If $\Big(\dfrac{p^{-1}q^{2}}{p^2q^{-4}}\Big)^7 ÷ \Big(\dfrac{p^3q^{-5}}{p^{-2}q^{3}}\Big)^{-5} = p^xq^y$, find x + y, where p and q are different positive primes.

Given,

$\Rightarrow \Big(\dfrac{p^{-1}q^{2}}{p^2q^{-4}}\Big)^7 ÷ \Big(\dfrac{p^3q^{-5}}{p^{-2}q^{3}}\Big)^{-5} = p^xq^y \\[1em] \Rightarrow \Big(\dfrac{p^{-1}q^{2}}{p^2q^{-4}}\Big)^7 ÷ \Big(\dfrac{p^{-2}q^3}{p^3q^{-5}}\Big)^5 = p^xq^y \\[1em] \Rightarrow \dfrac{p^{-7}q^{14}}{p^{14}q^{-28}} \times \Big(\dfrac{p^3q^{-5}}{p^{-2}q^3}\Big)^5 = p^xq^y \\[1em] \Rightarrow \dfrac{q^{14}.q^{28}}{p^{14}.p^7} \times \dfrac{p^{15}q^{-25}}{p^{-10}q^{15}} = p^xq^y \\[1em] \Rightarrow \dfrac{q^{14 +28}}{p^{14 + 7}} \times \dfrac{p^{15}.p^{10}}{q^{25}.q^{15}} = p^xq^y \\[1em] \Rightarrow \dfrac{q^{42}}{p^{21}} \times \dfrac{p^{25}}{q^{40}} = p^xq^y \\[1em] \Rightarrow p^{4}q^{2} = p^xq^y \\[1em] \therefore x = 4 \text{ and } y = 2.$

x + y = 4 + 2 = 6.

Hence, x + y = 6.