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

Answer

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

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