Function ^2 system: x^2 - y^2 + xy = 2, x^2 - 2xy = 2

[Adderall]

In the red portion is the problem's enunciation and down my try. I have 5x^2+2=0 and it's imposible. Where is my mistake ?

 

[HallsofIvy]

That looks to me like the hard way to do it! From $\displaystyle -y^2+ 3xy= 0$, instead of using the quadratic formula, factor:
y(3x- y)= 0 which gives y= 0 and y= 3x as you get.

If y= 0, then the first equation becomes $\displaystyle x^2= 2$ so that $\displaystyle x= \pm\sqrt{2}$, So one solution is $\displaystyle x= \sqrt{2}$, y= 0 and another is $\displaystyle x= -\sqrt{2}$, y= 0.

If y= 3, then the first equation becomes $\displaystyle x^2- 9^2+ 3x^2= -5x^2= 2$. As you say, that has no (real) solution. So there is no solution (in the real numbers) for y= 3. The only solutions to the system are the two given above.