# How can you solve (6x^-2y^7)/(3x^-7y^-3)?

Mar 28, 2018

See a solution process below:

#### Explanation:

First, rewrite the expression as:

$\frac{6}{3} \left({x}^{-} \frac{2}{x} ^ - 7\right) \left({y}^{7} / {y}^{-} 3\right) \implies$

$2 \left({x}^{-} \frac{2}{x} ^ - 7\right) \left({y}^{7} / {y}^{-} 3\right)$

Now, use this rule of exponents to simplify the expression:

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

$2 \left({x}^{\textcolor{red}{- 2}} / {x}^{\textcolor{b l u e}{- 7}}\right) \left({y}^{\textcolor{red}{7}} / {y}^{\textcolor{b l u e}{- 3}}\right) \implies$

$2 {x}^{\textcolor{red}{- 2} - \textcolor{b l u e}{- 7}} {y}^{\textcolor{red}{7} - \textcolor{b l u e}{- 3}} \implies$

$2 {x}^{\textcolor{red}{- 2} + \textcolor{b l u e}{7}} {y}^{\textcolor{red}{7} + \textcolor{b l u e}{3}} \implies$

$2 {x}^{5} {y}^{10}$

Mar 28, 2018

$2 {x}^{5} {y}^{10}$

#### Explanation:

any thing with a negative power will switch its place up and down. so, $\frac{6 {x}^{7} {y}^{7} {y}^{3}}{3 {x}^{2}}$
now as on the numerator we have ${y}^{7} \mathmr{and} {y}^{3} a \mathrm{dd} s u p = {y}^{10} \left({a}^{b} \cdot {a}^{c} = {a}^{c + b}\right)$
$\frac{6 {x}^{7} {y}^{10}}{3 {x}^{2}}$
we can separate${x}^{7} = {x}^{2} \mathmr{and} {x}^{5}$
now solve it: $\frac{\cancel{6} \textcolor{red}{2} \cancel{{x}^{2}} {x}^{5} {y}^{10}}{\cancel{3} \cancel{{x}^{2}}}$