# How do you simplify (-8/21 x^2y^3) times (-7/16xy^2)?

##### 1 Answer
Aug 15, 2017

See a solution process below:

#### Explanation:

First, rewrite the expression as:

$\left(- \frac{8}{21} \times - \frac{7}{16}\right) \left({x}^{2} \times x\right) \left({y}^{3} \times {y}^{2}\right) \implies$

$\left(- \frac{\textcolor{red}{\cancel{\textcolor{b l a c k}{8}}}}{\textcolor{b l u e}{\cancel{\textcolor{b l a c k}{21}} 3}} \times - \frac{\textcolor{b l u e}{\cancel{\textcolor{b l a c k}{7}}}}{\textcolor{red}{\cancel{\textcolor{b l a c k}{16}}} 2}\right) \left({x}^{2} \times x\right) \left({y}^{3} \times {y}^{2}\right) \implies$

$\left(- \frac{1}{3} \times - \frac{1}{2}\right) \left({x}^{2} \times x\right) \left({y}^{3} \times {y}^{2}\right) \implies$

$\frac{1}{6} \left({x}^{2} \times x\right) \left({y}^{3} \times {y}^{2}\right)$

Next, use this rule for exponents to rewrite the $x$ term:

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

$\frac{1}{6} \left({x}^{2} \times x\right) \left({y}^{3} \times {y}^{2}\right) \implies \frac{1}{6} \left({x}^{2} \times {x}^{\textcolor{red}{1}}\right) \left({y}^{3} \times {y}^{2}\right)$

Now, use this rule of exponents to simplify the $x$ and $y$ terms:

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

$\frac{1}{6} \left({x}^{\textcolor{red}{2}} \times {x}^{\textcolor{b l u e}{1}}\right) \left({y}^{\textcolor{red}{3}} \times {y}^{\textcolor{b l u e}{2}}\right) \implies$

$\frac{1}{6} {x}^{\textcolor{red}{2} + \textcolor{b l u e}{1}} {y}^{\textcolor{red}{3} + \textcolor{b l u e}{2}} \implies$

$\frac{1}{6} {x}^{3} {y}^{5}$

Or

$\frac{{x}^{3} {y}^{5}}{6}$