Solving Rational Equations and Inequalities

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Solving Rational Equations and Inequalities
5-5
Warm Up
Lesson Presentation
Lesson Quiz
Holt Algebra 2
Holt McDougal Algebra 2
2
Warm Up Find the least common multiple for each
pair.
1. 2x2 and 4x2 2x
2x2(2x 1)
2. x 5 and x2 x 30
(x 5)(x 6)
Add or subtract. Identify any x-values for which
the expression is undefined.
3.
x ? 0
4.
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Objective
Solve rational equations and inequalities.
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Vocabulary
rational equation extraneous solution rational
inequality
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A rational equation is an equation that contains
one or more rational expressions. The time t in
hours that it takes to travel d miles can be
determined by using the equation t , where
r is the average rate of speed. This equation is
a rational equation.
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To solve a rational equation, start by
multiplying each term of the equation by the
least common denominator (LCD) of all of the
expressions in the equation. This step eliminates
the denominators of the rational expression and
results in an equation you can solve by using
algebra.
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Example 1 Solving Rational Equations
Solve the equation x 3.
Multiply each term by the LCD, x.
x2 18 3x
Simplify. Note that x ? 0.
x2 3x 18 0
Write in standard form.
(x 6)(x 3) 0
Factor.
x 6 0 or x 3 0
Apply the Zero Product Property.
x 6 or x 3
Solve for x.
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Example 1 Continued
x 3
Check
x 3
3
3
3
6 3
3
3 6
3
3
?
3
3
?
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Check It Out! Example 1a
Multiply each term by the LCD, 3x.
10x 12 6x
Simplify. Note that x ? 0.
4x 12
Combine like terms.
x 3
Solve for x.
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Check It Out! Example 1b
Multiply each term by the LCD, 4x.
24 5x 7x
Simplify. Note that x ? 0.
24 12x
Combine like terms.
x 2
Solve for x.
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Check It Out! Example 1c
Solve the equation x 1.
Multiply each term by the LCD, x.
x2 6 x
Simplify. Note that x ? 0.
x2 x 6 0
Write in standard form.
(x 2)(x 3) 0
Factor.
x 2 0 or x 3 0
Apply the Zero Product Property.
x 2 or x 3
Solve for x.
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An extraneous solution is a solution of an
equation derived from an original equation that
is not a solution of the original equation. When
you solve a rational equation, it is possible to
get extraneous solutions. These values should be
eliminated from the solution set. Always check
your solutions by substituting them into the
original equation.
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Example 2A Extraneous Solutions
Solve each equation.
Multiply each term by the LCD, x 2.
Divide out common factors.
Simplify. Note that x ? 2.
5x 3x 4
x 2
Solve for x.
The solution x 2 is extraneous because it makes
the denominators of the original equation equal
to 0. Therefore, the equation has no solution.
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Example 2A Continued
Check Substitute 2 for x in the original equation.
Division by 0 is undefined.
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Example 2B Extraneous Solutions
Solve each equation.

Multiply each term by the LCD, 2(x 8).
Divide out common factors.
2(2x 5) x(x 8) 11(2)
Simplify. Note that x ? 8.
Use the Distributive Property.
4x 10 x2 8x 22
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Example 2B Continued
x2 4x 32 0
Write in standard form.
(x 8)(x 4) 0
Factor.
x 8 0 or x 4 0
Apply the Zero Product Property.
x 8 or x 4
Solve for x.
The solution x 8 us extraneous because it makes
the denominator of the original equation equal to
0. The only solution is x 4.
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Example 2B Continued
Check
Write
as
Graph the left side of the equation as Y1.
Identify the values of x for which Y1 0.
The graph intersects the x-axis only when x 4.
Therefore, x 4 is the only solution.
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Check It Out! Example 2a
Solve the equation .

Multiply each term by the LCD, (x 4)(x 4).
Divide out common factors.
Simplify. Note that x ? 4.
16 2x 8
x 4
Solve for x.
The solution x 4 is extraneous because it makes
the denominators of the original equation equal
to 0. Therefore, the equation has no solution.
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Check It Out! Example 2b
Multiply each term by the LCD, 6(x 1).
Divide out common factors.
6 6x x(x x)
Simplify. Note that x ? 1.
Use the Distributive Property.
6 6x x2 x
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Check It Out! Example 2b Continued
Write in standard form.
0 x2 5x 6
0 (x 6)(x 1)
Factor.
Apply the Zero Product Property.
x 6 0 or x 1 0
Solve for x.
x 6 or x 1
The solution x 1 us extraneous because it makes
the denominator of the original equation equal to
0. The only solution is x 6.
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Example 3 Problem-Solving Application
A jet travels 3950 mi from Chicago, Illinois, to
London, England, and 3950 mi on the return trip.
The total flying time is 16.5 h. The return trip
takes longer due to winds that generally blow
from west to east. If the jets average speed
with no wind is 485 mi/h, what is the average
speed of the wind during the round-trip flight?
Round to the nearest mile per hour.
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Example 3 Continued
The answer will be the average speed of the wind.

