System of two equations with two unknowns - can't get rid of $xy$ The system is:
$x^2 + 2y^2 + 3xy = 12$
$y^2 - 3y = 4$
I try to turn $x^2 + 2y^2 + 3xy$ into $(x + y)^2 + y^2 + xy$ , but it's a dead end from here. Can anyone please help?

If you look at the second part $$12 y^2 - 3y = 4$$ it is just a quadratic equations the roots of which being $$y_{\pm}=\frac{1}{24} \left(3\pm\sqrt{201}\right)$$ Now, consider $x^2 + 2y^2 + 3xy = 4$ where $y$ is a known parameter and you get another quadratic in $x$ the roots of which being $$x=\frac{1}{2} \left(\pm\sqrt{y^2+16}-3 y\right)$$ Replace $y$ by its values and get the corresponding $x$'s.
This will give you four solutions.
