Algebra: A Combined Approach (4th Edition)

Published by Pearson
ISBN 10: 0321726391
ISBN 13: 978-0-32172-639-1

Chapter 7 - Section 7.5 - Solving Equations Containing Rational Expressions - Exercise Set: 23

Answer

the solutions are -4 and 6

Work Step by Step

$\frac{2x}{x+2}$-2=$\frac{x-8}{x-2}$ (x+2)(x-2)[$\frac{2x}{x+2}$-2]=$\frac{x-8}{x-2}$(x+2)(x-2) (x+2)(x-2)$\frac{2x}{x+2}$-2(x+2)(x-2)=(x-8)(x+2) 2x(x-2)-2($x^{2}$-4)=$x^{2}$+2x-8x-16 2$x^{2}$-4x-2$x^{2}$+8=$x^{2}$-6x-16 -4x+8=$x^{2}$-6x-16 0=$x^{2}$-6x-16+4x-8 0=$x^{2}$-2x-24 0=$x^{2}$+4x-6x-24 0=x(x+4)-6(x+4) 0=(x-6)(x+4) x-6=0 or x+4=0 x=6 or x=-4 The solutions are -4 and 6 Check Let x=-4 $\frac{2(-4)}{-4+2}$-2=$\frac{-4-8}{-4-2}$ $\frac{-8}{-2}$-2=$\frac{-12}{-6}$ 4-2=2 2=2 Let x=6 $\frac{2*6}{6+2}$-2=$\frac{6-8}{6-2}$ $\frac{12}{8}$-2=$\frac{-2}{4}$ $\frac{6}{4}$-$\frac{8}{4}$=$\frac{-2}{4}$ $\frac{-2}{4}$=$\frac{-2}{4}$
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