Решите систему...

$\left \{ {{x^2+y^2+x+y=32} \atop {x^2+y^2-3x-3y=4}} \right.$ ⇔ $\left \{ {{x^2+y^2=32-(x+y)} \atop {x^2+y^2-3(x+y)=4}} \right.$

$\left \{ {{x^2+y^2=32-(x+y)} \atop {32-(x+y)-3(x+y)=4}} \right.$
32-(x+y)-3(x+y)=4
32-4(x+y)=4
4(x+y)=28
x+y=7
$\left \{ {{x^2+y^2=25} \atop {x+y=7}} \right.$ ⇔ $\left \{ {{x^2+(7-x)^2=25} \atop {y=7-x}} \right.$
x²+(7-x)²=25
x²+49-14x+x²=25
2x²-14x+24=0
D=196-4*2*24=4=2²
x₁=3 u x₂=4
$\left \{ {{y_1=4} \atop {x_1=3}} \right. \left \{ {{y_2=3} \atop {x_2=4}} \right.$

Ответ A(3;4),(4;3)