Your example is not an illuminating one, because you used two equations in two variables. You would therefore expect there, generically, to be a solution for most values of $(a,b,\ldots, l)$. I'll get back to your example, but let's start with the more informative case of two equations in one variable. Then I'll use this as an excuse to talk about properness.

Starting very small, when do $ax+b=0$ and $cx+d=0$ have a common root? Answer: We must have $ad-bc =0$. In addition, we must either have $a$ and $c \neq 0$, or else $a=b=0$, or else $c=d=0$. So, there is a polynomial $ad-bc$ which basically cuts out the location where the two polynomials have a common root, but we then have to remove some smaller strata by inequalities.

More generally, when do $f_m x^m + f_{m-1} x^{m-1} + \cdots + f_0=0$ and $g_n x^n + g_{n-1} x^{n-1} + \cdots + g_0=0$ have a common root? The polynomial which plays the role of $ad-bc$ is the determinant
$$R(f,g) := \det \left( \begin{matrix}
f_m & f_{m-1} & \cdots & f_0 & 0 & \cdots & 0 \\
0 & f_m & f_{m-1} & \cdots & f_0 & \cdots & 0 \\
\ddots \\
0 & \cdots & 0 & f_m & f_{m-1} & \cdots & f_0 \\
g_n & g_{n-1} & \cdots & g_0 & 0 & \cdots & 0 \\
0 & g_n & g_{n-1} & \cdots & g_0 & \cdots & 0 \\
\ddots \\
0 & \cdots & 0 & g_m & g_{m-1} & \cdots & g_0
\end{matrix} \right)$$

There are $n$ rows in the first block and $m$ in the second block. The determinant $R(f,g)$ vanishes if and only if there are polynomials $a$ and $b$, of degree $n-1$ and $m-1$, so that $af+bg=0$. It is easy to see that this is true if and only if either (1) $f$ and $g$ have a common root or (2) $f_m = g_n =0$. The little nuisance terms come up in ruling out case (2).

The polynomial $R$ is called the resultant. If you wnat to learn more about this, including generalizations to more polynomials in more variables, the terms to search for are "resultant" and "elimination theory". Standard references are *Using Algebraic Geometry* by Cox, Little and O'Shea and *Discriminants, resultants, and multidimensional determinants* by Gelfand, Kapranov and Zelevinsky.

The awkward thing about your chosen example is that the analogue of the resultant is $0$, and you only have correction terms. I don't really feel like working it out.

The theory looks much prettier if we use homogeneous polynomials. For example, $ax+by$ and $cx+dy$ have a common nonzero root if and only if $ad-bc=0$, with no special cases. Two homogeneous quadratics in three variables *always* have a common root.

Here is the big result about homogeneous polynomials: Let $S$ be the ring $\mathbb{C}[t_1, \ldots, t_k, x_0, x_1, \ldots, x_n]$. We think of the $t$'s as parameters and the $x$'s as the homogeneous coordinates. Let $F_1(x, t)$, ..., $F_N(x, t)$ be any collection of polynomials, homogeneous in the $x$'s. Then there are polynomials $G_1(t)$, ..., $G_M(t)$ such that, given a point $(s_1, \ldots, s_k) \in \mathbb{C}^k$, there is a nonzero common root of the $F_i(s, x)$ if and only if $G_1(s) = \cdots = G_M(s)=0$.

This theorem is better phrased in the language of the Zariski topology. I don't know how much algebraic geometry you've taken. A subset of $\mathbb{C}^k$ is called Zariski closed if it is the zero locus of a set of polynomials. So the statement here is the following:

If we have a Zariski closed subset of
$\mathbb{CP}^n \times \mathbb{C}^k$,
then its projection to $\mathbb{C}^k$
is Zariski closed.

In other words,

The mapping $\mathbb{CP}^n \times \mathbb{C}^k \to \mathbb{C}^k$ is closed.

The corresponding statement is NOT true for $\mathbb{C}^n$. Consider the subset of $\mathbb{C} \times \mathbb{C}$ cut out by $xy=1$. This hyperbola is closed. But the projection onto one of the coordinates is $\{ x \neq 0 \}$, which is not closed.

Now, in fact, a stronger statement is true:

For any variety $B$, the mapping $\mathbb{CP}^N \times B \to B$ is closed.

We express this by saying that $\mathbb{CP}^N$ is universally closed.

If you get far enough in algebraic geometry, you'll run across the theorem that projective space is proper. The definition of proper is universally closed, plus some other conditions which we can ignore for now. So this theorem is the highly abstract way of making the statement about $F$'s and $G$'s above. Hopefully, I have given you something to look forward to as you continue in algebraic geometry!

Historical Ramblings in Algebraic Geometry and Related Algebra, Abhyankar famously quotes Weil:"The device that follows, which, it may be hoped, ﬁnally eliminates from algebraic geometry the last traces of elimination theory [...]". This passage has been quoted many times since by those who wholeheartedly agree that elimination was put to rest prematurely. $\endgroup$ – Thierry Zell Dec 11 '10 at 14:35