Решить систему методом подстановки х+у=4 у+ху=6

Решите задачу:

$\left \{ {{x+y=4} \atop {y+xy=6}} \right. \; \; \left \{ {{x=4-y} \atop {y+y(4-y)=6}} \right. \; \; \left \{ {{x=4-y} \atop {y+4y-y^2=6}} \right. \; \; \left \{ {{x=4-y} \atop {y^2-5y+6=0}} \right. \\\\ \left \{ {{x=4-y} \atop {(y-2)(y-3)=0}} \right. \; \; \left \{ {{x_1=2\; ,\; x_2=1} \atop {y_1=2\; ,\; y_2=3}} \right. \\\\Otvet:\; \; (2,2)\; ,\; (1,3)\; .$