Решите систему уравнений y^2 = x + 4y { x + y = 4

Решите задачу:

$\left \{ {{y^2=x+4y} \atop {x+y=4}} \right. \left \{ {{y^2=x+4y} \atop {x=4-y}} \right. \left \{ {{y^2=4-y+4y} \atop {x=4-y}} \right. \left \{ {{y^2-3y-4=0}(1) \atop {x=4-y}(2)} \right.\\(1)y^2-3y-4=0\\a-b+c=0\\y_1=-1\\y_2=4\\ (2)x_1=4+1=5\\x_2=4-4=0\\................................................................................\\(x_1;y_1)=(0;4)\\(x_2;y_2)=(5;-1)$