Решите систему уравнений

Решите задачу:

$\left \{ {{2^{x} +2^{y} = 40} \atop {2^{x-1} -2^{y-1}=12}} \right. \\ \left \{ {{2^{x} =40-2^{y} } \atop { \frac{40-2^{y}}{2} - \frac{2^{y}}{2} = 12}} \right. \\ \left \{ {{2^{x} =40-2^{y}} \atop {40-2^{y}-2^{y}=24}} \right. \\ \left \{ {{2^{x} =40-2^{y}} \atop {2*2^{y}=16}} \right. \\ \left \{ {{2^{x} =40-2^{y}} \atop {2^{y}=8}} \right. \\ \left \{ {{2^{x} =40-2^{y}} \atop {2^{y}=2^{3}}} \right. \\ \left \{ {{2^{x} =40-2^{y}} \atop {y=3}} \right. \\$
$\left \{ {{2^{x} =40-8} \atop {y=3}} \right. \\ \left \{ {{2^{x} =32} \atop {y=3}} \right. \\ \left \{ {{2^{x} =2^{5}} \atop {y=3}} \right. \\ \left \{ {{x =5} \atop {y=3}} \right. \\$