6.207. $\left\{\begin{array}{l}x+y+z=6, \\ x(y+z)=5, \\ y(x+z)=8 .\end{array}\right.$
## Solution.

Let's rewrite the system of equations as

$\left\{\begin{array}{l}x+y+z=6, \\ x y+x z=5, \\ y x+y z=8 .\end{array} \Rightarrow z=6-x-y\right.$.

$\left\{\begin{array}{l}x y+x(6-x-y)=5, \\ y x+y(6-x-y)=8\end{array} \Leftrightarrow\left\{\begin{array}{l}x^{2}-6 x+5=0, \\ y^{2}-6 y+8=0,\end{array} \quad\right.\right.$ from which $\quad x_{1}=5, x_{2}=1 ;$

$y_{1}=2, y_{2}=4$.

Thus,

$$
\left\{\begin{array} { l } 
{ x _ { 1 } = 5 , } \\
{ y _ { 1 } = 2 , } \\
{ z _ { 1 } = - 1 ; }
\end{array} \left\{\begin{array} { l } 
{ x _ { 2 } = 5 , } \\
{ y _ { 2 } = 4 , } \\
{ z _ { 2 } = - 3 ; }
\end{array} \left\{\begin{array} { l } 
{ x _ { 3 } = 1 , } \\
{ y _ { 3 } = 2 , } \\
{ z _ { 3 } = 6 - 1 - 2 = 3 ; }
\end{array} \left\{\begin{array}{l}
x_{4}=1 \\
y_{4}=4 \\
z_{4}=6-1-4=1
\end{array}\right.\right.\right.\right.
$$

Answer: $(5 ; 2 ;-1),(5 ; 4 ;-3),(1 ; 2 ; 3),(1 ; 4 ; 1)$.