Помогите решить пожалуйста оба номера.

Решите задачу:

$1)\; \; \left \{ {{x-y=4} \atop {5^{x+y}=25}} \right. \; \left \{ {{x-y=4} \atop {5^{x+y}=5^2}} \right. \; \left \{ {{x-y=4} \atop {x+y=2}} \right. \; \left \{ {{2x=6} \atop {2y=-2}} \right. \; \left \{ {{x=3} \atop {y=-1}} \right.$

$2)\; \; \left \{ {{3^{x}\cdot 2^{y}=12} \atop {2^{y+1}-3^{x}=5}} \right. \; \left \{ {{2^{y}=\frac{12}{3^{x}} \atop {2^{y}\cdot 2-3^{x}-5=0}} \right. \; \left \{ {{2^{y}=\frac{12}{3^{x}} \atop {2\cdot \frac{12}{3^{x}}-3^{x}-5=0}} \right. \\\\t=3^{x}\ \textgreater \ 0\; ,\; \; \frac{24}{t}-t-5=0\; ,\; \; \frac{24-t^2-5t}{t}=0\\\\t^2+5t-24=0\; , \\\\t_1=-8\ \textless \ 0\; \; ne\; podxodit\\\\t_2=3\; \; \to \; \; 3^{x}=3\; ,\; \; x=1\\\\2^{y}=\frac{12}{3}=4\; ,\; \; 2^{y}=2^2\; ,\; \; y=2\\\\Otvet:\; \; (1,2).\\$