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: 36


$=\displaystyle \frac{x-2}{x-3}, \quad x\neq-2,3$

Work Step by Step

The fractions have a common denominator, we add the numerators... $\displaystyle \frac{x^{2}-4x}{x^{2}-x-6}+\frac{4x-4}{x^{2}-x-6}=\frac{x^{2}-4x+4x-4}{x^{2}-x-6}$ $=\displaystyle \frac{x^{2}-4}{x^{2}-x-6}$ ... factor what we can... $\left\{\begin{array}{lll} x^{2}-4 & =(x+2)(x-2) & \text{(diff. of squares)}\\ x^{2}-x-6 & =(x-3)(x+2) & \text{(factors of c whose sum is b)} \end{array}\right.$ $=\displaystyle \frac{(x+2)(x-2)}{(x-3)(x+2)}$ ... exclude values that yield 0 in the denominator: $x\neq-2,3$ ... and, cancel the common factor $=\displaystyle \frac{x-2}{x-3}, \quad x\neq-2,3$
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