• List the important information
• The jet spent 16.5 h on the round-trip.
• It went 3950 mi east and 3950 mi west.
• Its average speed with no wind is 485 mi/h.

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Example 3 Continued
Let w represent the speed of the wind. When the
jet is going east, its speed is equal to its
speed with no wind plus w. When the jet is going
west, its speed is equal to its speed with no
wind minus w.
Distance (mi) Average Speed (mi/h) Time (h)
East 3950 485 w
West 3950 485 w
total time

time east

time west
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The LCD is (485 w)(485 w).
Simplify. Note that x ? 485.
16.5(485 w)(485 w) 3950(485 w) 3950
(485 w)
Use the Distributive Property.
3,881,212.5 16.5w2 1,915,750 3950w
1,915,750 3950w
3,881,212.5 16.5w2 3,831,500
Combine like terms.
16.5w2 49,712.5
Solve for w.
w 55
The speed of the wind cannot be negative.
Therefore, the average speed of the wind is 55
mi/h.
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Example 3 Continued
If the speed of the wind is 55 mi/h, the jets
speed when going east is 485 55 540 mi/h. It
will take the jet approximately 7.3 h to travel
3950 mi east. The jets speed when going west is
485 55 430 mi/h. It will take the jet
approximately 9.2 h to travel 3950 mi west. The
total trip will take 16.5 h, which is the given
time.
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Check It Out! Example 3
On a river, a kayaker travels 2 mi upstream and 2
mi downstream in a total of 5 h. In still water,
the kayaker can travel at an average speed of 2
mi/h. Based on this information, what is the
average speed of the current of this river? Round
to the nearest tenth.
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Check It Out! Example 3 Continued
The answer will be the average speed of the
current.

• List the important information
• The kayaker spent 5 hours kayaking.
• She went 2 mi upstream and 2 mi downstream.
• Her average speed in still water is 2 mi/h.

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Check It Out! Example 3 Continued
Let c represent the speed of the current. When
the kayaker is going upstream, her speed is equal
to her speed in still water minus c. When the
kayaker is going downstream, her speed is equal
to her speed in still water plus c.
Distance (mi) Average Speed (mi/h) Time (h)
Up 2 2 c
Down 2 2 c
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The LCD is (2 c)(2 c).
Simplify. Note that x ? 2.
5(2 c)(2 c) 2(2 c) 2(2 c)
Use the Distributive Property.
20 5c2 4 2c 4 2c
20 5c2 8
Combine like terms.
5c2 12
Solve for c.
c 1.5
The speed of the current cannot be negative.
Therefore, the average speed of the current is
about 1.5 mi/h.
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Check It Out! Example 3 Continued
If the speed of the current is about 1.5 mi/h,
the kayakers speed when going upstream is 2
1.5 0.5 mi/h. It will take her about 4 h to
travel 2 mi upstream. Her speed when going
downstream is about 2 1.5 3.5 mi/h. It will
take her 0.5 h to travel 2 mi downstream. The
total trip will take about 4.5 hours which is
close to the given time of 5 h.
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Example 4 Work Application
Natalie can finish a 500-piece puzzle in about 8
hours. When Natalie and Renzo work together, they
can finish a 500-piece puzzle in about 4.5 hours.
About how long will it take Renzo to finish a
500-piece puzzle if he works by himself?
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Example 4 Continued

1

Multiply by the LCD,8h.
4.5h 36 8h
Simplify.
Solve for h.
36 3.5h
10.3 h
It will take Renzo about 10.3 hours, or 10 hours
17 minutes to complete a 500-piece puzzle working
by himself.
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Check It Out! Example 4
Julien can mulch a garden in 20 minutes. Together
Julien and Remy can mulch the same garden in 11
minutes. How long will it take Remy to mulch the
garden when working alone?
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Check It Out! Example 4 Continued

