Systems of equations over reals 
Solve over reals
  $$2x^2+xy+x-3y+3=0$$
$$3xy+y^2-2x+6y-6=0$$

Adding or subtracting these equations doesn't help, and completing the square also doesn't work. I'm considering substituting $x+y$ and $xy$, but it also doesn't show much promise. Thanks!

Another way to do it.
From the first equation, extract $y$
$$y=-\frac{2 x^2+x+3}{x-3}$$
Plug $y$ in the second equation to get
$$\frac{-2 x^4+5 x^3+49 x^2+51 x+9}{(x-3)^2}=0$$ By inspection $x=-1$ and $x=-3$ are roots of the numerator. So, we are left with
$$\frac{(x+1) (x+3) \left(2 x^2-13 x-3\right)}{(x-3)^2}=0$$ So, the four roots for $x$ and the corresponding $y$'s.
