Answer
$\dfrac{6}{7x^{2}y^{4}}$
Work Step by Step
$\bf{\text{Solution Outline:}}$
To simplify the given expression, $
\left( \dfrac{2x^3y^{-4}}{5xy^5} \right)\left( \dfrac{15xy^2}{7x^5y^{-3}} \right)
,$ use the laws of exponents.
$\bf{\text{Solution Details:}}$
Using the Quotient Rule of the laws of exponents, which states that $\dfrac{x^m}{x^n}=x^{m-n},$ the expression above simplifies to
\begin{array}{l}\require{cancel}
\left( \dfrac{2x^3y^{-4}}{5xy^5} \right)\left( \dfrac{15xy^2}{7x^5y^{-3}} \right)
\\\\
\left( \dfrac{2x^{3-1}y^{-4-5}}{5} \right)\left( \dfrac{15x^{1-5}y^{2-(-3)}}{7} \right)
\\\\=
\left( \dfrac{2x^{2}y^{-9}}{5} \right)\left( \dfrac{15x^{-4}y^{5}}{7} \right)
.\end{array}
Using the Product Rule of the laws of exponents, which is given by $x^m\cdot x^n=x^{m+n},$ the expression above is equivalent to
\begin{array}{l}\require{cancel}
\left( \dfrac{2x^{2}y^{-9}}{5} \right)\left( \dfrac{15x^{-4}y^{5}}{7} \right)
\\\\=
\dfrac{2(15)x^{2+(-4)}y^{-9+5}}{5(7)}
\\\\=
\dfrac{2(\cancel{15}^3)x^{2-4}y^{-9+5}}{\cancel{5}^1(7)}
\\\\=
\dfrac{6x^{-2}y^{-4}}{7}
.\end{array}
Using the Negative Exponent Rule of the laws of exponents, which states that $x^{-m}=\dfrac{1}{x^m}$ or $\dfrac{1}{x^{-m}}=x^m,$ the expression above is equivalent to
\begin{array}{l}\require{cancel}
\dfrac{6x^{-2}y^{-4}}{7}
\\\\=
\dfrac{6}{7x^{2}y^{4}}
.\end{array}
