\section*{Problem 1 - 161211}

Determine all triples of real numbers \((x, y, z)\) that satisfy the system of equations:

\[
\begin{array}{r}
x + yz = 7 \\
xy + z = 5 \\
x + y + z = 6
\end{array}
\]
}

Let \((x, y, z)\) be a solution of the system of equations (1), (2), (3). From (3) it follows that

\[
x=6-y-z
\]

Substituting this into (1), (2) yields

\[
6-y-z+yz=7 \quad ; 6y-y^2-yz+z=5
\]

Adding (5), (6) results in

\[
6+5y-y^2=12 \quad ; \quad y^2-5y+6=0
\]

Therefore, \(y\) can only be \(2\) or \(3\).

From \(y=2\) and (5);(4) it follows that \(4-z+2z=7 ; z=3, x=1\)

From \(y=3\) and (5);(4) it follows that \(3-z+3z=7 ; z=2, x=1\)

Thus, the given system can only have the solutions \((1,2,3)\) and \((1,3,2)\). Verification confirms that both triples are indeed solutions.

Taken from \([5]\)