Помогите решить 2 и 3 Ответ к 2 : (2;3) Ответ к 3 : (3;1)

1. $\left \{ {{5^x-3^{y-1}=16} \atop {5^{x-1}+3^y=32}} \right. \\ \left \{ {{5^x-3^{y-1}=16} \atop { \frac{5^x}{5}=32-3^y }} \right. \\ \left \{ {{5^x-3^{y-1}=16} \atop {5^x=160-15*3^{y-1}}} \right. \\ \left \{ {{160-15*3^{y-1}-3^{y-1}=16} \atop {5^x=160-15*3^{y-1}}} \right. \\ \left \{ {{10 - 3^{y-1}=1} \atop {5^x=160-15*3^{y-1}}} \right. \\ \left \{ {{3^{y-1}=3^2} \atop {5^x=160-15*3^{y-1}}} \right. \\ \left \{ {{y-1=2} \atop {5^x=160-15*3^{y-1}}} \right.$
$\left \{ {{y=3} \atop {5^x=160-15*3^2} \right. \\ \left \{ {{y=3} \atop {5^x=160-135}} \right. \\ \left \{ {{y=3} \atop {5^x=5^2}} \right. \\ \left \{ {{y=3} \atop {x=2}} \right.$

Ответ: (2;3)

2. $\left \{ {{2^x-2^y=6} \atop {2^{x+y}=16}} \right. \\ \left \{ {{2^x = 6+2^y} \atop {2^x*2^y=16}} \right. \\ \left \{ {{2^x=6+2^y} \atop {(6+2^y)2^y=16}} \right. \\ \left \{ {{2^x=6+2^y} \atop {(2^y)^2+6*2^y-16=0}} \right. \\ (2^y)^2+6*2^y-16=0 \\ D=36+64=100 \\ 2^y= \frac{-6+10}{2} =2 \\ y=1 \\ \left \{ {{2^x=6+2^1} \atop {y=1}} \right. \\ \left \{ {{2^x=8} \atop {y=1}} \right. \\ \left \{ {{2^x=2^3} \atop {y=1}} \right. \\ \left \{ {{x=3} \atop {y=1}} \right.$

Ответ: (3;1)