Inverse Variation
ALGEBRA 1 LESSON 12-1
(For help, go to Lesson 5-5.)
Suppose y varies directly with x. Find each constant of variation.
1. y = 5x
2. y = –7x
3. 3y = x
4. 0.25y = x
Write an equation of the direct variation that includes the given point.
5. (2, 4)
6. (3, 1.5)
7. (–4, 1)
12-1
8. (–5, –2)
Inverse Variation
ALGEBRA 1 LESSON 12-1
Solutions
1. y = 5x; constant of variation = 5
2. y = –7x; constant of variation = –7
3.
3y = x
1
1
•
3y
=
•x
3
3
y = 1 x; constant of variation = 1
3
3
4. 0.25y = x
1
y=x
4
4 •1 y = 4 • x
4
y = 4x; constant of variation = 4
12-1
Inverse Variation
ALGEBRA 1 LESSON 12-1
Solutions (continued)
5. Point (2, 4) in y = kx: 4 = k(2), so k = 2 and y = 2x.
6. Point (3, 1.5) in y = kx: 1.5 = k(3), so k = 0.5 and y = 0.5x.
7. Point (–4, 1) in y = kx: 1 = k(–4), so k = – 1 and y = – 1 x.
4
4
8. Point (–5, –2) in y = kx: –2 = k(–5), so k = 2 and y = 2 x.
5
12-1
5
Inverse Variation
ALGEBRA 1 LESSON 12-1
Suppose y varies inversely with x, and y = 9 when x = 85.
Write an equation for the inverse variation.
xy = k
(8)(9) = k
Use the general form for an inverse variation.
Substitute 8 for x and 9 for y.
72 = k
Multiply to solve for k.
xy = 72
Write an equation. Substitute 72 for k in xy = k.
72
The equation of the inverse variation is xy = 72 or y = x .
12-1
Inverse Variation
ALGEBRA 1 LESSON 12-1
The points (5, 6) and (3, y) are two points on the graph of an
inverse variation. Find the missing value.
x1 • y1 = x2 • y2
5(6) = 3y2
Use the equation x1 • y1 = x2 • y2 since you know
coordinates, but not the constant of variation.
Substitute 5 for x1, 6 for y1, and 3 for x2.
30 = 3y2
Simplify.
10 = y2
Solve for y2.
The missing value is 10. The point (3, 10) is on the graph of the
inverse variation that includes the point (5, 6).
12-1
Inverse Variation
ALGEBRA 1 LESSON 12-1
Jeff weighs 130 pounds and is 5 ft from the lever’s fulcrum. If
Tracy weighs 93 pounds, how far from the fulcrum should she sit in
order to balance the lever?
Relate: A weight of 130 lb is 5 ft from the fulcrum.
A weight of 93 lb is x ft from the fulcrum.
Weight and distance vary inversely.
Define: Let weight1 = 130 lb
Let weight2 = 93 lb
Let distance1 = 5 ft
Let distance2 = x ft
12-1
Inverse Variation
ALGEBRA 1 LESSON 12-1
(continued)
Write:
weight1 • distance1 = weight2 • distance2
130
•
5
=
93
•
x
Substitute.
650 = 93x
Simplify.
650
93 = x
Solve for x.
6.99 = x
Simplify.
Tracy should sit 6.99, or 7 ft, from the fulcrum to balance the lever.
12-1
Inverse Variation
ALGEBRA 1 LESSON 12-1
Decide if each data set represents a direct variation or an
inverse variation. Then write an equation to model the data.
a.
x
3
5
10
y
10
6
3
The values of y seem to vary
inversely with the values of x.
Check each product xy.
xy: 3(10) = 30
5(6) = 30
The product of xy is the same for all pairs of data.
So, this is an inverse variation, and k = 30.
The equation is xy = 30.
12-1
10(3) = 30
Inverse Variation
ALGEBRA 1 LESSON 12-1
(continued)
b.
y
x
x
2
4
8
y
3
6
12
3
= 1.5
2
The values of y seem to vary directly
with the values of x.
y
Check each ratio x .
6
= 1.5
4
y
The ratio x is the same for all pairs of data.
So, this is a direct variation, and k = 1.5.
The equation is y = 1.5x.
12-1
12
= 1.5
8
Inverse Variation
ALGEBRA 1 LESSON 12-1
Explain whether each situation represents a direct variation
or an inverse variation.
a. You buy several souvenirs for $10 each.
The cost per souvenir times the number of souvenirs equals the total
cost of the souvenirs.
cost
Since the ratio souvenirs is constant at $10 each, this is a direct variation.
b. The cost of a $25 birthday present is split among several friends.
The cost per person times the number of people equals the total
cost of the gift.
Since the total cost is a constant product of $25, this is an inverse variation.
12-1
Inverse Variation
ALGEBRA 1 LESSON 12-1
pages 640–642 Exercises
12. 6
24. direct variation; y = 0.5x
1. xy = 18
13. 7
25. inverse variation; xy = 60
2. xy = 2
14. 3
26. inverse variation; xy = 72
3. xy = 56
15. 130
4. xy = 1.5
16. 12
27. Direct variation; the ratio cost
pound
is constant at $1.79.
5. xy = 24
17. 96
6. xy = 7.7
18. 3125
7. xy = 2
19. 2
8. xy = 0.5
20. 1
9. xy = 0.06
21. 20
10. 8
22. 3 h
11. 15
23. 13.3 mi/h
28. Inverse variation; the total
number of slices is constant at 8.
6
29. Inverse variation; the product
of the length and width remains
constant with an area of
24 square units.
30. 32; xy = 32
31. 1.1; rt = 1.1
12-1
32. 2.5; xy = 2.5
Inverse Variation
ALGEBRA 1 LESSON 12-1
41. direct variation; y = 0.4x; 8
33. 1; ab = 1
42. direct variation; y = 70x; 0.9
34. 15.6; pq = 15.6
43. inverse variation; xy = 48; 0.5
35. 375; xy = 375
44. a. greater
b. greater
c. less
36. Direct variation; the ratio
of the perimeter to the side
length is constant at 3.
45. a. 16 h; 10 h; 8 h; 4 h
b. hr worked, rate of pay
c. rt = 80
37. Inverse variation; the product
of the rate and the time
is always 150.
46. Check students’ work.
38. Direct variation; the ratio of
the circumference to the
radius is constant at 2 .
47. 10.2 L
48. p: y = 0.5x; q: xy = 8
39. 121 ft
49. a. y is doubled.
b. y is halved.
40. 2.4 days
12-1
Inverse Variation
ALGEBRA 1 LESSON 12-1
2
50. 4; s 1 d = 1 sd2 = k, so s = 4 k 2 .
2
4
d
55. [4] a.
51. a. x4y = k
x4y
b. z = k
52. C
53. F
b. The variables speed and
time are inversely related.
54. [2] Direct variation: y = kx,
c. 2 1 h
2
[3] one computational error
[2] one part missing
[1] two parts missing
10 = 5k, k = 2. So when x = 8,
y = 2 • 8 = 16. Inverse variation:
xy = k, 5 • 10 = 50,
So when x = 8, y = 50, or 6.25.
8
56. 8
17
[1] no work shown OR
one computational error
15
57. 17
12-1
Inverse Variation
ALGEBRA 1 LESSON 12-1
58. 15
68. (3a – 1)(a + 4)
59. 8
69. (5x + 2)(3x + 7)
17
17
70. (2y – 3)(y + 8)
60. 8
15
61. 15
8
62. 5.1
63. 12.0
64. 12.0
65. 10.6
66. 2.2
67. 2.5
12-1
Inverse Variation
ALGEBRA 1 LESSON 12-1
1. The points (5, 1) and (10, y) are on the graph of an inverse variation. Find y.
0.5
2. Find the constant of variation k for the inverse variation where
a = 2.5 when b = 7.
17.5
y
3. Write an equation to model the data x
1
and complete the table.
1
2
1
xy = 3
3
6
3
1
6
1
9
1
18
4. Tell whether each situation represents a direct variation or an inverse variation.
a. You buy several notebooks for $3 each.
direct variation
b. The $45 cost of a dinner at a restaurant is split among several people.
Inverse variation
12-1
Graphing Rational Functions
ALGEBRA 1 LESSON 12-2
(For help, go to Lessons 5-3, 8-7, and 10-1.)
Evaluate each function for x = –2, 0, 3.
1. ƒ(x) = x – 8
2. g(x) = x2 + 4
3. y = 3x
5. g(x) = –x2
6. y = 2x
Graph each function.
4. ƒ(x) = 2x + 1
12-2
Graphing Rational Functions
ALGEBRA 1 LESSON 12-2
Solutions
1. ƒ(x) = x – 8 for x = –2, 0, 3:
ƒ(–2) = –2 – 8 = –10
ƒ(0) = 0 – 8 = –8
ƒ(3) = 3 – 8 = –5
2. g(x) = x2 + 4 for x = –2, 0, 3:
g(–2) = (–2)2 + 4 = 4 + 4 = 8
g(0) = 02 + 4 = 0 + 4 = 4
g(3) = 32 + 4 = 9 + 4 = 13
3. y = 3x for x = –2, 0, 3:
1
1
1
y = 3–2 = 32 = 3 • 3 = 9
y = 30 = 1
y = 33 = 3 • 3 • 3 = 9 • 3 = 27
12-2
Graphing Rational Functions
ALGEBRA 1 LESSON 12-2
Solutions (continued)
4. ƒ(x) = 2x + 1
5. g(x) = – x2
6. y = 2x
12-2
Graphing Rational Functions
ALGEBRA 1 LESSON 12-2
The function t = 70
r models the time it will take you to travel
70 miles at different rates of speed. Graph this function.
