Solution af a system of 2 quadratic equations I have a system of two quadratic equations with unknowns $x$ and $y$:
$$a_{1 1} x y + a_{1 2} x^2 + a_{1 3} y^2 + a_{1 4} x + a_{1 5} y + a_{1 6} = 0,\\
a_{2 1} x y + a_{2 2} x^2 + a_{2 3} y^2 + a_{2 4} x + a_{2 5} y + a_{2 6} = 0,$$
where $a_{i j}$ are arbitrary scalars.
Is there an algebraic solution of the above system?
 A: From Intersecting two conics:

The solutions to a two second degree equations system in two variables may be seen as the coordinates of the intersections of two generic conic sections. In particular two conics may possess none, two or four possibly coincident intersection points. The best method of locating these solutions exploits the homogeneous matrix representation of conic sections, i.e. a 3x3 symmetric matrix which depends on six parameters.
  The procedure to locate the intersection points follows these steps:
  
  
*
  
*given the two conics $C_1$ and $ C_2$ consider the pencil of conics given by their linear combination $\lambda C_1 + \mu C_2$
  
*identify the homogeneous parameters $(\lambda,\mu)$ which corresponds to the degenerate conic of the pencil. This can be done by imposing that $\det(\lambda C_1 + \mu C_2) = 0$, which turns out to be the solution to a third degree equation.
  
*given the degenerate conic $C_0$, identify the two, possibly coincident, lines constituting it
  
*intersects each identified line with one of the two original conic; this step can be done efficiently using the dual conic representation of $C_0$
  
*the points of intersection will represent the solution to the initial equation system
  

A: Each equation describes a conic. Thus, you are trying to compute the intersection $\cal C_1\cap\cal C_2$ of two conics. In general, you have to expect four intersection points (e.g., think of two ellipses meeting transversally).
There are special cases where the task is simplified. For instance, if one of the two conics splits as the union of two lines, or if one of the equations reduces easily to the form $y=f(x)$.
In the general case you can either try Elimination Theory or you can exploit the fact that conics are rational curves, i.e. that there is a parametrization
$$
\Bbb R\ni t\mapsto (x(t),y(t))\in\cal C_1\qquad(*)
$$
given by rational functions, i.e. quotient of polynomials in $t$. Then if you plug this "rational description" of $\cal C_1$ into the equation of $\cal C_2$ will eventually get a polynomial equation of degree 4 in the variable $t$ only, whose solutions correspond to the intersection points. In order to solve this degree 4 equation you need either some luck, or some patience to go browsing old Algebra books.
In order to get (*), the usual method is to find just one point $(x_o,y_o)\in\cal C_1$, consider the lines
$$
\ell_t: y=t(x-x_o)+y_o
$$
through it and describe the second intersection of $\cal C_1\cap\ell_t$ in terms of $t$.
