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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 $xyzuv$?

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4 Answers 4

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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.

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  • $\begingroup$ Very nicely done. $\endgroup$
    – Simon S
    Commented Nov 2, 2014 at 19:08
  • $\begingroup$ A very clean, concise solution. $\endgroup$
    – SimonT
    Commented Nov 2, 2014 at 20:20
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$$\begin{align*} y+u+x+v&=0\tag1\\ z+y+v+u&=1\tag2\\ x+y+z+u&=5\tag3\\ z+u+v+x&=2\tag4\\ v+x+y+z&=4\tag5 \end{align*}$$

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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    $\begingroup$ 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 as y+u+x+v=0\tag1. $\endgroup$
    – Ruslan
    Commented Nov 2, 2014 at 20:02
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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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  • $\begingroup$ That is the solution. $\endgroup$ Commented Nov 2, 2014 at 19:04
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    $\begingroup$ Strictly, it asked What is the value of xyzuv? namely 12 :) $\endgroup$
    – smci
    Commented Nov 2, 2014 at 23:47
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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 $\mathbf{s}$.

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