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!

$$2x^2+xy+x-3y+3=0\tag{1}$$
$$3xy+y^2-2x+6y-6=0\tag{2}$$
Take 2$\times$(1)+(2) to get,
$$4x^2+5xy+y^2=0 \implies (4x+y)(x+y)=0$$
which yields $y=-x$ and $y=-4x$. Plug them into (1)
$$x^2+4x+3=0\implies x=-1,\>x=-3$$
$$2x^2-13x-3=0\implies x=\frac{13\pm\sqrt{193}}{4}$$
Then, evaluated the corresponding $y$'s to obtain the follow solutions,
$$(-1,1),\>\>\>(-3,3),\>\>\>
\left(\frac{13+\sqrt{193}}{4},-13-\sqrt{193}\right),
\left(\frac{13-\sqrt{193}}{4},-13+\sqrt{193}\right)$$
