## Intermediate Algebra (6th Edition)

$\dfrac{1}{5}$
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}