000
x = = 1538.5
585
That is, we need 1539 part-time students and
4617 full-time students.
17. Demand: n = − 2.25 p + 3162.5
a. Revenue = demand ⋅ price
R = n ⋅ p
f ( p) = ( − 2.25 p + 3162.5) p
2
f ( p) = − 2.25 p + 3162.5 p
b. Set f ( p ) = 0
2
2.25 3162.5 0
− p + p =
−2.25 p( p − 1405.56) = 0
( )
p p − 1405.56 = 0
p = 0 or p = 1405.56
Thus p-intercepts are (0,0) and (1405.56,0)
If p = $0., we will give the guitars away; if
p = $1405.56, there will be no demand. In
either case, no revenue will be generated.
c. The revenue function is part of the parabola
2
y = − 2.25 p + 3162.5p
which has standard form
2
y = − 2.25( p − 1405.56 p + 493896.60)
+ 1111267.36
2
2.25( 702.78) 1111267.36
y = − p − +
The maximum R value will occur at the vertex of
the parabola, which is (702.78, 1111267.36).
Thus the maximum monthly revenue is
$1,111,267.36, which occurs when the price per
guitar is $702.78
19. a. Start by plotting the data.
A linear model seems appropriate. Using
linear regression we get
n = − 136.2 p + 5861.64
356

SSM: Intermediate Algebra
Section 11.5 Quiz
b. Revenue = demand ⋅ price
R = n ⋅ p
f ( p) = ( − 136.2 p + 5861.64) p
= − +
2
f ( p) 136.2 p 5861.64 p
c. Set f ( p ) = 0
2
136.2 p 5861.64 p 0
− + =
−136.2 p( p − 43.04) = 0
( )
p p − 43.04 = 0
p = 0 or p = 43.04
Thus the p-intercepts are (0,0) and
(43.04,0).
p = $0., we will give the CDs away; if
p = $43.04, there will be no demand. In
either case, no revenue will be generated.
d. The revenue function is part of the parabola
2
y = − 136.2 p + 5861.64 p
which has standard form
y = − p − p + +
2
136.2( 43.04 463.11) 63075.64
2
y = −136.2( p − 21.52) + 63075.64
The maximum R value will occur at the
vertex of the parabola, which is (21.52,
63075.64). Thus the maximum monthly
revenue is $63,075.64, which occurs when
the price per CD is $21.52.
21. a. Start by plotting the data.
A linear model seems appropriate. Using
linear regression we get
n = − 2.164 p + 2493.6
b. The variable n is the dependent variable
because the number of demands depends on
the price.
c. Revenue = demand ⋅ price
R = n ⋅ p
f ( p) = ( − 2.164 p + 2493.6) p
2
f ( p) 2.164 p 2493.6 p
= − +
If
d. Set f ( p ) = $240,000.
− + =
2
2.164 p 2493.6 p 240000
− + − =
2
2.164 p 2493.6 p 240000 0
Using the quadratic formula:
− ± − − −
p =
2( −2.164)
2
2493.6 (2493.6) 4( 2.164)( 240000)
− 2493.6 ± 2034.85
or p =
so
−4.328
p = 106 or p = 1046.31
Thus if we sell the printers for either $106
apiece or $1046.31 apiece, our revenue will
be $240,000.
e. The revenue function is part of the parabola
2
y = − 2.164p + 2493.6 p
which has standard form
Section 11.5 Quiz
y = − p − p + +
2
2.164( 1152.31 331954.89) 718350.39
y = − − +
2
2.164( p 576.16) 718350.39
The maximum R value will occur at the
vertex of the parabola, which is (576.16,
718350.39). Thus the maximum monthly
revenue is $718,350.39, which occurs when
the price per printer is $576.16.
1. x = dollars invested in CD
y = dollars invested in mutual fund
a. x + y = 8000 or y = 8000 − x
I = .029 x + .127y
f ( x) = .029 x + .127(8000 − x)
= .029 x + .127 ⋅8000 − .127x
f ( x) = − .098x + 1016 0≤ x ≤8000
b. f (500) = − .098(500) + 1016
f (500) = $967.
If $500 is invested in the CD and $7500 in
the mutual fund, $967 in interest will be
earned.
357

Homework 11.6
SSM: Intermediate Algebra
c. We want f ( x) = 500
− .098x
+ 1016 = 500
− .098x
= −516
−516
x = = $5265.31
−.098
You should invest $5265.31 in the CD and
$2734.69 in the mutual fund in order to earn
$500 interest.
2. x = number of $30 tickets sold
y = number of $50 tickets sold
a. x + y = 16000 or y = 16000 − x
R = 30x + 50y
f ( x) = 30x + 50(16000 − x)
= 30x
+ 50⋅16000 − 50x
f ( x) = − 20x + 800000 0≤ x ≤16000
b. The slope of f is -20; this means that for
every (additional) $30 ticket sold, the
revenue will decrease by $20.
c. We want R = $500,000.
− 20x
+ 800000 = 500000.
− 20x
= − 300000.
−300000
x = = 15000
−20
Thus we should sell 15,000 $30 tickets and
1000 $50 tickets to earn $500,000 in
revenue.
3. Demand: n = − 120 p + 65000
a. Revenue = demand ⋅ price
R = n ⋅ p
f ( p) = ( − 120 p + 65000) p
= − +
2
f ( p) 120 p 65000 p
b. Set f ( p ) = 0
− + =
2
120 p 65000 p 0
−120 p( p − 541.67) = 0
( )
p p − 541.67 = 0
p = 0 or p = 541.67
Thus the
p-intercepts are (0,0) and
(541.67,0).
If p = $0., we will give the DVD players
away; if p = $541.67, there will be no
demand. In either case, no revenue will be
generated.
c. The revenue function is part of the parabola
2
y = − 120 p + 65000 p
which has standard form
Homework 11.6
2
y = −120( p − 541.67 p + 73351.6) + 8802191.67
2
y = −120( p − 270.84) + 8802191.67
The maximum R value will occur at the
vertex of the parabola, which is (270.84,
8802191.67). Thus the maximum monthly
revenue is $8,802,191.67, which occurs
when the price per DVD player is $270.84.
1. 2 = 2 +
2
c a b
c
c
c
2 2 2
2
2
= 5 + 12
= 25 + 144
= 169
c = 13
3. 2 = 2 +
2
c a b
c
c
c
2 2 2
2
2
= 4 + 5
= 16 + 25
= 41
c =
41
5. 2 + 2 =
2
a b c
2 2 2
3 + b = 8
2
9 + b = 64
b
2
= 55
b =
7. 2 + 2 =
2
a b c
a
a
2 2 2
2
+ 5 = 7
2
55
+ 25 = 49
a
= 24
a =
24
a = 2 6
358

