Question

Решите систему уравнений 1. x^2 +xy +y^2=13 2. x +y =4

Answer 1

$\left \{ {{x^{2} +xy+y^{2}=13 } \atop {x+y=4}} \right.$

$(x+y)^{2} =x^{2} +2xy+y^{2}$

$x^{2} +2xy+y^{2}=16$

$x^{2} +y^{2} =16-2xy$

$\left \{ {{x^{2} +y^{2}=13-xy } \atop {x^{2} +y^{2}=16-2xy }} \right.$

13-xy=16-2xy

xy=16-13

xy=3

x=4-y

(4-y)y=3

$4y-y^{2} =3$

$y^{2}$-4y+3=0

D=16-12=4

$y_{1}$=$\frac{4+2}{2} =\frac{6}{2}=3$     $y_{2} =\frac{4-2}{2}=1$

$x_{1} =4-3=1$       $x_{2}= 4-1=3$

Ответ:($x_{1}$;$y_{1}$)=(1;3)   ($x_{2}$;$y_{2}$)=(3;1)