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Holt Algebra 1

12-7 Solving Rational Equations12-7 Solving Rational Equations

Holt Algebra 1

ObjectivesObjectives

NotesNotes

PracticePractice

Holt Algebra 1

12-7 Solving Rational Equations

Solve rational equations.

Identify extraneous solutions.

Objectives

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A rational equation is an equation that contains one or more rational expressions. If a rational equation is a proportion, it can be solved using the Cross Product Property.

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Example 1: Solving Rational Equations by Using Cross Products

Use cross products.

5x = (x – 2)(3)

5x = 3x – 6

2x = –6

Solve . Check your answer.

x = –3

Check

Distribute 3 on the right side.

Subtract 3x from both sides.

–1 –1

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Check It Out! Example 2

Solve . Check your answer.

21x = (x – 7)(3)

21x = 3x –21

18x = –21

x =

Use cross products.

Distribute 3 on the right side.

Subtract 3x from both sides.

Check

Divide both sides by 18.

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Some rational equations contain sums or differences of rational expressions. To solve these, you must find the LCD of all the rational expressions in the equation.

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Example 3: Solving Rational Equations by Using the LCD

Solve the equation. Check your answer.

Step 1 Find the LCD

2x(x + 1) Include every factor of the denominator.

Step 2 Multiply both sides by the LCD

Distribute on the left side.

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Example 3 Continued

Step 3 Simplify and solve.

(2x)(2) +6(x +1) = 5(x +1)

4x + 6x + 6 = 5x + 5

10x + 6 = 5x + 5

5x = –1

Divide out common factors.

Simplify.

Distribute and multiply.

Combine like terms.

Subtract 5x and 6 from both sides.


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Example 3 Continued

Check Verify that your solution is not extraneous.

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Example 4: Solving Rational Equations by Using the LCD

Solve the equation. Check your answer.

Step 1 Find the LCD

(x2) Include every factor of the denominator.


Distribute on the left side.

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Example 4 Continued

Step 3 Simplify and solve.

Divide out common factors.

4x – 3 = x2

0 = x2 – 4x + 3

(x – 3)(x – 1) = 0

x = 3, 1

Simplify.Subtract 4x and -3

from both sides.

Factor.

Solve.

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Example 4 Continued


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Example 5

Solve each equation. Check your answer.

Step 1 Find the LCD

t(t +3) Include every factor of the denominator.


Distribute on the right side.

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Example 5 Continued

Solve each equation. Check your answer.

Divide out common terms.

8t = (t + 3) + t(t + 3)

8t = t + 3 + t2 + 3t

0 = t2 – 4t + 3

0 = (t – 3)(t – 1)

t = 3, 1

Simplify.

Distribute t.

Combine like terms.

Factor.

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Example 5 Continued


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Example 6: Problem-Solving Application

Copy machine A can make 200 copies in 60 minutes. Copy machine B can make 200 copies in 10 minutes. How long will it take both machines working together to make 200 copies?

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List the important information:

• Machine A can print the copies in 60 minutes, which is of the job in 1 minute.

• Machine B can print the copies in 10 minutes, which is of the job in 1 minute.

11 Understand the Problem

The answer will be the number of minutes m machine A and machine B need to print the copies.

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The part of the copies that machine A can print plus the part that machine B can print equals the complete job. Machine A’s rate times the number of minutes plus machine B’s rate times the number of minutes will give the complete time to print the copies.

22 Make a Plan

(machine A’s rate)

m (machine B’s rate)

m+ complete job

=

m m+ = 1

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Solve33

Multiply both sides by the LCD, 60.

1m + 6m = 60 Distribute 60 on the left side.

7m = 60 Combine like terms.


Machine A and Machine B working together can print the copies in a little more than 8.5 minutes.

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Look Back44

Machine A prints of the copies per minute and machine B prints of the copies per minute. So in minutes, machine A prints of the copies and machine B prints

of the copies. Together, they print

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When you multiply each side of an equation by the LCD, you may get an extraneous solution. An extraneous solution is a solution to a resulting equation that is not a solution to the original equation.

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Extraneous solutions may be introduced by squaring both sides of an equation or by multiplying both sides of an equation by a variable expression.

Helpful Hint

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Example 7: Extraneous Solutions

Solve . Identify any extraneous solutions.

Step 1 Solve.

2(x2 – 1) = (x + 1)(x – 6)

2x2 – 2 = x2 – 5x – 6

x2 + 5x + 4 = 0

(x + 1)(x + 4) = 0

x = –1 or x = –4

Use cross products.

Distribute 2 on the left side. Multiply the right side.

Subtract x2 from both sides. Add 5x and 6 to both sides.

Factor the quadratic expression.Use the Zero Product Property.Solve.

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Example 7 Continued

Solve . Identify any extraneous solutions.

Step 2 Find extraneous solutions.

Because and

are undefined –1 is

not a solution.

The only solution is – 4, so – 1 is an extraneous solution.

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Example 8

Solve. Identify any extraneous solutions.

Step 1 Solve.

(x – 2)(x – 7) = 3(x – 7)

Use cross products.

Distribute 3 on the right side. Multiply the left side.

2x2 – 9x + 14 = 3x – 21

X2 – 12x + 35 = 0

Subtract 3x from both sides. Add 21 to both sides.

(x – 7)(x – 5) = 0

x = 7 or x = 5

Factor the quadratic expression.Use the Zero Product Property.Solve.

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Example 8 Continued

Step 2 Find extraneous solutions.

The only solution is 5, so 7 is an extraneous solution.

Because and

are undefined 7 is

not a solution.

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• EXIT TICKET:– Solve the equation.

2

122 xx

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