Precalculus: Concepts Through Functions, A Unit Circle Approach to Trigonometry (3rd Edition)

Published by Pearson
ISBN 10: 0-32193-104-1
ISBN 13: 978-0-32193-104-7

Appendix A - Review - A.6 Rational Expressions - A.6 Assess Your Understanding - Page A53: 47


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

Work Step by Step

Make the expressions similar using their LCD which is $(x-1)(x+1)$, to obtain: $$ \frac{x}{x+1}+\frac{2x-3}{x-1}=\frac{x(x-1)}{(x-1)(x+1)}+\frac{(2x-3)(x+1)}{(x-1)(x+1)}, \quad x\ne-1, 1.$$ Now, since the denominators are the same, add the numerators to obtain: \begin{align*} \frac{x(x-1)}{(x-1)(x+1)}+\frac{(2x-3)(x+1)}{(x-1)(x+1)}&=\frac{x^2-x}{(x-1)(x+1)}+\frac{2x^2+2x-3x-3}{(x-1)(x+1)}\\ \\&=\frac{x^2-x}{(x-1)(x+1)}+\frac{2x^2-x-3}{(x-1)(x+1)}\\ \\&=\frac{x^2-x+2x^2-x-3}{(x-1)(x+1)} \\ \\&=\frac{3x^2-2x-3}{(x-1)(x+1)},\quad x\ne-1, 1\end{align*}
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