## College Algebra (6th Edition)

Published by Pearson

# Chapter P - Prerequisites: Fundamental Concepts of Algebra - Exercise Set P.6 - Page 86: 52

#### Answer

$\frac{20x-6}{(5x+2)(5x-2)}; x \ne \frac{-2}{5}, \frac{2}{5}$

#### Work Step by Step

$\frac{3}{5x+2} + \frac{5x}{25x^{2}-4}$ $25x^{2}-4$ is in the form of $(a^{2}-b^{2})$ $(a^{2}-b^{2}) = (a+b)(a-b)$ So, $25x^{2}-4 = (5x+2)(5x-2)$ The given expression can be written as $=\frac{3}{5x+2} + \frac{5x}{ (5x+2)(5x-2)}; x \ne \frac{-2}{5}, \frac{2}{5}$ Taking Least Common Denominator, $=\frac{3(5x-2)+5x}{ (5x+2)(5x-2)}; x \ne \frac{-2}{5}, \frac{2}{5}$ $=\frac{15x-6+5x}{ (5x+2)(5x-2)}; x \ne \frac{-2}{5}, \frac{2}{5}$ Combine like terms $=\frac{20x-6}{(5x+2)(5x-2)}; x \ne \frac{-2}{5}, \frac{2}{5}$

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