1

Multiply by the LCD,20m.
11m 220 20m
Simplify.
Solve for m.
220 9m
24 m
It will take Remy about 24 minutes to mulch the
garden working by himself.
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A rational inequality is an inequality that
contains one or more rational expressions. One
way to solve rational inequalities is by using
graphs and tables.
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Example 5 Using Graphs and Tables to Solve
Rational Equations and Inequalities
(9, 3)
The graph of Y1 is at or below the graph of Y2
when x lt 6 or when x 9.
Vertical asymptote x 6
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Example 5 Continued
Use a table. The table shows that Y1 is
undefined when x 6 and that Y1 Y2 when x 9.
The solution of the inequality is x lt 6 or x 9.
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Check It Out! Example 5a
(4, 4)
The graph of Y1 is at or below the graph of Y2
when x lt 3 or when x 4.
Vertical asymptote x 3
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Check It Out! Example 5a continued
Use a table. The table shows that Y1 is
undefined when x 3 and that Y1 Y2 when x 4.
The solution of the inequality is x lt 3 or x 4.
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Check It Out! Example 5b
(5, 2)
The graph of Y1 is at or below the graph of Y2
when x 5.
Vertical asymptote x 1
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Check It Out! Example 5b continued
Use a table. The table shows that Y1 is
undefined when x 1 and that Y1 Y2 when x
5.
The solution of the inequality is x 5.
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You can also solve rational inequalities
algebraically. You start by multiplying each term
by the least common denominator (LCD) of all the
expressions in the inequality. However, you must
consider two cases the LCD is positive or the
LCD is negative.
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Example 6 Solving Rational Inequalities
Algebraically
Case 1 LCD is positive.
Step 1 Solve for x.
Multiply by the LCD.
Simplify. Note that x ? 8.
6 3x 24
Solve for x.
30 3x
10 x
Rewrite with the variable on the left.
x 10
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Example 6 Continued
Step 2 Consider the sign of the LCD.
LCD is positive.
x 8 gt 0
Solve for x.
x gt 8
For Case 1, the solution must satisfy x 10 and
x gt 8, which simplifies to x 10.
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Example 6 Solving Rational Inequalities
Algebraically
Case 2 LCD is negative.
Step 1 Solve for x.
Multiply by the LCD. Reverse the inequality.
Simplify. Note that x ? 8.
6 3x 24
Solve for x.
30 3x
10 x
Rewrite with the variable on the left.
x 10
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Example 6 Continued
Step 2 Consider the sign of the LCD.
LCD is positive.
x 8 gt 0
Solve for x.
x gt 8
For Case 2, the solution must satisfy x 10 and
x lt 8, which simplifies to x lt 8.
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Check It Out! Example 6a
Case 1 LCD is positive.
Step 1 Solve for x.
Multiply by the LCD.
Simplify. Note that x ? 2.
6 4x 8
Solve for x.
2 4x
Rewrite with the variable on the left.
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Check It Out! Example 6a Continued
Step 2 Consider the sign of the LCD.
LCD is positive.
x 2 gt 0
Solve for x.
x gt 2
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Check It Out! Example 6a Continued
Case 2 LCD is negative.
Step 1 Solve for x.
Multiply by the LCD. Reverse the inequality.
Simplify. Note that x ? 2.
6 4x 8
Solve for x.
2 4x
Rewrite with the variable on the left.
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Check It Out! Example 6a Continued
Step 2 Consider the sign of the LCD.
LCD is negative.
x 2 lt 0
Solve for x.
x lt 2
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Check It Out! Example 6b
Case 1 LCD is positive.
Step 1 Solve for x.
Multiply by the LCD.
Simplify. Note that x ? 3.
9 lt 6x 18
Solve for x.
9 lt 6x
Rewrite with the variable on the left.
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Check It Out! Example 6b Continued
Step 2 Consider the sign of the LCD.
LCD is positive.
x 3 gt 0
Solve for x.
x gt 3
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Check It Out! Example 6b Continued
Case 2 LCD is negative.
Step 1 Solve for x.
Multiply by the LCD. Reverse the inequality.
Simplify. Note that x ? 3.
9 gt 6x 18
Solve for x.
9 gt 6x
Rewrite with the variable on the left.
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Check It Out! Example 6b Continued
Step 2 Consider the sign of the LCD.
LCD is negative.
x 3 lt 0
Solve for x.
x lt 3
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Lesson Quiz
Solve each equation or inequality.
1.
x 1 or x 4
2.
no solution
3.
x 5

4.
2
3 lt x 5
5. A college basketball player has made 58 out of
82 attempted free-throws this season. How many
additional free-throws must she make in a row to
raise her free-throw percentage to 90?
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