Question

развязать систему уравнений
x +y =2
2x^2 +xy+y^2=16

Answer 1

$displaystyle left { {{x+y=2} atop {2x^2+xy+y^2=16}} right. Rightarrow left { {{x+y=2} atop {x^2+(x+y)^2-xy=16}} right. Rightarrow\ \ \ Rightarrow left { {{x+y=2} atop {x^2-xy+2^2=16}} right. Rightarrow left { {{y=2-x} atop {x^2-x(2-x)=12}} right. \ x^2-2x+x^2=12\ 2x^2-2x-12=0|:2\ x^2-x-6=0$

По т. Виета: $x_1=-2;,,,,,,,,,, x_2=3$

$y_1=2-x_1=2+2=4\ y_2=2-x_2=2-3=-1$