# How do you simplify (x^3y)/(xy^5)*(x^2y^9)/x^8 and write it using only positive exponents?

Mar 4, 2018

See a solution process below:

#### Explanation:

First, rewrite the expression as:

$\left(\frac{{x}^{2} \times {x}^{2}}{x \times {x}^{8}}\right) \left(\frac{y \times {y}^{9}}{y} ^ 5\right)$

Next, use this rule of exponents to rewrite the expression again:

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

$\left(\frac{{x}^{2} \times {x}^{2}}{{x}^{\textcolor{red}{1}} \times {x}^{8}}\right) \left(\frac{{y}^{\textcolor{red}{1}} \times {y}^{9}}{y} ^ 5\right)$

Then, use this rule of exponents to combine the common exponents in the numerator and denominator:

${x}^{\textcolor{red}{a}} \times {x}^{\textcolor{b l u e}{b}} = {x}^{\textcolor{red}{a} + \textcolor{b l u e}{b}}$

$\left(\frac{{x}^{\textcolor{red}{2}} \times {x}^{\textcolor{b l u e}{2}}}{{x}^{\textcolor{red}{1}} \times {x}^{\textcolor{b l u e}{8}}}\right) \left(\frac{{y}^{\textcolor{red}{1}} \times {y}^{\textcolor{b l u e}{9}}}{y} ^ 5\right)$

$\left({x}^{\textcolor{red}{2} + \textcolor{b l u e}{2}} / \left({x}^{\textcolor{red}{1} + \textcolor{b l u e}{8}}\right)\right) \left({y}^{\textcolor{red}{1} + \textcolor{b l u e}{9}} / {y}^{5}\right)$

$\left({x}^{4} / {x}^{9}\right) \left({y}^{10} / {y}^{5}\right)$

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

${x}^{\textcolor{red}{a}} / {x}^{\textcolor{b l u e}{b}} = {x}^{\textcolor{red}{a} - \textcolor{b l u e}{b}}$

$\left({x}^{4} / {x}^{9}\right) \left({y}^{\textcolor{red}{10}} / {y}^{\textcolor{b l u e}{5}}\right)$

$\left({x}^{4} / {x}^{9}\right) {y}^{\textcolor{red}{10} - \textcolor{b l u e}{5}}$

$\left({x}^{4} / {x}^{9}\right) {y}^{5}$

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

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

$\left({x}^{\textcolor{red}{4}} / {x}^{\textcolor{b l u e}{9}}\right) {y}^{5}$

$\frac{1}{x} ^ \left(\textcolor{b l u e}{9} - \textcolor{red}{4}\right) \times {y}^{5}$

$\frac{1}{x} ^ 5 \times {y}^{5}$

${y}^{5} / {x}^{5}$

Or

${\left(\frac{y}{x}\right)}^{5}$

We can do this problem easily be breaking it down step by step.

#### Explanation:

Let's handle each expression first, then multiply them both together.
For $\frac{{x}^{3} y}{x {y}^{5}}$, we first start by looking at the x terms. Remember that for division of exponents with the same base, we subtract the exponents. Therefore, ${x}^{3} / x = {x}^{3 - 1} = {x}^{2}$.
Now we do the same thing for the y terms. For this part, recall that to write using only positive exponents that means that all terms with negative exponents are written as their reciprocal, in other words, they go on the bottom of the fraction. Therefore, $\frac{y}{y} ^ 5 = {y}^{1 - 5} = {y}^{-} 4 = \frac{1}{y} ^ 4$.
So the simplification of the first expression is ${x}^{2} / {y}^{4}$.
Now we so the same thing for the second expression.
${x}^{2} / {x}^{8} = {x}^{2 - 8} = {x}^{-} 6 = \frac{1}{x} ^ 6$.
Therefore, the simplification of the second expression is ${y}^{9} / {x}^{6}$.
Now we multiply both expressions together and simplify just as we did before.
${x}^{2} / {y}^{4} \cdot {y}^{9} / {x}^{6} = \frac{{x}^{2} {y}^{9}}{{y}^{4} {x}^{6}} = {y}^{9 - 4} {x}^{2 - 6} = {y}^{5} {x}^{-} 4 = {y}^{5} / {x}^{4}$.