Given 2^x = 8^y+1 and 9^y = 3^x-9 , then value of x+y is = ?

Explanation:-

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