Step 1: Make a table of values.
r
t
15
20
30
40
60
4.67
3.5
2.33
1.75
1.17
Step 2: Plot the points.
12-2
Graphing Rational Functions
ALGEBRA 1 LESSON 12-2
4
Identify the vertical asymptote of y = x – 3 . Then graph the
function.
Step 1: Find the vertical asymptote.
x–3=0
The numerator and denominator have no common
factors. Find any value(s) where the denominator
equals zero.
x=3
This is the equation of the vertical asymptote.
12-2
Graphing Rational Functions
ALGEBRA 1 LESSON 12-2
(continued)
Step 2: Make a table of values.
Use values of x near 3,
the asymptote.
x
y
2
1
0
–1
4
5
6
–4
–2
4
–3
–1
4
2
Step 3: Graph the function.
4
3
12-2
Graphing Rational Functions
ALGEBRA 1 LESSON 12-2
Identify the asymptotes of y =
4
+ 3. Then graph the
x+4
function.
Step 1: From the form of the function, you can see that there is a
vertical asymptote at x = –4 and a horizontal asymptote at
y = 3. Sketch the asymptotes.
x
y
Step 2: Make a table of values
using values of x near –4.
12-2
–10
–8
–6
–5
–3
–2
–1
2
2 13
2
1
–1
7
5
4 13
3 2
3
Graphing Rational Functions
ALGEBRA 1 LESSON 12-2
(continued)
Step 3: Graph the function.
12-2
Graphing Rational Functions
ALGEBRA 1 LESSON 12-2
Describe the graph of each function.
x
a. y = 6
1
The graph is a line with slope 6 and y-intercept 0.
b. y = 6x
The graph is of exponential growth.
c. y = 6x2
The graph is a parabola with axis of symmetry at x = 0.
d. y = | x – 6|
The graph is an absolute value function with a vertex at (6, 0).
12-2
Graphing Rational Functions
ALGEBRA 1 LESSON 12-2
(continued)
6
e. y = x
The graph is a rational function with vertical asymptote at x = 0 and
horizontal asymptote at y = 0.
f. y = x – 6
The graph is the radical function y =
x shifted right 6 units.
g. y = 6x
The graph is a line with slope 6 and y-intercept 0.
12-2
Graphing Rational Functions
ALGEBRA 1 LESSON 12-2
pages 648–650 Exercises
1.
5.
3.
6. 0
2.
7. 2
4.
8. –2
9. 2
10. x = 2, y = 0
11. x = –1, y = 0
12. x = 1, y = –1
13. x = 0, y = 2
12-2
Graphing Rational Functions
ALGEBRA 1 LESSON 12-2
14. x = 0;
16. x = –1;
18. x = –4;
17. x = 5;
19. x = –4;
15. x = 0;
12-2
Graphing Rational Functions
ALGEBRA 1 LESSON 12-2
20. x = 0, y = –5;
22. x = 0, y = –6;
24. x = 3, y = –5;
21. x = 0, y = 5;
23. x = –1, y = 4;
25. x = 1, y = –2;
12-2
Graphing Rational Functions
ALGEBRA 1 LESSON 12-2
26. line with slope 4, y-int. 1
36. moves graph 3 units to the right
27. absolute value function with vertex (4, 0)
37. lowers graph 15 units
28. exponential decay
38. moves graph 12 units left
29. line with slope 1 , y-int. 0
39. moves the graph up 12 units
4
30. rational function, with asymptotes
x = 0, y = 1
31. radical function; y = x
shifted right 4, up 1
32. parabola with axis of
symmetry x = 0
33. rational function with
asymptotes x = –4, y = –1
34. parabola with axis of symmetry x = – 1
4
35. moves graph 1 unit to the left
12-2
40. moves the graph left 3 units
41. moves the graph down 2 units
42. moves the graph 3 units left
and 2 units down
Graphing Rational Functions
ALGEBRA 1 LESSON 12-2
43. x = 0, y = 0;
45. x = –4, y = 0;
47. x = –1, y = 4;
46. x = 0, y = 1;
48. x = –1, y = –3;
44. x = 0, y = 0;
12-2
Graphing Rational Functions
ALGEBRA 1 LESSON 12-2
49. x = 1, y = 3;
51. x = 3, y = –2;
1
50. x = –5, y = 1;
1
52. Answers may vary. Sample: ƒ(x) = + 3, g(x) =
x
x
53. 17.8 lumens; 1.97 lumens
54. a.
b. x = 0, y = 0; x = 0, y = 0
c. y is any real number except 0; y > 0.
12-2
Graphing Rational Functions
ALGEBRA 1 LESSON 12-2
55. a.
57.
d>
– 40
b. 16; 1600; 160,000
c. The signal is extremely strong
when you are in the immediate
vicinity of a transmitter and it will
interfere with the other station.
58.
56. The graph of y = 3 and y = – 3 are both composed of
x
x
two curves with asymptotes x = 0 and y = 0. The graph of
y = – 3 is a reflection of the graph of y = 3 over the y-axis.
x
x
12-2
Graphing Rational Functions
ALGEBRA 1 LESSON 12-2
59.
62.
60.
No; ƒ(x) = (x + 2)(x + 1) is equivalent to
x+2
g(x) = x + 1 for all values except x = –2.
63. C
64. I
61. a. x = –3, y = –2
1
b. y =
–2
x+3
65. A
66. C
12-2
Graphing Rational Functions
ALGEBRA 1 LESSON 12-2
4
+5
x–1
is a translation of ƒ(x) = 4
x
67. [2] a. The graph of g(x) =
5 units up and 1 unit right.
b. x = 1 and y = 5
[1] one part answered correctly
68. xy = 21
69. xy = 16
70. xy = 22
71. xy = 21.08
72. 0
73. 1
74. 2
12-2
75. 3(d – 6)(d + 6)
76. 2(m – 12)(m + 5)
77. (t 2 + 3)(t – 1)
Graphing Rational Functions
ALGEBRA 1 LESSON 12-2
3
1. Identify the vertical asymptote of y = x + 2 . Then graph the function.
1
2. Identify the vertical and horizontal asymptotes of y = x – 2 – 3.
Then graph the function.
3. Describe the graph of each function.
a. y =
x+3
b. y = x
3
c. y = 3 + 2
x
d. y = 3x2
12-2
Graphing Rational Functions
ALGEBRA 1 LESSON 12-2
3
1. Identify the vertical asymptote of y = x + 2 . Then graph the function.
x = –2
1
2. Identify the vertical and horizontal asymptotes of y = x – 2 – 3.
Then graph the function.
x = 2; y = –3
12-2
Graphing Rational Functions
ALGEBRA 1 LESSON 12-2
3. Describe the graph of each function.
a. y =
x+3
radical function y =
x shifted left 3 units
b. y = x
3
1
line with slope 3 and y-intercept 0
c. y = 3 + 2
x
rational function with vertical asymptote at x = 0 and
horizontal asymptote at y = 2
d. y = 3x2
parabola with axis of symmetry at x = 0.
12-2
Simplifying Rational Expressions
ALGEBRA 1 LESSON 12-3
(For help, go to Lessons 9-5 and 9-6.)
Write each fraction in simplest form.
1.
8
2
15
2. –
24
3.
25
35
Factor each quadratic expression.
4. x2 + x – 12
5. x2 + 6x + 8
6. x2 – 2x – 15
7. x2 + 8x + 16
8. x2 – x – 12
9. x2 – 7x + 12
12-3
Simplifying Rational Expressions
ALGEBRA 1 LESSON 12-3
Solutions
8
15
1. 2 = 8 ÷ 2 = 4
25
5•5
3•5
5
2. – 24 = – 3 • 8 = – 8
5
3. 35 = 5 • 7 = 7
4. Factors of –12 with a sum of 1: 4 and –3.
x2 + x – 12 = (x + 4)(x – 3)
5. Factors of 8 with a sum of 6: 2 and 4.
x2 + 6x + 8 = (x + 2)(x + 4)
6. Factors of –15 with a sum of –2: 3 and –5.
x2 – 2x – 15 = (x + 3)(x – 5)
12-3
Simplifying Rational Expressions
ALGEBRA 1 LESSON 12-3
Solutions (continued)
7. Factors of 16 with a sum of 8: 4 and 4.
x2 + 8x + 16 = (x + 4)(x + 4) or (x + 4)2
8. Factors of –12 with a sum of –1: 3 and –4.
x2 – x – 12 = (x + 3)(x – 4)
9. Factors of 12 with a sum of –7: –3 and –4.
x2 – 7x + 12 = (x – 3)(x – 4)
12-3
Simplifying Rational Expressions
ALGEBRA 1 LESSON 12-3
Simplify 3x + 9 .
x+3
Factor the numerator. The denominator
cannot be factored.
