Intermediate Algebra (6th Edition)

Published by Pearson
ISBN 10: 0321785045
ISBN 13: 978-0-32178-504-6

Chapter 6 - Review - Page 404: 35

Answer

$\dfrac{5-2x}{2(x-1)}$

Work Step by Step

Factoring the given expression, $ \dfrac{2}{x-1}-\dfrac{3x}{3x-3}+\dfrac{1}{2x-2} ,$ results to \begin{array}{l}\require{cancel} \dfrac{2}{x-1}-\dfrac{3x}{3(x-1)}+\dfrac{1}{2(x-1)} .\end{array} Using the $LCD= 6(x-1) $, the expression above simplifies to \begin{array}{l}\require{cancel} \dfrac{6(2)-2(3x)+3(1)}{6(x-1)} \\\\= \dfrac{12-6x+3}{6(x-1)} \\\\= \dfrac{15-6x}{6(x-1)} \\\\= \dfrac{3(5-2x)}{6(x-1)} \\\\= \dfrac{\cancel{3}(5-2x)}{\cancel{3}\cdot2(x-1)} \\\\= \dfrac{5-2x}{2(x-1)} .\end{array}
Update this answer!

You can help us out by revising, improving and updating this answer.

Update this answer

After you claim an answer you’ll have 24 hours to send in a draft. An editor will review the submission and either publish your submission or provide feedback.