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 $xyzuv$?
$$\begin{align*} y+u+x+v&=0\tag1\\ z+y+v+u&=1\tag2\\ x+y+z+u&=5\tag3\\ z+u+v+x&=2\tag4\\ v+x+y+z&=4\tag5 \end{align*}$$
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, $$ x\cdot y\cdot z \cdot u\cdot v = 2\times 1 \times 3 \times (-1) \times (-2)$$ $$ \therefore \boxed{\ x\,y\,z\,u\,v = 12\ } $$
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7$\begingroup$ Instead of using these ugly points, you can use
\tag1
,\tag2
etc.. Then the tags will appear right-justified, while the equations will be centered. E.g. your first equation can be entered asy+u+x+v=0\tag1
. $\endgroup$ – Ruslan Nov 2 '14 at 20:02
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.
Did you try and solve the equations, using Gaussian elimination, or some other method? I found $(x,y,z,u,v) = (2,1,3,-1,-2)$
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1$\begingroup$ Strictly, it asked What is the value of xyzuv? namely 12 :) $\endgroup$ – smci Nov 2 '14 at 23:47
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 $\mathbf{s}$.