Question

Решите систему уравнений:
x+y=2
2x^2+xy+y^2=8

Answer 1

$\left \{ {{x+y=2} \atop {2x^2+xy+y^2=8}} \right. \\ \\ 1. \\ x+y=2 \\ x=2-y \\ \\ 2. \\ 2(2-y)^2+y(2-y)+y^2=8 \\2(4-4y+y^2)+2y-y^2+y^2=8 \\ 8-8y+2y^2+2y-y^2+y^2-8=0 \\ 2y^2-6y=0 \\ 2y(y-3)=0 \\ y=0;y=3 \\ \\ 3. \\ \left \{ {{y=3} \atop {x=-1}} \right. \\ \left \{ {{y=0} \atop {x=2}} \right.$

Ответ:(-1;3)(2;0)