Precalculus (6th Edition)

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

Chapter R - Review of Basic Concepts - R.5 Rational Expressions - R.5 Exercises: 75

Answer

$\dfrac{-b^2+b+2}{-b^2+b}$

Work Step by Step

Simplify by multiplying the LCD (which is $(1-b)(1+b)$) to both the numerator and the denominator to obtain: $\require{cancel} =\dfrac{(1-b)(1+b)\left(1+\frac{1}{1-b}\right)}{(1-b)(1+b)(1-\frac{1}{1+b})} \\=\dfrac{(1-b)(1+b)(1)+(1-b)(1+b)(\frac{1}{1-b})}{(1-b)(1+b)(1)-(1-b)(1+b)(\frac{1}{1+b})} \\=\dfrac{(1-b)(1+b)+(1+b)}{(1-b)(1+b)-(1-b)}$ Use the formula $(x-y)(x+y)=x^2-y^2$ to obtain: $=\dfrac{(1^2-b^2)+1+b}{(1^2-b^2)-1+b} \\=\dfrac{1-b^2+1+b}{1-b^2-1+b} \\=\dfrac{2+b-b^2}{b-b^2} \\=\dfrac{-b^2+b+2}{-b^2+b}$
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