3x + 9
3(x + 3)
=
x+3
x+3
= 3(x + 3)
1 x+3
=3
1
Divide out the common factor x + 3.
Simplify.
12-3
Simplifying Rational Expressions
ALGEBRA 1 LESSON 12-3
Simplify
x2
4x – 20 .
x2 – 9x + 20
4x – 20
4(x – 5)
=
– 9x + 20
(x – 4) (x – 5)
1
4(x
–
5)
=
(x – 4) (x – 5) 1
=
4
x–4
Factor the numerator and the denominator.
Divide out the common factor x – 5.
Simplify.
12-3
Simplifying Rational Expressions
ALGEBRA 1 LESSON 12-3
Simplify 3x – 272 .
81 – x
3x – 27
81 – x2
=
3(x – 9)
(9 – x) (9 + x)
3(x – 9)
= – 1 (x – 9) (9 + x)
Factor the numerator and the denominator.
Factor –1 from 9 – x.
=
1
3(x – 9)
– 1 (x – 9) (9 + x)
1
Divide out the common factor x – 9.
=–
3
9+x
Simplify.
12-3
Simplifying Rational Expressions
ALGEBRA 1 LESSON 12-3
The baking time for bread depends, in part, on its size and
shape. A good approximation for the baking time, in minutes, of a
60 • volume
30rh
cylindrical loaf is surface
,
or
, where the radius r and the length
area
r+h
h of the baked loaf are in inches. Find the baking time for a loaf that is
8 inches long and has a radius of 3 inches. Round your answer to the
nearest minute.
30rh
30 (3) (8)
=
r+h
3+8
=
Substitute 8 for r and 3 for h.
720
11
Simplify.
65
Round to the nearest whole number.
The baking time is approximately 65 minutes.
12-3
Simplifying Rational Expressions
ALGEBRA 1 LESSON 12-3
pages 654–656 Exercises
1. 2a + 3
4
2. 1
7x
3. 1
3
4. 1
2
20. –
14. b + 3
24. 13 min
1
m–2
5. 3x
15.
6. x + 2
16. –1
7. 2
17.
2
b+4
9. 1
m–7
18. –2
x2
3
8.
1
v+5
21. – 1
w–4
w
w–7
11. a + 1
5
12. m + 3
m+2
13. c – 4
c+3
10.
–4
t+1
19. – 1
2
12-3
22. 36 min
23. 13 min
2r – 1
r+5
7z + 2
26.
z–1
5t – 4
27.
3t – 1
4a2
28. 2a – 1
25.
29. 3(z + 4)
z3
Simplifying Rational Expressions
ALGEBRA 1 LESSON 12-3
30. 2s + 1
38.
31. – 2a + 1
39.
s2
a+3
32. 4 + 3m
m–7
33. –c(3c + 5)
5c + 4
40.
41.
34. Answers may vary. Sample:
3
(x – 2)(x + 3)
35. a. i. 2b + 4h
bh
42.
43.
ii. 2h + 2r
rh
b. 4 ; 4
9 9
36. The student canceled terms
instead of factors.
2
37. –3 is not in the domain of x – 9 .
x+3
44.
5w
5w + 6
1
4
3y
4(y + 4)
t+3
3(t + 2)
m–n
m + 10n
a – 3b
a + 4b
6v – 7w
3v – 2w
45. sometimes
46. sometimes
47. never
48. C
12-3
49. I
50. B
51. D
52. C
53. [2] The student put the
4 in the numerator
rather than in
the denominator.
x–5
=
4x – 20
x–5 = 1
4(x – 5)
4
[1] no explanation OR
incorrectly simplified
expression
Simplifying Rational Expressions
ALGEBRA 1 LESSON 12-3
54. vertical asymptote: x = 0;
horizontal asymptote: y = 2;
56. vertical asymptote: x = 0;
horizontal asymptote: y = –4;
57. 10
55. vertical asymptote: x = 4;
horizontal asymptote: y = 0;
2
58. a2b3c4 b
59. 2
60.
2
2
5m2
61. y = x2, y = –2x2, y = 3x2
62. y = 1 x2, y = 1 x2, y = 2 x2
4
3
5
63. y = 0.5x2, y = 2x2, y = –4x2
64. y = –x2, y = 2.3x2, y = –3.8x2
12-3
Simplifying Rational Expressions
ALGEBRA 1 LESSON 12-3
Simplify each expression.
1. 6 – 2x
x–3
–2
4.
6x2 – x – 12
8x2 – 10x – 3
3x + 4
4x + 1
2.
x2 + 8x
x2 – 64
3.
x–6
36 – x2
–1
x+6
x
x–8
5. 4x2 + x
x3
4x + 1
x2
12-3
Multiplying and Dividing Rational Expressions
ALGEBRA 1 LESSON 12-4
(For help, go to Lessons 8-3 and 9-6.)
Simplify each expression.
1. r 2 • r 8
2. b3 • b4
3. c7 ÷ c2
4. 3x4 • 2x5
5. 5n2 • n2
6. 15a3 (–3a2)
Factor each polynomial.
7. 2c2 + 15c + 7
8. 15t2 – 26t + 11
12-4
9. 2q2 + 11q + 5
Multiplying and Dividing Rational Expressions
ALGEBRA 1 LESSON 12-4
Solutions
1. r 2 • r 8 – r (2 + 8) = r 10
3. c7 ÷ c2 = c(7 – 2) = c5
2. b3 • b4 = b(3+4) = b7
4.
3x4 • 2x5 = (3 • 2)(x4 • x5) = 6x(4 + 5) = 6x9
5. 5n2 • n2 = 5(n2 • n2) = 5n(2 + 2) = 5n4
6. 15a3(–3a2) = 15(–3)(a3 • a2) = –45a(3 + 2) = –45a5
7. 2c2 + 15c + 7 = (2c + 1)(c + 7)
Check: (2c + 1)(c + 7) = 2c2 + 14c + 1c + 7
= 2c2 + 15c + 7
8. 15t2 – 26t + 11 = (15t – 11)(t – 1)
Check: (15t – 11)(t – 1) = 15t2 – 15t – 11t + 11
= 15t2 – 26t + 11
9. 2q2 + 11q + 5 = (2q + 1)(q + 5)
Check: (2q + 1)(q + 5) = 2q2 + 10q + 1q + 5
= 2q2 + 11q + 5
12-4
Multiplying and Dividing Rational Expressions
ALGEBRA 1 LESSON 12-4
Multiply.
7
8
a. y • y2
7
8
56
•
=
2
y
y
y3
x
Multiply the numerators and multiply
the denominators.
x–2
b. x + 5 • x – 6
x
x–2
x(x – 2)
•
=
x+5
x–6
(x + 5) (x – 6)
Multiply the numerators and multiply the
denominators. Leave the answer in
factored form.
12-4
Multiplying and Dividing Rational Expressions
ALGEBRA 1 LESSON 12-4
Multiply 3x +1 and
4
8x
.
9x2 – 1
3x + 1
8x
3x + 1
8x
•
=
•
4
9x2 – 1
4
(3x – 1) (3x + 1)
2
1
8x
3x +1 •
=
(3x – 1) (3x + 1)
4
1
1
=
2x
3x – 1
Factor denominator.
Divide out the common
factors (3x +1) and 4.
Simplify.
12-4
Multiplying and Dividing Rational Expressions
ALGEBRA 1 LESSON 12-4
Multiply
5x + 1
and x2 + 7x + 12.
3x + 12
5x + 1
5x + 1
(x + 3) (x + 4)
2 + 7x + 12) =
•
(x
•
3x + 12
3 (x + 4)
1
=
=
5x + 1
3 (x + 4)
1
1
(x + 3) (x + 4)
•
1
(5x +1) (x + 3)
3
12-4
Factor.
Divide out the
common factor x + 4.
Leave in factored form.
Multiplying and Dividing Rational Expressions
ALGEBRA 1 LESSON 12-4
Divide
x2 + 13x +40
by
x–7
x+8
.
x2 – 49
x2 + 13x +40
x2 + 13x +40
x+8
÷ x2 – 49 =
x–7
x–7
=
(x + 5) (x + 8)
x–7
1
(x + 8) (x + 5)
=
x–7
1
x2 – 49
• x+8
2
Multiply by x – 49 , the
x+8
reciprocal of x2 + 8 .
x – 49
•
(x + 7) (x – 7)
x+8
Factor.
•
1
(x + 7) (x – 7)
x+81
Divide out the common
factors x + 8 and x – 7.
= (x + 5) (x + 7)
Leave in factored form.
12-4
Multiplying and Dividing Rational Expressions
ALGEBRA 1 LESSON 12-4
2
Divide x + 9x + 14
by (8x2 + 16x).
11x
1
x2 + 9x + 14
8x2 + 16x
x2 + 9x + 14
=
÷
•
2
8x + 16x
11x
1
11x
=
1
(x + 7) (x + 2)
•
8x (x + 2)
11x
1
1
(x + 7) (x + 2)
• 8x (x + 2)
=
11x
1
=
x+7
88x2
12-4
Multiply by the
reciprocal of 8x2 + 16x
Factor.
Divide out the common
factor x + 2.
Simplify.
Multiplying and Dividing Rational Expressions
ALGEBRA 1 LESSON 12-4
pages 659–661 Exercises
1. 35x
2.
3.
4.
5.
