## Elementary Algebra

$\dfrac{2x}{5}$
Cancelling the common factor between the numerator and the denominator, the given expression, $\dfrac{14xy}{35y} ,$ simplifies to \begin{array}{l}\require{cancel} \dfrac{\cancel{7y}\cdot2x}{\cancel{7y}\cdot5} \\\\= \dfrac{2x}{5} \end{array}