Intermediate Algebra for College Students (7th Edition)

Published by Pearson
ISBN 10: 0-13417-894-7
ISBN 13: 978-0-13417-894-3

Chapter 11 - Cumulative Review Exercises - Page 872: 16


The answer is $\frac{2x^2+5x-2}{(x-5)(x+2)}$.

Work Step by Step

The given expression is $\frac{2x+1}{x-5}-\frac{4}{x^2-3x-10}$ Factor the denominator in the second terms. $=x^2-3x-10$ $=x^2-5x+2x-10$ $=x(x-5)+2(x-5)$ $(x-5)(x+2)$ Plug into the expression. $=\frac{2x+1}{x-5}-\frac{4}{(x-5)(x+2)}$ The least common denominator for both terms is $=(x-5)(x+2)$ Multiply and divide the first terms by $x+2$. $=\frac{(2x+1)(x+2)}{(x-5)(x+2)}-\frac{4}{(x-5)(x+2)}$ Clear the parentheses in the first term. $=\frac{2x^2+4x+x+2}{(x-5)(x+2)}-\frac{4}{(x-5)(x+2)}$ Both denominators are equal therefore, Subtract both numerators. $=\frac{2x^2+4x+x+2-4}{(x-5)(x+2)}$ Simplify. $=\frac{2x^2+5x-2}{(x-5)(x+2)}$.
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