A Five Equations problem? If,
$$\begin{align*}
    y+u+x+v&=0\\
    z+y+v+u&=1\\
    x+y+z+u&=5\\
    z+u+v+x&=2\\
    v+x+y+z&=4\,,
\end{align*}$$
What is the value of $xyzuv$?
 A: Did you try and solve the equations, using Gaussian elimination, or some other method? I found $(x,y,z,u,v) = (2,1,3,-1,-2)$
A: Let $S=x+y+z+u+v$. Then, for example, your first equation can be written
$$
S-z=0
$$
You can do the same thing with the other four equations, since each of them is missing a different variable. Add the resulting equations to get $5S-S=0+1+5+2+4=12$, so $S=3$. From this and the transformed equations, we get the values of each of the variables, and hence their product.
A: Hint : Solve for $$\mathbf{A} \mathbf{s} = \mathbf{b},$$
where $\mathbf{s} = [x \ y \  z \  u \ v]^T$
Result should be the product of all the elements of $\mathbf{s}$.
A: $$\begin{align*}
    y+u+x+v&=0\tag1\\
    z+y+v+u&=1\tag2\\
    x+y+z+u&=5\tag3\\
    z+u+v+x&=2\tag4\\
    v+x+y+z&=4\tag5
\end{align*}$$                                                            
Adding these five equations, we get:
$$\begin{align*}
    4(x+y+z+u+v)&=12\\
    x+y+z+u+v&=3\tag{6}
\end{align*}$$
$$
\begin{align}
(6) - (1) &\implies z=3\\
(6) - (2) &\implies x=2\\
(6) - (3) &\implies v=-2\\
(6) - (4) &\implies y=1\\
(6) - (5) &\implies u=-1
\end{align}
$$
So, from these results,
$$
x\cdot y\cdot z \cdot u\cdot v = 2\times 1 \times 3 \times (-1) \times (-2)$$
$$
\therefore  \boxed{\ x\,y\,z\,u\,v = 12\ }
$$
