2^х*2^у=16 log3х (три снизу)+log3у (три снизу)=1 Решить систему уравнений

$\left \{ {{ 2^{x }* 2^{y} =16 } \atop { log_{3}x+log_{3}y=1 }} \right. \\ \\ \left \{ {{ 2^{x+y}= 2^{4} } \atop {log_{3}(xy)=log_{3}3}} \right. \\ \\ \left \{ {{xy=3} \atop {x+y=4}} \right. \\ \\ \left \{ {{yx=3} \atop {x=4-y}} \right. \\ \\ \left \{ {{y(4-y)=3} \atop {x=4-y}} \right. \\ \\ \left \{ {{x=4-y} \atop {- y^{2}+4y-3=0 }} \right. \\ \\ \left \{ {{ y^{2}-4y+3=0 } \atop {x=4-y}} \right.$

$1. \left \{ {{y=3} \atop {x=4-3}} \right. \\ \\ \left \{ {{y=3} \atop {x=1}} \right. \\ \\ (1;3)$

$2. \left \{ {{y=1} \atop {x=4-1}} \right. \\ \\ \left \{ {{y=1} \atop {x=3}} \right. \\ \\ (3;1)$

Ответ: (3;1);(1;3)