If [tex]3x-y=12[/tex], then [tex]y=3x-12[/tex]. Substituting that into [tex]\dfrac{8^x}{2^y}[/tex] would give us...
[tex]\begin{aligned}\dfrac{8^x}{2^y} &= \dfrac{8^x}{2^{3x-12}} \\[0.9em]&= \dfrac{8^x}{2^{3x}\cdot2^{-12}}\\[0.9em]&= \dfrac{8^x}{(2^3)^x\cdot2^{-12}}\\[0.9em]&= \dfrac{8^x}{8^{x}\cdot2^{-12}}\\[0.9em]&= \dfrac{1}{1\cdot2^{-12}}\\[0.9em]&= \dfrac{1}{2^{-12}}\\[0.9em]&= 2^{12}\end{aligned}[/tex]
