College Algebra (6th Edition)

Published by Pearson
ISBN 10: 0-32178-228-3
ISBN 13: 978-0-32178-228-1

Chapter P - Summary, Review, and Test - Review Exercises - Page 91: 117

Answer

$\frac{2x^{2}-3}{(x+3)(x-3)(x-2)} ; x \ne -3,3,2;$

Work Step by Step

$\frac{x}{x^{2}-9} + \frac{x-1}{x^{2}-5x+6}$ Factors of $x^{2}-5x+6$ are $(x-3)(x-2)$ $x^{2}-9 = (x+3)(x-3) $ using the formula $a^{2}-b^{2} = (a+b)(a-b)$ $= \frac{x}{(x+3)(x-3)} + \frac{x-1}{(x-3)(x-2)} ; x \ne -3,3,2;$ Take LCD, $= \frac{x(x-2)+(x-1)(x+3)}{(x+3)(x-3)(x-2)} ; x \ne -3,3,2;$ $= \frac{x^{2}-2x+x^{2}+3x-x-3}{(x+3)(x-3)(x-2)} ; x \ne -3,3,2;$ Combine like terms. $=\frac{2x^{2}-3}{(x+3)(x-3)(x-2)} ; x \ne -3,3,2;$
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