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: 37


The simplified form of a complex rational expression $\frac{\frac{3}{x-7}}{\frac{4x+3}{x+1}}$ is $\frac{3\left( x+1 \right)}{\left( x-7 \right)\left( 4x+3 \right)}$.

Work Step by Step

$\frac{\frac{3}{x-7}}{\frac{4x+3}{x+1}}$ Divide the numerator by the denominator, $\frac{\frac{3}{x-7}}{\frac{4x+3}{x+1}}=\frac{3}{x-7}\div \frac{4x+3}{x+1}$ $\begin{align} & \frac{A}{B}\div \frac{C}{D}=\frac{A}{B}\cdot \frac{D}{C} \\ & =\frac{AD}{BC} \end{align}$ where $\frac{A}{B},\left( B\ne 0 \right)$, $\frac{C}{D},\left( D\ne 0 \right)$ are rational expressions with $\frac{C}{D}\ne 0$ The reciprocal of $\frac{4x+3}{x+1}$ is $\frac{x+1}{4x+3}$ So, multiply the reciprocal of the divisor, $\frac{\frac{3}{x-7}}{\frac{4x+3}{x+1}}=\frac{3}{x-7}\cdot \frac{x+1}{4x+3}$ Simplify the terms, $\begin{align} & \frac{\frac{3}{x-7}}{\frac{4x+3}{x+1}}=\frac{3\cdot \left( x+1 \right)}{\left( x-7 \right)\cdot \left( 4x+3 \right)} \\ & =\frac{3\left( x+1 \right)}{\left( x-7 \right)\left( 4x+3 \right)} \end{align}$
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