Question

решите систему уравнения:
{x^2+y^2+x+y=32
{x^2+y^2-3x-3y=4

A(3;4),(4;3)
B(1,5;2,5),(3,5;-3,5)
C(-1;-2),(-2;-1)
D(-3;-4),(-4;-3)

Answer 1

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

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