Intermediate Algebra for College Students (7th Edition)

Published by Pearson
ISBN 10: 0-13417-894-7
ISBN 13: 978-0-13417-894-3

Chapter 6 - Section 6.6 - Rational Equations - Exercise Set - Page 463: 39


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

Work Step by Step

Factor the denominators $x^{2}-x=x(x-1)$ $x^{2}-1=(x-1)(x+1)$ LCD = $x(x-1)(x+1)$ $\displaystyle \frac{x+2}{x^{2}-x}-\frac{6}{x^{2}-1}= \displaystyle \frac{x+2}{x(x-1)}\cdot\frac{x+1}{x+1}-\frac{6}{(x-1)(x+1)}\cdot\frac{x}{x}$ $= \displaystyle \frac{(x+2)(x+1)-6x}{x(x-1)(x+1)}$ $= \displaystyle \frac{x^{2}+3x+2-6x}{x(x-1)(x+1)}$ $= \displaystyle \frac{x^{2}-3x+2}{x(x-1)(x+1)}$ ... factors of 2 whose sum is -3 ... are -1 and -2 $= \displaystyle \frac{(x-1)(x-2)}{x(x-1)(x+1)}\qquad$... cancel the common factor, $(x-1)$ = $\displaystyle \frac{x-2}{x(x+1)}$
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