6.
7.
8.
9.
36
12
t2
40
3a5
m(m – 2)
(m + 2)(m – 1)
2x(x – 1)
3(x + 1)
12x2
5(x + 1)
2c
c–1
5x4
2
9
t
20. x – 1
10. 1
3
11. 1
2
12. 3(4x + 1)
x–1
x+3
21. 6
22. – 1
13. 4(t + 1)(t + 2)
23.
14. 3(2m + 1)(m + 2)
24.
15. (x – 1)(x – 2)
25.
16. x + 1
26.
3
2
17. – 2d –25
6d
18. 2 1
c –1
19. 1
s+4
12-4
27.
28.
2
–1
3
2(x + 2)
x–1
n–3
4n + 5
3
x
11
7k – 15
1
x+1
29. t + 3
Multiplying and Dividing Rational Expressions
ALGEBRA 1 LESSON 12-4
30. c + 1
39. $132.96
31.
40. a.
b.
c.
d.
32.
33.
34.
c–1
3t – 5
7t 2
5(2x – 5)
x–5
x–2
x–3
x–5
x
x–2
4(x + 7)
2m2(m + 2)
42.
(m – 1)(m + 4)
43. 2
a+5
r+3
44.
(r – 1)(r + 1)2
41.
35. The student forgot to rewrite
the expression using the
reciprocal before canceling.
36. Answers may vary.
Sample: 9(m +2 1) • m + 2 ; m +2 2
37. 0, 4, –4
$100,000
360 payments
$599.55
$215,838
3m
3(m + 1)
45. She wrote w5 as a fraction so
she could easily see what
she could cancel.
m
38. $88.71
12-4
Multiplying and Dividing Rational Expressions
ALGEBRA 1 LESSON 12-4
x2
46. a.
4(2x + 1)2
b. x(3x + 2)2
4(2x + 1)
47. 9m 2(m + 1)
2
56. D
57. G
2
58. [2] x – 1 •
3x
=
x
x–1
(x + 1)(x – 1)(3x) = 3(x + 1)
x(x – 1)
48. 1
49.
50.
51.
52.
53.
x
y+5
–(2a + 3b)(a + 2b)
(5a + b)(2a – 3b)
m–2
2m(m – 1)
x(x – 2)
2(x – 1)
1
(w + 2)(w + 3)
54. B
55. G
[1] one computational error
OR answer with no work shown
59. b – 5
2
60. 3
4k
61. 7
3
4
q
62.
4
2
63. 5t – 9
8
12-4
Multiplying and Dividing Rational Expressions
ALGEBRA 1 LESSON 12-4
m2
2 – 3m
1
65.
2a2 – 3
64.
74.
76.
66. 2z + 3
z+1
67. 2c – 9
2c + 8
68. 8.2
69. 5.3
70. 5
75.
71. 11
72. 0.2
73. 7.1
12-4
Multiplying and Dividing Rational Expressions
ALGEBRA 1 LESSON 12-4
Multiply or divide.
1.
3.
5.
6.
7x 2
15
•
5
14x
2. 6x + 3
3x
2
•
x+6
x+3
1
2 + 5x + 6)
(x
÷
x+1
(x + 1)(x + 2)
2x + 4
x+1
÷
x2 + 11x + 18
x2 + 14x + 45
(x2 + 12x + 11)
4. 4x + 8
3x
2(x + 5)
x+1
x+9
• x2 + 20x + 99
12-4
x+1
x2 + 9x + 18
2x + 1
9x 2
• x+2
12x
3(x + 3)
Dividing Polynomials
ALGEBRA 1 LESSON 12-5
(For help, go to Lessons 9-1 and 9-3.)
Write each polynomial in standard form.
1. 9a – 4a2 + 1
2. 3x2 – 6 + 5x – x3
3. –2 + 8t
Find each product.
4. (2x + 4)(x + 3)
5. (–3n – 4)(n – 5)
6. (3a2 + 1)(2a – 7)
12-5
Dividing Polynomials
ALGEBRA 1 LESSON 12-5
Solutions
1. 9a – 4a2 + 1 = –4a2 + 9a + 1
2. 3x2 – 6 + 5x – x3 = –x3 + 3x2 + 5x – 6
3. –2 + 8t = 8t – 2
4. (2x + 4)(x + 3) = (2x)(x) + (2x)(3) + (4)(x) + (4)(3)
= 2x2 + 6x + 4x + 12 = 2x2 + 10x + 12
5. (–3n – 4)(n – 5) = (–3n)(n) + (–3n)(–5) + (–4)(n) + (–4)(–5)
= –3n2 + 15n – 4n + 20 = –3n2 + 11n + 20
6. (3a2 + 1)(2a – 7) = (3a2)(2a) + (3a2)(–7) + (1)(2a) + (1)(–7)
= 6a3 – 21a2 + 2a – 7
12-5
Dividing Polynomials
ALGEBRA 1 LESSON 12-5
Divide (18x3 + 9x2 – 15x) by 3x2.
1
(18x3 + 9x2 – 15x) ÷ 3x2 = (18x3 + 9x2 – 15x) • 3x2 .
18x3
9x2
=
+
3x2
3x2
–
15x
3x2
5
= 6x1 + 3x0 – x
Multiply by the
reciprocal of 3x2.
Use the Distributive Property.
Use the division rules for exponents.
5
= 6x + 3 – x
Simplify.
12-5
Dividing Polynomials
ALGEBRA 1 LESSON 12-5
Divide (5x2 + 2x – 3) by (x + 2)
Step 1: Begin the long division process.
Align terms by their degree. So put 5x
above 2x of the dividend.
5x
x + 2 5x2 + 2x – 3
5x2 + 10x
– 8x – 3
Divide: Think 5x2 ÷ x = 5x.
Multiply: 5x(x + 2) = 5x2 + 10x. Then subtract.
Bring down – 3.
12-5
Dividing Polynomials
ALGEBRA 1 LESSON 12-5
(continued)
Step 2: Repeat the process: Divide, multiply, subtract, bring down.
5x – 8
x + 2 5x2 + 2x – 3
5x2 + 10x
– 8x – 3
Divide: –8x ÷ x = – 8
– 8x – 16
Multiply: – 8(x + 2) = – 8x – 16. Then subtract.
13
The remainder is 13.
13
The answer is 5x – 8 + x + 2 .
12-5
Dividing Polynomials
ALGEBRA 1 LESSON 12-5
The width and area of a rectangle are shown in the figure
below. What is the length?
Since A = w, divide the area by the width
to find the length.
2x – 3
3x2 + 2x + 3
6x3 – 5x2 + 0x – 9
6x3 – 9x2
4x2 + 0x
4x2 + 6x
–6x – 9
–6x – 9
0
Rewrite the dividend with 0x.
The length of the rectangle is (3x2 + 2x + 3) in.
12-5
Dividing Polynomials
ALGEBRA 1 LESSON 12-5
Divide (–8x – 2 + 6x2) by (–1 + x).
Rewrite –8x – 2 + 6x2 as 6x2 – 8x – 2 and –1 + x as x – 1.
Then divide.
6x – 2
x – 1 6x2 – 8x – 2
6x2 – 6x
–2x –2
–2x + 2
–4
4
The answer is 6x – 2 – x – 1 .
12-5
Dividing Polynomials
ALGEBRA 1 LESSON 12-5
pages 664–666 Exercises
11. 3x – 1
1. x4 – x3 + x2
12. –2q – 10 +
2. 3x4 – 2
13. 5t – 50
3. 3c2 + 2c – 1
14. 2w2 + 2w + 5 –
x
3
4.
n2
– 18n + 3
5. 4 – 16
q
6. –t 3 + 2t 2 – 4t + 5
7. x – 3
8. 2t + 9 +
10
w–1
15. b2 – 3b – 1 + 3
3b – 1
16. c2 – 1
c–1
17. t 2 – 2t – 2
18. n2 – 2n – 21 –
16
t–3
8
n+2
19. (r 2 + 5r + 1) cm
20. (4c2 – 8c + 16) ft
9. n – 1
10. y – 3 +
22
2q + 1
8
y+2
21. b + 12 +
12-5
1
b+4
Dividing Polynomials
ALGEBRA 1 LESSON 12-5
22. a – 1 –
33. k 2 – 0.3k – 0.4
2
a+4
23. 10w – 681 + 49,046
34. 3s – 8 +
w + 72
24. t –
29
2s + 3
5
z+1
36. 6m2 – 24m + 99 – 326
m+4
35. –2z2 + 3z – 4 +
9
t+4
25. 2x2 + 5x + 2
26. 3q 2 + 2q + 3 + 12
37. –16c2 – 20c – 25
27. 3x + 2 – 1
38. 2r 4 + r 2 – 7
28. c2 + 11c – 15 + 8
39. t – 1 +
10
b–1
30. y2 + 5y + 29 + 138
y–5
40. z3 – 3z2 + 10z – 30 +
q–2
2x
c
29. 2b2 + 2b + 10 +
2t
2t 3 + 1
88
z+3
41. a. Answers may vary. Sample:
(c3 + 3c2 – 2c – 4); (c + 1)
31. 28a – 12
32. 5t 3 – 25t 2 + 115t – 575 + 2881
t+5
b. (c3 + 3c2 – 2c – 4) ÷ (c + 1)
= c2 + 2c – 4
12-5
Dividing Polynomials
ALGEBRA 1 LESSON 12-5
42. a. y = 2 – 1
x+3
b. Answers may vary. Sample:
x –2 –1 0 1 2
y
1
3
2
5
3
7
4
44. m2 + 5m + 4
45. a. d – 2 +
3
d+1
b. d 2 – 2d + 3 –
9
5
4
d+1
c. d 3 – 2d 2 + 3d – 4 +
5
d+1
d. Answers may vary. Sample:
6
d+1
4
3
2
e. d – 2d + 3d – 4d + 5 – 6
d+1
d 4 – 2d 3 + 3d 2 – 4d + 5 –
46. 12
c. vertical asymptote: x = –3
horizontal asymptote: y = 2
43. The binomial is a factor of the
polynomial if there is no
remainder from the division.
