Precalculus (6th Edition)

Published by Pearson
ISBN 10: 013421742X
ISBN 13: 978-0-13421-742-0

Chapter 1 - Equations and Inequalities - 1.6 Other Types of Equations and Applications - 1.6 Exercises - Page 146: 24


The solution is $x=3$

Work Step by Step

$\dfrac{3}{x^{2}+x-2}-\dfrac{1}{x^{2}-1}=\dfrac{7}{2x^{2}+6x+4}$ Factor all rational expressions completely: $\dfrac{3}{(x+2)(x-1)}-\dfrac{1}{(x-1)(x+1)}=\dfrac{7}{2(x+2)(x+1)}$ Multiply the whole equation by $2(x+2)(x-1)(x+1)$: $2(x+2)(x-1)(x+1)\Big[\dfrac{3}{(x+2)(x-1)}-\dfrac{1}{(x-1)(x+1)}=\dfrac{7}{2(x+2)(x+1)}\Big]$ $3(2)(x+1)-(2)(x+2)=7(x-1)$ Evaluate the indicated operations: $6x+6-2x-4=7x-7$ Take all terms with $x$ to the left side and all terms without $x$ to the right side: $6x-2x-7x=-7-6+4$ Simplify both sides: $-3x=-9$ Solve for $x$: $x=\dfrac{-9}{-3}$ $x=3$
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