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8.9: Complex Rational Expressions

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Simple And Complex Fractions

Simple Fraction

In section 8.2 we saw that a simple fraction was a fraction of the form PQ, where P and Q are polynomials and Q0.

Complex Fraction

A complex fraction is a fraction in which the numerator or denominator, or both, is a fraction. The fractions

81523 and 11x11x2

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

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 8.9.1

x38x512

Steps 1 and 2 are not necessary so we proceed with step 3:

x38x512=x3812x5=x382312x52=32x2

Example 8.9.2

11x11x2

Step 1: Combine the terms of the numerator: LCD = x.

11x=xx1x=x1x

Step 2: Combine the terms of the denominator: LCD = x2.

11x2=x2x21x2=x21x2

Step 3: Divide the numerator by the denominator.

x1xx21x2=x1xx2x21=x1xx2(x+1)(x+1)=xx+1

Thus,

11x11x2=xx+1

Example 8.9.3

213m7m22+3m+1m2

Step 1: Combine the terms of the numerator: LCD = m2.

213m7m2=2m2m213mm27m2=2m213m7m2

Step 2: Combine the terms of the denominator: LCD = m2

2+3m+1m2=2m2m2+3mm2+1m2=2m2+3m+1m2

Step 3: Divide the numerator by the denominator:

2m213m7m22m2+3m1m2=2m213m7m2m22m2+3m+1=(2m+1)(m7)m2m2(2m+1)(m+1)=m7m+1

Thus,

213m7m22+3m+1m2=m7m+1

Practice Set A

Use the combine-divide method to simplify each expression.

Practice Problem 8.9.1

27x2615x38

Answer

125x

Practice Problem 8.9.2

31x3+1x

Answer

3x13x+1

Practice Problem 8.9.3

1+xyxy2x

Answer

xy(xy)

Practice Problem 8.9.4

m3+2mm4+3m

Answer

m2m3

Practice Problem 8.9.5

1+1x111x1

Answer

xx2

The LCD-Multiply-Divide Method

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 8.9.4

14a21+2a

Step 1: The LCD =a2.

Step 2: Multiply both the numerator and denominator by a2.

a2(14a2)a2(1+2a)=a21a24a2a21+a22a=a24a2+2a.

Step 3: Reduce:

a24a2+2a=(a+2)(a2)a(a+2)=a2a

Thus,

14a21+2a=a2a

Example 8.9.5

15x6x21+6x+5x2

Step 1: The LCD is x2.

Step 2: Multiply the numerator and denominator by x2.

x2(15x6x2)x2(1+6x+5x2)=x21x25xx26x2x21+x26x+x25x2=x25x6x2+6x+5

Step 3: Reduce:

x25x6x2+6x+5=(x6)(x+1)(x+5)(x+1)=x6x+5

Thus,

15x6x21+6x+5x2=x6x+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.

Practice Problem 8.9.6

27x2615x38

Answer

125x

Practice Problem 8.9.7

31x3+1x

Answer

3x13x+1

Practice Problem 8.9.8

1+xyxy2x

Answer

xy(xy)

Practice Problem 8.9.9

m3+2mm4+3m

Answer

m2m3

Practice Problem 8.9.10

1+1x111x1

Answer

xx2

Exercises

For the following problems, simplify each complex rational expression.

Exercise 8.9.1

1+14114

Answer

53

Exercise 8.9.2

1131+13

Exercise 8.9.3

11y1+1y

Answer

y1y+1

Exercise 8.9.4

a+1xa1x

Exercise 8.9.5

ab+cbabcb

Answer

a+cac

Exercise 8.9.6

5m+4m5m4m

Exercise 8.9.7

3+1x3x+1x2

Answer

x

Exercise 8.9.8

1+xx+y1xx+y

Exercise 8.9.9

2+5a+125a+1

Answer

2a+72a3

Exercise 8.9.10

11a11+1a1

Exercise 8.9.11

41m22+1m

Answer

2m1m

Exercise 8.9.12

91x231x

Exercise 8.9.13

k1kk+1k

Answer

k1

Exercise 8.9.14

mm+11m+12

Exercise 8.9.15

2xy2xyy2xy3

Answer

3y2(2xy)2

Exercise 8.9.16

1a+b1ab1a+b+1ab

Exercise 8.9.17

5x+35x35x+3+5x3

Answer

3x

Exercise 8.9.18

2+1y+11y+23

Exercise 8.9.19

1x21y21x+1y

Answer

yxxy

Exercise 8.9.20

1+5x+6x211x12x2

Exercise 8.9.21

1+1y2y21+7y+10y2

Answer

y1y+5

Exercise 8.9.22

3nm2mn3nm+4+mn

Answer

3x4

Exercise 8.9.23

yx+yxxyxx+y+yxy

Exercise 8.9.24

aa2aa+22aa2+a2a+2

Answer

4a2+4

Exercise 8.9.25

3211m+1

Exercise 8.9.26

x111xx+11+1x

Answer

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

Exercise 8.9.27

In electricity theory, when two resistors of resistance R1 and R2 ohms are connected in parallel, the total resistance R is:

R=11R1+1R2

Write this complex fraction as a simple fraction.

Exercise 8.9.28

According to Einstein's theory of relativity, two velocities v1 and v2 are not added according to v=v1+v2, but rather by

v=v1+v21+v1v2c2

Write this complex fraction as a simple fraction.

Einstein's formula is really only applicable 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=v1+v2 provides an extremely good approximation.

Answer

c2(V1+V2)c2+V1V2

Exercises For Review

Exercise 8.9.30

Supply the missing word. Absolute value speaks to the question of how ____ and not “which way.”

Exercise 8.9.31

Find the product. (3x+4)2

Answer

9x2+24x+16

Exercise 8.9.32

Factor x4y4

Exercise 8.9.33

Solve the equation 3x15x+3=0.

Answer

x=7

Exercise 8.9.34

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?


This page titled 8.9: Complex Rational Expressions is shared under a CC BY 4.0 license and was authored, remixed, and/or curated by Denny Burzynski & Wade Ellis, Jr. (OpenStax CNX) .

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