Решите систему уравнений {у2+2х=2 Х+у=1

$\left \{ {{ y^{2}+2x =2} \atop {x+y=1}} \right. \ \textgreater \ \left \{ {{ y^{2}+2-2y=2(1) } \atop {x=1-y}} \right. 1) y^{2}-2y+2-2=0 y^{2}-2y=0 y(y-2)=0 y=0(1), y-2=0, y=2(2) 1\left \{ {{y=0} \atop {x=1-0}} \right. \ \textgreater \ \left \{ {{y=0} \atop {x=1}} \right. 2 \left \{ {{y=2} \atop {x=1-2}} \right. \ \textgreater \ \left \{ {{y=2} \atop {x=-1}} \right.$

Ответ:(1;0);(-1;2)