System of two equations with two unknowns - can't get rid of $xy$

Question

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?

Answer 1

HINT:

Solve $12y^2-3y-4=0$ for $y$

Set those two values one by one in the first equation

Answer 2

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.

Answer 3

The system consists the two equations $x^2 + 2y^2 + 3xy = 12$ and (!) $y^2 - 3y = 4.$

From the second equation we get $y=-1, y=4.$ Then from the first equation we find $x.$