SSM: Intermediate Algebra Homework 11.6
9.
2 2 2
c = a + b
c
c
c
2
2
2
( 2) ( 5)
= 2 + 5
= 7
c =
= +
7
11. c 2 = a 2 + b
2
c
c
c
2 2 2
2
2
= 21 + 7
= 441+
49
= 490
c =
490
c = 7 10
2 2
13. ( 2,5 ) and ( 6,7 )
( 6 2) ( 7 5)
2 2
= 4 + 2
= 16 + 4
20
≈ 4.47
2 2
d = − + −
=
15. ( − 3,5 ) and ( 4,2)
( 4 ( 3)
) ( 2 5)
2
( )
= 7 + −3
= 49 + 9
58
≈ 7.62
2
2 2
d = − − + −
=
17. ( −4, −1 ) and ( −1, − 5)
( 1 ( 4)
) ( 5 ( 1)
)
2
( )
= 3 + −4
= 9 + 16
= 5
25
2
2 2
d = − − − + − − −
=
19. ( 2.1,8.9 ) and ( 5.6,1.7 )
( 5.6 2.1) ( 1.7 8.9)
2
( )
= 3.5 + −7.2
= 12.25 + 51.84
64.09
≈ 8.01
2 2
d = − + −
=
21. ( −2.18, − 5.74 ) and ( 3.44,6.29)
2
( 3.44 ( 2.18)
) ( 6.29 ( 5.74)
)
2 2
= 5.62 + 12.03
= 31.5844 + 144.7209
176.3053
≈ 13.28
2 2
d = − − + − −
=
23. C ( 0,0 ) and r = 5
( ) ( )
2 2 2
x − h + y − k = r
( x ) ( y )
2 2 2
− 0 + − 0 = 5
x
2 2
+ y = 25
25. C ( 0,0 ) and r = 6.7
( ) ( )
2 2 2
x − h + y − k = r
( x ) ( y )
2 2 2
− 0 + − 0 = 6.7
x
2 2
27. C ( 5,3 ) and r = 2
( ) ( )
( x ) ( y )
( x ) ( y )
+ y = 44.89
2 2 2
x − h + y − k = r
29. C ( )
2 2 2
− 5 + − 3 = 2
2 2
− 5 + − 3 = 4
− 2,1 and r = 4
( ) ( )
( x ( )) ( y )
2 2 2
x − h + y − k = r
2 2 2
− − 2 + − 1 = 4
( x ) ( y )
2 2
+ 2 + − 1 = 16
359

Homework 11.6
SSM: Intermediate Algebra
31. C ( )
−7, − 3 and r = 3
( ) ( )
2 2 2
x − h + y − k = r
( x ( )) y ( )
( ) ( )
2 2
− − 7 + − − 3 = 3
33. x
2 + y
2 = 25
( x ) ( y )
2 2
+ 7 + + 3 = 3
The equation has the form
Therefore, C = ( 0,0)
and
r
2
= 25
r =
r = 5
−4
25
4
−4
y
4
x
2
2 2 2
x + y = r .
r
2
= 16
r = 16
r = 4
y
8
4
x
4 8
2 2
39. ( x + 6) + ( y − 1)
= 7
2 2
( x −( − 6)
) + ( y − 1) = ( 7 )
The equation is in the form
( x h) ( y k)
r
The center is ( 6,1)
2 2 2
− + − = .
y
8
2
C − and the radius is r = 7 .
35. x
2 + y
2 = 8
The equation has the form
Therefore, C = ( 0,0)
and
r
2
= 8
r =
8
r = 2 2
−4
4
−4
y
4
x
2 2 2
x + y = r .
x
−8
8
−8
2 2
41. ( x + 3) + ( y + 2)
= 1
( x ( )) ( y ( ))
2 2 2
− − 3 + − − 2 = 1
The equation is in the form
2 2 2
( x − h) + ( y − k)
= r .
The center is ( 3, 2)
4
y
C − − and the radius is r = 1 .
2 2
37. ( x ) ( y )
− 3 + − 5 = 16
The equation is in the form
2 2 2
( x − h) + ( y − k)
= r .
The center is ( h,
k ) or ( 3,5)
C and
−4
−4
4
x
360

SSM: Intermediate Algebra Homework 11.6
43. a = 5, b = 20, c = length of ladder
2 2 2
c = a + b
c
c
c
2 2 2
2
2
= 5 + 20
= 25 + 400
= 425
c =
425
c ≈ 20.62
The ladder must be approximately 20.62 feet
long.
45. a = 465, c = 964, b = distance
2 2 2
a + b = c
2 2 2
465 + b = 964
b
2
= 713071
b =
713071
b ≈ 844.44
a+ b+ c = 465 + 844.44 + 964
= 2273.44
The total distance of the road trip would be
approximately 2273.44 miles.
47. C ( 1, − 2)
and r = 3 .
( ) ( )
( x ) ( y ( ))
2 2 2
x − h + y − k = r
2 2 2
− 1 + − − 2 = 3
( x ) ( y )
2 2
− 1 + + 2 = 9
49. The radius is the distance from the center to any
point on the circle. The distance between
3,2 5,6 is given by:
C ( ) and ( )
( 5 3) ( 6 2)
2 2
= 2 + 4
= 4 + 16
20
2 2
d = − + −
=
The radius is r = 20 .
The equation of the circle is
( ) ( )
2 2 2
x − h + y − k = r
2 2
( x − 3) + ( y − 2) = ( 20)
( x ) ( y )
2 2
− 3 + − 2 = 20
2
51. Answers may vary. One possible answer:
( x ) ( y )
2 2
− 2 + − 3 = 9
( x ) ( y )
2 2
− 5 + − 6 = 9
y
8
4
(5, 3)
4 8
53. Find the equation of the circle that has center
C 3,2 and radius r = 4 .
( )
( ) ( )
2 2 2
x − h + y − k = r
( x ) ( y )
2 2 2
− 3 + − 2 = 4
x
2 2
( x − 3) + ( y − 2)
= 16
Find the coordinates of five points, ( x,
y ) , that
satisfy this equation. Two possible answers are:
7,2 .
( 3,6 ) and ( )
55. No. The graph of the relation is a circle with
radius 7 and centered at the origin. The graph
fails the vertical line test.
57. a. The square root of a nonnegative number is
a nonnegative real number and the square
root of a negative number is an imaginary
number. Therefore, y ≥ 0 for real number
values of y.
b.
−4
4
−4
y
59. a. Sketches may vary. One example:
8
4
x
8
361

