How to solve a system of equations of two quadratics So, i have such general system
$$\begin{cases} ax^2 + bx + c = dy^2 + ey + f \\ hx^2 + ix + j = ky^2 + ly + m \end{cases}$$
Where $x$ and $y$ are variables and 12 alphabet letters are parameters.
I need to get a formula for at least one of the variables, depending on those 12 parameters. I've tried hard to solve this myself but i either fall into recursive problem or just can't do it.
Two systems, two variables, there should be from 0 to 2 answers for one variable, but my knowledge isn't enough for this.
 A: Rewriting the system as
$$\begin{cases} a_1x^2 + a_2x + a_3 = a_4y^2 + a_5y + a_6 \\ b_1x^2 + b_2x + b_3 = b_4y^2 + b_5y + b_6 \end{cases}$$
one obtains in the generic case by using discriminants,
$$
x= \frac{- a_1b_3 + a_1b_4y^2 + a_1b_5y + a_1b_6 + a_3b_1 - a_4b_1y^2 - a_5b_1y - a_6b_1}{a_1b_2 - a_2b_1},
$$
where $y$ is a root of some polynomial $f$ with coefficients $a_i,b_i$.
A: Your answer for $y$ is a fourth order equation in $y$. Equivalently, your answer for $x$ is a fourth order equation in $x$.
Here is a way how to proceed for $y$, avoiding roots:


*

*Replace only $x^2 = z$ in both equations (leave the linear $x$-terms unchanged).

*Solve the second equation for $z$ and insert into the first one. 

*That modified first equation is now linear in $x$ - solve for $x$.

*Insert that $x$ into the second equation. You get:


$$
(b - \frac{a i}{h})^2  (j - m - ly - ky^2) + (b - \frac{a i}{h}) (i(f - c + ey + dy^2 - \frac{a (k y^2 + l y - j + m)}{h}))+ h(f - c + ey + d y^2 - \frac{a (k y^2 + l y - j + m)}{h})^2 = 0
$$
or, in slightly deconvoluted typing:
$$
(b - \frac{a i}{h})^2 (j - m - ly - ky^2) + (b - \frac{a i }{h}) (i r)+ h r^2 = 0 \quad \rm{\bf (1)}
$$
with
$$
r = f - c + ey + d y^2 - \frac{a (k y^2 + l y - j + m)}{h}
$$
which is a fourth order equation in $y$. This should also be clear from the start since solving each equation for $x$ gives two roots each, and equating each root of the first equation with each root of the second one gives four equations, however with roots.
How many solutions are obtained depends on the 12 variables (a..m). 
In particular, if $bh = a i$, we can solve directly to obtain $r=0$ which is quadratic in $y$ and hence directly produces two solutions.
Similarly, if $dh = ak$, we have that $r$ will be linear in $y$ and, upon inserting $r$ in (1),again there is only a quadratic equation in $y$ to solve.
