Solution (By Examveda Team)
$$\eqalign{
& {{\text{2}}^x}{\text{ = }}{{\text{8}}^{y + 1}} \cr
& \Leftrightarrow {{\text{2}}^x}{\text{ = }}{\left( {{2^3}} \right)^{y + 1}} = {2^{\left( {3y + 3} \right)}} \cr
& \Leftrightarrow x = 3y + 3 \cr
& \Leftrightarrow x - 3y = 3.......(i) \cr
& {9^y} = {3^{x - 9}} \cr
& \Leftrightarrow {\left( {{3^2}} \right)^y}{\text{ = }}{{\text{3}}^{x - 9}} \cr
& \Leftrightarrow 2y = x - 9 \cr
& \Leftrightarrow x - 2y = 9......({\text{ii}}) \cr
& {\text{Subtracting (i) from (ii),}} \cr
& {\text{we}}\,{\text{get}}\,y = 6 \cr
& {\text{Putting }}y\,{\text{ = 6 in (i),}} \cr
& {\text{we get }}x{\text{ = 21}} \cr
& \therefore x + y = 21 + 6 = 27 \cr} $$
Join The Discussion