47. a. t = d
r
b.
t2
– 7t + 12
48. 2a2b2 – 3ab3 + 5ab2
49. 3x + 2y
12-5
Dividing Polynomials
ALGEBRA 1 LESSON 12-5
50. 10r 5 + 2r 4 + 5r 2
56. [2]
51. 2b3 – 2b2 + 3
52. a. x – 3 + 10
x+5
b. ƒ(x) = x – 3 +
10
x+5
c. y = x – 3
d.
53. C
54. G
55. B
x2 + 3x + 2
2x – 1 2x3 + 5x2 + x – 2
2x3 – x2
6x2 + x
6x2 – 3x
4x – 2
4x – 2
x+1
0
x + 2 x2 + 3x + 2
x2 + 2x
x+2
x+2
0
The width is x + 1.
[1] one computational error OR
correct answer with no work shown
12-5
Dividing Polynomials
ALGEBRA 1 LESSON 12-5
57. [4] a.
3
x + 4 3x + 10
3x + 12
–2
y=3– 2
58. n + 2
59. (t – 5)(3t + 1)(2t + 11)
(2t – 55)(t + 1)(3t)
60. 3c + 8
2c + 7
2
61. (x + 5)(x + 4)
(x + 7)(x + 8)2
x+4
b. Tables may vary. Sample:
62.
63.
64.
65.
66.
67.
vertical asymptote: x = –4
68.
horizontal asymptote: y = 3;
69.
[3] appropriate methods, but with one computational error 70.
[2] no asymptotes OR no graph
71.
[1] one part only
72.
12-5
5
9.4
15
17.9
12.0
16.2
5.38
17
–12.7
63.25
6.32
Dividing Polynomials
ALGEBRA 1 LESSON 12-5
Divide.
1. (x8 – x6 + x4) ÷ x2
2. (4x2 – 2x – 6) ÷ (x + 1)
x6 – x4 + x2
4x – 6
3. (6x3 + 5x2 + 11) ÷ (2x + 3)
4. (29 + 64x3) ÷ (4x + 3)
2
2
16x2 – 12x + 9 + 4x + 3
3x2 – 2x + 3 + 2x + 3
12-5
Adding and Subtracting Rational Expressions
ALGEBRA 1 LESSON 12-6
(For help, go to Lessons 1-5 and 906.)
Simplify each expression.
1. 4 + 2
9
9
2. 3 – 5
7
7
3. 1 + – 5
2
2
4. 5 + 2
6
9
5. 1 – 1
4
3
6. 5 – 3
12
4
7. 4x + 2x
9
9
8. 7x – x
12
12
9.
7
1
–
12y
12y
Factor each quadratic expression.
10. x2 + 3x + 2
11. y2 + 7y + 12
12-6
12. t 2 – 4t + 4
Adding and Subtracting Rational Expressions
ALGEBRA 1 LESSON 12-6
Solutions
1. 4 + 2 = 4 + 2 = 6 = 3 • 2 = 2
9
9
3•3
3
9
9
2. 3 – 5 = 3 – 5 = –2 = – 2
7
7
7
7
7
3. 1 + – 5
2
2
=
1 + (– 5)
–4
=
= –2
7
2
4. 5 + 2 = 5 • 3 + 2 • 2 = 15 + 4 = 19 or 1 1
6
9
6•3
9•2
18
18
18
18
5. 1 – 1 = 1 • 3 – 1 • 4 = 3 – 4 = 3 – 4 = –1 = – 1
12
12
12
4
3
4•3
3•4
12
12
6. 5 – 3 = 5 – 3 • 3 = 5 – 9 = 5 – 9 = –4 = – 1
12
4
12
4•3
12
12
12
12
3
12-6
Adding and Subtracting Rational Expressions
ALGEBRA 1 LESSON 12-6
Solutions (continued)
7. 4x + 2x = 4x + 2x = 6x = 3 • 2 • x = 2x or 2 x
9
9
9
9
3•3
3
3
8. 7x – x = 7x – x = 6x = 6 • x = x or 1 x
12
12
12
12
6•2
2
2
9.
7
1
7–1
6
6•1
1
–
=
=
=
=
12y
12y
12y
12y
6 • 2y
2y
10. Factors of 2 with a sum of 3: 1 and 2.
x2 + 3x + 2 = (x + 1)(x + 2)
11. Factors of 12 with a sum of 7: 3 and 4.
y2 + 7y + 12 = (y + 3)(y + 4)
12. Factors of 4 with a sum of –4: –2 and –2.
t2 – 4t + 4 = (t – 2)(t – 2) or (t – 2)2
12-6
Adding and Subtracting Rational Expressions
ALGEBRA 1 LESSON 12-6
Add
4
x+3
2
4
and 2 .
x+3
x+3
4+2
+ x+3 = x+3
=
6
x+3
Add the numerators.
Simplify the numerator.
12-6
Adding and Subtracting Rational Expressions
ALGEBRA 1 LESSON 12-6
3x + 5
4x + 7
Subtract 3x2 + 2x – 8
4x + 7
–
3x2 + 2x – 8
3x + 5
3x2 + 2x – 8
=
from 3x2 + 2x – 8 .
4x + 7 – (3x + 5)
3x2 + 2x – 8
=
4x + 7 – 3x + 5
3x2 + 2x – 8
=
3x2
x+2
+ 2x – 8
Subtract the
numerators.
Use the Distributive Property.
Simplify the numerator.
1
=
=
x+2
(3x – 4) (x + 2)
1
Factor the denominator. Divide
out the common factor x + 2.
1
3x – 4
Simplify.
12-6
Adding and Subtracting Rational Expressions
ALGEBRA 1 LESSON 12-6
3
1
Add 4x + .
8
3
1
Step 1: Find the LCD of 4x and 8 .
4x = 2 • 2 • x
Factor each denominator.
8=2•2•2
LCD = 2 • 2 • 2 • x = 8x
Step 2: Rewrite using the LCD and add.
3
1
2•3
1•x
+
=
+
4x
8
2 • 4x
8•x
6
x
= 8x + 8x
=
Rewrite each fraction using the LCD.
Simplify numerators and denominators.
6+x
8x
Add the numerators.
12-6
Adding and Subtracting Rational Expressions
ALGEBRA 1 LESSON 12-6
Add
7
3
and
.
x+4
x–5
Step 1: Find the LCD of x + 4 and x – 5. Since there are no common
factors, the LCD is (x + 4)(x – 5).
Step 2: Rewrite using the LCD and add.
7
3
7 (x – 5)
3 (x + 4)
+
=
+
x+4
x–5
(x + 4) (x – 5)
(x + 4) (x – 5)
7x – 35
3x + 12
= (x + 4) (x – 5) + (x + 4) (x – 5)
=
7x – 35 + 3x + 12
(x + 4) (x – 5)
Rewrite the fractions using
the LCD.
Simplify each numerator.
Add the numerators.
10x – 23
= (x + 4) (x – 5)
Simplify the numerator.
12-6
Adding and Subtracting Rational Expressions
ALGEBRA 1 LESSON 12-6
The distance between Seattle, Washington, and Miami,
Florida, is about 5415 miles. The ground speed for jet traffic from
Seattle to Miami can be about 14% faster than the ground speed from
Miami to Seattle. Use r for the jet’s ground speed. Write and simplify
an expression for the round-trip air time.
Miami to Seattle time:
5415
r
5415
Seattle to Miami time: 1.14r
time =
distance
rate
time =
distance
rate
12-6
14% more than a number
is 114% of the number.
Adding and Subtracting Rational Expressions
ALGEBRA 1 LESSON 12-6
(continued)
An expression for the total time is
5415
5415
+
r
1.14r
6173
5415
+
1.14r
1.14r
11588
5415 5415
+ 1.14r
r
Rewrite using the LCD, 1.14r.
= 1.14r
Add the numerators.
= 10165
Simplify.
r
12-6
Adding and Subtracting Rational Expressions
ALGEBRA 1 LESSON 12-6
pages 669–671 Exercises
1.
2.
3.
4.
5.
6.
7.
8.
9.
9
2m
7
6t – 1
n+2
n+3
14
c–5
2s2 + 1
4s2 + 2
6c – 28
2c + 7
–3
2–b
– 1
t2+1
–2t
2t – 3
10. 1
21. 189 –3 9n
11. 2
22.
12. –1
23.
13. 2x2
24.
14. 18
25.
