## Intermediate Algebra for College Students (7th Edition)

$\displaystyle \frac{3\sqrt{3y}}{xy}$
Simpify the denominator using $\sqrt[n]{a}\times\sqrt[n]{b}=\sqrt[n]{ab},\qquad$ and $\sqrt[n]{a^{n}}=(\sqrt[n]{a})^{n}=a$ (for positive a). $\sqrt{3x^{2}y}=\sqrt{x^{2}\times 3y}=\sqrt{x^{2}}\times\sqrt{3y} = x\sqrt{3y}$ $... =\displaystyle \frac{9}{x\sqrt{3y}}$ $\displaystyle \color{red}{ \cdot\frac{\sqrt{3y}}{\sqrt{3y}} }\qquad$ (rationalize) $=\displaystyle \frac{9\sqrt{3y}}{x(\sqrt{3y})^{2}}$ $=\displaystyle \frac{9\sqrt{3y}}{x(3y)}$ $\qquad$ ... reduce 3, the common factor = $\displaystyle \frac{3\sqrt{3y}}{xy}$