Решите систему уравнений: х^2+y^2-6y=0 y+2x=0 И второе: x^2-xy+y^2=63 x+y=-3

1)
$\left \{ {{x^2+y^2-6y=0} \atop {y+2x=0}} \right. \\y=-2x \\x^2+(-2x)^2-6(-2x)=0 \\x^2+4x^2+12x=0 \\5x^2+12x=0 \\x(5x+12)=0 \\x_1=0 \\5x+12=0 \\5x=-12 \\x_2= -\frac{12}{5} =-2,4 \\y_1=-2*0=0 \\y_2=-2*-2,4=4,8$
Ответ: (0;0), (-2,4;4,8)
2)
$\left \{ {{x^2-xy+y^2=63} \atop {x+y=-3}} \right. \\ \left \{ {{(x^2+2xy+y^2)-3xy=63} \atop {x+y=-3}} \right. \\ \left \{ {{(x+y)^2-3xy=63} \atop {x+y=-3}} \right. \\\left \{ {{(-3)^2-3xy=63} \atop {x+y=-3}} \right. \\\left \{ {{-3xy=63-9} \atop {x+y=-3}} \right. \\\left \{ {{xy=-18} \atop {x+y=-3}} \right. \\x=-3-y \\(-3-y)*y=-18 \\y^2+3y=18 \\y^2+3y-18=0 \\D=9+72=81=9^2 \\y_1= \frac{-3+9}{2} =3 \\y_2= \frac{-12}{2} =-6 \\x_1=-3-3=-6 \\x_2=-3-(-6)=-3+6=3$
Ответ: (-6;3), (3;-6)