Question

${\left(\begin{array}{c}\frac{x^{-1}{y}^2}{x^2{y}^{-4}}\end{array}\right)}^7÷{\left(\begin{array}{c}\frac{x^3{y}^{-5}}{x^{-2}{y}^3}\end{array}\right)}^{-6}=x^p.y^q$. Then $p+q=$

Answer 1

${\left(\frac{x^{-1}{y}^2}{x^2{y}^{-4}}\right)}^7÷{\left(\frac{x^3{y}^{-5}}{x^{-2}{y}^3}\right)}^{-6}=x^p{y}^q$

$\Rightarrow \left(x^{-21}{y}^{42}\right)÷\left(x^{-30}{x}^{48}\right)\Rightarrow x^{-21+30}{y}^{42-48}$

$\Rightarrow x^9{y}^{-6}\therefore p+q=(9-6)=3$