How do you simplify (10x^4 y^9) /( 30x^12 y^-3)?

Mar 10, 2018

See a solution process below:

Explanation:

First, rewrite the expression as:

$\frac{10}{30} \left({x}^{4} / {x}^{12}\right) \left({y}^{9} / {y}^{-} 3\right) \implies$

$\frac{1 \textcolor{red}{\cancel{\textcolor{b l a c k}{0}}}}{3 \textcolor{red}{\cancel{\textcolor{b l a c k}{0}}}} \left({x}^{4} / {x}^{12}\right) \left({y}^{9} / {y}^{-} 3\right) \implies$

$\frac{1}{3} \left({x}^{4} / {x}^{12}\right) \left({y}^{9} / {y}^{-} 3\right)$

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

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

$\frac{1}{3} \left({x}^{\textcolor{red}{4}} / {x}^{\textcolor{b l u e}{12}}\right) \left({y}^{9} / {y}^{-} 3\right) \implies$

$\frac{1}{3} \left(\frac{1}{x} ^ \left(\textcolor{b l u e}{12} - \textcolor{red}{4}\right)\right) \left({y}^{9} / {y}^{-} 3\right) \implies$

$\frac{1}{3} \left(\frac{1}{x} ^ 8\right) \left({y}^{9} / {y}^{-} 3\right) \implies$

$\frac{1}{3 {x}^{8}} \left({y}^{9} / {y}^{-} 3\right)$

Now, use this rule for exponents to simplify the $y$ term:

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

$\frac{1}{3 {x}^{8}} \left({y}^{\textcolor{red}{9}} / {y}^{\textcolor{b l u e}{- 3}}\right) \implies$

$\frac{1}{3 {x}^{8}} \left({y}^{\textcolor{red}{9} - \textcolor{b l u e}{- 3}}\right) \implies$

$\frac{1}{3 {x}^{8}} \left({y}^{\textcolor{red}{9} + \textcolor{b l u e}{3}}\right) \implies$

$\frac{1}{3 {x}^{8}} \left({y}^{12}\right) \implies$

${y}^{12} / \left(3 {x}^{8}\right)$