# Simplify and express the answer using positive exponents: (\frac{2x^{-1}}

Simplify and express the answer using positive exponents:
$$\displaystyle{\left({\frac{{{2}{x}^{{-{1}}}}}{{{3}{y}}}}\right)}^{{-{2}}}$$

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yunitsiL
Given expression:
$$\displaystyle{\left({\frac{{{2}{x}^{{-{1}}}}}{{{3}{y}}}}\right)}^{{-{2}}}$$
We know that $$\displaystyle{\left({a}^{{m}}\right)}^{{n}}={a}^{{{m}\times{n}}}$$
So, we have
$$\displaystyle{\left({\frac{{{2}{x}^{{-{1}}}}}{{{3}{y}}}}\right)}^{{-{2}}}={\frac{{{2}^{{-{2}}}\cdot{\left({x}\right)}^{{-{1}{x}-{2}}}}}{{{\left({3}{y}\right)}^{{-{2}}}}}}$$
$$\displaystyle{\left({\frac{{{2}{x}^{{-{1}}}}}{{{3}{y}}}}\right)}^{{-{2}}}={\frac{{{\left({x}\right)}^{{2}}}}{{{2}^{{2}}\cdot{3}^{{-{2}}}\cdot{y}^{{-{2}}}}}}$$
$$\displaystyle{\left({\frac{{{2}{x}^{{-{1}}}}}{{{3}{y}}}}\right)}^{{-{2}}}={\frac{{{\left({x}\right)}^{{2}}\cdot{y}^{{2}}\cdot{3}^{{2}}}}{{{2}^{{2}}}}}$$
$$\displaystyle{\left({\frac{{{2}{x}^{{-{1}}}}}{{{3}{y}}}}\right)}^{{-{2}}}={\frac{{{9}{x}^{{2}}{y}^{{2}}}}{{{4}}}}$$