College Algebra (10th Edition)

Published by Pearson
ISBN 10: 0321979478
ISBN 13: 978-0-32197-947-6

Chapter R - Section R.7 - Rational Expressions - R.7 Assess Your Understanding - Page 71: 49


$\dfrac{3x^2-2x-3}{(x-1)(x+1)}; x\ne -1, 1$

Work Step by Step

The LCM of $x+1$ and $x-1$ is $(x+1)(x-1)$. Use the LCM as the expressions' LCD to make them similar: $=\dfrac{x\color{blue}{(x-1)}}{(x+1)\color{blue}{(x-1)}} +\dfrac{(2x-3)\color{blue}{(x+1)}}{(x-1)\color{blue}{(x+1)}} \\=\dfrac{x^2-x}{(x+1)(x-1)}+\dfrac{2x(x+1)-3(x+1)}{(x-1)(x+1)} \\=\dfrac{x^2-x}{(x+1)(x-1)}+\dfrac{2x^2+2x-3x-3}{(x-1)(x+1)} \\=\dfrac{x^2-x}{(x+1)(x-1)}+\dfrac{2x^2-x-3}{(x-1)(x+1)}$ The expressions are similar so add the numerators and copy the denominator to obtain: $=\dfrac{x^2-x+2x^2-x-3}{(x-1)(x+1)} \\=\dfrac{3x^2-2x-3}{(x-1)(x+1)}; x\ne -1, 1$
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