# 8.6 Rational equations

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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 identify rational equations, understand and be able to use the method of solving rational expressions, be able to recognize extraneous solutions.</para>

## Overview

• Rational Equations
• The Logic Behind The Process
• The Process
• Extraneous Solutions

## Rational equations

When one rational expression is set equal to another rational expression, a rational equation results.

Some examples of rational equations are the following (except for number 5):

$\frac{3x}{4}=\frac{15}{2}$

$\frac{x+1}{x-2}=\frac{x-7}{x-3}$

$\frac{5a}{2}=10$

$\frac{3}{x}+\frac{x-3}{x+1}=\frac{6}{5x}$

$\frac{x-6}{x+1}$ is a rational expression , not a rational equation.

## The logic behind the process

It seems most reasonable that an equation without any fractions would be easier to solve than an equation with fractions. Our goal, then, is to convert any rational equation to an equation that contains no fractions. This is easily done.

To develop this method, let’s consider the rational equation

$\frac{1}{6}+\frac{x}{4}=\frac{17}{12}$

The LCD is 12. We know that we can multiply both sides of an equation by the same nonzero quantity, so we’ll multiply both sides by the LCD, 12.

$12\left(\frac{1}{6}+\frac{x}{4}\right)=12\text{\hspace{0.17em}}·\text{\hspace{0.17em}}\frac{17}{12}$

Now distribute 12 to each term on the left side using the distributive property.

$12\text{\hspace{0.17em}}·\text{\hspace{0.17em}}\frac{1}{6}+12\text{\hspace{0.17em}}·\text{\hspace{0.17em}}\frac{x}{4}=12\text{\hspace{0.17em}}·\text{\hspace{0.17em}}\frac{17}{12}$

Now divide to eliminate all denominators.

$\begin{array}{ccc}2\text{\hspace{0.17em}}·\text{\hspace{0.17em}}1+3\text{\hspace{0.17em}}·\text{\hspace{0.17em}}x& =& 17\\ \hfill 2+3x& =& 17\end{array}$

Now there are no more fractions, and we can solve this equation using our previous techniques to obtain 5 as the solution.

## The process

We have cleared the equation of fractions by multiplying both sides by the LCD. This development generates the following rule.

## Clearing an equation of fractions

To clear an equation of fractions, multiply both sides of the equation by the LCD.

When multiplying both sides of the equation by the LCD, we use the distributive property to distribute the LCD to each term. This means we can simplify the above rule.

## Clearing an equation of fractions

To clear an equation of fractions, multiply every term on both sides of the equation by the LCD.

can someone help me with some logarithmic and exponential equations.
20/(×-6^2)
Salomon
okay, so you have 6 raised to the power of 2. what is that part of your answer
I don't understand what the A with approx sign and the boxed x mean
it think it's written 20/(X-6)^2 so it's 20 divided by X-6 squared
Salomon
I'm not sure why it wrote it the other way
Salomon
I got X =-6
Salomon
ok. so take the square root of both sides, now you have plus or minus the square root of 20= x-6
oops. ignore that.
so you not have an equal sign anywhere in the original equation?
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