Download Rational Equations and Inequalities in One Variables

SECT
ION
7.5
7.5
OBJECTIVES
Rational Equations and
Inequalities in One Variable
“One person’s constant is another person’s variable.”
–Susan Gerhart
1. Solve rational
equations in one
variable algebraically
2. Solve literal
equations involving a rational expression
3. Solve rational inequalities in one
variable algebraically
Example 1
Rational Equations
Applications of your work in algebra will often result in equations involving rational
expressions. Our objective in this section is to develop methods to find solutions for
such equations.
The usual technique for solving such equations is to multiply both sides of the
equation by the lowest common denominator (LCD) of all the rational expressions appearing in the equation. The resulting equation will be cleared of fractions, and we can
then proceed to solve the equation as before. Example 1 illustrates the process.
Clearing Equations of Fractions
Solve.
x
2x
13
5
3
The LCD for 3 and 5 is 15. Multiplying both sides of the equation by 15, we have
2x
x
15 15 13
3
5
Distribute 15 on the left.
2x
x
15 15 15 13
3
5
10x 3x 195
13x 195
x 15
The solution set is {15}.
528
Simplify. The equation is now
cleared of fractions.
Section 7.5
■
Rational Equations and Inequalities in One Variable
529
To check, substitute 15 in the original equation.
2 15
15
13
3
5
10 3 13
13 13
Caution
!
A true statement.
So 15 is the solution for the equation.
Be Careful! A common mistake is to confuse an equation such as
2x
x
13
3
5
and an expression such as
2x
x
3
5
Let’s compare.
Equation:
2x
x
13
3
5
Here we want to solve the equation for x, as in Example 1. We multiply both sides by
the LCD to clear fractions and proceed as before.
Expression:
2x
x
3
5
Here we want to find a third fraction that is equivalent to the given expression. We
write each fraction as an equivalent fraction with the LCD as a common denominator.
2x
x
2x 5
x3
3
5
35
53
10x
3x
10x 3x
15
15
15
13x
15
✓ CHECK YOURSELF 1
■
Solve.
3x
x
7
2
3
530
Chapter 7
■
Rational Expressions
The process is similar when variables are in the denominators. Consider Example 2.
Example 2
Solving an Equation Involving Rational Expressions
Solve.
We assume that x cannot
have the value 0. Do you see
why?
7
3
1
2 2
4x
x
2x
The LCD of 4x, x2, and 2x2 is 4x2. So, multiplying both sides by 4x2, we have
7
3
1
4x2 2 4x2 2
4x
x
2x
7
3
1
4x2 4x2 2 4x2 4x
x
2x2
Distribute 4x2 on the left side.
Simplify.
7x 12 2
7x 14
x2
The solution set is {2}.
We leave the check of the solution, x 2, to you. Be sure to return to the original equation and substitute 2 for x.
✓ CHECK YOURSELF 2
■
Solve.
5
4
7
2 2
2x
x
2x
Example 3 illustrates the same solution process when there are binomials in the
denominators.
Example 3
Solving an Equation Involving Rational Expressions
Solve.
Here we assume that x
cannot have the value 2 or
3.
4
3x
3 x2
x3
Section 7.5
■
531
Rational Equations and Inequalities in One Variable
The LCD is (x 2)(x 3). Multiplying by that LCD, we have
Note that multiplying each
term by the LCD is the same
as multiplying both sides of
the equation by the LCD.
4
3x
(x 2)(x 3) (x 2)(x 3)(3) (x 2)(x 3) x2
x3
Or, simplifying each term, we have
4(x 3) 3(x 2)(x 3) 3x(x 2)
We now clear the parentheses and proceed as before.
4x 12 3x2 3x 18 3x2 6x
3x2 x 30 3x2 6x
x 30 6x
5x 30
x 6
The solution set is {6}.
Again, we leave the check of this solution to you.
✓ CHECK YOURSELF 3
■
Solve.
5
2x
2 x4
x3
Factoring plays an important role in solving equations containing rational
expressions.
