340
(6-38)
Chapter 6
Rational Expressions
6.6
In this
section
●
Equations with Rational
Expressions
●
Extraneous Solutions
SOLVING EQUATIONS WITH
RATIONAL EXPRESSIONS
Many problems in algebra can be solved by using equations involving rational
expressions. In this section you will learn how to solve equations that involve rational expressions, and in Section 6.7 and 6.8 you will solve problems using these
equations.
Equations with Rational Expressions
We solved some equations involving fractions in Section 2.3. In that section the
equations had only integers in the denominators. Our first step in solving those
equations was to multiply by the LCD to eliminate all of the denominators.
E X A M P L E
1
Integers in the denominators
Solve 1 2
helpful
hint
Note that it is not necessary
to convert each fraction into
an equivalent fraction with a
common denominator here.
Since we can multiply both
sides of an equation by any
expression we choose, we
choose to multiply by the
LCD. This tactic eliminates
the fractions in one step.
x2
3
1
6.
Solution
The LCD for 2, 3, and 6 is 6. Multiply each side of the equation by 6:
1 x2 1
3
2
6
1 x2
6 2
3
2
1
x2
6 6 2
3
3 2(x 2) 3 2x 4 2x x
1
6 6
1
6 6
1
1
6
3
Original equation
Multiply each side by 6.
Distributive property
Simplify.
Distributive property
Subtract 7 from each side.
Divide each side by 2.
Check x 3 in the original equation:
1 32 1 1 3 2 1
2
3
2 3 6 6 6
The solution to the equation is 3.
■
When a numerator contains a binomial, as in Example 1, the
CAUTION
numerator must be enclosed in parentheses when the denominator is eliminated.
To solve an equation involving rational expressions, we usually multiply each
side of the equation by the LCD for all the denominators involved, just as we do for
an equation with fractions.
6.6
E X A M P L E
2
Solving Equations with Rational Expressions
(6-39)
341
Variables in the denominators
1
1
1
Solve x 6 4.
Solution
We multiply each side of the equation by 12x, the LCD for 4, 6, and x:
1 1 1
Original equation
x
6 4
1
1 1
12x 12x Multiply each side by 12x.
x
6
4
2
3
1
1
1
12x 12 x 12 x x
6
4
12 2x 3x
12 x
Distributive property
Simplify.
Subtract 2x from each side.
Check that 12 satisfies the original equation:
1
1
1
2
3
1
12 6 12 12 12 4
E X A M P L E
3
An equation with two solutions
0
Solve the equation 10
x
study tip
Your mood for studying
should match the mood in
which you are tested. Being
too relaxed during studying
will not match the increased
level of activation you attain
during a test. Likewise, if you
get too tensed-up during a
test, you will not do well because your test-taking mood
will not match your studying
mood.
■
100
x5
9.
Solution
The LCD for the denominators x and x 5 is x(x 5):
100
100
Original equation
9
x5
x
100
100
Multiply each side by
x(x 5) x(x 5) x(x 5)9
x(x 5).
x5
x
All denominators are
(x 5)100 x(100) (x2 5x)9
eliminated.
2
100x 500 100x 9x 45x
Simplify.
2
500 200x 9x 45x
0 9x2 155x 500 Get 0 on one side.
0 (9x 25)(x 20) Factor.
9x 25 0
or x 20 0
Zero factor property
25
x or
x 20
9
A check will show that both 295 and 20 satisfy the original equation.
■
Extraneous Solutions
In a rational expression we can replace the variable only by real numbers that do not
cause the denominator to be 0. When solving equations involving rational expressions, we must check every solution to see whether it causes 0 to appear in a
denominator. If a number causes the denominator to be 0, then it cannot be a solution to the equation. A number that appears to be a solution but causes 0 in a denominator is called an extraneous solution.
342
(6-40)
Chapter 6
E X A M P L E
4
Rational Expressions
An equation with an extraneous solution
Solve the equation 1 x 1.
x2
2x 4
Solution
Because the denominator 2x 4 factors as 2(x 2), the LCD is 2(x 2).
1
x
2(x 2) 2(x 2) 2(x 2) 1
x2
2(x 2)
2 x 2x 4
2 3x 4
6 3x
2x
Multiply each side of the
original equation by 2(x 2).
Simplify.
Check 2 in the original equation:
1
2
1
22 224
The denominator 2 2 is 0. So 2 does not satisfy the equation, and it is an extra■
neous solution. The equation has no solutions.
E X A M P L E
5
Another extraneous solution
Solve the equation 1 x
1
x3
x 2
.
x3
Solution
The LCD for the denominators x and x 3 is x(x 3):
x2
1
1
x
x3 x3
x2
1
1
x(x 3) x(x 3) x(x 3) x3
x
x3
x 3 x x(x 2)
2x 3 x2 2x
0 x2 4x 3
0 (x 3)(x 1)
x30
or
x10
x3
or
x1
Original equation
Multiply each side by x(x 3).
If x 3, then the denominator x 3 has a value of 0. If x 1, the original equa■
tion is satisfied. The only solution to the equation is 1.
CAUTION
Always be sure to check your answers in the original equation
to determine whether they are extraneous solutions.
