Precalculus (6th Edition)

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

Chapter 1 - Equations and Inequalities - Summary Exercises on Solving Equations - Exercises - Page 149: 6



Work Step by Step

Write the denominators in factored form: $\dfrac{5}{x+3}-\dfrac{6}{x-2}=\dfrac{3}{(x+3)(x-2)}$ Multiply $(x+3)(x-2)$ to both sides of the equation to obtain: $\require{cancel} (x+3)(x-2)\left[\dfrac{5}{x+3}-\dfrac{6}{x-2}\right]=(x+3)(x-2)\left[\dfrac{3}{(x+3)(x-2)}\right]$ $\require{cancel} \\\cancel{(x+3)}(x-2)\left[\dfrac{5}{\cancel{x+3}}\right]-(x+3)\cancel{(x-2)}\left[\dfrac{6}{\cancel{x-2}}\right]=\cancel{(x+3)(x-2)}\left[\dfrac{3}{\cancel{(x+3)(x-2)}}\right]$ $5(x-2)-6(x+3)=3 \\5(x)-5(2)-6(x)-6(3)=3 \\5x-10-6x-18=3 \\-x-28=3 \\-x=3+28 \\-x=31 \\-1(-x)=-1(31) \\x=-31$
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