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: 60

Answer

$\frac{1}{2y-x}$.

Work Step by Step

Factor each term of the given expression: $\frac{x+2y}{x^2+4xy+4y^2}-\frac{2x}{x^2-4y^2}=\frac{x+2y}{(x+2y)(x+2y)}-\frac{2x}{(x+2y)(x-2y)}$ Cancel common terms. $=\frac{1}{(x+2y)}-\frac{2x}{(x+2y)(x-2y)}$ LCD is $=(x+2y)(x-2y)$. Multiply each numerator and denominator by the extra factor required to form the LCD. $=\frac{1}{x+2y}\times \frac{x-2y}{x-2y}-\frac{2x}{(x+2y)(x-2y)}$ Simplify. $=\frac{x-2y}{(x+2y)(x-2y)}-\frac{2x}{(x+2y)(x-2y)}$ Add the numerators. $=\frac{x-2y-2x}{(x+2y)(x-2y)}$ Simplify. $=\frac{-2y-x}{(x+2y)(x-2y)}$ Or we can write. $=\frac{-(x+2y)}{(x+2y)(x-2y)}$ Cancel common terms. $=-\frac{1}{x-2y}$. Or we can write. $=\frac{1}{2y-x}$.
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