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 71: 27


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

Work Step by Step

$...=\displaystyle \frac{6x}{x^{2}-4}\div\frac{3x-9}{2x+4}=\frac{6x}{x^{2}-4}\cdot\frac{2x+4}{3x-9}$ Factoring, $6=2\cdot 3$ $ x^{2}-4=(x+2)(x-2)\quad$ (a difference of squares) $2x+4=2(x+2)$ $3x-9=3(x-3)$ ... = $\displaystyle \frac{2\cdot 3\cdot x\cdot 2(x+2)}{(x+2)(x-2)\cdot 3(x-3)}$ ... cancel the following from both sides of the fraction line: $3$, $(x+2)$ ... = $\displaystyle \frac{4x}{(x-2)(x-3)}$
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