Problem source

Consider the system of equations
\begin{alignat*}{1}
2yz+zx-5xy & =2\\
yz-zx+2xy & =1\\
yz-2zx+6xy & =3
\end{alignat*}             
Show that 
\[xyz=\pm 6\] and find
the possible values of $x$, $y$ and $z$.

Solution source

Consider the linear $3\times 3$ system in $yz, zx, xy$, then

\begin{align*}
\left(\begin{array}{ccc|c}
  2 & 1 & -5 & 2 \\
  1 & -1 & 2 & 1 \\
  1 & -2 & 6 & 3 \\
\end{array}\right) \\ 
\left(\begin{array}{ccc|c}
  1 & -1 & 2 & 1 \\
  1 & -2 & 6 & 3 \\
  2 & 1 & -5 & 2 \\
\end{array}\right) \\ 
\left(\begin{array}{ccc|c}
  1 & -1 & 2 & 1 \\
  0 & -1 & 4 & 2 \\
  0 & 3 & -9 & 0 \\
\end{array}\right) \\ \left(\begin{array}{ccc|c}
  1 & -1 & 2 & 1 \\
  0 & -1 & 4 & 2 \\
  0 & 0 & 3 & 6 \\
\end{array}\right) \\ 
\end{align*}

Therefore $yz = 2, zx = 6, xy = 3 \Rightarrow  (xyz)^2 = 36 \Rightarrow xyz = \pm 6$. If $xyz = 6, x = 3, y = 1, z = 2$, if $xyz = -6, x = -3, y = -1, z = -2$