Elementary and Intermediate Algebra: Concepts & Applications (6th Edition)

Published by Pearson
ISBN 10: 0-32184-874-8
ISBN 13: 978-0-32184-874-1

Chapter R - Elementary Algebra Review - R.6 Rational Expressions and Equations - R.6 Exercise Set - Page 980: 33


The simplified form of the expression $\frac{x-2}{x+5}-\frac{x+3}{x-4}$ is $\frac{-7\left( 2x+1 \right)}{\left( x-4 \right)\left( x+5 \right)}$.

Work Step by Step

$\frac{x-2}{x+5}-\frac{x+3}{x-4}$ Solve to obtain the LCD of the denominator of the fractions that is $\left( x+5 \right)$ and $\left( x-4 \right)$. On solving the LCD is $\left( x-4 \right)\left( x+5 \right)$. The rational expression is written as a fraction in such a way that the LCD is the denominator of the fraction: $\frac{x-2}{x+5}=\frac{\left( x-2 \right)\left( x-4 \right)}{\left( x+5 \right)\left( x-4 \right)}$ And $\frac{x+3}{x-4}=\frac{\left( x+3 \right)\left( x+5 \right)}{\left( x-4 \right)\left( x+5 \right)}$ Since the fractions have a common denominator, add the numerators and the LCD is the denominator: $\begin{align} & \frac{x-2}{x+5}-\frac{x+3}{x-4}=\frac{\left( x-2 \right)\left( x-4 \right)}{\left( x+5 \right)\left( x-4 \right)}-\frac{\left( x+3 \right)\left( x+5 \right)}{\left( x+5 \right)\left( x-4 \right)} \\ & =\frac{\left( x-2 \right)\left( x-4 \right)-\left( x+3 \right)\left( x+5 \right)}{\left( x+5 \right)\left( x-4 \right)} \end{align}$ Simplify the above expression as: $\begin{align} & \frac{x-2}{x+5}-\frac{x+3}{x-4}=\frac{x\cdot \left( x-4 \right)-2\cdot \left( x-4 \right)-x\cdot \left( x+5 \right)-3\cdot \left( x+5 \right)}{\left( x+5 \right)\left( x-4 \right)} \\ & =\frac{{{x}^{2}}-4x-2x+8-{{x}^{2}}-5x-3x-15}{\left( x+5 \right)\left( x-4 \right)} \\ & =\frac{{{x}^{2}}-{{x}^{2}}-4x-2x-5x-3x+8-15}{\left( x+5 \right)\left( x-4 \right)} \\ & =\frac{-14x-7}{\left( x+5 \right)\left( x-4 \right)} \end{align}$ Take out $-7$ from the numerator. $\frac{x-2}{x+5}-\frac{x+3}{x-4}=\frac{-7\left( 2x+1 \right)}{\left( x-4 \right)\left( x+5 \right)}$
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