The example of three linear equations in two variables is typical. The equations $ax+by+c=0$, $dx+ey+f=0$ and $gx+hy+i=0$ have a common root if
- $\det \begin{pmatrix} a & b & c \\ d & e & f \\ g & h & i \end{pmatrix} =0$ AND one of the following two conditions is met
- EITHER at least one of $\det \begin{pmatrix} a & b \\ d & e \end{pmatrix}$, $\det \begin{pmatrix} a & b \\ g & h \end{pmatrix}$, $\det \begin{pmatrix} d & e \\ g & h\end{pmatrix}$ is nonzero OR
- all nine $2 \times 2$ minors of $\begin{pmatrix} a & b & c \\ d & e & f \\ g & h & i \end{pmatrix}$ vanish.
Note that we cannot give a single polynomial $R(a,b,c,d,e,f,g,h,i)$ which vanishes if and only if the equations have a common root, because the property of these equations having a common root is not closed. For example, for all $\epsilon \neq 0$, the equations $y=0$, $y+\epsilon x=1$ and $y+2\epsilon x = 2$ have a common root at $(0, 1/\epsilon)$, but the equations $y=0$, $y=1$, $y=2$, obtained in the limit as $\epsilon \to 0$, have no common root. Note that this also shows that no list of equations could do the job.
In general, a constructible set is a subset of an algebraically closed field formed from finitely many polynomial equalities and the boolean connectors AND, OR and NOT. For example, the set
$$\{ (a,b,c,d,e,f,g,h,i,x,y) : ax+by+c=dx+ey+f=gx+hy+i=0 \}.$$
A theorem of Chevalley says that the projection of a constructible set onto a subset of its coordinates is again constructible. So the projection of the above set onto the $(a,b,c,d,e,f,g,h,i)$ coordinates must be an constructible set, and I've given a boolean description above.