Решите систему уравнения: x^2+4y=8 x+y=2

$\left \{ {{x^2+4y=8} \atop {x+y=2}} \right.$

$\left \{ {{y=2-x} \atop {x^2+4y=8}} \right.$

$\left \{ {{y=2-x} \atop {x^2+4(2-x)=8}} \right.$

$\left \{ {{y=2-x} \atop {x^2+8-4x-8=0}} \right.$

$\left \{ {{y=2-x} \atop {x^2-4x=0}} \right.$

$\left \{ {{y=2-x} \atop {x(x-4)=0}} \right.$

$\left \{ {{y=2-x} \atop {x=0}} \right.$ или $\left \{ {{y=2-x} \atop {x=4}} \right.$

$\left \{ {{y=2 \atop {x=0}} \right.$ или $\left \{ {{y=-2 \atop {x=4}} \right.$

Ответ: $(0;2)$ ; $(4;-2)$