Solving set of two quadratic equations Hi I am trying to simultaneously solve these two equations (it is a chemistry problem but I tried to write in in pure mathematical terms):
$$a_1x + a_2xy + 2a_3x^2y + a_4xy^2+a_5 = 0$$
$$a_1x + a_2xy + 2a_3x^2y + 2a_4xy^2+a_6 = 0$$
I am to be honest stuck in terms of how to solve for $x$ (or $y$) without ending up with the opposite variable $y$ (or $x$) in that in solution. I am sure there some trick I am missing here to simplify this? Or can one only solve this sort of problem using the method of successive approximation? Apologies if there is an answer somewhere already but I couldn't find any.

In general you use one of the equations and solve for $x$ or $y$ and accept that the solution is in terms of the other variable.  That is supposed to happen.  Then take those two solutions and put them into the othe equation and solve.
Example:
$a_1x + a_2xy + 2a_3x^2y + a_4xy^2+a_5 = 0$
$(2a_3y)x^2 + (a_1 + a_2y + a_4y^2)x + a_5 = 0$
$x = \frac {-(a_1 + a_2y + a_4y^2) \pm \sqrt {(a_1 + a_2y + a_4y^2)^2 - 4(2a_3y)a_5}}{2(2a_3y)}$
So plug that into:
$a_1\frac {-(a_1 + a_2y + a_4y^2) \pm \sqrt {(a_1 + a_2y + a_4y^2)^2 - 4(2a_3y)a_5}}{2(2a_3y)} + a_2\frac {-(a_1 + a_2y + a_4y^2) \pm \sqrt {(a_1 + a_2y + a_4y^2)^2 - 4(2a_3y)a_5}}{2(2a_3y)}y + 2a_3(\frac {-(a_1 + a_2y + a_4y^2) \pm \sqrt {(a_1 + a_2y + a_4y^2)^2 - 4(2a_3y)a_5}}{2(2a_3y)})^2y + 2a_4(\frac {-(a_1 + a_2y + a_4y^2) \pm \sqrt {(a_1 + a_2y + a_4y^2)^2 - 4(2a_3y)a_5}}{2(2a_3y)})y^2+a_6 = 0$
And solve for $y$.
But in this case we can make it easier.
$a_1x + a_2xy + 2a_3x^2y + a_4xy^2+a_5 = 0$
$a_1x + a_2xy + 2a_3x^2y + 2a_4xy^2+a_6 = 0$ so subtract and get
$a_4xy^2 + a_6 - a_5 = 0$
$a_4xy^2 = a_5- a_6$ so
$a_1x + a_2xy + 2a_3x^2y + 2a_5 - a_6 = 0$ 
And $y = \pm \sqrt{\frac {a_5-a_6}{a_4x}}$
So $a_1x + a_2x(\pm \sqrt{\frac {a_5-a_6}{a_4x}}) + 2a_3x^2(\pm \sqrt{\frac {a_5-a_6}{a_4x}}) + 2a_5 - a_6 = 0$
Solve for that to get $x$.
