College Algebra (11th Edition)

Published by Pearson
ISBN 10: 0321671791
ISBN 13: 978-0-32167-179-0

Chapter 1 - Section 1.6 - Other Types of Equations and Applications - 1.6 Exercises - Page 134: 18

Answer

Both 1 and -1 are extraneous solutions. There is no solution.

Work Step by Step

$$\frac{-x}{x+1}-\frac{1}{x-1}=\frac{-2}{x^2-1}$$ $$((x+1)(x-1))(\frac{-x}{x+1}-\frac{1}{x-1})=((x+1)(x-1))(\frac{-2}{x^2-1})$$ $$-x(x-1)-(x+1)=-2$$ $$-x^2+x-x-1=-2$$ $$-x^2+1=0$$ $$x^2-1=0$$ $$(x+1)(x-1)=0$$ $$x=-1,1$$ Both 1 and -1 are extraneous solutions. There is no solution.
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