Example 4
Solving an Equation Involving Rational Expressions
Solve.
3
7
2
x3
x3
x2 9
In factored form, the denominator on the right side is (x 3)(x 3), which forms the
LCD, and we multiply each term by that LCD.
3
7
2
(x 3)(x 3) (x 3)(x 3) (x 3)(x 3) x3
x3
(x 3)(x 3)
532
Chapter 7
■
Rational Expressions
Again, simplifying each term on the right and left sides, we have
3(x 3) 7(x 3) 2
3x 9 7x 21 2
4x 28
x7
The solution set is {7}.
Be sure to check this result by substitution in the original equation.
✓ CHECK YOURSELF 4
■
4
3
5
Solve .
x4
x1
x2 3x 4
Whenever we multiply both sides of an equation by an expression containing a
variable, there is the possibility that a proposed solution may make that multiplier 0.
As we pointed out earlier, multiplying by 0 does not give an equivalent equation, and
therefore verifying solutions by substitution serves not only as a check of our work but
also as a check for extraneous solutions. Consider Example 5.
Example 5
Solving an Equation Involving Rational Expressions
Solve.
x
2
7 x2
x2
Note that we must assume
that x 2.
The LCD is x 2, and multiplying, we have
(x 2) 7(x 2) (x 2)
x 2
x2
x
Note that each of the three
terms gets multiplied by
(x 2).
2
Simplifying yields
x 7(x 2) 2
x 7x 14 2
6x 12
x2
Section 7.5
■
Rational Equations and Inequalities in One Variable
533
To check this result, by substituting 2 for x, we have
Caution
!
Because division by 0 is
undefined, we conclude that
2 is not a solution for the
original equation. It is an
extraneous solution. The
original equation has no
solution.
2
2
7 22
22
2
2
7 0
0
The solution set is empty. The set is written {} or .
✓ CHECK YOURSELF 5
■
x3
1
Solve 4 .
x4
x4
Equations involving rational expressions may also lead to quadratic equations, as
illustrated in Example 6.
Example 6
Solving an Equation Involving Rational Expressions
Solve.
Assume x 3 and x 4.
x
15
2x
x4
x3
x2 7x 12
After factoring the trinomial denominator on the right, the LCD of x 3, x 4,
and x2 7x 12 is (x 3)(x 4). Multiplying by that LCD, we have
x
15
2x
(x 3)(x 4) (x 3)(x 4) (x 3)(x 4) x4
x3
(x 3)(x 4)
Simplifying yields
x(x 3) 15(x 4) 2x
Remove the parentheses.
x 3x 15x 60 2x
Write in standard form and factor.
2
x2 16x 60 0
(x 6)(x 10) 0
So
x6
or
x 10
Verify that 6 and 10 are both solutions for the original equation. The solution set is
{6, 10}.
534
Chapter 7
■
Rational Expressions
✓ CHECK YOURSELF 6
■
3x
2
36
Solve .
x2
x3
x2 5x 6
The following algorithm summarizes our work in solving equations containing rational expressions.
Solving Equations Containing Rational Expressions
Step 1 Clear the equation of fractions by multiplying both sides of the equation
by the LCD of all the fractions that appear.
Step 2 Solve the equation resulting from step 1.
Step 3 Check all solutions by substitution in the original equation.
The method in this section may also be used to solve certain literal equations for
a specified variable. Consider Example 7.
Example 7
A parallel electric circuit. The
symbol for a resistor is
Solving a Literal Equation
R1
R2
Recall that the numbers 1 and
2 are subscripts. We read R1
as “R sub 1” and R2 as “R
sub 2.”
If two resistors with resistances R1 and R2 are connected in parallel, the combined resistance R can be found from
1
1
1
R
R1
R2
Solve the formula for R.
First, the LCD is RR1R2, and we multiply:
1
1
1
RR1R2 RR1R2 RR1R2 R
R1
R2
Section 7.5
■
Rational Equations and Inequalities in One Variable
535
Simplifying yields
R1R2 RR2 RR1
Factor out R on the right.
