College Algebra (10th Edition)

Published by Pearson
ISBN 10: 0321979478
ISBN 13: 978-0-32197-947-6

Chapter R - Section R.7 - Rational Expressions - R.7 Assess Your Understanding - Page 72: 80


$\displaystyle \frac{-5x+4}{(x-2)(x+1)(x+3)}$

Work Step by Step

$\displaystyle \frac{\frac{x-2}{x+1}-\frac{x}{x-2}}{x+3}=(\frac{x-2}{x+1}-\frac{x}{x-2})\div(x+3)$ $=(\displaystyle \frac{x-2}{x+1}\cdot\frac{x-2}{x-2}-\frac{x}{x-2}\cdot\frac{x+1}{x+1})\div(x+3)$ $=(\displaystyle \frac{x^{2}-4x+4}{(x-2)(x+1)}-\frac{x^{2}-x}{(x-2)(x+1)})\div(x+3)$ $=(\displaystyle \frac{-5x+4}{(x-2)(x+1)})\div(x+3)$ ... division = multiplication with the reciprocal, $=\displaystyle \frac{-5x+4}{(x-2)(x+1)}\cdot\frac{1}{(x+3)}$ $=\displaystyle \frac{-5x+4}{(x-2)(x+1)(x+3)}$
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