Algebra: A Combined Approach (4th Edition)

Published by Pearson
ISBN 10: 0321726391
ISBN 13: 978-0-32172-639-1

Chapter 7 - Section 7.7 - Simplifying Complex Fractions - Exercise Set - Page 549: 22

Answer

$\dfrac{x-\dfrac{1}{2x+1}}{1-\dfrac{x}{2x+1}}=2x-1$

Work Step by Step

$\dfrac{x-\dfrac{1}{2x+1}}{1-\dfrac{x}{2x+1}}$ Evaluate the substractions indicated in the numerator and the denominator: $\dfrac{x-\dfrac{1}{2x+1}}{1-\dfrac{x}{2x+1}}=\dfrac{\dfrac{x(2x+1)-1}{2x+1}}{\dfrac{2x+1-x}{2x+1}}=\dfrac{\dfrac{2x^{2}+x-1}{2x+1}}{\dfrac{x+1}{2x+1}}=...$ Evaluate the division: $...=\dfrac{2x^{2}+x-1}{2x+1}\div\dfrac{x+1}{2x+1}=\dfrac{(2x^{2}+x-1)(2x+1)}{(x+1)(2x+1)}=...$ Factor the first parentheses in the numerator and simplify: $...=\dfrac{(x+1)(2x-1)(2x+1)}{(x+1)(2x+1)}=2x-1$
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