Answer
$\dfrac{3(y-2)}{4(2y+3)}$
Work Step by Step
The given expression, $
\dfrac{\dfrac{y-2}{16}}{\dfrac{2y+3}{12}}
,$ simplifies to
\begin{array}{l}\require{cancel}
\dfrac{y-2}{16}\div\dfrac{2y+3}{12}
\\\\=
\dfrac{y-2}{16}\cdot\dfrac{12}{2y+3}
\\\\=
\dfrac{y-2}{\cancel{4}\cdot4}\cdot\dfrac{\cancel{4}\cdot3}{2y+3}
\\\\=
\dfrac{3(y-2)}{4(2y+3)}
.\end{array}