6.6
WARM-UPS
Solving Equations with Rational Expressions
(6-41)
343
True or false? Explain your answers.
1. The LCD is not used in solving equations with rational expressions. False
2. To solve the equation x 2 8x, we divide each side by x. False
3. An extraneous solution is an irrational number. False
Use the following equations for Questions 4–10.
3
5
2
a) x x2 3
4.
5.
6.
7.
8.
9.
10.
6. 6
1 1 3
b) x 2 4
1
1
c) 2 x1
x1
To solve Eq. (a), we must add the expressions on the left-hand side. False
Both 0 and 2 satisfy Eq. (a). False
To solve Eq. (a), we multiply each side by 3x2 6x. True
The only solution to Eq. (b) is 4. True
Equation (b) is equivalent to 4 2x 3x. True
To solve Eq. (c), we multiply each side by x2 1. True
The numbers 1 and 1 do not satisfy Eq. (c). True
EXERCISES
Reading and Writing After reading this section, write out the
answers to these questions. Use complete sentences.
1. What is the typical first step for solving an equation involving rational expressions?
The first step is usually to multiply each side by the LCD.
2. What is the difference in procedure for solving an equation
involving rational expressions and adding rational expressions?
In adding rational expressions we build up each expression
to get the LCD as the common denominator.
3. What is an extraneous solution?
An extraneous solution is a number that appears when
we solve an equation, but it does not check in the original
equation.
4. Why do extraneous solutions sometimes occur for equations with rational expressions?
Extraneous solutions occur because we multiply by an expression involving a variable and that variable expression
might have a value of zero.
Solve each equation. See Example 1.
x
x
x x x
5. 5 7 12
6. 11 30
3
2
3 2 5
y 2 y 1
z 5 z 3
8. 6
7. 30
5 3 6 3
6 4 2 4
t
3
t4
4 v1 v5
9. 5 10. 8
5
10
30
12
4
3
1
1 w 10
w1
11. 4
10
5
15
6
q q 1 13 q 1
12. 2
5
2
20
4
Solve each equation. See Example 2.
1 1 3
3 1 5
13. 4
14. 8
x 2 4
x 4 8
2
1
7
1
1
1
16. 3
15. 4
3x 2x 24
6x 8x 72
1 a2 a2
17. 3
2
a
2a
1 1 b1
3
18. 4
5b
b 5
10
1 k3
k1
1
19. 2
3
6k
3k
2k
3 p 3 2p 1 5
20. 5
p
3p
2p
6
Solve each equation. See Example 3.
x
5
x
4
21. 5, 2
22.
4, 3
2 x3
3 x1
1 1
2
1
1
3 5
24. , 4
23. 2, 3
x1 x 6
w 1 2w 40 3
a4
a1
1
25. 3, 3
a2 4 a 2 a 2
b 17
1
b2
26. 4, 5
2
b 1
b1 b1
Solve each equation. Watch for extraneous solutions. See
Examples 4 and 5.
2
1
x
27. 2
x1 x x1
344
(6-42)
Chapter 6
Rational Expressions
3
x
4
1
28. 6
x x3 x3 3
x1
5
2
29. No solution
x2 x3 x3
y1
6
7
30. No solution
y2 y8 y8
3y
6
31. 1 No solution
y2 y2
5
y7
32. 1 No solution
y 3 2y 6
z
1
2z 5
33. 3
z 1 z 2 z2 3z 2
z
1
7
34. 1
2
z 2 z 5 z 3z 10
In Exercises 35–56, solve each equation.
a 5
35. 10
4 2
y 6 18
36. 3 5 5
w 3w
37. 0
6
11
2m 3m
38. 0
3
2
x
5
39. 5, 5
x 5
3
x
40. 3, 3
x
3
x3 x3
41. 3, 5
5
x
a4 a4
42. 4, 2
2
a
1
x
43. 1
x2 x2
3
w
44. 3
w2 w2
1
1
3
45. 3
2x 4 x 2 2
7
1
4
46. 4
3x 9 x 3 3
3
2
47. 0
a2 a 6 a2 4
8
6
48. 6
a2 a 6 a2 9
4
1
25
49. 4
c2 2c c6
3
1
10
50. 3
x 1 1 x x2 1
1
3
4
51. 20
2
x 9 x3 x3
3
5
1
52. 9
x 2 x 3 x2 x 6
3
1
1
53. 3
2x 4 x 2 3x 1
5
1
1
54. 3
2m 6 m 1 m 3
2t 1 3t 1
t
55. 3
3t 3 6t 6 t 1
4w 1 w 1 w 1
56. 2
3w 6
3
w2
Solve each problem.
57. Lens equation. The focal length f for a camera lens is related to the object distance o and the image distance i by the
formula
1 1 1
.
f
o
i
See the accompanying figure. The image is in focus at distance i from the lens. For an object that is 600 mm from a
50-mm lens, use f 50 mm and o 600 mm to find i.
6
54 mm
11
o
i
FIGURE FOR EXERCISE 57
58. Telephoto lens. Use the formula from Exercise 57 to find
the image distance i for an object that is 2,000,000 mm
from a 250-mm telephoto lens.
250.03 mm
FIGURE FOR EXERCISE 58