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

$\displaystyle \frac{6\sqrt{xy}+9x+y}{y-9x}$
We lose the square roots in the denominator by applying the difference of squares formula: $(a\sqrt{x}+b\sqrt{y})(a\sqrt{x}-b\sqrt{y})=(a\sqrt{x})^{2}-(b\sqrt{y})^{2}=a^{2}x-b^{2}y$ $\displaystyle \frac{3\sqrt{x}+\sqrt{y}}{\sqrt{y}-3\sqrt{x}}\color{red}{ \cdot\frac{\sqrt{y}+3\sqrt{x}}{\sqrt{y}+3\sqrt{x}} }\qquad$ (rationalize) $=\displaystyle \frac{(3\sqrt{x}+\sqrt{y})(\sqrt{y}+3\sqrt{x})}{(\sqrt{y})^{2}-(3\sqrt{x})^{2}}$ ... use FOIL for the numerator $=\displaystyle \frac{3\sqrt{xy}+9(\sqrt{x})^{2}+(\sqrt{y})^{2}+3\sqrt{xy}}{y-3^{2}x}$ $=\displaystyle \frac{6\sqrt{xy}+9x+y}{y-9x}$