Solve the system of equation 2\(^{x + y}\) = 32, \(3^{3y - x}\) = 27
\(2^{x + y} = 32, 3^{3y - x}\) = 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
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