Развязать систему уравнений x +y =2 2x^2 +xy+y^2=16

$\displaystyle \left \{ {{x+y=2} \atop {2x^2+xy+y^2=16}} \right. \Rightarrow \left \{ {{x+y=2} \atop {x^2+(x+y)^2-xy=16}} \right. \Rightarrow\\ \\ \\ \Rightarrow \left \{ {{x+y=2} \atop {x^2-xy+2^2=16}} \right. \Rightarrow \left \{ {{y=2-x} \atop {x^2-x(2-x)=12}} \right. \\ x^2-2x+x^2=12\\ 2x^2-2x-12=0|:2\\ x^2-x-6=0$

По т. Виета: $x_1=-2;\,\,\,\,\,\,\,\,\,\, x_2=3$

$y_1=2-x_1=2+2=4\\ y_2=2-x_2=2-3=-1$