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 - Page 314: 11



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: $\frac{x}{x-4}-\frac{2x}{x+4}=-1$ $\frac{x(x+4)-2x(x-4)}{(x+4)(x-4)}=-1$ $\frac{x^{2}+4x-2x^{2}+8x}{(x+4)(x-4)}=-1$ $\frac{-x^{2}+12x}{(x+4)(x-4)}=-1$ Now, we cross multiply: $\frac{-x^{2}+12x}{(x+4)(x-4)}=-1$ $-x^{2}+12x=-1(x-4)(x+4)$ $-x^{2}+12x=-1(x^{2}-4^{2})$ $-x^{2}+12x=-1(x^{2}-16)$ $-x^{2}+12x=-x^{2}+16$ $12x=16$ $x=\frac{16}{12}$ $x=\frac{4}{3}$ Therefore, the solution set is {$\frac{4}{3}$}.
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