# How do you rationalize the denominator and simplify (sqrtx-sqrty)/(sqrt(5x)+sqrt(6y))?

Jan 21, 2018

See a solution process below:

#### Explanation:

To rationalize the denominator and simplify, first, multiply the fraction by the appropriate form of $1$:

$\frac{\sqrt{5 x} - \sqrt{6 y}}{\sqrt{5 x} - \sqrt{6 y}} \times \frac{\sqrt{x} - \sqrt{y}}{\sqrt{5 x} + \sqrt{6 y}} \implies$

$\frac{\left(\sqrt{5 x} - \sqrt{6 y}\right) \left(\sqrt{x} - \sqrt{y}\right)}{\left(\sqrt{5 x} - \sqrt{6 y}\right) \left(\sqrt{5 x} + \sqrt{6 y}\right)} \implies$

$\frac{\sqrt{5 x} \sqrt{x} - \sqrt{5 x} \sqrt{y} - \sqrt{6 y} \sqrt{x} + \sqrt{6 y} \sqrt{y}}{{\left(\sqrt{5 x}\right)}^{2} + \sqrt{5 x} \sqrt{6 y} - \sqrt{5 x} \sqrt{6 y} - {\left(\sqrt{6 y}\right)}^{2}} \implies$

$\frac{\sqrt{5 x \cdot x} - \sqrt{5 x \cdot y} - \sqrt{6 y \cdot x} + \sqrt{6 y \cdot y}}{5 x + 0 - 6 y} \implies$

$\frac{\sqrt{5 {x}^{2}} - \sqrt{5 x y} - \sqrt{6 x y} + \sqrt{6 {y}^{2}}}{5 x - 6 y} \implies$

$\frac{\sqrt{5} x - \sqrt{5 x y} - \sqrt{6 x y} + \sqrt{6} y}{5 x - 6 y}$

If required, we can factor the numerator as:

$\frac{\sqrt{5} x - \sqrt{5} \sqrt{x y} - \sqrt{6} \sqrt{x y} + \sqrt{6} y}{5 x - 6 y} \implies$

$\frac{\sqrt{5} \left(x - \sqrt{x y}\right) - \sqrt{6} \left(\sqrt{x y} - y\right)}{5 x - 6 y}$

Or, we can also factor the numerator as:

$\frac{\sqrt{5} x - \sqrt{5} \sqrt{x y} - \sqrt{6} \sqrt{x y} + \sqrt{6} y}{5 x - 6 y} \implies$

$\frac{\sqrt{5} x - \sqrt{x y} \left(\sqrt{5} + \sqrt{6}\right) + \sqrt{6} y}{5 x - 6 y}$