6.193. $\left\{\begin{array}{l}x+y z=2, \\ y+z x=2, \\ z+x y=2 .\end{array}\right.$
## Solution.

From the third equation of the system, we find $z=2-x y$ and substitute it into the first and second equations:

$$
\begin{aligned}
& \left\{\begin{array} { l } 
{ x + y ( 2 - x y ) = 2 , } \\
{ y + ( 2 - x y ) x = 2 , } \\
{ z = 2 - x y }
\end{array} \Leftrightarrow \left\{\begin{array} { l } 
{ x + 2 y - x y ^ { 2 } = 2 , } \\
{ y + 2 x - x ^ { 2 } y = 2 } \\
{ z = 2 - x y }
\end{array} \Leftrightarrow \left\{\begin{array}{l}
x+2 y-x y^{2}=2, \\
y\left(1-x^{2}\right)-(2-2 x)=0, \\
z=2-x y,
\end{array}\right.\right.\right. \\
& \Leftrightarrow\left\{\begin{array}{l}
x+2 y-x y^{2}=2, \\
(1-x)(y(1+x)-2)=0, \\
z=2-x y .
\end{array}\right.
\end{aligned}
$$

The second equation is equivalent to two equations: $1-x=0$ or $y(1+x)-2=0$. Therefore, the system is equivalent to two systems of equations

1) $\left\{\begin{array}{l}x+2 y-x y^{2}=2, \\ 1-x=0, \\ z=2-x y\end{array}\right.$
or 2) $\left\{\begin{array}{l}x+2 y-x y^{2}=2, \\ y(1+x)-2=0, \\ z=2-x y .\end{array}\right.$
2) From the second equation of the system

$$
\begin{aligned}
& x=1 \Rightarrow\left\{\begin{array} { l } 
{ 1 + 2 y - y ^ { 2 } = 2 , } \\
{ x = 1 , } \\
{ z = 2 - y }
\end{array} \Leftrightarrow \left\{\begin{array} { l } 
{ y ^ { 2 } - 2 y + 1 = 0 , } \\
{ x = 1 , } \\
{ z = 2 - y }
\end{array} \Leftrightarrow \left\{\begin{array}{l}
(y-1)^{2}=0, \\
x=1, \\
z=2-y
\end{array},\right.\right.\right. \\
& \Leftrightarrow\left\{\begin{array} { l } 
{ y = 1 , } \\
{ x = 1 , } \\
{ z = 2 - y . }
\end{array} \left\{\begin{array}{l}
y_{1}=1, \\
x_{1}=1, \\
z_{1}=1 .
\end{array}\right.\right.
\end{aligned}
$$

2) Write the system in the form

$$
\begin{aligned}
& \left\{\begin{array}{l}
\left(x-x y^{2}\right)-(2-2 y)=0, \\
y=\frac{2}{1+x}, \\
z=2-x y
\end{array}\right. \\
& \Leftrightarrow\left\{\begin{array}{l}
x(1-y)(1+y)-2(1-y)=0, \\
y=\frac{2}{1+x} \\
z=2-x y
\end{array}\right. \\
& \Leftrightarrow\left\{\begin{array}{l}
(1-y)(x(1+y)-2)=0, \\
y=\frac{2}{1+x} \\
z=2-x y
\end{array}\right.
\end{aligned}
$$

The first equation of the last system is equivalent to two equations:

$1-y=0$ or $x(1+y)=2$.

Thus, the system is equivalent to two systems of equations:

$$
\left\{\begin{array} { l } 
{ 1 - y = 0 , } \\
{ y = \frac { 2 } { 1 + x } , } \\
{ z = 2 - x y }
\end{array} \left\{\begin{array} { l } 
{ x ( 1 + y ) = 2 , } \\
{ y = \frac { 2 } { 1 + x } , } \\
{ z = 2 - x y . }
\end{array} \Rightarrow \left\{\begin{array} { l } 
{ x _ { 2 } = 1 , } \\
{ y _ { 2 } = 1 , } \\
{ z _ { 2 } = 1 ; }
\end{array} \left\{\begin{array} { l } 
{ x _ { 3 } = - 2 , } \\
{ y _ { 3 } = - 2 , } \\
{ z _ { 3 } = - 2 ; }
\end{array} \left\{\begin{array}{l}
x_{4}=1, \\
y_{4}=1 \\
z_{4}=1
\end{array}\right.\right.\right.\right.\right.
$$

Answer: $(1 ; 1 ; 1),(-2 ;-2 ;-2)$.