R1R2 R(R2 R1)
Divide by R2 R1 to isolate R.
R R2
1
R
R2 R1
R R2
R 1
R1 R2
or
✓ CHECK YOURSELF 7
■
Solve for D1.
1
1
1
F
D1
D2
Note: This formula involves
the focal length of a convex
lens.
The Distance Formula
drt
Distance equals rate times time.
Sometimes use of this formula leads to a rational equation, as in Example 8.
Example 8
Solving a Motion Problem
A boat can travel 16 miles per hour in still water. If the boat can travel 5 miles downstream in the same time it takes to travel 3 miles upstream, what is the rate of the river’s
current?
Step 1
We want to find the rate of the current.
Step 2
Let c be the rate of the current. Then 16 c is the rate of the boat going
downstream and 16 c is the rate going upstream.
Step 3
We’ll use a chart to help us set up an appropriate equation.
d
r
t
Downstream
5 miles
16 c
Upstream
3 miles
16 c
5
16 c
3
16 c
536
Chapter 7
■
Rational Expressions
d
We found the time by dividing distance by rate. If d r t, then t . The key to
r
finding our equation is noting that the time is the same upstream and downstream, so
5
3
16 c
16 c
Step 4
Multiplying both sides by the LCD (16 c)(16 c) yields
5(16 c) 3(16 c)
80 5c 48 3c
32 8c
c4
The current is moving at 4 miles per hour.
Step 5
5
3
To check, verify that .
16 4
16 4
✓ CHECK YOURSELF 8
■
A plane flew 540 miles into a steady 30 mi/h wind. The pilot then returned along the
same route with a tailwind. If the entire trip took seven and one-half hours, what would
his speed have been in still air?
Another application that frequently results in rational equations is the work problem. Example 9 illustrates.
Solving a Work Problem
Example 9
One computer printer can print a company’s paychecks in 40 minutes (min). A second
printer can print them in 80 min. If both printers are working, how long will it take to
print the paychecks?
Before we learn how to solve work problems, we should look at a couple of common
errors made in attempting to solve such problems.
Caution
!
Error 1: Students sometimes try adding the two times together. If we add the 40
min and the 80 min we get 120 min. Is this a reasonable answer? Certainly not!
If one printer does the job in 40 min, why would it take 120 min for two printers
to do it? It wouldn’t.
Section 7.5
Caution
!
■
Rational Equations and Inequalities in One Variable
537
Error 2: A more reasonable approach would be to give half the job to each printer.
The first printer would finish its half of the job in 20 min. The second would finish its half in 40 min. The first printer would be idle for the final 20 min, so we
know the job could have been finished faster.
Although error 2 does not solve the problem, it does give us guidelines for a reasonable answer. When the printers work together, it will take them somewhere between
20 and 40 min to finish the job. To find the exact amount of time, we use the work
principle.
The Work Principle: Given an object, A, which completes a task in time a, and
an object, B, which completes the same task in time b, we find the amount of time
it takes them to complete the task together by solving the following equation for
variable t.
t
t
1
a
b
Now we can solve the problem.
Step 1
We are looking for the time it takes to print the paychecks.
Step 2
Our variable, t is the time it will take the printers to complete the task.
t
t
We have the equation 1
40
80
Multiply by LCD 80.
Step 3
Step 4
2t t 80
3t 80
80
t min
3
2
26 min
3
26 min 40 sec
Step 5
To verify this answer, we find what fraction of the job each printer does in
80
80
3
3
2
1
this time. The first printer does — of the job. The second printer does — 40
3
80
3
2
1
of the job. Together, they do the entire job 1 .
3
3
✓ CHECK YOURSELF 9
■
It would take Sasha 48 days to paint the house. Natasha could do it in 36 days. How
long would it take them to paint the house if they worked together?
