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Complex Rational Expressions

Module by: Wade Ellis, Denny Burzynski. E-mail the authors

Summary:

This module is from Elementary Algebra by Denny Burzynski and Wade Ellis, Jr.

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.

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.

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.

Overview

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

Simple And Complex Fractions

Simple Fraction

In Section (Reference) we saw that a simple fraction was a fraction of the form P Q , P Q , where P P and Q Q are polynomials and Q0 Q0 .

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 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.

Example 1

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 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

Example 2

1 1 x 1 1 x 2 Step 1:       Combine the terms of the numerator: LCD=x. 1 1 x = x x 1 x = x1 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. x1 x x 2 1 x 2 = x1 x · x 2 x 2 1 = x1 x x 2 ( x+1 ) ( x1 ) = x x+1 Thus, 1 1 x 1 1 x 2 = x x+1 1 1 x 1 1 x 2 Step 1:       Combine the terms of the numerator: LCD=x. 1 1 x = x x 1 x = x1 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. x1 x x 2 1 x 2 = x1 x · x 2 x 2 1 = x1 x x 2 ( x+1 ) ( x1 ) = x x+1 Thus, 1 1 x 1 1 x 2 = x x+1

Example 3

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 13m m 2 7 m 2 = 2 m 2 13m7 m 2 Step 2:        Combine the terms of the denominator: LCD= m 2 2+ 3 m + 1 m 2 = 2 m 2 m 2 + 3m m 2 + 1 m 2 = 2 m 2 +3m+1 m 2 Step 3:       Divide the numerator by the denominator. 2 m 2 13m7 m 2 2 m 2 +3m1 m 2 = 2 m 2 13m7 m 2 · m 2 2 m 2 +3m+1 = ( 2m+1 ) ( m7 ) m 2 · m 2 ( 2m+1 ) ( m+1 ) = m7 m+1 Thus, 2 13 m 7 m 2 2+ 3 m + 1 m 2 = m7 m+1 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 13m m 2 7 m 2 = 2 m 2 13m7 m 2 Step 2:        Combine the terms of the denominator: LCD= m 2 2+ 3 m + 1 m 2 = 2 m 2 m 2 + 3m m 2 + 1 m 2 = 2 m 2 +3m+1 m 2 Step 3:       Divide the numerator by the denominator. 2 m 2 13m7 m 2 2 m 2 +3m1 m 2 = 2 m 2 13m7 m 2 · m 2 2 m 2 +3m+1 = ( 2m+1 ) ( m7 ) m 2 · m 2 ( 2m+1 ) ( m+1 ) = m7 m+1 Thus, 2 13 m 7 m 2 2+ 3 m + 1 m 2 = m7 m+1

Practice Set A

Use the combine-divide method to simplify each expression.

Exercise 1

27 x 2 6 15 x 3 8 27 x 2 6 15 x 3 8

Solution

12 5x 12 5x

Exercise 2

3 1 x 3+ 1 x 3 1 x 3+ 1 x

Solution

3x1 3x+1 3x1 3x+1

Exercise 3

1+ x y x y 2 x 1+ x y x y 2 x

Solution

x y( xy ) x y( xy )

Exercise 4

m3+ 2 m m4+ 3 m m3+ 2 m m4+ 3 m

Solution

m2 m3 m2 m3

Exercise 5

1+ 1 x1 1 1 x1 1+ 1 x1 1 1 x1

Solution

x x2 x x2

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.

Example 4

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 +2a Step 3:        Reduce. a 2 4 a 2 +2a = ( a+2 ) ( a2 ) a ( a+2 ) = a2 a Thus, 1 4 a 2 1+ 2 a = a2 a 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 +2a Step 3:        Reduce. a 2 4 a 2 +2a = ( a+2 ) ( a2 ) a ( a+2 ) = a2 a Thus, 1 4 a 2 1+ 2 a = a2 a

Example 5

1 5 x 6 x 2 1+ 6 x + 5 x 2 Step1:TheLCDis x 2 . Step2:Multiplythenumeratoranddenominatorby 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 5x6 x 2 +6x+5 Step3:Reduce. x 2 5x6 x 2 +6x+5 = ( x6 )( x+1 ) ( x+5 )( x+1 ) = x6 x+5 Thus, 1 5 x 6 x 2 1+ 6 x + 5 x 2 = x6 x+5 1 5 x 6 x 2 1+ 6 x + 5 x 2 Step1:TheLCDis x 2 . Step2:Multiplythenumeratoranddenominatorby 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 5x6 x 2 +6x+5 Step3:Reduce. x 2 5x6 x 2 +6x+5 = ( x6 )( x+1 ) ( x+5 )( x+1 ) = x6 x+5 Thus, 1 5 x 6 x 2 1+ 6 x + 5 x 2 = x6 x+5

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.

