X^2+x+xy=8 y^2+y+xy=4 решить систему уравнений

Решите задачу:

$\left \{ {{x^2+x+xy=8} \atop {y^2+y+xy=4}} \right. \oplus \left \{ {{x^2+y^2+x+y+2xy=12} \atop {x^2+x+xy=8}} \right. \; \left \{ {{(x+y)^2+(x+y)-12=0} \atop {x^2+x+xy=8}} \right. \\\\ t=x+y\; ,\; \; t^2+t-12=0\; ,\; \; t_1=-4\; ,\; t_2=3\; \; (teorema\; Vieta)\\\\a)\; \; \left \{ {{x+y=-4} \atop {x^2+x+xy=8}} \right. \; \left \{ {{y=-x-4} \atop {x^2+x+x(-x-4)=8}} \right. \; \left \{ {{y=-x-4} \atop {x^2+x-x^2-4x=8}} \right. \; \left \{ {{y=-x-4} \atop {-3x=8}} \right.$

$\left \{ {{y=-\frac{4}{3}} \atop {x=-\frac{8}{3}}} \right. \\\\b)\; \; \left \{ {{x+y=3} \atop {x^2+x+xy=8}} \right. \; \left \{ {{y=3-x} \atop {x^2+x+x(3-x)=8}} \right. \; \left \{ {{y=3-x} \atop {x^2+x+3x-x^2=8}} \right. \; \left \{ {{y=3-x} \atop {4x=8}} \right. \\\\\left \{ {{y=1} \atop {x=2}} \right. \\\\Otvet:\; \; (-\frac{8}{3}\, ,\, -\frac{4}{3})\; ,\; \; (2,1)\; .$