Algebra: A Combined Approach (4th Edition)

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

Chapter 7 - Cumulative Review: 48

Answer

$\frac{30}{(x+3)}$

Work Step by Step

$\frac{6x^{2}-18x}{3x^{2}-2x}\times\frac{15x-10}{x^{2}-9}$ =$\frac{6x(x-3)}{x(3x-2)}\times\frac{5(3x-2)}{x^{2}-3^{2}}$ =$\frac{6x(x-3)}{x(3x-2)}\times\frac{5(3x-2)}{(x+3)(x-3)}$ =$\frac{6x}{x(3x-2)}\times\frac{5(3x-2)}{(x+3)}$ =$\frac{6x}{x}\times\frac{5}{(x+3)}$ =$\frac{6}{1}\times\frac{5}{(x+3)}$ =$\frac{30}{(x+3)}$
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