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: 26


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

Work Step by Step

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