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: 42


$=\displaystyle \frac{2(5x-14)}{(x-2)(x-3)}, \quad x\neq 2,3$

Work Step by Step

The denominators are different. Find a common denominator. LCD=$(x-2)(x-3)$ $\displaystyle \frac{8}{x-2}+\frac{2}{x-3}= \displaystyle \frac{8(x-3)}{(x-2)(x-3)}+\frac{2(x-2)}{(x-2)(x-3)}$ $=\displaystyle \frac{8x-24+2x-4}{(x-2)(x-3)}$ $=\displaystyle \frac{10x-28}{(x-2)(x-3)}$ $=\displaystyle \frac{2(5x-14)}{(x-2)(x-3)}$ ... exclude values that yield 0 in the denominator: $x\neq 2,3$ ... and, cancel common factors (here, there are none) $=\displaystyle \frac{2(5x-14)}{(x-2)(x-3)}, \quad x\neq 2,3$
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.