## Intermediate Algebra: Connecting Concepts through Application

$\dfrac{6}{7x^{2}y^{4}}$
$\bf{\text{Solution Outline:}}$ 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}