Section 11.6 Quiz
SSM: Intermediate Algebra
b. a = k,
b = k
2 2 2
c = a + b
2 2 2
c = k + k
c
2 2
= 2k
c =
c =
c = k
2k
2
c. c = k 2
= 3 2
2
d. c = k 2
5 = k 2
k =
5
2
2
k
2
5 2
k = ⋅
2 2
k =
5 2
2
Section 11.6 Quiz
1. a = 4, c = 8
2 2 2
a + b = c
2 2 2
4 + b = 8
2
16 + b = 64
b
2
= 48
b =
48
b = 4 3
The other leg is 4 3 inches.
2. b = 16, c = 19
2 2 2
a + b = c
a
a
2 2 2
2
+ 16 = 19
+ 256 = 361
a
2
= 105
a =
105
a ≈ 10.25
The height of the screen is about 10.25 inches.
3. ( −2, − 5)
and ( 3, − 1)
( 3 ( 2)
) ( 1 ( 5)
)
2 2
= 5 + 4
= 25 + 16
41
2 2
d = − − + − − −
=
4. ( − 3,2 ) and ( 4, 2)
( 4 ( 3)
) ( 2 2)
2 2
= 7 + 0
= 7
49
2 2
d = − − + −
=
5. C ( )
− 3,2 and r = 6
( ) ( )
( x ( )) ( y )
2 2 2
x − h + y − k = r
2 2 2
− − 3 + − 2 = 6
( x ) ( y )
2 2
+ 3 + − 2 = 36
6. C ( 0,0 ) and r = 2.8
( ) ( )
2 2 2
x − h + y − k = r
( x ) ( y )
2 2 2
− 0 + − 0 = 2.8
x
7. x
2 + y
2 = 12
2 2
+ y = 7.84
The equation is in the form
center is C ( 0,0)
and
r
2
= 12
r =
12
r = 2 3
−4
4
−4
y
4
x
2 2 2
x + y = r . The
362

SSM: Intermediate Algebra Homework 11.7
2 2
8. ( x ) ( y )
+ 4 + − 3 = 25
( x ( )) ( y )
2 2 2
− − 4 + − 3 = 5
The equation is in the form
2 2 2
( x − h) + ( y − k)
= r . The center is C ( − 4,3)
and the radius is r = 5 .
y
12
Homework 11.7
1.
2 2
x y
+ = 1
36 9
a
2
= 36, a = 6
x-intercepts: ( − 6,0 ),6,0
( )
b
2
= 9, b = 3
y-intercepts: ( 0, − 3 ),( 0,3)
y
−12 4
−4
9. C ( 2, − 1)
x
The radius is the distance from the center,
2, 1
4,7 that lies on the circle.
( − ) , to the point ( )
( 4 2) 2
( 7 ( 1)
) 2
d = − + − −
2 2
= 2 + 8
= 4 + 64
= 68
The equation of the circle is
( ) ( )
2 2 2
x − h + y − k = r
2
( x ) y ( )
2
( ) ( )
− 2 + − − 1 = 68
( x ) ( y )
2 2
− 2 + + 1 = 68
10. Answers may vary. One possible answer:
( x )
( x )
2 2
+ 2 + y = 4
2 2
− 3 + y = 9
−4
4
−4
y
(0, 0)
4
x
2
3.
5.
−4
2 2
4
−4
x y
+ = 1
4 36
a
2
= 4, a = 2
x-intercepts: ( − 2,0 ),( 2,0)
b
2
= 36, b = 6
y-intercepts: ( 0, − 6,0,6 ) ( )
−4
4
−4
2 2
x y
+ = 1
100 16
a
2
y
= 100, a = 10
x-intercepts: ( − 10,0 ),( 10,0)
b
2
= 16, b = 4
y-intercepts: ( 0, − 4,0,4 ) ( )
y
8
4
4
x
x
−4
−8
4
x
363

Homework 11.7
SSM: Intermediate Algebra
7.
9.
2 2
25x
+ 4 y = 100
2 2
25x
4y
100
+ =
100 100 100
a
2
2 2
x y
+ = 1
4 25
= 4, a = 2
x-intercepts: ( − 2,0 ),( 2,0)
b
2
= 25, b = 5
y-intercepts: ( 0, − 5 ),( 0,5)
−4
y
2
−2
2 2
2 2
2 2
4
9x
+ 100y
= 900
9x
100y
900
+ =
900 900 900
a
2
x y
+ = 1
100 9
= 100, a = 10
x-intercepts: ( − 10,0 ),( 10,0)
b
2
= 9, b = 3
y-intercepts: ( 0, − 3 ),( 0,3)
y
8
−4 4
−8
11. x
2 + y
2 = 36
2 2
x y 36
+ =
36 36 36
2 2
x y
+ = 1
36 36
a
2
= 36, a = 6
x-intercepts: ( − 6,0 ),6,0
( )
b
2
= 36, b = 6
x
x
y-intercepts: ( 0, − 6,0,6 ) ( )
−4
4
−4
13. x
2 + 25y
2 = 25
15.
y
2 2
x 25y
25
+ =
25 25 25
a
2
2 2
x y
+ = 1
25 1
= 25, a = 5
x-intercepts: ( − 5,0 ),( 5,0)
b
2
= 1, b = 1
y-intercepts: ( 0, − 1 ),( 0,1)
−4
4
−4
y
2 2
5x
+ 16y
= 80
2 2
5x
16y
80
+ =
80 80 80
a
2
2 2
x y
+ = 1
16 5
= 16, a = 4
x-intercepts: ( − 4,0 ),( 4,0)
b
2
= 5, b = 5
y-intercepts: ( 0, − 5 ),( 0, 5)
−4
y
4
−2 2
4
4
x
x
x
364

