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: 23


The solution is $x=-9$

Work Step by Step

$\dfrac{4}{x^{2}+x-6}-\dfrac{1}{x^{2}-4}=\dfrac{2}{x^{2}+5x+6}$ Factor all rational expressions completely: $\dfrac{4}{(x+3)(x-2)}-\dfrac{1}{(x-2)(x+2)}=\dfrac{2}{(x+3)(x+2)}$ Multiply the whole equation by $(x+3)(x-2)(x+2)$: $(x+3)(x-2)(x+2)\Big[\dfrac{4}{(x+3)(x-2)}-\dfrac{1}{(x-2)(x+2)}=\dfrac{2}{(x+3)(x+2)}\Big]$ $4(x+2)-(x+3)=2(x-2)$ Evaluate the indicated operations: $4x+8-x-3=2x-4$ Take all terms with $x$ to the left side and all terms without $x$ to the right side: $4x-x-2x=-4-8+3$ Simplify both sides: $x=-9$
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