Elementary Algebra

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

Chapter 7 - Algebraic Fractions - 7.6 - More Fractional Equations and Problem Solving - Problem Set 7.6: 12

Answer

{$6$}

Work Step by Step

First, we find the LCM of the denominators of the two fractions on the left-hand side of the equation in order to develop a single fraction: $\frac{2x}{x+2}+\frac{x}{x-2}=3$ $\frac{2x(x-2)+x(x+2)}{(x+2)(x-2)}=3$ $\frac{2x^{2}-4x+x^{2}+2x}{(x+2)(x-2)}=3$ $\frac{3x^{2}-2x}{(x+2)(x-2)}=3$ Now, we cross multiply: $\frac{3x^{2}-2x}{(x+2)(x-2)}=3$ $3x^{2}-2x=3(x-2)(x+2)$ $3x^{2}-2x=3(x^{2}-2^{2})$ $3x^{2}-2x=3(x^{2}-4)$ $3x^{2}-2x=3x^{2}-12$ $-2x=-12$ $x=\frac{-12}{-2}$ $x=6$ Therefore, the solution set is {$6$}.
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