Question

# How do you simplify \frac { 3x ^ { - 2} y ^ { 3} } { 9x ^ { 4} y ^ { 5} }?

How do you simplify \frac { 3x ^ { - 2} y ^ { 3} } { 9x ^ { 4} y ^ { 5} }?

Answer 1

See a solution process below:

#### Explanation:

First, rewrite this expression as:

$\frac{3}{9} \left({x}^{-} \frac{2}{x} ^ 4\right) \left({y}^{3} / {y}^{5}\right) \implies \frac{1}{3} \left({x}^{-} \frac{2}{x} ^ 4\right) \left({y}^{3} / {y}^{5}\right)$

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

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

$\frac{1}{3} \left({x}^{\textcolor{red}{- 2}} / {x}^{\textcolor{b l u e}{4}}\right) \left({y}^{3} / {y}^{5}\right) \implies \frac{1}{3} \left(\frac{1}{x} ^ \left(\textcolor{b l u e}{4} - \textcolor{red}{- 2}\right)\right) \left({y}^{3} / {y}^{5}\right) \implies$

$\frac{1}{3} \left(\frac{1}{x} ^ \left(\textcolor{b l u e}{4} + \textcolor{red}{2}\right)\right) \left({y}^{3} / {y}^{5}\right) \implies \frac{1}{3} \left(\frac{1}{x} ^ \left(\textcolor{b l u e}{6}\right)\right) \left({y}^{3} / {y}^{5}\right) \implies \frac{1}{3 {x}^{6}} \left({y}^{3} / {y}^{5}\right)$

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

$\frac{1}{3 {x}^{6}} \left({y}^{\textcolor{red}{3}} / {y}^{\textcolor{b l u e}{5}}\right) \implies \frac{1}{3 {x}^{6}} \left(\frac{1}{y} ^ \left(\textcolor{b l u e}{5} - \textcolor{red}{3}\right)\right) \implies \frac{1}{3 {x}^{6}} \left(\frac{1}{y} ^ 2\right) \implies$

$\frac{1}{3 {x}^{6} {y}^{2}}$

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