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: 36


The simplified form of a complex rational expression $\frac{\frac{x}{3}-\frac{3}{x}}{\frac{1}{x}+\frac{1}{3}}$ is $x-3$.

Work Step by Step

$\frac{\frac{x}{3}-\frac{3}{x}}{\frac{1}{x}+\frac{1}{3}}$ The denominators within the complex rational expression are $x\text{ and }3$. Thus, the LCD is $3x$. Multiply the expression with $\frac{3x}{3x}$: $\frac{\frac{x}{3}-\frac{3}{x}}{\frac{1}{x}+\frac{1}{3}}=\frac{\frac{x}{3}-\frac{3}{x}}{\frac{1}{x}+\frac{1}{3}}\cdot \frac{3x}{3x}$ Apply the distributive property, $\begin{align} & \frac{\frac{x}{3}-\frac{3}{x}}{\frac{1}{x}+\frac{1}{3}}=\frac{\frac{x}{3}\cdot 3x-\frac{3}{x}\cdot 3x}{\frac{1}{x}\cdot 3x+\frac{1}{3}\cdot 3x} \\ & =\frac{x\cdot x-3\cdot 3}{1\cdot 3+1\cdot x} \end{align}$ Simplify the terms, $\begin{align} & \frac{\frac{x}{3}-\frac{3}{x}}{\frac{1}{x}+\frac{1}{3}}=\frac{{{x}^{2}}-9}{3+x} \\ & =\frac{{{x}^{2}}-9}{x+3} \end{align}$ $\begin{align} & \frac{\frac{x}{3}-\frac{3}{x}}{\frac{1}{x}+\frac{1}{3}}=\frac{\left( x+3 \right)\left( x-3 \right)}{x+3} \\ & =x-3 \end{align}$
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