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: 21


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

Work Step by Step

$\frac{2-x}{5{{x}^{2}}}\div \frac{{{x}^{2}}-4}{3x}$ $\frac{A}{B}\div \frac{C}{D}=\frac{A}{B}\cdot \frac{D}{C}$ The reciprocal of $\frac{{{x}^{2}}-4}{3x}$ is$\frac{3x}{{{x}^{2}}-4}$. So, multiply the reciprocal of the divisor, $\begin{align} & \frac{2-x}{5{{x}^{2}}}\div \frac{{{x}^{2}}-4}{3x}=\frac{2-x}{5{{x}^{2}}}\cdot \frac{3x}{{{x}^{2}}-4} \\ & =\frac{\left( 2-x \right)\left( 3x \right)}{5{{x}^{2}}\left( {{x}^{2}}-4 \right)} \end{align}$ Factor the terms, $\begin{align} & \left( 2-x \right)\left( 3x \right)=-\left( x-2 \right)\left( 3x \right) \\ & 5{{x}^{2}}\left( {{x}^{2}}-4 \right)=5{{x}^{2}}\left( {{x}^{2}}-{{2}^{2}} \right) \\ & =5{{x}^{2}}\left( x+2 \right)\left( x-2 \right) \end{align}$ So, the rational expression becomes, $\frac{2-x}{5{{x}^{2}}}\div \frac{{{x}^{2}}-4}{3x}=\frac{-\left( x-2 \right)\left( 3x \right)}{5{{x}^{2}}\left( x+2 \right)\left( x-2 \right)}$ Regroup and remove the factor equal to 1, $\begin{align} & \frac{2-x}{5{{x}^{2}}}\div \frac{{{x}^{2}}-4}{3x}=\frac{-\left( x-2 \right)\left( 3x \right)}{\left( x-2 \right)\left( x+2 \right)5{{x}^{2}}} \\ & =\frac{-\left( x-2 \right)}{\left( x-2 \right)}\cdot \frac{\left( 3x \right)}{\left( x+2 \right)5{{x}^{2}}} \\ & =-1\cdot \frac{3x}{\left( x+2 \right)5{{x}^{2}}} \\ & =-\frac{3}{5x\cdot \left( x+2 \right)} \end{align}$
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