15. 7z
16. 35b3c
17. 35 + 6a
15a
18. 12 – 2x
3x
2
19. 18 + 20x
15x8
20. 9 + 2m
24m3
12-6
26.
27.
28.
29.
7n
45 + 36x2
20x2
17m – 47
(m + 2)(m – 7)
a2 + 9a + 12
(a + 3)(a + 5)
a2 + 12a + 15
4(a + 3)
2
c + 7c + 20
(c + 5)(c + 3)
4t 2 + 5t + 5
t 2(t + 1)
18a + 3
(2a + 1)(2a – 1)
a. 1 + 1
r
0.7r
b. 17
7r
c. about 0.8 h
Adding and Subtracting Rational Expressions
ALGEBRA 1 LESSON 12-6
30.
31.
32.
33.
34.
35.
36.
37.
38.
39.
–y 2 +
2y + 2
3y + 1
h2+ h + 1
2t 2 – 7
r – 2k – 6
9 + p3
–3 – x – z
xy 2z
k + 3km
2m2
12c – 15a
abc
3
c – a3
abc
10x + 15
x+2
–21t + 33
2t – 3
6x
(x – 3)(x + 3)2
45. 8x 2 – 1
40. k – 1
k–6
x
41. The student added the
terms in the denominators.
46. 8
47. – 3x – 5
42. a. 2 + 2 ; 18
r
1.25r 5r
b. 2 + 2 ; 9
d
0.8d 2d
48.
c. Yes; they both represent
the time it takes to make
a round trip.
43. Answers may vary. Sample:
Not always; the numerator
may contain a factor of the LCD.
50.
44. Answers may vary.
2
3
2
Sample: 2w , 3w ; 3w + 11w – 6w
(w + 3)(w – 3)
w+3 w–3
12-6
49.
51.
52.
53.
54.
x(x – 5)
32x
x–5
x–5
4x
1
2x(x – 5)
– d+3
d+4
3
–x + 6x2 + 35x – 50
(a + 12)(a – 5)
x
x+4
5a – 8
(a + 2)(a – 5)
55. D
Adding and Subtracting Rational Expressions
ALGEBRA 1 LESSON 12-6
56. H
2
59. x + 2x – 1
57. D
60. 5b2 – 10b + 30 – 60
58. [2] a. 10 + 10 = total time;
61. 6
2
b+2
r+3
r
10(r + 3) + 10r = 10r + 30 + 10r =
r(r + 3)
r(r + 3)
20r + 30 = 10(2r + 3)
r(r + 3)
r(r + 3)
b. 10(2(12) + 3) = 10(24 + 3) =
12(12 + 3)
12(15)
10(27) = 270 = 3
180
180 2
The ride took 1 1 hours.
2
[1] one computational error
OR no work shown
12-6
62. 3
63. no solution
64. ± 6.9
65. ± 3.9
66. no solution
Adding and Subtracting Rational Expressions
ALGEBRA 1 LESSON 12-6
2
1
2
1. Add 5x + 9 .
18 + 5x
45x
4(5x + 7)
(3x + 4)(x + 2)
8
3
3. Subtract x2 – 4 – x2 – 4 .
x2
6
2. Add 3x + 4 + x + 2 .
4. Subtract
4(y – 1)
y+2
5
–4
x–4
5. Add 8 + x + 3 .
9x + 20
x+3
12-6
6y – 7
2y – 3
–
.
y+2
y+2
Solving Rational Equations
ALGEBRA 1 LESSON 12-7
(For help, go to Lessons 4-1 and 12-6.)
Solve each proportion.
1. 1 = 3
x
5
2.
3
5
=
t
2
3. m = 27
3
m
Find the LCD of each group of expressions.
4.
3
;
4n
1
;
2
2
n
5.
1
;
3x
2
;
5
12-7
4
3x
6.
1
1
5
;
;
2
8y
y
6
Solving Rational Equations
ALGEBRA 1 LESSON 12-7
Solutions
1. 1
2.
3
= 5
x
3x = 5
5
3.
3
5
=
t
2
5t = 6
2
x = 3 or 1 3
6
m
27
=
3
m
m2 = 81
1
t = 5 or 1 5
m=±
81 = ± 9
4. 4n = 2 • 2 • n; 2 = 2; n = n; LCD = 2 • 2 • n = 4n
5. 3x = 3 • x; 5 = 5; 3x = 3 • x; LCD = 3 • 5 • x = 15x
6. 8y = 2 • 2 • 2 • y; y2 = y • y; 6 = 2 • 3; LCD = 2 • 2 • 2 • 3 • y • y = 24y2
12-7
Solving Rational Equations
ALGEBRA 1 LESSON 12-7
Solve 3 + 4 = 14 .
8
8x
1
8x
x
2x
3
4
14
+
=
8
x
2x
The denominators are 8, x, and 2x.
The LCD is 8x.
3
4
= 8x 14
+
8
x
2x
Multiply each side by the LCD.
1 4
3
+ 8x x
8
1
1
4
= 8x 14
2x
1
3x + 32 = 56
3x = 24
x=8
Use the Distributive Property.
No rational expressions.
Now you can solve.
Subtract 32 from each side.
Divide each side by 3, then simplify.
12-7
Solving Rational Equations
ALGEBRA 1 LESSON 12-7
(continued)
3
4
14
2(8)
Check: 8 + 8
7
8
7
8
14
16
=
7
8
12-7
Solving Rational Equations
ALGEBRA 1 LESSON 12-7
6
5
Solve x2 = x – 1. Check the solution.
5
x2
6
x2
1
x
5
6
2
x
=
x
x2
1
1
x2
= x2 x – 1
– x2 (1)
6 = 5x – x2
Use the Distributive Property.
Simplify.
x2 – 5x + 6 = 0
Collect like terms on one side.
(x – 3)(x – 2) = 0
Factor the quadratic expression.
x – 3 = 0 or x – 2 = 0
x = 3 or
Multiply each side by the LCD, x2.
x=2
Use the Zero-Product Property.
Solve.
12-7
Solving Rational Equations
ALGEBRA 1 LESSON 12-7
(continued)
Check:
6
32
5
3
2
2
=
3
3
–1
6
22
5
2
3
3
=
2
2
12-7
–1
Solving Rational Equations
ALGEBRA 1 LESSON 12-7
Renee can mow the lawn in 20 minutes. Joanne can do the
same job in 30 minutes. How long will it take them if they work together?
Define: Let n = the time to complete the job if they work together (in minutes).
Person
Renee
Joanne
Work Rate
(part of job/min.)
1
20
1
30
Time Worked
(min)
n
n
Part of
Job Done
n
20
n
30
Relate: Renee’s part done + Joanne’s part done = complete job.
12-7
Solving Rational Equations
ALGEBRA 1 LESSON 12-7
(continued)
Relate: Renee’s part done + Joanne’s part done = complete job.
n
20
Write:
60
n
n
+
20
30
+
= 60(1)
3n + 2n = 60
5n = 60
n = 12
n
30
=
1
Multiply each side by the LCD, 60.
Use the Distributive Property.
Simplify.
Simplify.
It will take two of them 12 minutes to mow the lawn working together.
1
Check: Renee will do 12 • 20 = 3 of the job, and Joanne will do 12 • 1 = 2
1
5
1 30 5
of the job. Together, they will do 3 + 2 = 1, or the whole job.
5
12-7
5
Solving Rational Equations
ALGEBRA 1 LESSON 12-7
Solve
4
1
=
. Check the solution.
2
x
x+8
4
1
=
2
x
x+8
4(x + 8) = x2(1)
Write cross products.
4x + 32 = x2
Use the Distributive Property.
x2 – 4x – 32 = 0
Collect terms on one side.
(x – 8)(x + 4) = 0
Factor the quadratic expression.
x – 8 = 0 or x + 4 = 0
Use the Zero-Product Property.
x = 8 or
x = –4
Solve.
12-7
Solving Rational Equations
ALGEBRA 1 LESSON 12-7
(continued)
4
1
Check: x2 = x + 8
4
82
1
8+8
4
=
64
1
16
4
(–4)2
4
16
1
–4 + 8
=
1
4
12-7
Solving Rational Equations
ALGEBRA 1 LESSON 12-7
Solve
x+3
4
=
.
x–1
x–1
x+3
4
=
x–1
x–1
(x + 3) (x – 1) = 4(x – 1)
x2 – x + 3x – 3 = 4x – 4
Write the cross products.
Use the Distributive Property.
x2 + 2x – 3 = 4x – 4
Combine like terms.
x2 – 2x + 1 = 0
Subtract 4x – 4 from each side.
(x – 1) (x – 1) = 0
x–1=0
x=1
Factor.
Use the Zero-Product Property.
Simplify.
12-7
Solving Rational Equations
ALGEBRA 1 LESSON 12-7
(continued)
Check:
1+3
1–1
4
0
4
1–1
=
4
0
Undefined! There is no division by 0.
The equation has no solution because 1 makes a denominator equal 0.
12-7
Solving Rational Equations
ALGEBRA 1 LESSON 12-7
pages 675–677 Exercises
12. –4
24. –14
13. –2
25. 1 , 2
2. 3
14. –1
26. 3
3. –1
15. – 2
27. –5, 2
1. –2
4. 6, –1
5. – 1
3
6. –2, 4
7. 1, 4
8. 5
9. 1, 3
10. 1
3
11. 16
3
2
3
16. 1 5 h
7
17.
