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: 48


$\displaystyle \frac{2(x^{2}+9)}{(x-3)(x+3)},\quad x\neq-3,3$

Work Step by Step

The denominators are different. Find a common denominator. LCD=$(x+3)(x-3)$ $\displaystyle \frac{x+3}{x-3}+\frac{x-3}{x+3}= \displaystyle \frac{x+3}{x-3}\times\frac{x+3}{x+3}+\frac{x-3}{x+3}\times\frac{(x-3)}{(x-3)}=$ $=\displaystyle \frac{(x+3)(x+3)+(x-3)(x-3)}{(x-3)(x+3)}\quad $ ... apply$: (a\pm b)^{2}=a^{2}\pm 2ab+b^{2}$ $ =\displaystyle \frac{x^{2}+6x+9+x^{2}-6x+9}{(x-3)(x+3)} $ $ =\displaystyle \frac{2x^{2}+18}{(x-3)(x+3)} $ $=\displaystyle \frac{2(x^{2}+9)}{(x-3)(x+3)},\quad x\neq-3,3$
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