SSM: Intermediate Algebra Homework 11.7
17.
2 2
16x
+ 25y
= 1
2 2
x y
+ = 1
1/16 1/ 25
2 1 1
a = , a =
16 4
1 1
x-intercepts:
⎛ ⎜ − ,0 ⎞ ⎟, ⎛ ⎜ ,0
⎞
⎟
⎝ 4 ⎠ ⎝ 4 ⎠
2 1 1
b = , b =
25 5
1 1
y-intercepts:
⎛ ⎜0, −
⎞ ⎟, ⎛ ⎜0,
⎞
⎟
⎝ 5 ⎠ ⎝ 5 ⎠
−1
1
−1
y
1
x
23.
then sketch the inclined asymptotes.
4
2 2
2
y
−2
x y
− = 1
25 81
a
2
= 25, a = 5
−4
x-intercepts: ( − 5,0 ),( 5,0)
2
x
b = 81, b = 9
Sketch a dashed rectangle that contains the
−5,0 , 5,0 , 0, − 9 , and 0,9 , and
points ( ) ( ) ( ) ( )
then sketch the inclined asymptotes.
y
8
19.
2 2
x y
− = 1
16 4
a
2
= 16, a = 4
x-intercepts: ( − 4,0 ),( 4,0)
−8
−8
8
x
21.
2
b = 4, b = 2
Sketch a dashed rectangle that contains the
−4,0 , 4,0 , 0, − 2 , and 0,2 , and
points ( ) ( ) ( ) ( )
then sketch the inclined asymptotes.
4
−3 3
2 2
−4
y
y x
− = 1
16 25
b
2
= 16, b = 4
y-intercepts: ( 0, − 4,0,4 ) ( )
x
25.
2 2
16x
− 4y
= 64
2 2
16x
4y
64
− =
64 64 64
a
2
2 2
x y
− = 1
4 16
= 4, a = 2
x-intercepts: ( − 2,0 ),( 2,0)
2
b = 16, b = 4
Sketch a dashed rectangle that contains the
−2,0 , 2,0 , 0, − 4 , and 0,4 , and
points ( ) ( ) ( ) ( )
then sketch the inclined asymptotes.
y
8
2
a = 25, a = 5
Sketch a dashed rectangle that contains the
−5,0 , 5,0 , 0, − 4 , and 0,4 , and
points ( ) ( ) ( ) ( )
−8
−8
8
x
365

Homework 11.7
SSM: Intermediate Algebra
27. x
2 − 9y
2 = 9
2 2
x 9y
9
− =
9 9 9
a
2 2
x y
− = 1
9 1
2
= 9, a = 3
x-intercepts: ( − 3,0 ),3,0
( )
2
b = 1, b = 1
Sketch a dashed rectangle that contains the
−3,0 ,3,0,0, − 1, and 0,1 , and then
points ( ) ( ) ( ) ( )
sketch the inclined asymptotes.
−5
4
−4
29. y
2 − x
2 = 4
2 2
2 2
y
y x 4
− =
4 4 4
y x
− = 1
4 4
b
2
= 4, b = 2
y-intercepts: ( 0, − 2,0,2 ) ( )
2
5
x
a = 4, a = 2
Sketch a dashed rectangle that contains the
−2,0 , 2,0 , 0, − 2 , and 0,2 , and
points ( ) ( ) ( ) ( )
then sketch the inclined asymptotes.
y
33.
y-intercepts: ( 0, − 1 ),( 0,1)
2
a = 16, a = 4
Sketch a dashed rectangle that contains the
−4,0 , 4,0 , 0, − 1 , and 0,1 , and
points ( ) ( ) ( ) ( )
then sketch the inclined asymptotes.
−5
4
−4
y
2 2
25x
− 7 y = 175
2 2
25x
7 y 175
− =
175 175 175
a
2
2 2
x y
− = 1
7 25
= 7, a = 7
x-intercepts: ( − 7,0 ),( 7,0)
2
5
b = 25, b = 5
Sketch a dashed rectangle that contains the
x
points ( − 7,0 ),( 7,0 ),( 0, −5 ), and ( 0, − 5)
,
and then sketch the inclined asymptotes.
−8
−8
y
8
8
x
4
−4
4
x
−4
31.
2 2
16y
− x = 16
2 2
16y
x 16
− =
16 16 16
b
2
2 2
y x
− = 1
1 16
= 1, b = 1
366

SSM: Intermediate Algebra Homework 11.7
35.
2 2
x y
+ = 1
64 4
Ellipse
a
2
= 64, a = 8
x-intercepts: ( − 8,0 ),( 8,0)
b
2
= 4, b = 2
y-intercepts: ( 0, − 2,0,2 ) ( )
y
y-intercepts: ( 0, − 9,0,9 ) ( )
y
8
4
x
−8
8
−4
−8
−8
8
−8
37. x
2 − y
2 = 1
39.
2 2
x y
− = 1
1 1
Hyperbola
a
2
= 1, a = 1
x-intercepts: ( − 1,0 ),( 1,0 )
2
8
x
b = 1, b = 1
Sketch a dashed rectangle that contains the
−1,0 , 1,0 , 0, − 1 , and 0,1 , and then
points ( ) ( ) ( ) ( )
sketch the inclined asymptotes.
y
4
−4 4
−4
2 2
81x
+ 49y
= 3969
2 2
81x
49y
3969
+ =
3969 3969 3969
2 2
x y
+ = 1
49 81
Ellipse
a
2
= 49, a = 7
x-intercepts: ( − 7,0 ),( 7,0)
b
2
= 81, b = 9
x
41. x
2 + y
2 = 1
43.
2 2
x y
+ = 1
1 1
Circle
a
2
= 1, a = 1
x-intercepts: ( − 1,0 ),( 1,0 )
b
2
= 1, b = 1
y-intercepts: ( 0, − 1 ),( 0,1)
y
4
−4 4
−4
2 2
9y
− 4x
= 144
2 2
9y
4x
144
− =
144 144 144
2 2
y x
− = 1
16 36
Hyperbola
2
b = 16, b = 4
y-intercepts: ( 0, − 4,0,4 ) ( )
2
x
a = 36, a = 6
Sketch a dashed rectangle that contains the
−6,0 ,6,0,0, − 4, and 0, 4 , and
points ( ) ( ) ( ) ( )
then sketch the inclined asymptotes.
−8
8
−8
y
8
x
367

