Intermediate Algebra (6th Edition)

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

Chapter 6 - Review - Page 403: 8

Answer

$\dfrac{1}{5}$

Work Step by Step

Factoring the expressions and then cancelling the common factor/s between the numerator and the denominator, the given expression, $ \dfrac{5x-15}{25x-75} ,$ simplifies to \begin{array}{l}\require{cancel} \dfrac{5(x-3)}{25(x-3)} \\\\= \dfrac{5(x-3)}{5\cdot5(x-3)} \\\\= \dfrac{\cancel{5(x-3)}}{5\cdot\cancel{5(x-3)}} \\\\= \dfrac{1}{5} .\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.