## Elementary Algebra

$\dfrac{5b}{7}$
Cancelling the common factor between the numerator and the denominator, the given expression, $\dfrac{65abc}{91ac} ,$ simplifies to \begin{array}{l}\require{cancel} \dfrac{\cancel{13ac}\cdot5b}{\cancel{13ac}\cdot7} \\\\= \dfrac{5b}{7} \end{array}