# How do you simplify \frac { - 16r ^ { 3} y ^ { 2} } { - 4r ^ { 2} y ^ { 7} }?

May 4, 2017

See a solution process below:

#### Explanation:

First, rewrite the expression as:

$\frac{- 16}{-} 4 \left({r}^{3} / {r}^{2}\right) \left({y}^{2} / {y}^{7}\right) \implies 4 \left({r}^{3} / {r}^{2}\right) \left({y}^{2} / {y}^{7}\right)$

Next, use these two rule of exponents to simplify the $r$ and $y$ terms:

${x}^{\textcolor{red}{a}} / {x}^{\textcolor{b l u e}{b}} = {x}^{\textcolor{red}{a} - \textcolor{b l u e}{b}}$ and ${x}^{\textcolor{red}{a}} / {x}^{\textcolor{b l u e}{b}} = \frac{1}{x} ^ \left(\textcolor{b l u e}{b} - \textcolor{red}{a}\right)$

$4 \left({r}^{\textcolor{red}{3}} / {r}^{\textcolor{b l u e}{2}}\right) \left({y}^{\textcolor{red}{2}} / {y}^{\textcolor{b l u e}{7}}\right) \implies 4 \left({r}^{\textcolor{red}{3} - \textcolor{b l u e}{2}}\right) \left(\frac{1}{y} ^ \left(\textcolor{b l u e}{7} - \textcolor{red}{2}\right)\right) = 4 \left({r}^{1}\right) \left(\frac{1}{y} ^ 5\right) = \frac{4 {r}^{1}}{y} ^ 5$

Now, use this rule of exponents to complete the simplification of the $r$ term:

${a}^{\textcolor{red}{1}} = a$

$\frac{4 {r}^{\textcolor{red}{1}}}{y} ^ 5 = \frac{4 r}{y} ^ 5$