# Question #131a1

Jul 22, 2016

Here's what I got.

#### Explanation:

Your starting expression looks like this

$12 \frac{\sqrt{x - y}}{y} + 12 \frac{x}{\sqrt{x - y}} - 12 \frac{\sqrt{{\left(x - y\right)}^{2}}}{x - y}$

The first thing to do here is identify the common denominator by taking a look at the three fractions you have in your expression.

Your goal when finding the common denominator is to get the denominators of the fractions to match, i.e. you need the three fractions to have the same denominator.

The first fraction has $y$ as the denominator. The second has $\sqrt{x - y}$. The third has $\left(x - y\right)$. To get all three to have the same value, simply multiply each by the product of the other two.

In other words, the common denominator here will be the product of the three denominators. In this case, you have

$y \cdot \left(x - y\right) \cdot \sqrt{x - y} \to$ the common denominator

To get to this point, multiply the first fraction by

$\textcolor{b l u e}{1} = \frac{\left(x - y\right) \cdot \sqrt{x - y}}{\left(x - y\right) \cdot \sqrt{x - y}}$

to get

$12 \frac{\sqrt{x - y}}{y} \cdot {\overbrace{\frac{\left(x - y\right) \cdot \sqrt{x - y}}{\left(x - y\right) \cdot \sqrt{x - y}}}}^{\textcolor{b l u e}{= 1}} = \frac{12 \cdot \left(x - y\right) \cdot \sqrt{x - y} \cdot \sqrt{x - y}}{y \cdot \left(x - y\right) \cdot \sqrt{x - y}}$

Notice that because you're essentially multiplying by $1$, the value of the fraction does not change. You thus have

$12 \frac{\sqrt{x - y}}{y} = \frac{12 \cdot \left(x - y\right) \cdot {\left(\sqrt{x - y}\right)}^{2}}{y \cdot \left(x - y\right) \cdot \sqrt{x - y}}$

$= \frac{12 \cdot \left(x - y\right) \left(x - y\right)}{y \cdot \left(x - y\right) \cdot \sqrt{x - y}}$

Now do the same for the second fraction. You must multiply it by

$\textcolor{b l u e}{1} = \frac{y \cdot \left(x - y\right)}{y \cdot \left(x - y\right)}$

to get

$12 \frac{x}{\sqrt{x - y}} = 12 \frac{x}{\sqrt{x - y}} \cdot {\overbrace{\frac{y \cdot \left(x - y\right)}{y \cdot \left(x - y\right)}}}^{\textcolor{b l u e}{= 1}}$

$= \frac{12 \cdot x \cdot y \cdot \left(x - y\right)}{y \cdot \left(x - y\right) \cdot \sqrt{x - y}}$

Finally, do the same for the third fraction. You must multiply it by

$\textcolor{b l u e}{1} = \frac{y \cdot \sqrt{x - y}}{y \cdot \sqrt{x - y}}$

to get

$12 \frac{\sqrt{{\left(x - y\right)}^{3}}}{x - y} = 12 \frac{\sqrt{{\left(x - y\right)}^{3}}}{x - y} \cdot {\overbrace{\frac{y \cdot \sqrt{x - y}}{y \cdot \sqrt{x - y}}}}^{\textcolor{b l u e}{= 1}}$

$= \frac{12 y \cdot \sqrt{{\left(x - y\right)}^{3} \cdot \left(x - y\right)}}{\left(y \cdot \left(x - y\right) \cdot \sqrt{x - y}\right)}$

$= \frac{12 y \cdot {\left(x - y\right)}^{2}}{\left(y \cdot \left(x - y\right) \cdot \sqrt{x - y}\right)}$

Now you're ready to put all this together and simply the expression

$\frac{12 {\left(x - y\right)}^{2}}{\left(y \cdot \left(x - y\right) \cdot \sqrt{x - y}\right)} + \frac{12 x y \left(x - y\right)}{\left(y \cdot \left(x - y\right) \cdot \sqrt{x - y}\right)} - \frac{12 y {\left(x - y\right)}^{2}}{\left(y \cdot \left(x - y\right) \cdot \sqrt{x - y}\right)}$

Focus on the numerator first

$12 {\left(x - y\right)}^{2} + 12 x y \left(x - y\right) - 12 y {\left(x - y\right)}^{2}$

$12 \left(x - y\right) \left[\left(x - y\right) + x y - y \left(x - y\right)\right]$

$12 \left(x - y\right) \left(x - y + \textcolor{red}{\cancel{\textcolor{b l a c k}{x y}}} - \textcolor{red}{\cancel{\textcolor{b l a c k}{x y}}} + {y}^{2}\right)$

$12 \left(x - y\right) \left({y}^{2} + x - y\right)$

The expression now becomes

$\frac{12 \left(x - y\right) \left({y}^{2} + x - y\right)}{y \left(x - y\right) \sqrt{x - y}}$

Notice that the $\left(x - y\right)$ can be simplified

$\frac{12 \textcolor{red}{\cancel{\textcolor{b l a c k}{\left(x - y\right)}}} \left({y}^{2} + x - y\right)}{y \textcolor{red}{\cancel{\textcolor{b l a c k}{\left(x - y\right)}}} \sqrt{x - y}}$

to get

$\textcolor{g r e e n}{| \overline{\underline{\textcolor{w h i t e}{\frac{a}{a}} \textcolor{b l a c k}{\frac{12 \left({y}^{2} + x - y\right)}{y \cdot \sqrt{x - y}}} \textcolor{w h i t e}{\frac{a}{a}} |}}}$

Note that

$\sqrt{x - y} = {\left(x - y\right)}^{\frac{1}{2}}$

Because $x > y$, you will always have $x - y \ne 0$, but you do need to have $y \ne 0$.