College Algebra (6th Edition)

Published by Pearson
ISBN 10: 0-32178-228-3
ISBN 13: 978-0-32178-228-1

Chapter P - Prerequisites: Fundamental Concepts of Algebra - Exercise Set P.6: 36

Answer

$\frac{x-2}{x-3}; x \ne 3,-2$

Work Step by Step

$ =\frac{x^{2}-4x}{x^{2}-x-6} + \frac{4x-4}{x^{2}-x-6}$ Same denominator - add numerators. $ =\frac{x^{2}-4x+4x-4}{x^{2}-x-6}$ Combine like terms. $ =\frac{x^{2}-4}{x^{2}-x-6}$ Factor $ =\frac{(x+2)(x-2)}{(x+2)(x-3)}; x \ne 3,-2;$ $=\frac{x-2}{x-3}; x \ne 3,-2$
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