Homework 11.7
SSM: Intermediate Algebra
45.
2 2
x y
− = 1
25 25
Hyperbola
a
2
= 25, a = 5
x-intercepts: ( − 5,0 ),( 5,0)
2
b = 25, b = 5
Sketch a dashed rectangle that contains the
−5,0 , 5,0 , 0, − 5 , and 0,5 , and
points ( ) ( ) ( ) ( )
then sketch the inclined asymptotes.
−8
8
−8
47. x
2 + y
2 = 16
2 2
y
x y
+ = 1
16 16
Circle
a
2
= 16, a = 4
x-intercepts: ( − 4,0 ),( 4,0)
b
2
= 16, b = 4
y-intercepts: ( 0, − 4,0,4 ) ( )
y
8
x
51.
y
2
−2
−2
2 2
2 2
2 2
2
9x
− 9y
= 81
9x
9y
81
− =
81 81 81
x y
− = 1
9 9
Hyperbola
a
2
= 9, a = 3
x-intercepts: ( − 3,0 ),3,0
( )
2
b = 9, b = 3
Sketch a dashed rectangle that contains the
−3,0 ,3,0,0, − 3, and 0,3 , and
points ( ) ( ) ( ) ( )
then sketch the inclined asymptotes.
y
4
−4 4
−4
x
x
49.
2
−2
−2
2 2
2 2
2 2
2
4x
+ 4y
= 64
4x
4y
64
+ =
64 64 64
x y
+ = 1
16 16
Circle
a
2
= 16, a = 4
x-intercepts: ( − 4,0 ),( 4,0)
x
53. a. i.
x
c
2 2
y
+ = 1
d
2 2
x y
+ = 1
4 16
Ellipse
a
2
= 4, a = 2
x-intercepts: ( − 2,0 ),( 2,0)
b
2
= 16, b = 4
y-intercepts: ( 0, − 4,0,4 ) ( )
y
4
b
2
= 16, b = 4
y-intercepts: ( 0, − 4,0,4 ) ( )
−4 4
−4
x
368

SSM: Intermediate Algebra Homework 11.7
ii.
iii.
x
c
2 2
y
+ = 1
d
2 2
x y
− = 1
4 16
Hyperbola
a
2
= 4, a = 2
x-intercepts: ( − 2,0 ),( 2,0)
2
b = 16, b = 4
Sketch a dashed rectangle that contains
−2,0 , 2,0 , 0, − 4 , and
the points ( ) ( ) ( )
( 0, 4 ) , and then sketch the inclined
asymptotes.
x
c
y
5
−4 4
2 2
−5
y
+ = 1
d
2 2
y x
− = 1
16 4
Hyperbola
b
2
= 16, b = 4
y-intercepts: ( 0, − 4,0,4 ) ( )
2
a = 4, a = 2
Sketch a dashed rectangle that contains
−2,0 , 2,0 , 0, − 4 , and
the points ( ) ( ) ( )
( 0, 4 ) , and then sketch the inclined
asymptotes.
y
5
x
Circle
a
2
= 4, a = 2
x-intercepts: ( − 2,0 ),( 2,0)
b
2
= 4, b = 2
y-intercepts: ( 0, − 2,0,2 ) ( )
4
y
−4 4
−4
b. If c > 0 and d > 0 and c ≠ d , then the
graph is an ellipse.
If c > 0 and d < 0 , then the graph is a
hyperbola with x-intercepts.
If c < 0 and d > 0 , then the graph is a
hyperbola with y-intercepts.
If c = d and c > 0 , then the graph is a
circle.
2 2
x y
55. No. The graph of + = 1 is an ellipse. The
4 81
graph fails the vertical line test.
57. a.
b.
5 4
2
y = − x
2
The square root of a nonnegative number is
a nonnegative real number and the square
root of a negative number is an imaginary
number. Since 5 2 is positive, 5 4
2
− x is
2
nonnegative for real values of y. Thus,
y ≥ 0 for real number values of y.
y
x
iv.
x
c
−4 4
−5
2 2
y
+ = 1
d
2 2
x y
+ = 1
4 4
x
4
−4 4
−4
x
369

Section 11.7 Quiz
SSM: Intermediate Algebra
59. Answers may vary. Possible answer:
2 2
x y
+ = 1
2 2
a b
To keep the ellipses from intersecting, we
increase the value of a and b for each equation.
Some possible equations:
2 2 2 2
x y x y
+ = 1, + = 1,
1 4 4 9
2 2 2 2
x y x y
+ = 1, + = 1,
9 16 16 25
2 2
x y
+ = 1
25 36
8
y
2.
2 2
y x
− = 1
49 9
b
2
= 49, b = 7
y-intercepts: ( 0, − 7,0,7 ) ( )
2
a = 9, a = 3
Sketch a dashed rectangle that contains the
−3,0 ,3,0,0, − 7, and 0,7 , and
points ( ) ( ) ( ) ( )
then sketch the inclined asymptotes.
−8
y
4
−4
8
x
−8
−8
61. x 2 + y 2 = r
2
2 2 2
x y r
+ =
r r r
2 2 2
2 2
x y
+ = 1
2 2
r r
Ellipse
8
2 2 2
x
Since a = b = r , it is also a circle. (Note that
all circles are ellipses just as all squares are
rectangles.)
Section 11.7 Quiz
3.
2 2
4x
− y = 16
2 2
4x
y 16
− =
16 16 16
a
2 2
x y
− = 1
4 16
2
= 4, a = 2
x-intercepts: ( − 2,0 ),( 2,0)
2
b = 16, b = 4
Sketch a dashed rectangle that contains the
−2,0 , 2,0 , 0, − 4 , and 0,4 , and
points ( ) ( ) ( ) ( )
then sketch the inclined asymptotes.
y
5
1.
2 2
x y
+ = 1
9 25
a
2
= 9, a = 3
x-intercepts: ( − 3,0 ),3,0
( )
−4 4
−5
x
b
2
= 25, b = 5
y-intercepts: ( 0, − 5 ),( 0,5)
y
4
−4 4
x
−4
370

