Решите систему уравнений методом подстановки: ( х^2+ху-х-у=2 (х-у=2

$\left \{ {{x^{2}+xy-x-y=2}\atop {x-y=2}} \right.\\\left \{ {{y=x-2} \atop {x^{2}+x(x-2)-x-(x-2)=2} } \right. \\\left \{ {{y=x-2} \atop {x^{2}+x^{2}-2x-x-x+2=2}} \right.\\\left \{ {{y=x-2} \atop {2x^{2}-4x=0} } \right.\\\left \{ {{y=x-2} \atop {x^{2}-2x=0} } \right.\\\left \{ {{y=x-2} \atop {x(x-2)=0}} \right.\\\left \{ {{y=x-2} \atop {x=2;x=0}} \right.\\1)\left \{ {x_{1}= 2}} \atop {y_{1}=0} } \right. \\2)\left \{ {{x_{2}=0} \atop {y_{2}=-2} } \right.$

Ответ: (2 ; 0) , ( 0 ; - 2)