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 6.2 - Adding and Subtracting Rational Expressions - Exercise Set - Page 427: 51


$\displaystyle \frac{x^{2}+2x-14}{3(x+2)(x-2)} $

Work Step by Step

$\displaystyle \frac{x+7}{3x+6}+\frac{x}{-(x^{2}-4)}=\frac{x+7}{3x+6}-\frac{x}{(x^{2}-4)}$ 1. Find the LCD . 1st denominator = $3(x+2)$ 2nd denominator = $(x-2)(x+2)$ List factors of the 1st denominator. From each next denominator, add only those factors that do not yet appear in the list. LCD = $3(x+2)(x-2)$ 2. Rewrite each rational expression with the the LCDas the denominator = $\displaystyle \frac{(x+7)(x-2)}{3(x+2)(x-2)}-\frac{x\cdot 3}{(x-2)(x+2)\cdot 3}$ 3. Add or subtract numerators, placing the resulting expression over the LCD. = $\displaystyle \frac{(x+7)(x-2)-3x}{3(x+2)(x-2)}$ 4. If possible, simplify. = $\displaystyle \frac{x^{2}-2x+7x-14-3x}{3(x+2)(x-2)} $ = $\displaystyle \frac{x^{2}+2x-14}{3(x+2)(x-2)} $
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