Four Equations in Four Unknowns 
Four Equations in Four Unknowns

  Completely solve the following equation!
$$\begin{eqnarray}
x&+&y&+&z&+&w&=&10 \\
x^2&+&y^2&+&z^2&+&w^2&=&30 \\
x^3&+&y^3&+&z^3&+&w^3&=&100\\
&&&&&&xyzw&=&24
\end{eqnarray}$$

This is problem 3 on page 3  of

Mathematical Quickies

270 Stimulating Problems with Solutions

by Charles W. Trigg

Dover Publications, Inc., New York

ISBN 0-486-24949-2
Here is the solution from page 78 of the book:

By inspection $(1,2,3,4)$ is a solution if the first and fourth equation and satisfies the second and third equations. Since the equations are symmetrical in $x$, $y$, $z$, $w$ the other $23$ permutations of $1,2,3,4$ are solutions also.
  But these are all the solutions, since the product of the degrees of the equations is 4!

I assume that the exclamation mark at the end of the last sentence is a factorial symbol because the product of the degree of the equation is 24. 
This seems to be a property that is similar to the fact that a univariate polynomial of degree $n$ has at most $n$ zeroes. But I can't see how to generalize this to multivariate equations.
Why does this system of equations where the product of the degree of the equations is 24 has at most 24 solutions? 
 A: I. Yes, you are correct that a univariate polynomial of degree $n$ has $n$ zeros (counting multiplicity). However, your system are just the roots of a univariate in disguise. 
The clue is the elementary symmetric polynomials $x+y+z+w$ and $xyzw$. If these unknowns are indeed the roots of the quartic,
$$F(u)=u^4+au^3+bu^2+cu+d=0$$ then,

$$\begin{aligned}
x+y+z+w &=10 = -a\\
x^2+y^2+z^2+w^2 &=30 = a^2-2b\\
x^3+y^3+z^3+w^3 &=100=-a^3 + 3 a b - 3 c\\
xyzw &=24=d
\end{aligned}\tag1$$

It is easy to solve for $a,b,c,d\,$ above and we get,
$$F(u)=u^4 - 10u^3 + 35u^2 - 50u + 24 = 0$$
$$F(u)=(u - 1)(u - 2)(u - 3)(u - 4)=0$$
hence these, including their permutations, are all the solutions.
II. Let $n_r$ be the number of roots and $n_d$ be the product of the degrees.  By Bezout's theorem, then $n_r$ is at most equal to $n_d$. The observation that your example has $n_r=n_d=24$ is just a peculiarity of the system. If we tweak it slightly,

$$\begin{aligned}
x+y+z+w &=10 = -a\\
x^2+y^2+z^2+w^2 &=30 = a^2-2b\\
\color{blue}{x^5+y^5+z^5+w^5} &=100=-a^5 + 5 a^3 b - 5 a b^2 - 5 a^2 c + 5 b c + 5 a d\\
xyzw &=24=d
\end{aligned}\tag2$$

Solving for $a,b,c,d,\,$ they turn out to be rational so,
$$F(u)=u^4 - 10u^3 + 35u^2 - \color{blue}{\tfrac{602}{13}}u + 24 = 0$$
though $F(u)$ is no longer rationally factorable. Like the previous, by including the permutations of the $u_i$, there are again $n_r=24$ roots, but the product of the degrees is different now as $n_d=1\times2\times5\times4 = 40$ so $n_r\neq n_d$.
