Elementary Algebra

Published by Cengage Learning
ISBN 10: 1285194055
ISBN 13: 978-1-28519-405-9

Chapter 7 - Algebraic Fractions - Chapter 7 Review Problem Set: 14

Answer

$= \frac{5x-4}{(x+2)(x+5)(x-5)}$

Work Step by Step

$\frac{2}{x^{2}+7x+10} + \frac{3}{x^{2}-25}$ 1. Factor the denominator of the first fraction $ = \frac{2}{x^{2}+5x+2x+10} + \frac{3}{x^{2}-25}$ $ = \frac{2}{x(x+5)+2(x+5)} + \frac{3}{x^{2}-25}$ $ = \frac{2}{(x+2)(x+5)} + \frac{3}{x^{2}-25}$ 2. Factor the denominator of the second fraction: $ = \frac{2}{(x+2)(x+5)} + \frac{3}{x^{2}-5x+5x-25}$ $ = \frac{2}{(x+2)(x+5)} + \frac{3}{x(x-5)+5(x-5)}$ $ = \frac{2}{(x+2)(x+5)} + \frac{3}{(x-5)(x+5)}$ 3. Find a common denominator: Multiply the first fraction by $(x-5)$ and the second fraction by $(x+2)$ to obtain a common denominator of $(x+2)(x+5)(x-5)$ $ = \frac{2(x-5)}{(x+2)(x+5)(x-5)} + \frac{3(x+2)}{(x-5)(x+5)(x+2)}$ 4. Add the fractions $= \frac{2(x-5)+3(x+2)}{(x+2)(x+5)(x-5)}$ $= \frac{2x-10+3x+6}{(x+2)(x+5)(x-5)}$ $= \frac{5x-4}{(x+2)(x+5)(x-5)}$
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