## Algebra 1

$\frac{\sqrt {3n}}{n}$
We rationalize the denominator. $\frac{\sqrt 6}{\sqrt {2n}}$ $\longrightarrow$ multiply top and bottom by $\sqrt {2n}$ $\frac{\sqrt 6}{\sqrt {2n}}\times\frac{\sqrt {2n}}{\sqrt {2n}}$ $=\frac{\sqrt {12n}}{2n}$ $=\frac{\sqrt {4\times3n}}{2n}$ $=\frac{2\sqrt {3n}}{2n}$ $=\frac{\sqrt {3n}}{n}$