# How do you simplify \frac { 4y ^ { 2} } { x ^ { 0} y ^ { 3} }?

Mar 15, 2018

See a solution process below:

#### Explanation:

First, use this rule of exponents to simplify the $x$ term:

${a}^{\textcolor{red}{0}} = 1$

$\frac{4 {y}^{2}}{{x}^{\textcolor{red}{0}} {y}^{3}} \implies \frac{4 {y}^{2}}{1 \cdot {y}^{3}} \implies \frac{4 {y}^{2}}{y} ^ 3$

Next, use these rule of exponents to simplify the $y$ term:

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

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

Mar 15, 2018

It becomes $\frac{4}{y}$

#### Explanation:

First of all, anything raised to the zero power is going to be 1, so that ${x}^{0}$ just is one, and won't affect the problem. Second of all, you can cancel out exponents when working with fractions, so the ${y}^{2} / {y}^{3}$ will become $\frac{1}{y} ^ 1$ and you can simplify that to just $\frac{1}{y}$. We can't forget the 4 either, so in all actuality, your expression simplifies to $\frac{4 \cdot 1}{y}$ or just $\frac{4}{y}$