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<para>This module is from<link document="col10614">Elementary Algebra</link>by Denny Burzynski and Wade Ellis, Jr.</para><para>A detailed study of arithmetic operations with rational expressions is presented in this chapter, beginning with the definition of a rational expression and then proceeding immediately to a discussion of the domain. The process of reducing a rational expression and illustrations of multiplying, dividing, adding, and subtracting rational expressions are also included. Since the operations of addition and subtraction can cause the most difficulty, they are given particular attention. We have tried to make the written explanation of the examples clearer by using a "freeze frame" approach, which walks the student through the operation step by step.</para><para>The five-step method of solving applied problems is included in this chapter to show the problem-solving approach to number problems, work problems, and geometry problems. The chapter also illustrates simplification of complex rational expressions, using the combine-divide method and the LCD-multiply-divide method.</para><para>Objectives of this module: be able to distinguish between simple and complex fractions, be able to simplify complex fractions using the combine-divide and the LCD-multiply-divide method.</para>

Overview

  • Simple And Complex Fractions
  • The Combine-Divide Method
  • The LCD-Multiply-Divide Method

Simple and complex fractions

Simple fraction

In Section [link] we saw that a simple fraction was a fraction of the form P Q , where P and Q are polynomials and Q 0 .

Complex fraction

A complex fraction is a fraction in which the numerator or denominator, or both, is a fraction. The fractions 8 15 2 3 and 1 1 x 1 1 x 2 are examples of complex fractions, or more generally, complex rational expressions.

There are two methods for simplifying complex rational expressions: the combine-divide method and the LCD-multiply-divide method.

The combine-divide method

  1. If necessary, combine the terms of the numerator together.
  2. If necessary, combine the terms of the denominator together.
  3. Divide the numerator by the denominator.

Sample set a

Simplify each complex rational expression.

x 3 8 x 5 12 Steps 1 and 2 are not necessary so we proceed with step 3 . x 3 8 x 5 12 = x 3 8 · 12 x 5 = x 3 8 2 · 12 3 x 5 2 = 3 2 x 2

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1 1 x 1 1 x 2 Step 1:       Combine the terms of the numerator: LCD = x . 1 1 x = x x 1 x = x 1 x Step 2:       Combine the terms of the denominator: LCD = x 2 1 1 x 2 = x 2 x 2 1 x 2 = x 2 1 x 2 Step 3:       Divide the numerator by the denominator . x 1 x x 2 1 x 2 = x 1 x · x 2 x 2 1 = x 1 x x 2 ( x + 1 ) ( x 1 ) = x x + 1 Thus, 1 1 x 1 1 x 2 = x x + 1

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2 13 m 7 m 2 2 + 3 m + 1 m 2 Step 1:        Combine the terms of the numerator: LCD = m 2 . 2 13 m 7 m 2 = 2 m 2 m 2 13 m m 2 7 m 2 = 2 m 2 13 m 7 m 2 Step 2:        Combine the terms of the denominator: LCD = m 2 2 + 3 m + 1 m 2 = 2 m 2 m 2 + 3 m m 2 + 1 m 2 = 2 m 2 + 3 m + 1 m 2 Step 3:       Divide the numerator by the denominator . 2 m 2 13 m 7 m 2 2 m 2 + 3 m 1 m 2 = 2 m 2 13 m 7 m 2 · m 2 2 m 2 + 3 m + 1 = ( 2 m + 1 ) ( m 7 ) m 2 · m 2 ( 2 m + 1 ) ( m + 1 ) = m 7 m + 1 Thus, 2 13 m 7 m 2 2 + 3 m + 1 m 2 = m 7 m + 1

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Practice set a

Use the combine-divide method to simplify each expression.

27 x 2 6 15 x 3 8

12 5 x

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3 1 x 3 + 1 x

3 x 1 3 x + 1

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1 + x y x y 2 x

x y ( x y )

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m 3 + 2 m m 4 + 3 m

m 2 m 3

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1 + 1 x 1 1 1 x 1

x x 2

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The lcd-multiply-divide method

  1. Find the LCD of all the terms.
  2. Multiply the numerator and denominator by the LCD.
  3. Reduce if necessary.

Sample set b

Simplify each complex fraction.

