College Algebra (6th Edition)

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

Chapter P - Prerequisites: Fundamental Concepts of Algebra - Exercise Set P.6 - Page 87: 65

Answer

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

Work Step by Step

$\frac{x- \frac{x}{x+3}}{x+2}$ $= \frac{ \frac{x(x+3)-x}{x+3}}{x+2} ; x \ne -2,-3;$ $= \frac{ \frac{x^{2}+3x-x}{x+3}}{x+2} ; x \ne -2,-3;$ $= \frac{ \frac{x^{2}+2x}{x+3}}{x+2} ; x \ne -2,-3;$ $ = \frac{x^{2}+2x}{x+3} \times \frac{1}{x+2} ; x \ne -2,-3;$ $ = \frac{x(x+2)}{x+3} \times \frac{1}{x+2} ; x \ne -2,-3;$ $= \frac{x}{(x+3)} ; x \ne -2,-3;$
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