Exercise 6

27 x 2 6 15 x 3 8 27 x 2 6 15 x 3 8

Solution

12 5x 12 5x

Exercise 7

3 1 x 3+ 1 x 3 1 x 3+ 1 x

Solution

3x1 3x+1 3x1 3x+1

Exercise 8

1+ x y x y 2 x 1+ x y x y 2 x

Solution

x y( xy ) x y( xy )

Exercise 9

m3+ 2 m m4+ 3 m m3+ 2 m m4+ 3 m

Solution

m2 m3 m2 m3

Exercise 10

1+ 1 x1 1 1 x1 1+ 1 x1 1 1 x1

Solution

x x2 x x2

Exercises

For the following problems, simplify each complex rational expression.

Exercise 11

1+ 1 4 1 1 4 1+ 1 4 1 1 4

Solution

5 3 5 3

Exercise 12

1 1 3 1+ 1 3 1 1 3 1+ 1 3

Exercise 13

1 1 y 1+ 1 y 1 1 y 1+ 1 y

Solution

y1 y+1 y1 y+1

Exercise 14

a+ 1 x a 1 x a+ 1 x a 1 x

Exercise 15

a b + c b a b c b a b + c b a b c b

Solution

a+c ac a+c ac

Exercise 16

5 m + 4 m 5 m 4 m 5 m + 4 m 5 m 4 m

Exercise 17

3+ 1 x 3x+1 x 2 3+ 1 x 3x+1 x 2

Solution

x x

Exercise 18

1+ x x+y 1 x x+y 1+ x x+y 1 x x+y

Exercise 19

2+ 5 a+1 2 5 a+1 2+ 5 a+1 2 5 a+1

Solution

2a+7 2a3 2a+7 2a3

Exercise 20

1 1 a1 1+ 1 a1 1 1 a1 1+ 1 a1

Exercise 21

4 1 m 2 2+ 1 m 4 1 m 2 2+ 1 m

Solution

2m1 m 2m1 m

Exercise 22

9 1 x 2 3 1 x 9 1 x 2 3 1 x

Exercise 23

k 1 k k+1 k k 1 k k+1 k

Solution

k1 k1

Exercise 24

m m+1 1 m+1 2 m m+1 1 m+1 2

Exercise 25

2xy 2xy y 2xy 3 2xy 2xy y 2xy 3

Solution

3 y 2 ( 2xy ) 2 3 y 2 ( 2xy ) 2

Exercise 26

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

Exercise 27

5 x+3 5 x3 5 x+3 + 5 x3 5 x+3 5 x3 5 x+3 + 5 x3

Solution

3 x 3 x

Exercise 28

2+ 1 y+1 1 y + 2 3 2+ 1 y+1 1 y + 2 3

Exercise 29

1 x 2 1 y 2 1 x + 1 y 1 x 2 1 y 2 1 x + 1 y

Solution

yx xy yx xy

Exercise 30

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

Exercise 31

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

Solution

y1 y+5 y1 y+5

Exercise 32

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

Exercise 33

x 4 3x1 1 2x2 3x1 x 4 3x1 1 2x2 3x1

Solution

3x4 3x4

Exercise 34

y x+y x xy x x+y + y xy y x+y x xy x x+y + y xy

Exercise 35

a a2 a a+2 2a a2 + a 2 a+2 a a2 a a+2 2a a2 + a 2 a+2

Solution

4 a 2 +4 4 a 2 +4

Exercise 36

3 2 1 1 m+1 3 2 1 1 m+1

Exercise 37

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

Solution

( x2 )( x+1 ) ( x1 )( x+2 ) ( x2 )( x+1 ) ( x1 )( x+2 )

Exercise 38

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

R= 1 1 R 1 + 1 R 2 R= 1 1 R 1 + 1 R 2

Write this complex fraction as a simple fraction.

Exercise 39

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

v= v 1 + v 2 1+ v 1 v 2 c 2 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=186,000miles per second ). ( c=186,000miles per second ). At very much lower velocities, such as 500 miles per hour, the formula v= v 1 + v 2 v= v 1 + v 2 provides an extremely good approximation.

Solution

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

Exercises For Review

Exercise 40

((Reference)) Supply the missing word. Absolute value speaks to the question of how

          
and not “which way.”

Exercise 41

((Reference)) Find the product. ( 3x+4 ) 2 . ( 3x+4 ) 2 .

Solution

9 x 2 +24x+16 9 x 2 +24x+16

Exercise 42

((Reference)) Factor x 4 y 4 . x 4 y 4 .

Exercise 43

((Reference)) Solve the equation 3 x1 5 x+3 =0. 3 x1 5 x+3 =0.

Solution

x=7 x=7

Exercise 44

((Reference)) 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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