Precalculus: Mathematics for Calculus, 7th Edition

Published by Brooks Cole
ISBN 10: 1305071751
ISBN 13: 978-1-30507-175-9

Chapter 1 - Section 1.5 - Equations - 1.5 Exercises - Page 56: 28



Work Step by Step

$\dfrac{4}{x-1}+\dfrac{2}{x+1}=\dfrac{35}{x^{2}-1}$ The denominator $x^{2}-1$ is a difference of two squares. We can factor it like this. $\dfrac{4}{x-1}+\dfrac{2}{x+1}=\dfrac{35}{(x-1)(x+1)}$ Now, we can take $(x-1)(x+1)$ to multiply to the left side of the equation: $(x-1)(x+1)(\dfrac{4}{x-1}+\dfrac{2}{x+1})=35$ Evaluate the products: $\dfrac{4(x-1)(x+1)}{x-1}+\dfrac{2(x-1)(x+1)}{x+1}=35$ Simplify: $4(x+1)+2(x-1)=35$ $4x+4+2x-2=35$ Solve for $x$: $6x=35-4+2$ $6x=33$ $x=\dfrac{33}{6}$
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