28. –1
12.7 min
29. 1
2
18. 3
30. – 6 , –1
19. 10, –10
31. 0, 2
20. – 3 , 4
32. 12 h
21. 4
33. a. 32
b. Answers may vary.
Sample: Cross-multiplying;
I think it's quicker.
5
2
22. no solution
23. 6
12-7
Solving Rational Equations
ALGEBRA 1 LESSON 12-7
33 (continued)
c. No; it only works for rational
equations that are proportions.
34. a = 4, b = 7 , c = 11, d = – 1
27
3
2
35. a. y1 = 6 + 1, y2 = (x + 7)
x2
6
35. (continued)
c. Yes; the x-values are solutions
to the original equation since
both sides are equal.
36. 66.6Ω
37. 20Ω
38. 3.75Ω
39. 20Ω
40. Answers may vary.
Sample: 2b
b+2
41. 40 mi/h
42. 9
b. (–9.53, 1.07), (–4.16, 1.35),
(–1.12, 5.76), (0.81, 10.16)
43. 0, 1
2
44. –1
12-7
=
6b
4b + 3
Solving Rational Equations
ALGEBRA 1 LESSON 12-7
52. [2]
45. 1
46.
2(3 + x) = 16 + x
23.5 min
47. a. 0.80s
6 + 2x = 16 + x
b. 50 – s
x = 10
c. 0.30(50 – s)
3 + 10
1
16 + 10
2
13 = 1
26
2
d. 0.8s + 0.3(50 – s) = (0.62)(50)
e. 32
[1] appropriate method,
but with one computational error
f. 32 L of 80% solution
and 18 L of 30% solution
48. 11 1 h
3
49. B
50. G
51. C
3+x
= 1;
16 + x
2
53. –
3
x 2y 2z
2
54. 3h + 2ht + 4h
2(t – 2)(t + 2)
–4k – 61
55.
(k – 4)(k + 10)
12-7
Solving Rational Equations
ALGEBRA 1 LESSON 12-7
56.
62. –18, –5
59.
63. 8, 11
64. –23, 1
65. –8, 6
60.
66. –13, –4
57.
67. –91, –1
61.
58.
12-7
Solving Rational Equations
ALGEBRA 1 LESSON 12-7
x
2
2
x+1
= x .
1. Solve 4x + 3 = 5 .
2. Solve
1
3. Solve
–2
1
= 3(2 + x) .
3
4. Solve
4
1
1
+
= x+3 .
2x
2
–2
1
–4
1, 3
5. Juanita can wash the car in 30 minutes. Gabe can wash the car in
40 minutes. Working together, how long will it take?
1
17 7 min
12-7
Counting Methods and Permutations
ALGEBRA 1 LESSON 12-8
(For help, go to Lessons 12-8 and 4-6.)
You roll a number cube. Find each probability.
1. P(even number)
2. P(prime number)
3. P(a number greater than 5)
4. P(a negative number)
You roll a blue number cube and a yellow number cube.
Find each probability.
5. P(blue 1 and yellow 2)
6. P(blue even and yellow odd)
12-8
Counting Methods and Permutations
ALGEBRA 1 LESSON 12-8
Solutions
3
1
1. P(even number) = P(2, 4, or 6) = 6 = 2
2. P(prime number) = P(2, 3, or 5) = 3 = 1
6
3. P(a number greater than 5) = P(6) = 1
6
4. P(a negative number) = 0 = 0
6
2
5. P(blue 1 and yellow 2) = P(blue 1) • P(yellow 2) =
1
1
1
=
•
6
6
36
6. P(blue even and yellow odd) = P(blue even) • P(yellow odd) =
3
3
9
1
•
=
=
6
6
36
4
12-8
Counting Methods and Permutations
ALGEBRA 1 LESSON 12-8
Suppose you have three shirts and three ties that coordinate.
Make a tree diagram to find the number of possible outfits you have.
Shirts
Ties
Outfits
Shirt 1
Tie1
Tie 2
Tie 3
Shirt 1, Tie 1
Shirt 1, Tie 2
Shirt 1, Tie 3
Shirt 2
Tie1
Tie 2
Tie 3
Shirt 2, Tie 1
Shirt 2, Tie 2
Shirt 2, Tie 3
Shirt 3
Tie1
Tie 2
Tie 3
Shirt 3, Tie 1
Shirt 3, Tie 2
Shirt 3, Tie 3
There are nine possible outfits.
12-8
Counting Methods and Permutations
ALGEBRA 1 LESSON 12-8
Suppose there are two routes you can drive to get from
Austin, Texas to Dallas, Texas, and four routes from Dallas, Texas to
Tulsa, Oklahoma. How many possible routes are there from Austin to
Tulsa through Dallas?
2•4=8
Routes from
Austin to Dallas
Routes from Austin to Tulsa through Dallas
Routes from Dallas
to Tulsa
There are eight possible routes from Austin to Tulsa.
12-8
Counting Methods and Permutations
ALGEBRA 1 LESSON 12-8
In how many ways can 11 students enter a classroom?
There are 11 choices for the first student, 10 for the second, 9 for the
third, and so on.
11 • 10 • 9 • 8 • 7 • 6 • 5 • 4 • 3 • 2 • 1 = 39,916,800
Use a calculator.
There are 39,916,800 possible ways in which the students can enter
the classroom.
12-8
Counting Methods and Permutations
ALGEBRA 1 LESSON 12-8
Simplify 8P5.
Method 1: Use pencil and paper,
8P5 =
8•7•6•5•4
= 6720
The first factor is 8 and there are 5 factors.
Simplify.
Method 2: Use a graphing calculator.
Enter the first factor,8.
Use MATH to select nPR in the PRB screen.
Input 5, since there are 5 factors. Press enter.
8P5 =
6720
12-8
Counting Methods and Permutations
ALGEBRA 1 LESSON 12-8
Suppose you use five different letters from the 26 letters of
the alphabet to make a password. Find the number of possible fiveletter passwords.
There are 26 letters in the alphabet. You are finding the number of
permutations of 26 letter arranged 5 at a time.
8P5 =
26 • 25 • 24 • 23 • 22
Use a calculator.
= 7,893,600
There are 7,893,600 five-letter passwords in which letters do not repeat.
12-8
Counting Methods and Permutations
ALGEBRA 1 LESSON 12-8
pages 682–685 Exercises
1. 10 choices
Shirt 1
Tie 1
Tie 2
Tie 3
Tie 4
Tie 5
Shirt 2
Tie 1
Tie 2
Tie 3
Tie 4
Tie 5
2.
S1, T1
S1, T2
S1, T3
S1, T4
S1, T5
S2, T1
S2, T2
S2, T3
S2, T4
S2, T5
12 menus
Main
Salad Soup Course
C
V
B
S
S
C
C
B
S
C
V
B
C
S
C
C
B
S
12-8
Menu
SVC
SVB
SVS
SCC
SCB
SCS
CVC
CVB
CVS
CCC
CCB
CCS
Counting Methods and Permutations
ALGEBRA 1 LESSON 12-8
5.
14. 42
a. 8, 10, 10, 10
b. 8,000,000 telephone numbers 15. 5040
a. 6
16. 12,144
b. 12
17. 8P6
3,628,800 orders
6.
120 arrangements
18. 9P7
7.
1680
19. 8P4
8.
3024
9.
360
20. a. 2; 2
b. 4
c. No; number of consonants •
number of vowels = number of vowels •
number of consonants.
3.
4.
10. 120
11. 5040
12. 5040
21. a. 24
b. 1
24
13. 2520
12-8
Counting Methods and Permutations
ALGEBRA 1 LESSON 12-8
22. a. 24
b.
27. a. 260,000 license plates
b. 23,920,000 license plates
1
14,950
28. a. Check students’ work.
b. Check students’ work.
c. Answers may vary. Sample:
No; if someone tries to guess
your password, they’ll probably
try your name or initials first.
29. a. 17,576 codes
b. 17,526 codes
23. 3
30. a. 35,152 call letters
b. 913,952 call letters
24. 2
31. a. 18,278 companies
b. 12,338,352 companies
c. NASDAQ; 12,320,074
more companies
25. 5
26. a. 10,000
b. With repetition; there are more
permutations when repetition
is allowed.
32. a. 2
b. 6
c. (n – 1)!
12-8
Counting Methods and Permutations
ALGEBRA 1 LESSON 12-8
33. 7
41. 3
53. – 7 ±
34. a. 60 numbers
b. 6 numbers
42. –8, 6
54. –2 ±
43. –10, 1
55. 5, 13
c. 1
10
b 9
10
35. False; if a = 3 and b = 2,
then (a – b)! = 1! = 1,
but a! – b! = 3! – 2! = 4.
2
44. –8, 5
45. 1 , –5
2
46. AB = 7, AC = 5
47. BC
34, AC
54
36. 72
48. AB
21, BC
5
37. 16
49. AB
48, AC
7
38. 4
50. –6 ±
39. 24
51. no solution
40. 3
52. –4 ± 3
10
35
12-8
3
97
2
5
Counting Methods and Permutations
ALGEBRA 1 LESSON 12-8
1. Jeff has five shirts, three pairs of pants, and two ties. How many
possible outfits can he make?
