Intermediate Algebra (6th Edition)

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

Chapter 6 - Review - Page 403: 8



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}
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