Solutions to a system of equations I know that Bezout's theorem says that if you take two plane curves, then their maximal number of intersection points is the product of their degrees. However, assume that I have two irreducible polynomials in say 6 variables and I plug in values for 4 of them. I'm left with two plane curves. What can be said about their number of intersections?
I assume the original two polynomials are irreducible. I know that there are cases like
$$u+f(x,y),\;\; v+f(x,y),$$
where I can plug in zero for $u,v$ and then I end up with just one curve. Therefore, is it possible to say something like ''for most values of the other 3 variables'' the number of intersection points is the product of the degrees?
 A: Yes, what you want to say is true.
Let's stick to projective space and homogeneous equations to keep things tidy. (There's a process called "homogenisation" that allows us to pass from inhomogeneous polynomials in $n$ variables to homogeneous polynomials in $n+1$ variables of the same degree which then define hypersurfaces in projective space $\mathbf{P}^n.$ So we can answer the projective version first, and then go back to the original version.)
Suppose your two original polynomials homogenise to give $F$ and $G$. Then they define two hypersurfaces $\{F=0\}$ and $\{G=0\}$ in projective space $\mathbf P^6$. Since $F$ and $G$ are irreducible and distinct, these two hypersurfaces intersect in a subset $X$ of dimension 4 and degree $(\operatorname{deg} F \cdot \operatorname{deg} G)$.
Now "plugging in values" for 4 variables in your setting corresponds, in the projective setting, to intersecting with a linear subspace $L$ of codimension 4, in other words a plane.
Then the statement you want is that for "almost all" choices of the plane $L$, the intersection $X \cap L$ will consist of  $(\operatorname{deg} F \cdot \operatorname{deg} G)$ points in $\mathbf P^6$. Of course, there are some choices of $L$ for which this will fail --- it could be that $\{F=0\} \cap \{G=0\}$ contains a line, for example, in which case any plane $L$ containing that line will intersect in an infinite set of points. But those "bad" $L$ are a smaller-dimensional set of the set of all possible choices of $L$.
Of course you asked about inhomogeneous polynomials, so let's go back to the affine picture. For a slightly smaller set of choices of $L$ than in the previous paragraph, none of the points of $X \cap L$ will be "points at infinity" in $\mathbf P^6$, so your original equations have exactly  $(\operatorname{deg} F \cdot \operatorname{deg} G)$ common solutions.