1 4 a 2 1 + 2 a Step 1:      The LCD = a 2 . Step 2:      Multiply both the numerator and denominator by  a 2 . a 2 ( 1 4 a 2 ) a 2 ( 1 + 2 a ) = a 2 · 1 a 2 · 4 a 2 a 2 · 1 + a 2 · 2 a = a 2 4 a 2 + 2 a Step 3:        Reduce . a 2 4 a 2 + 2 a = ( a + 2 ) ( a 2 ) a ( a + 2 ) = a 2 a Thus, 1 4 a 2 1 + 2 a = a 2 a

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1 5 x 6 x 2 1 + 6 x + 5 x 2 Step 1 : The LCD is x 2 . Step 2 : Multiply the numerator and denominator by x 2 . x 2 ( 1 5 x 6 x 2 ) x 2 ( 1 + 6 x + 5 x 2 ) = x 2 · 1 x 2 · 5 x x 2 · 6 x 2 x 2 · 1 + x 2 · 6 x + x 2 · 5 x 2 = x 2 5 x 6 x 2 + 6 x + 5 Step 3 : Reduce . x 2 5 x 6 x 2 + 6 x + 5 = ( x 6 ) ( x + 1 ) ( x + 5 ) ( x + 1 ) = x 6 x + 5 Thus, 1 5 x 6 x 2 1 + 6 x + 5 x 2 = x 6 x + 5

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Practice set b

The following problems are the same problems as the problems in Practice Set A. Simplify these expressions using the LCD-multiply-divide method. Compare the answers to the answers produced in Practice Set A.

27 x 2 6 15 x 3 8

12 5 x

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3 1 x 3 + 1 x

3 x 1 3 x + 1

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1 + x y x y 2 x

x y ( x y )

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m 3 + 2 m m 4 + 3 m

m 2 m 3

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1 + 1 x 1 1 1 x 1

x x 2

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Exercises

For the following problems, simplify each complex rational expression.

1 1 y 1 + 1 y

y 1 y + 1

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a b + c b a b c b

a + c a c

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1 + x x + y 1 x x + y

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2 + 5 a + 1 2 5 a + 1

2 a + 7 2 a 3

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1 1 a 1 1 + 1 a 1

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4 1 m 2 2 + 1 m

2 m 1 m

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k 1 k k + 1 k

k 1

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2 x y 2 x y y 2 x y 3

3 y 2 ( 2 x y ) 2

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1 a + b 1 a b 1 a + b + 1 a b

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5 x + 3 5 x 3 5 x + 3 + 5 x 3

3 x

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1 x 2 1 y 2 1 x + 1 y

y x x y

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1 + 5 x + 6 x 2 1 1 x 12 x 2

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1 + 1 y 2 y 2 1 + 7 y + 10 y 2

y 1 y + 5

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3 n m 2 m n 3 n m + 4 + m n

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x 4 3 x 1 1 2 x 2 3 x 1

3 x 4

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y x + y x x y x x + y + y x y

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a a 2 a a + 2 2 a a 2 + a 2 a + 2

4 a 2 + 4

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x 1 1 1 x x + 1 1 + 1 x

( x 2 ) ( x + 1 ) ( x 1 ) ( x + 2 )

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In electricity theory, when two resistors of resistance R 1 and R 2 ohms are connected in parallel, the total resistance R is

R = 1 1 R 1 + 1 R 2

Write this complex fraction as a simple fraction.

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According to Einstein’s theory of relativity, two velocities v 1 and v 2 are not added according to v = v 1 + v 2 , but rather by

v = v 1 + v 2 1 + v 1 v 2 c 2

Write this complex fraction as a simple fraction.

Einstein's formula is really only applicale for velocities near the speed of light ( c = 186 , 000 miles per second ) . At very much lower velocities, such as 500 miles per hour, the formula v = v 1 + v 2 provides an extremely good approximation.

c 2 ( V 1 + V 2 ) c 2 + V 1 V 2

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Exercises for review

( [link] ) Supply the missing word. Absolute value speaks to the question of how and not “which way.”

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( [link] ) Find the product. ( 3 x + 4 ) 2 .

9 x 2 + 24 x + 16

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( [link] ) Solve the equation 3 x 1 5 x + 3 = 0.

x = 7

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( [link] ) One inlet pipe can fill a tank in 10 minutes. Another inlet pipe can fill the same tank in 4 minutes. How long does it take both pipes working together to fill the tank?

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Source:  OpenStax, Elementary algebra. OpenStax CNX. May 08, 2009 Download for free at http://cnx.org/content/col10614/1.3
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