SSM: Intermediate Algebra
Section 11.7 Quiz
4.
5.
6.
2 2
16x
+ 3y
= 48
2 2
16x
3y
48
+ =
48 48 48
a
2
2 2
x y
+ = 1
3 16
= 3, a = 3
x-intercepts: ( − 3,0 ),( 3,0)
b
2
= 16, b = 4
y-intercepts: ( 0, − 4,0,4 ) ( )
y
5
−4 4
2 2
−5
x y
− = 1
8 3
a
2
= 8, a = 2 2
x-intercepts: ( − 2 2,0 ),( 2 2,0)
2
b = 3, b = 3
Sketch a dashed rectangle that contains the
points ( −2 2,0 ),( 2 2,0 ),( 0, − 3 ),
and
( 0, 3 ) , and then sketch the inclined
asymptotes.
y
4
−4 4
−4
2 2
4y
− 4x
= 16
2 2
4y
4x
16
− =
16 16 16
b
2
2 2
y x
− = 1
4 4
= 4, b = 2
y-intercepts: ( 0, − 2,0,2 ) ( )
a
2
= 4, a = 2
x
x
7.
Sketch a dashed rectangle that contains the
−2,0 , 2,0 , 0, − 2 , and 0,2 , and
points ( ) ( ) ( ) ( )
then sketch the inclined asymptotes.
y
4
−4 4
2 2
−4
x y
+ = 1
5 14
a
2
= 5, a = 5
x-intercepts: ( − 5,0 ),( 5,0)
b
2
= 14, b = 14
y-intercepts: ( 0, − 14 ),( 0, 14)
y
4
−4 4
−4
+ 9 = 81
8. x
2 y
2
2 2
x 9y
81
+ =
81 81 81
a
2 2
x y
+ = 1
81 9
2
= 81, a = 9
x-intercepts: ( − 9,0 ),( 9,0)
b
2
= 9, b = 3
y-intercepts: ( 0, − 3 ),( 0,3)
−8
y
8
−8
8
x
x
x
371

Homework 11.8
SSM: Intermediate Algebra
2 2
x y
9. No. The graph of − = 1 is a hyperbola.
9 4
The graph fails the vertical line test.
10. Answers will vary. The points ( 0,3 ) and ( 0, − 3)
are y-intercepts. An ellipse with these y-
intercepts has the form
possibilities: let a = 1,2, and 4 .
2 2
x y
+ = 1
1 9
2 2
x y
+ = 1
4 9
2 2
x y
+ = 1
16 9
Homework 11.8
1. x
2 y
2
+ = 25
2 2
4x
+ 25y
= 100
−2
4
−4
y
(−5, 0) (5, 0)
2
2 2
x y
1
2
a + 9
= . Three
The two intersections points ( − 5,0)
and ( 5,0 )
are the solutions to the system.
Solve using elimination:
2 2
−4x
− 4y
= −100
2 2
4x
+ 25y
= 100
2
21y
= 0
y
2
= 0
y = 0
Let y = 0 in
x
2 2
+ 0 = 25
x
2
= 25
x = ± 5
x
2 2
x
+ y = 25 and solve for x.
The solutions are ( − 5,0)
and ( )
5,0 .
3.
5.
2
y = x + 1
y = − x + 3
(−2, 5)
−4
y
8
−2 2
(1, 2)
4
x
The two intersection points ( − 2,5)
and ( 1,2 )
are the solutions of the system.
Solve using substitution.
Substitute
x
2
x
2
2
( x )( x )
x + 1 for y in the second equation.
y = − x + 3
+ 1 = − x + 3
+ x − 2 = 0
+ 2 − 1 = 0
x + 2 = 0 or x − 1 = 0
x = − 2 or x = 1
Let x = − 2 and x = 1 in y = − x + 3 and solve
for y.
y = − − 2 + 3 y = − 1 + 3
( )
= 2 + 3
= 5
( )
= − 1+
3
= 2
2,5
1,2 .
The solutions are ( − ) and ( )
2
y = x − 2
2
y = − x + 6
4
(−2, 2) (2, 2)
−4
−4
y
4
x
The two intersection points ( − 2, 2)
and ( 2, 2 )
are the solutions of the system.
Solve using substitution.
2
Substitute x − 2 for y in the second equation.
372

SSM: Intermediate Algebra Homework 11.8
x
2
y = − x + 6
2 2
− 2 = − x + 6
2
2x
= 8
x
2
= 4
x = ± 2
Let x = − 2 and x = 2 in
for y.
( ) 2
y = −2 − 2
= 4 − 2
= 2
( ) 2
y = 2 − 2
= 4 − 2
= 2
2, 2
2
y = x − 2 and solve
The solutions are ( − ) and ( )
7. x
2 y
2
x
+ = 49
2 2
+ y = 16
y
2, 2 .
x
2
2 2
2
2 2
( x )
2
x
+ y = 25
+ − − 1 = 25
x + x + 2x
+ 1 = 25
2x
+ 2x
− 24 = 0
x
2
+ x − 12 = 0
( x )( x )
+ 4 − 3 = 0
x + 4 = 0 or x − 3 = 0
x = − 4 or x = 3
Let x = − 4 and x = 3 in y = −x
− 1 and solve
for y.
y = − −4 −1
y = − 3 −1
( )
= 4 −1
= 3
( )
= −3 −1
= −4
4,3
The solutions are ( − ) and ( 3, 4)
− .
−8
8
−8
8
x
11. y
2 x
2 2
− = 16
y + x = 4
4
y
(0, 4)
The graphs do not intersect. The solution set is
the empty set, ∅ .
Solve using elimination.
x
2 2
+ y = 49
2 2
−x
− y = −16
0 = 33 False
There is no solution.
9. x
2 + y
2 = 25
(−4, 3)
−4
y = −x
−1
4
−4
y
4
x
(3, −4)
The two intersection points ( − 4,3)
and ( 3, − 4)
are the solutions of the system.
Solve using substitution.
Substitute −x
− 1 for y in the first equation.
−4
(−3, −5)
−2
4
(3, −5)
The three intersection points ( −3, − 5)
, ( )
and ( 3, − 5)
are the solutions to the system.
Solve using elimination.
y
2 2
2
− x = 16
2
y + x = 4
y
+ y = 20
y
2
+ y − 20 = 0
( y )( y )
+ 5 − 4 = 0
y + 5 = 0 or y − 4 = 0
y = − 5 or y = 4
x
2
0, 4 ,
Let y = − 5 and y = 4 in y + x = 4 and solve
for x.
2
2
− 5 + x = 4 4 + x = 4
x
2
= 9
x = ± 3
x
2
= 0
x = 0
3, 5
The solutions are ( − − ) , ( 3, − 5)
, and ( )
0, 4 .
373

