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: 85


The required solution is $2$.

Work Step by Step

We have the given algebraic expression: $\left( 2-\frac{6}{x+1} \right)\left( 1+\frac{3}{x-2} \right)$. 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: $\begin{align} & 2-\frac{6}{x+1}=2\times \frac{\left( x+1 \right)}{\left( x+1 \right)}-\frac{6}{\left( x+1 \right)} \\ & =\frac{2\left( x+1 \right)}{\left( x+1 \right)}-\frac{6}{\left( x+1 \right)} \\ & =\frac{2x+2-6}{\left( x+1 \right)} \\ & =\frac{2x-4}{\left( x+1 \right)} \end{align}$. And factorize the numerator of the obtained fraction to get a simplified form: $2-\frac{6}{x+1}=\frac{2\left( x-2 \right)}{\left( x+1 \right)}$. Solve the second bracket of the given rational expression: $\begin{align} & 1+\frac{3}{x-2}=1\times \frac{\left( x-2 \right)}{\left( x-2 \right)}+\frac{3}{\left( x-2 \right)} \\ & =\frac{x-2}{\left( x-2 \right)}+\frac{3}{\left( x-2 \right)} \\ & =\frac{x-2+3}{\left( x-2 \right)} \\ & =\frac{x+1}{\left( x-2 \right)} \end{align}$. Therefore, the given expression becomes $\begin{align} & \left( 2-\frac{6}{x+1} \right)\left( 1+\frac{3}{x-2} \right)=\frac{2\left( x-2 \right)}{\left( x+1 \right)}\times \frac{\left( x+1 \right)}{\left( x-2 \right)} \\ & =2 \end{align}$. Hence, $\left( 2-\frac{6}{x+1} \right)\left( 1+\frac{3}{x-2} \right)=$ $2$.
Update this answer!

You can help us out by revising, improving and updating this answer.

Update this answer

After you claim an answer you’ll have 24 hours to send in a draft. An editor will review the submission and either publish your submission or provide feedback.