Precalculus (6th Edition) Blitzer

Published by Pearson
ISBN 10: 0-13446-914-3
ISBN 13: 978-0-13446-914-0

Chapter P - Section P.6 - Rational Expressions - Exercise Set - Page 85: 86


The required solution is $\frac{\left( 4x+5 \right)\left( x+4 \right)}{\left( x+2 \right)\left( x-1 \right)}$.

Work Step by Step

We have the given algebraic expression: $\left( 4-\frac{3}{x+2} \right)\left( 1+\frac{5}{x-1} \right)$. For an algebraic expression, a rational expression is an expression that can be expressed in the form $\frac{p}{q}$, where, both $p\ \text{and }q$ are polynomials and the denominator $q\ne 0$. Now, solve the first bracket of the given rational expression: $\begin{align} & 4-\frac{3}{x+2}=4\times \frac{\left( x+2 \right)}{\left( x+2 \right)}-\frac{3}{\left( x+2 \right)} \\ & =\frac{4\left( x+2 \right)}{\left( x+2 \right)}-\frac{3}{\left( x+2 \right)} \\ & =\frac{4x+8-3}{\left( x+2 \right)} \\ & =\frac{4x+5}{\left( x+2 \right)} \end{align}$. Also, solve the second bracket of the given rational expression: $\begin{align} & 1+\frac{5}{x-1}=1\times \frac{\left( x-1 \right)}{\left( x-1 \right)}+\frac{5}{\left( x-1 \right)} \\ & =\frac{x-1}{\left( x-1 \right)}+\frac{5}{\left( x-1 \right)} \\ & =\frac{x-1+5}{\left( x-1 \right)} \\ & =\frac{x+4}{\left( x-1 \right)} \end{align}$. Therefore, the given expression becomes $\begin{align} & \left( 4-\frac{3}{x+2} \right)\left( 1+\frac{5}{x-1} \right)=\frac{\left( 4x+5 \right)}{\left( x+2 \right)}\times \frac{\left( x+4 \right)}{\left( x-1 \right)} \\ & =\frac{\left( 4x+5 \right)\left( x+4 \right)}{\left( x+2 \right)\left( x-1 \right)} \end{align}$. Hence, $\left( 4-\frac{3}{x+2} \right)\left( 1+\frac{5}{x-1} \right)=$ $\frac{\left( 4x+5 \right)\left( x+4 \right)}{\left( x+2 \right)\left( x-1 \right)}$.
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