Question

Решите систему уравнений (2х^2 + y^2=9 ( x-y=3

Answer 1

$\left \{{{2x^{2}+y^{2}=9}\atop{x-y=3}} \right$

$\left \{{{y=x-3}\atop{2x^{2}+(x-3)^{2}=9}} \right$

$2x^{2}+x^{2}-6x+9=9$

$2x^{2}+x^{2}-6x+9-9=0$

$(2x^{2}+x^{2})-6x+(9-9)=0$

$3x^{2}-6x=0$

$x(3x-6)=0$

$x_{1}=0$

$3x-6=0$

$3x=6$

$x=6:3$

$x_{2}=2$

$y_{1}=x_{1}-3=0-3=-3$

$y_{2}=x_{2}-3=2-3=-1$

ответ: (0;-3), (2;-1)

Answer 2

х=3+у

2(3+у)^2+y^2=9

2(9+6y+y^2)+y^2-9=0

3y^2+12y+9=0

y^2+4y+3=0

y1=-3

y2=-1

х1=0

х2=2

вроде так.