\(2^{x + y} = 32, 3^{3y - x}\) = 27

\(2^{x + y} = 2^5\)

\(3^{3y - x} = 3^3\)

equating powers, we have

x + y = 5 ------------1

3y - x = 3 ------------2

solving simultaneously(any method)

x =  3 and y = 2