538
Chapter 7
■
Rational Expressions
The techniques we have discussed can also be used to find the zeros of rational
functions. Remember that a zero of a function is a value of x for which f(x) 0.
Example 10
Finding the Zeros of a Function
Find the zeros of
1
3
4
f(x) x
7x
21
Set the function equal to 0, and solve the resulting equation for x.
1
3
4
f(x) 0
x
7x
21
The LCD for x, 7x, and 21 is 21x. Multiplying both sides by 21x, we have
1
3
4
21x 21x 0
x
7x
21
21 9 4x 0
12 4x 0
Distribute 21x on the left side.
Simplify.
12 4x
3x
So 3 is the value of x for which f(x) 0, that is, 3 is a zero of f(x).
✓ CHECK YOURSELF 10
■
Find the zeros of the function.
5x 2
11
f(x) x5
4
Rational Inequalities
To solve inequalities involving rational expressions, we need some properties of division over the real numbers. Recall that
1.
The quotient of two positive numbers is always positive.
2.
The quotient of two negative numbers is always positive.
3.
The quotient of a positive number and a negative number is always negative.
We solve rational inequalities by using sign graphs, as Example 11 illustrates.
Section 7.5
■
Rational Equations and Inequalities in One Variable
539
Solving a Rational Inequality
Example 11
Solve.
x3
0
x2
x3
The graph of y is
x2
y
x
The inequality states that the quotient of x 3 and x 2 must be negative (less than
0). This means that the numerator and denominator must have opposite signs.
We start by finding the critical points. These are points where either the numerator or denominator is 0. In this case, the critical points are 3 and 2.
The solution depends on determining whether the numerator and denominator are
positive or negative. To visualize the process, start with a number line and label it as
shown below.
Sign of x 3 Sign of x 2 Sign of quotient For what values of x is y
less than 0?
x3
x2
2
3
Examining the sign of the numerator and denominator, we see that
For any x less than 2, the quotient is positive (quotient of two negatives).
For any x between 2 and 3, the quotient is negative (quotient of a negative and
a positive).
For any x greater than 3, the quotient is positive (quotient of two positives).
We return to the original inequality
x3
0
x2
This inequality is true only when the quotient is negative, that is, when x is between
2 and 3. This solution set can be written as {x
2 x 3} and represented on a
graph as follows.
2
3
540
Chapter 7
■
Rational Expressions
✓ CHECK YOURSELF 11
■
Solve and graph the solution set.
x4
0
x2
The solution process illustrated in Example 11 is valid only when the rational expression is isolated on one side of the inequality and is related to 0. If this is not the
case, we must write an equivalent inequality as the first step, as Example 12 illustrates.
Example 12
Solving a Rational Inequality
Solve.
2x 3
1
x1
Since the rational expression is not related to 0, we use the following procedure.
We have subtracted 1 from
both sides.
2x 3
1 0
x1
2x 3
x1
0
x1
x1
2x 3 (x 1)
0
x1
Form a common denominator on the left side.
Combine the expressions on the left side.
Simplify.
x4
0
x1
We can now proceed as before since the rational expression is related to 0. The critical points are 4 and 1, and the sign graph is formed as shown below.
Sign of x 4 Sign of x 1 Sign of quotient x4
x1
1
4
Section 7.5
We want a positive quotient.
■
Rational Equations and Inequalities in One Variable
From the sign graph the solution is
{x
x 1 or x 4}
The graph is shown below.
Note that 4 is included, but
1 cannot be included in the
solution set. Why?
[
1
4
✓ CHECK YOURSELF 12
■
Solve and graph the solution set.
2x 3
1
x2
✓ CHECK YOURSELF ANSWERS
■
1. {6}.
2. {3}.
8
6. 5, .
3
3. {9}.
4. {11}.
FD2
7. .
8. 150 mi/h.
D2 F
11. {x
x 2 or x 4}
12. {x
1 x 2}
2
4
[
1
2
5. No solution or .
4
9. 20 days.
10. 7.