Homework 11.8
SSM: Intermediate Algebra
13.
2 2
25x
− 9 y = 225
2 2
4x
+ 9y
= 36
4
y
(−3, 0) (3, 0)
−4
4
x
−4
The two intersection points ( − 3,0)
and ( 3,0 )
are the solutions to the system.
Solve using elimination.
2 2
25x
− 9 y = 225
2 2
4x
+ 9y
= 36
2
29x
= 261
x
2
= 9
x = ± 3
Let x = − 3 and x = 3 in
solve for y.
( ) 2 2
y
4 − 3 + 9 = 36
2
36 + 9y
= 36
2
9y
= 0
y
2
= 0
y = 0
2 2
4x
+ 9y
= 36 and
( ) 2 2
y
4 3 + 9 = 36
2
36 + 9y
= 36
2
9y
= 0
The solutions are ( − 3,0)
and ( )
15. 9x
2 + y
2 = 9
(−1, 0)
−4
y = 3x
+ 3
4
y
−4
(0, 3)
4
x
y
2
= 0
y = 0
3,0 .
The two intersection points ( − 1,0 ) and ( 0,3 )
are the solutions to the system.
Solve using substitution.
Substitute 3x + 3 for y in the first equation.
17.
2
2 2
2 2
9x
+ y = 9
( x )
9x
+ 3 + 3 = 9
9x + 9x + 18x
+ 9 = 9
2
2
18x
+ 18x
= 0
( )
18x x + 1 = 0
18x
= 0 or x + 1 = 0
x = 0 or x = −1
Let x = 0 and x = − 1 in y = 3x
+ 3 and solve
for y.
y = 3 0 + 3 y = 3 − 1 + 3
( )
= 0 + 3
= 3
( )
= − 3 + 3
= 0
The solutions are ( 0,3 ) and ( 1,0 )
2 2
4x
+ 9y
= 36
2 2
16x
+ 25y
= 225
−4
4
y
−4
4
x
− .
The graphs do not intersect. The solution set is
the empty set, ∅ .
Solve using elimination.
2 2
−16x
− 36y
= −144
2 2
16x
+ 25y
= 225
2
− 11y
= 81
2 81
y = −
11
2
Since y cannot be negative (in the real number
system), there is no solution.
19. y = x − 3
y = −x
−1
−2
4
y
−4
2
(1, −2)
x
374

SSM: Intermediate Algebra Homework 11.8
The intersection point ( 1, − 2)
is the solution to
the system.
Solve using substitution.
Substitute x − 3 for y in the second equation.
y = −x
−1
x − 3 = −x
−1
solve for y.
y = −
( ) 2
= 3 − 2
= 1
3 − 2
y =
( ) 2
= 3 − 2
= 1
The solutions are ( 3,1)
3 − 2
3,1 .
− and ( )
21.
x
2
( x ) = ( − x + 2)
( x )( x )
x = − x + 2
− 5x
+ 4 = 0
− 4 − 1 = 0
2 2
2
x = x − 4x
+ 4
x − 4 = 0 or x − 1 = 0
x = 4 or x = 1
Let x = 4 and x = 1 in y = x − 3 and solve for
y.
y =
4 − 3
= 2 − 3
y =
1 − 3
= 1−
3
= −1
= −2
Check each result in y = −x
− 1 .
( )
− 1 = − 4 −1
− 1 = −5 False
− 2 = −1−1
− 2 = −2 True
1, − 2 .
The only solution is ( )
2
2
y = 2x
− 5
y = x − 2
(− 3 , 1)
−4
4
y
( 3 , 1)
4
x
The two intersection points ( − 3,1)
and ( 3,1 )
are the solutions to the system.
Solve using substitution.
2
Substitute 2x − 5 for y in the second equation.
2
y = x − 2
2 2
2x
− 5 = x − 2
x
2
= 3
x = ±
3
Let x = − 3 and x = 3 in
2
y = x − 2 and
23.
2 2
25y
− 4x
= 100
2 2
9x
+ y = 9
(−0.74, 2.02)
−4
4
(−0.74, −2.02)
y
(0.74, 2.02)
4
(0.74, −2.02)
The four intersection points ( −0.74, − 2.02)
,
( − 0.74, 2.02)
, ( 0.74, − 2.02)
, and ( 0.74, 2.02 )
are the solutions of the system.
Solve using substitution.
2 2
9x
+ y = 9
y
Substitute
2 2
( x )
= 9 − 9x
2
9 − 9x for
2 2
25y
− 4x
= 100
2 2
25 9 − 9 − 4x
= 100
2 2
225 − 225x
− 4x
= 100
2
− 229x
= −125
x
2
=
x
2
y in the first equation.
125
229
125
x = ± ≈ ± 0.74
229
Let x = − 0.74 and x = 0.74 in
solve for y.
( ) 2 2
y
9 − 0.74 + = 9
y
( ) 2 2
y
9 0.74 + = 9
y
2
2
= 4.07
y = ± 4.07 ≈ ± 2.02
= 4.07
y = ± 4.07 ≈ ± 2.02
2 2
9x
+ y = 9 and
The solutions are ( −0.74, − 2.02)
, ( 0.74,2.02)
( 0.74, − 2.02)
, and ( 0.74, 2.02 ) .
− ,
375

