College Algebra (6th Edition)

Published by Pearson
ISBN 10: 0-32178-228-3
ISBN 13: 978-0-32178-228-1

Chapter P - Prerequisites: Fundamental Concepts of Algebra - Exercise Set P.6: 70

Answer

$\dfrac{\dfrac{6}{x^{2}+2x-15}-\dfrac{1}{x-3}}{\dfrac{1}{x+5}+1}=-\dfrac{x-1}{(x+6)(x-3)}$

Work Step by Step

$\dfrac{\dfrac{6}{x^{2}+2x-15}-\dfrac{1}{x-3}}{\dfrac{1}{x+5}+1}$ Factor the rational expression completely: $\dfrac{\dfrac{6}{x^{2}+2x-15}-\dfrac{1}{x-3}}{\dfrac{1}{x+5}+1}=\dfrac{\dfrac{6}{(x+5)(x-3)}-\dfrac{1}{x-3}}{\dfrac{1}{x+5}+1}=...$ Evaluate the operations indicated in the numerator and in the denominator: $...=\dfrac{\dfrac{6-(x+5)}{(x+5)(x-3)}}{\dfrac{1+x+5}{x+5}}=\dfrac{\dfrac{6-x-5}{(x+5)(x-3)}}{\dfrac{x+6}{x+5}}=\dfrac{\dfrac{-x+1}{(x+5)(x-3)}}{\dfrac{x+6}{x+5}}=...$ Simplify: $...=\dfrac{(-x+1)(x+5)}{(x+6)(x+5)(x-3)}=\dfrac{-x+1}{(x+6)(x-3)}=-\dfrac{x-1}{(x+6)(x-3)}$
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