Intermediate Algebra: Connecting Concepts through Application

Published by Brooks Cole
ISBN 10: 0-53449-636-9
ISBN 13: 978-0-53449-636-4

Chapter 1-8 - Cumulative Review: 8

Answer

$x = -7$

Work Step by Step

$\frac{6}{x+2} + 4 = \frac{4x}{x-3}$ $6 + 4(x+2) = \frac{4x(x+2)}{x-3}$ $6 + 4x + 8 = \frac{4x^{2}+8x}{x-3}$ $6(x-3) + 4x(x-3) + 8(x-3) = 4x^{2} + 8x$ $6x - 18 + 4x^{2} - 12x + 8x - 24 = 4x^{2} + 8x$ $4x^{2} - 6x + 8x - 42 = 4x^{2} + 8x$ $-6x - 42 = 0$ $-6x = 42$ $x = -7$ Check: $\frac{6}{x+2} + 4 \overset{?}{=} \frac{4x}{x-3}$ $\frac{6}{(-7)+2} + 4 \overset{?}{=} \frac{4(-7)}{-7-3}$ $\frac{6}{-5} + 4 \overset{?}{=} \frac{-28}{-10}$ $-\frac{6}{5} + 4 \overset{?}{=} \frac{28}{10}$ $\frac{14}{5} \overset{?}{=} \frac{28}{10}$ $\frac{14}{5} = \frac{14}{5}$
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