7
541
Exercises
1. Equation, {36}
2. Equation, {28}
3x
3. Expression, 10
x
4. Expression, 24
5. Equation, {5}
21x 25
6. Expression, 20
■
7.5
In Exercises 1 to 8, decide whether each of the following is an expression or an equation. If it is an equation, find a solution. If it is an expression, write it as a single
fraction.
x
x
1. 6
2
3
x
x
2. 3
4
7
x
x
3. 2
5
x
x
4. 6
8
3x 1
5. x 1
4
3x 1
x
x3
6. 2
5
4
x
x
1
7. 4
12
2
2x 1
x
8. 3
2
7. Equation, {3}
7x 2
8. Expression, 6
9. {5}
10. {7}
11. {8}
3
13. 2
12. {6}
15. {1}
16. {3}
9
17. 5
2
18. 3
2
19. 3
20. {3}
14. {2}
21. No solution or 22. {7}
23. {23}
24. {5}
25. {6}
26. x is any real except
x 0 and x 2
27. {4}
28. {6}
29. {8}
30. {5}
31. {4}
32. {4}
542
In Exercises 9 to 50, solve each equation.
x
3
x
7
9. 3
2
6
3
x
1
x
1
10. 10
5
5
2
4
3
10
11. x
4
x
3
5
7
12. x
3
x
5
1
1
13. 4x
2
2x
7
1
1
14. 6x
3
2x
3
2
15. x4
x3
5
4
16. x2
x1
9
2x
17. 2 x
x3
6
3x
18. 3 x
x1
3
5
13
19. x2
x
x2
7
2
6
20. x
x3
x
3
2
1
21. 2
2x 4
x2
2
5
3
22. x1
2x 2
4
x
x1
5
23. 3x 12
x4
3
x
x4
1
24. 4x 12
x3
8
x1
x3
3
25. x3
x
x2 3x
x1
x3
6
26. x2
x
x2 2x
1
2
2
27. x2
x2
x2 4
1
1
12
28. x4
x4
x2 16
7
1
x
29. x5
x5
x2 25
2
3
x
30. x2
x2
x2 4
11
5
1
31. x2
x2 x 6
x3
5
1
2
32. x4
x2
x2 2x 8
Section 7.5
3
33. 2
11
34. 2
35. No solution or 36. No solution or 37. {7}
38. {3}
39. {5}
40. {3}
5
41. 2
42. No solution or 1
43. , 6
2
■
Rational Equations and Inequalities in One Variable
5
3
24
33. x2
x3
x2 x 6
3
5
2
34. x1
x6
x2 7x 6
x
3
35. 2 x3
x3
x
5
36. 2 x5
x5
2
1
2
37. x2 3x
x2 2x
x2 x 6
543
2
4
3
38. x2 x
x2 5x 6
x2 6x
2
3
2
39. 2
2 2
x 4x 3
x 9
x 2x 3
2
1
3
40. x2 4
x2 x 2
x2 3x 2
44. {4, 3}
1
45. 2
2
46. 3
1
47. , 7
3
1
48. , 6
2
49. {8, 9}
9
50. , 10
2
ab
bx
51. 52. ba
bx
RR2
53. R2 R
D F
54. 1
D1 F
y1
55. y1
2y 3
56. y1
A
57. 1 rt
IR
58. E Ir
7
3
40
41. x5
x5
x2 25
3
18
5
42. x3
x2 9
x3
2x
2
3x
43. x3
x5
x2 8x 15
x
5x
3
44. x4
x2 x 12
x3
2x
5
1
45. 2
x2
x x6
x3
3x
2
2
46. 2
x1
x2
x 3x 2
7
16
47. 3
x2
x3
5
6
48. 2
x2
x2
11
10
49. 1 x3
x3
17
10
50. 2 x4
x2
In Exercises 51 to 58, solve each equation for the indicated variable.