30 outfits
2. Three of 18 students are to be selected for the student government
positions of president, vice-president, and treasurer. In how many
different ways can the three positions be filled?
4896 ways
3. Calculate 6P3.
120
4. An Italian restaurant offers four different choices of pasta, three
choices of sauce, and two choices of meat. How many different
dishes can you order that each consist of one pasta choice, one
sauce, and one meat?
24
12-8
Combinations
ALGEBRA 1 LESSON 12-9
(For help, go to Lessons 12-8 and 4-6.)
Evaluate each expression.
1. 5P3
2. 6P3
3. 7P3
4. 7P4
A and B are independent events. Find P(A and B) for the given
probabilities.
5. P(A) = 1 , P(B) = 3
3
7. P(A) =
4
9
5
, P(B) =
10
6
6. P(A) =
1
5
, P(B) =
8
9
8. P(A) = 0.35, P(B) = 0.2
12-9
Combinations
ALGEBRA 1 LESSON 12-9
(For help, go to Lessons 12-8 and 4-6.)
Solutions
1.
5P3
= 5 • 4 • 3 = 60
2.
6P3
= 6 • 5 • 4 = 120
3.
7P3
= 7 • 6 • 5 = 210
4.
7P4
= 7 • 6 • 5 • 4 = 840
5. P(A and B) = P(A) • P(B) = 1 • 3 = 1 • 3 = 1
3
4
3•4
4
6. P(A and B) = P(A) • P(B) = 1 • 5 = 5
8
9
72
3
3
7. P(A and B) = P(A) • P(B) = 9 • 5 = 3 • 3 • 5
=
=
10
6
5•2•3•2
2•2
4
8. P(A and B) = P(A) • P(B) = (0.35)(0.2) = 0.07
12-9
Combinations
ALGEBRA 1 LESSON 12-9
Simplify 9C6.
9C6 =
=
9P6
6P6
9•8•7•6•5•4
6•5•4•3•2•1
= 84
Write using permutation notation.
Write the product represented by
the notation.
Simplify.
12-9
Combinations
ALGEBRA 1 LESSON 12-9
Eighteen people enter a talent contest. Awards will be given
to the top ten finishers. How many different groups of ten winners can
be chosen?
The order in which the top ten winners are listed once they are chosen
does not distinguish one group of winners from another. You need the
number of combinations of 18 potential winners chosen 10 at a time.
Evaluate 18C10.
Method 1: Use pencil and paper.
18C10 =
=
18P10
10P10
Write using permutation notation.
18 • 17 • 16 • 15 • 14 • 13 • 12 • 11 • 10 • 9
10 • 9 • 8 • 7 • 6 • 5 • 4 • 3 • 2 • 1
= 43,758
Use a calculator.
12-9
Combinations
ALGEBRA 1 LESSON 12-9
(continued)
Method 2: Use a graphing calculator.
Enter the first factor, 18.
Use
to select nCr in the PRB screen.
Input 10, since there are ten factors. Press
18C10
.
= 43,758.
There are 43,758 different ten-person groups of winners that can be
chosen from a group of 18 people.
12-9
Combinations
ALGEBRA 1 LESSON 12-9
Suppose you have eight new CDs (three rock, two jazz, and
three country). If you choose the CDs at random, what is the
probability that the first one is country and the second one is jazz?
3
Three of the eight CDs
are country.
2
Two of the remaining CDs
are jazz.
probability the first choice is country = 8
probability the second choice is jazz = 7
3
2
P(country, then jazz) = 8 • 7
Multiply the probabilities.
6
Multiply.
= 56
3
= 28
Simplify.
3
The probability that you choose country, then jazz is 28 .
12-9
Combinations
ALGEBRA 1 LESSON 12-9
Suppose you have eight red pens and four black pens in a
box. You choose five pens without looking. What is the probability that
all the pens you choose are red?
There are 12 pens in all. Eight of the pens are red.
number of favorable outcomes = 8C5
number of ways to choose 5
pens from the 8 red pens
number of possible outcomes = 12C5
number of ways to choose 5 red
pens from 12 possible pens
12-9
Combinations
ALGEBRA 1 LESSON 12-9
(continued)
P(5 red pens) = number of favorable outcomes
total number of outcomes
=
8C5
Use the definition
of probability.
Substitute.
12C5
56
= 792
Simplify each expression.
7
= 99
Simplify.
7
The probability that you will chose five red pens is 99 , or about 7%.
12-9
Combinations
ALGEBRA 1 LESSON 12-9
pages 689–691 Exercises
12. 56
18. 35
1
5040
1. 1
13.
2. 6
14. a. 45
3. 15
b. 3
4. 20
c. 1
5. 15
6. 6
15
d. 2
15
b. 1
8. 28
c. 1
10. 21
11. 220
20. combination, since order
is not important
21. permutation, since order
is important
22. a.
15. a. 56
7. 28
9. 21
19. 4
56
d. 5
28
16. 10
17. 1
12-9
b. 6
c. 6
d. 45
Combinations
ALGEBRA 1 LESSON 12-9
23. a. Answers may vary. Sample:
It is a combination problem
because order is not important.
b. 45
c. Yes; Each line segment joins
two points and each handshake
connects two people.
25. Answers may vary. Sample:
Both permutations and combinations
are arrangements of some or all of a
group of objects. However, permutations
take into account order, and
combinations do not.
26. 4
24. a. 59,280 sequences
27. 8
b. 1482 sequences
28. a. 24
b. 1
c. 1
40
d. Answers may vary. Sample:
It is unlikely someone will guess
the right sequence with more
than 59,000 possibilities.
4
29. Check students’ work.
30. a. 792
b. 36
c. 1
22
12-9
Combinations
ALGEBRA 1 LESSON 12-9
31. always
36. (continued)
which uses the combination
32. sometimes
x!
formula xC2 = 2!(x – 2)! for x.
This represents the number of
combination groups of 2.
33. always
34. 1
28
35. a.
b.
c.
d.
36. a.
15
6; 3
20
300
c.
Groups can only be made from sets of
objects, which means they must be
integers.
37. B
38. F
39. C
b. Answers may vary. Sample:
It is the function ƒ(x) = x(x – 1) ,
2
40. [2] There are 10 possible pizzas;
combination; order does
not matter;
12-9
Combinations
ALGEBRA 1 LESSON 12-9
40. [2] (continued)
5P2
5•4
5C2 = 5P2 = 2 • 1 = 10
[1] incorrect explanation OR
minor error
49.
1
10
50. 5
51. –2 1
2
52. 0, 16
41. 12
53. –6.81, 0.81
42. 5040
54. 0.30, 6.70
43. 840
55. –5.46, 1.46
44. 9
56. –1.24, 1.35
45. 23
57. –6, 3
46. 92,610,000 license plates
58. –0.05, 13.38
47. 2
48. no solution
12-9
Combinations
ALGEBRA 1 LESSON 12-9
1. Simplify 30C4.
27,405
2. You have just received a box of chocolates. There are three turtles,
four caramels, and three chocolate-covered cherries. If you choose
two chocolates at random, what is the probability that the first one is
caramel and the second is a turtle?
2
15
3. There are twelve congressmen who want to be on the same
committee. Eight are Republicans and four are Democrats. If three
people are chosen for the committee, what is the probability that
three Republicans are chosen?
14
55
4. A friend has loaned you five books and you plan to read all of them.
In how many ways could you read the five books?
120 ways
12-9
Rational Expressions and Functions
ALGEBRA 1 CHAPTER 12
1. 30
8. x = 0; y = 0
2. 7.8
3. –24
4. 48.23
5. D
6. 1.2 h
7. x = 0; y = 0
9. x = 0; y = 3
12-A
10. x = 0; y = 3
Rational Expressions and Functions
ALGEBRA 1 CHAPTER 12
11. Answers may vary. Sample:
In direct variation and in inverse
variation, the variables are related
to each other by a constant. But in
direct variation, that number is the
ratio of any corresponding pair of
input and output values. Distance
traveled varies directly with average
speed. In inverse variation, that
number is the product of any
corresponding pair of input and
output values. The cost per person
of splitting a $14 pizza varies inversely
with the number of people who
are sharing it.
13. 1
12. x + 2
22. –6, –2
4
6x2
14. 14
3w – 5
2c(3c – 1)
15.
3(c + 5)(c – 3)
16. Answers may vary. Sample:
x+2
(x – 6)(x – 3)
17. 4x + 3 – 10
3x
18. x3 – 2x2 + 4x – 8
7
2x – 1
20. 2x2 – 5x – 2 + 17
3x + 2
21. 11 h
2
19. 2x3 – 2x2 – x +
23. 4
12-A
Rational Expressions and Functions
ALGEBRA 1 CHAPTER 12
24. no solution
34. 5
2
25. t + 5t + 5
35. 35
t(t + 1)
26. n + 9
n(n + 1)
27. 1
y+3
28. 7b 2 + 6b – 4
3b(b + 2)
36. 4
37. 20,160
38. 604,800
39. 10
29. Permutation; order is important;
870 different pairs.
40. 20 different combinations
30. Combination; order is not important;
15 different ways.
31. Combination; order is not important;
15 different pairs of toppings.
32. 15
33. 70
12-A
41. 33
91