HEALP!!!!!!!!!!!!!!!Решите систему х+у+ху=-1=х^2+xy+y^2=3

Решите задачу:

$\left \{ {{x+y+xy=-1} \atop {x^2+xy+y^2=3}} \right. \; \left \{ {{x+y=-xy-1} \atop {x^2+y^2=3-xy}} \right. \; \left \{ {{(x+y)^2=(-xy-1)^2} \atop {x^2+2xy+y^2=3-xy+2xy}} \right. \\\\ \left \{ {{(x+y)^2=x^2y^2+2xy+1} \atop {(x+y)^2=3+xy}} \right. \; \; \; \to \; \; \; x^2y^2+2xy+1=3+xy\\\\x^2y^2+xy-2=0\\\\t=xy\; ,\; \; \; t^2+t-2=0\; \; \to \; \; \; t_1=-2,\; t_2=1\; (Viet)$

$a)\; \; \left \{ {{xy=-2} \atop {x+y=-xy-1}} \right. \; \left \{ {{xy=-2} \atop {x+y=1}} \right. \; \left \{ {{xy=-2} \atop {y=1-x}} \right.$

$x(1-x)=-2\; ,\; \; x-x^2=-2\; ,\; \; x^2-x-2=0\\\\x_1=-1\; ,\; \; x_2=2\\\\b)\; \; \left \{ {{xy=1} \atop {x+y=-2}} \right. \; \left \{ {{x(-x-2)=1} \atop {y=-x-2}} \right. \; \; \to \; \; \; -x^2-2x=1\\\\x^2+2x+1=0\; ,\; \; (x+1)^2=0\; ,\; \; x=-1\; \; \to \; \; y=-1\\\\Otvet:\; \; (-1,2)\; ,\; \; (-1,-1)\; .$