Homework 11.8
SSM: Intermediate Algebra
25.
27.
2 2
25x
+ 9y
= 225
x
2 2
+ y = 16
(−2.23, 3.31) 4
(2.25, 3.31)
−5
(−2.25, −3.31) −4
y
2
−2
5
x
(2.25, −3.31)
The four intersection points ( −2.25, − 3.31)
,
( − 2.25,3.31)
, ( 2.25, − 3.31)
, and ( 2.25,3.31 )
are the solutions to the system.
Solve using elimination.
2 2
25x
+ 9y
= 225
2 2
−9x
− 9y
= −144
2
16x
= 81
2 81
x =
16
9
x = ±
4
x = ± 2.25
Let x = − 2.25 and x = 2.25 in
solve for y.
( ) 2 2
y
− 2.25 + = 16
y
( ) 2 2
y
2
2
= 10.9375
y = ± 3.31
2.25 + = 16
y
= 10.9375
y = ± 3.31
x
2 2
+ y = 16 and
The solutions are ( −2.25, − 3.31)
, ( 2.25,3.31)
( 2.25, − 3.31)
, and ( 2.25,3.31 ) .
2 2
9x
+ y = 85
2 2
2x
− 3y
= 6
Solve using elimination.
2 2
27x
+ 3y
= 255
2 2
2x
− 3y
= 6
2
29x
= 261
x
2
= 9
x = ± 3
Let x = − 3 and x = 3 in
2 2
− ,
9x
+ y = 85 and
29.
31.
solve for y.
( ) 2 2
y
9 − 3 + = 85
2
81+ y = 85
y
2
= 4
y = ± 2
( ) 2 2
y
9 3 + = 85
2
81+ y = 85
y
2
= 4
y = ± 2
The four solutions are ( −3, − 2)
, ( 3,2)
( 3, − 2)
, and ( )
equations.
2
2
y = 2x − 5x
−11
y = x − 3x
+ 4
Solve using substitution.
2
− ,
3,2 . Each result satisfies both
Substitute 2x
− 5x
− 11 for y in the second
equation.
2
y = x − 3x
+ 4
2 2
2x − 5x − 11 = x − 3x
+ 4
x
2
− 2x
− 15 = 0
( x )( x )
− 5 + 3 = 0
x − 5 = 0 or x + 3 = 0
x = 5 or x = −3
Let x = 5 and x = − 3 in
solve for y.
2
( ) ( )
y = 5 − 3 5 + 4
= 25 − 15 + 4
= 14
2
y = x − 3x
+ 4 and
2
( ) ( )
y = −3 − 3 − 3 + 4
= 9 + 9 + 4
= 22
The solutions are ( 5,14 ) and ( − 3,22)
result satisfies both equations.
2
y = x − 3x
+ 2
. Each
y = 2x
− 4
Solve using substitution.
Substitute 2x − 4 for y in the first equation.
x
2
( x )( x )
2
y = x − 3x
+ 2
2x − 4 = x − 3x
+ 2
− 5x
+ 6 = 0
− 3 − 2 = 0
x − 3 = 0 or x − 2 = 0
2
x = 3 or x = 2
Let x = 3 and x = 2 in y = 2x
− 4 and solve for
y.
376

SSM: Intermediate Algebra
Section 11.8 Quiz
33.
( )
y = 2 3 − 4
= 6 − 4
= 2
( )
y = 2 2 − 4
= 4 − 4
= 0
The solutions are ( 3,2 ) and ( )
satisfies both equations.
x
2 2
+ y = 25
2 2
4x
− 25y
= 100
2 2
4x
+ 25y
= 100
y
2,0 . Each result
Section 11.8 Quiz
1. x
2 y
2
9 + = 81
x
2 2
+ y = 9
y
8
(−3,0) (3,0)
−8
8
−8
x
4
(−5,0) (5,0)
−4
The two intersection points for all three graphs
5,0 5,0 . These are the solutions to
are ( − ) and ( )
the system. Each result satisfies all three
equations.
35. Answers may vary. One possible answer:
2 2
x + y = 16
x
2 2
− y = 16
37. 2x
2 + cy
2 = 82
2
y = x + dx + 5
Substitute ( 1,4 ) into each equation.
( ) c( )
2 2
2 1 + 4 = 82
2 + 16c
= 82
16c
= 80
c = 5
x
2
( ) d ( )
4 = 1 + 1 + 5
4 = 1+ d + 5
d = −2
39. Written response. Answers may vary.
2.
The two intersection points ( − 3,0)
and ( 3,0 )
are the solutions to the system.
Solve using elimination.
2 2
9x
+ y = 81
2 2
−9x
− 9y
= −81
2
− 8y
= 0
y
2
= 0
y = 0
Let y = 0 in
x
x
2 2
+ y = 9
2 2
+ 0 = 9
x
2
= 9
x = ± 3
x
2 2
+ y = 9 and solve for x.
The solutions are ( − 3,0)
and ( )
2
y = x − 2
y = − 2x
+ 1
(−3,7)
y
8
3,0 .
−4
−2 2 4
(1, −1)
x
The two intersection points ( 1, − 1)
and ( − 3,7)
are the solutions of the system.
Solve using substitution.
2
Substitute x − 2 for y in the second equation.
377

Section 11.8 Quiz
SSM: Intermediate Algebra
3.
x
2
x
2
( x )( x )
y = − 2x
+ 1
− 2 = − 2x
+ 1
+ 2x
− 3 = 0
+ 3 − 1 = 0
x + 3 = 0 or x − 1 = 0
x = − 3 or x = 1
Let x = − 3 and x = 1 in y = − 2x
+ 1 and solve
for y.
y = −2 − 3 + 1 y = − 2 1 + 1
( )
= 6 + 1
= 7
( )
= − 2 + 1
= −1
3,7
The solutions are ( − ) and ( 1, 1)
2
y = x + 3
2
y = x − 6x
+ 9
y
8
(1, 4)
− .
4.
2 2
25x
− 4y
= 100
2 2
9x
+ y = 9
−4
4
−4
y
4
The graphs do not intersect. There are no
solutions to the system.
Solve using elimination.
2 2
225x
− 36y
= 900
2 2
−225x
− 25y
= −225
2
− 61y
= 675
x
2 675
y = −
61
2
Since y must be nonnegative (in the real
number system), there are no solutions to the
system. The solution set is the empty set, ∅ .
−4
−2 2
4
x
The intersection point ( 1,4 ) is the solution to the
system.
Solve using substitution.
Substitute
2
x + 3 for y in the second equation.
2
y = x − 6x
+ 9
2 2
x + 3 = x − 6x
+ 9
6x
− 6 = 0
x − 1 = 0
x = 1
Let x = 1 in
( ) 2
y = 1 + 3
= 1+
3
= 4
The solution is ( )
2
y = x + 3 and solve for y.
1,4 .
5.
x
x
2 2
− y = 16
2 2
+ y = 16
( x 4)
y = +
y
8
x
(−4, 0) 8
−8
2
The intersection point of all three graphs is
− 4,0 . This is the solution of the system.
( )
6. Answers may vary. One possible answer:
2 2
x + y = 25
2
y = x + 5
378