1
1
1
51. for x
x
a
b
1
1
1
52. for a
x
a
b
1
1
1
53. for R1
R R1
R2
1
1
1
54. for D2
F
D1
D2
x1
55. y for x
x1
x3
56. y for x
x2
AP
57. t for P
Pr
nE
58. I for n
R nr
544
Chapter 7
■
Rational Expressions
59. {x
1 x 2}
In Exercises 59 to 72, solve each inequality, and graph the solution set.
60. {x
x 3 or x 2}
x2
59. 0
x1
x3
60. 0
x2
x4
61. 0
x2
61. {x
x 2 or x 4}
x6
62. 0
x3
x5
63. 0
x3
x3
64. 0
x2
62. {x
6 x 3}
2x 1
65. 0
x3
3x 2
66. 0
x4
x
2
67. 0
x3
x3
x
3
68. 0
x5
x5
x
4
69. x3
x3
x
2
70. x5
x5
2x 5
71. 1
x2
2x 3
72. 1
x4
63. {x
3 x 5}
64. {x
3 x 2}
1
65. x
x 3 or x 2
2
66. x
x 4
3
In Exercises 73 to 80, find the zeros of each function.
x
12
73. f(x) 10
5
4x
x
74. f(x) 3
6
12
5
75. f(x) x5
x
1
3
76. f(x) x2
x
67. {x
2 x 3}
68. {x
x 5 or x 3}
69. {x
3 x 4}
70. {x
x 2 or x 5}
72. {x
x 4 or x 1}
73. 24
74. 0
71. {x
x 2 or x 3}
25
75. 7
76. 3
Section 7.5
■
Rational Equations and Inequalities in One Variable
3
77. 4
2
1
5
77. f(x) x
x3
3x
2
3
1
78. f(x) x
x1
x2 x
78. 1
39
16
79. f(x) 1 x2
x
72
80. f(x) x 1
x
79. 3, 13
80. 9, 8
81. 6, 24
82. 4, 6
83. 9
84. 5
85. 4 mi/h
86. 4 mi/h
87. 150 mi/h
88. 40 mi/h
89. 36 min
90. 30 h
545
Solve the following problems.
81. Number analysis. One number is 4 times another number. The sum of the recip5
rocals of the numbers is . Find the two numbers.
24
82. Number analysis. The sum of the reciprocals of two consecutive even integers is
equal to 10 times the reciprocal of the product of those integers. Find the two
integers.
83. Number analysis. If the same number is subtracted from the numerator and deno11
1
minator of , the result is . Find that number.
15
3
8
84. Number analysis. The numerator of is multiplied by a number. That same num9
ber is subtracted from the denominator, and the result is 10. What is that number?
85. Motion. A motorboat can travel 20 mi/h in still water. If the boat can travel 3 mi
downstream on a river in the same time it takes to travel 2 mi upstream, what is
the rate of the river’s current?
86. Motion. Janet and Michael took a canoeing trip, traveling 6 mi upstream along a
river, against a 2 mi/h current. They then returned downstream to the starting point
of their trip. If their entire trip took 4 h, what was their rate in still water?
87. Motion. A plane flew 720 mi with a steady 30-mi/h tailwind. The pilot then returned to the starting point, flying against the same wind. If the round-trip flight
took 10 h, what was the plane’s airspeed?
88. Motion. A small jet has an airspeed (the rate in still air) of 300 mi/h. During one
day’s flights, the pilot noted that the plane could fly 85 mi with a tailwind in the
same time it took to fly 65 mi against the same wind. What was the rate of the
wind?
89. Work. One computer printer can print a company’s weekly payroll checks in 60
min. A second printer would take 90 min to complete the job. How long would it
take the two printers, operating together, to print the checks?
90. Work. An electrician can wire a house in 20 h. If she works with an apprentice,
the same job can be completed in 12 h. How long would it take the apprentice,
working alone, to wire the house?
91. What special considerations must be made when an equation contains rational expressions with variables in the denominator?
x
92. In solving the inequality 5, is it incorrect to find the solution by multix1
plying both sides by x 1? Why? What technique should be used?