Answer
$33|y|\sqrt{6}$
Work Step by Step
Factor each radicand so that one factor is a perfect square:
$4\sqrt{36y^2\cdot6}+3\sqrt{9y^2\cdot6}$
Recall the property (pg. 367):
$\sqrt[n]{a}\cdot\sqrt[n]{b}=\sqrt[n]{ab}$ (if $\sqrt[n]{a}$ and $\sqrt[n]{b}$ are real numbers)
Applying this property, we get:
$4\sqrt{36y^2\cdot6}+3\sqrt{9y^2\cdot6}$
$=4\sqrt{36y^2}\cdot \sqrt{6}+3\sqrt{9y^2}\cdot \sqrt{6}$
Recall that $6^2=36$, $\sqrt{y^2}=|y|$, and $3^2=9$.
Thus, the expression above simplifies to:
$4\cdot 6|y|\sqrt{6}+3\cdot 3|y|\sqrt{6}$
$=24|y|\sqrt{6}+9|y|\sqrt{6}$
$=(24|y|+9|y|)\sqrt{6}$
$=33|y|\sqrt{6}$
