6.206. $\left\{\begin{array}{l}x+y+x y=7 \\ x^{2}+y^{2}+x y=13 .\end{array}\right.$
Solution.

From the condition, we get $\left\{\begin{array}{l}(x+y)+x y=7, \\ (x+y)^{2}-x y=13 .\end{array}\right.$

Let $\left\{\begin{array}{l}x+y=u, \\ x y=v,\end{array}\right.$ then we have $\left\{\begin{array}{l}u+v=7, \\ u^{2}-v=13 .\end{array}\right.$

By adding the equations of the system, we get $u^{2}-7+u=13, u^{2}+u-20=0$, from which $u_{1}=-5, u_{2}=4$; then $v_{1}=12, v_{2}=3$.

This system is equivalent to two systems of equations

$$
\left\{\begin{array} { l } 
{ x + y = - 5 , } \\
{ x y = 1 2 }
\end{array} \text { or } \left\{\begin{array} { l } 
{ x + y = 4 , } \\
{ x y = 3 . }
\end{array} \Rightarrow \left\{\begin{array} { l } 
{ x _ { 1 } = 3 , } \\
{ y _ { 1 } = 1 ; }
\end{array} \left\{\begin{array}{l}
x_{2}=1 \\
y_{2}=3
\end{array}\right.\right.\right.\right.
$$

Answer: $(3 ; 1),(1 ; 3)$.