Answer
$\dfrac{2x}{5}$
Work Step by Step
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}
