6.067. $\left\{\begin{array}{l}(x+0.2)^{2}+(y+0.3)^{2}=1, \\ x+y=0.9 .\end{array}\right.$
## Solution.

Rewrite the system as

$$
\left\{\begin{array} { l } 
{ ( x + 0.2 ) ^ { 2 } + ( y + 0.3 ) ^ { 2 } = 1 , } \\
{ x + 0.2 - 0.2 + y + 0.3 - 0.3 = 0.9 }
\end{array} \Leftrightarrow \left\{\begin{array}{l}
(x+0.2)^{2}+(y+0.3)^{2}=1 \\
(x+0.2)+(y+0.3)=1.4
\end{array}\right.\right.
$$

Let $\left\{\begin{array}{l}x+0.2=u, \\ y+0.3=v .\end{array}\right.$

Then $\left\{\begin{array}{l}u^{2}+v^{2}=1, \\ u+v=1.4\end{array} \Leftrightarrow\left\{\begin{array}{l}(u+v)^{2}-2 u v=1, \\ u+v=1.4\end{array} \Leftrightarrow\right.\right.$

$\Leftrightarrow\left\{\begin{array}{l}(1.4)^{2}-2 u v=1, \\ u+v=1.4\end{array} \Leftrightarrow\left\{\begin{array}{l}1.96-2 u v=1, \\ u+v=1.4\end{array} \Leftrightarrow\left\{\begin{array}{l}u v=0.48, \\ u+v=1.4 .\end{array}\right.\right.\right.$

By Vieta's theorem, the only possible solutions are:

$\left\{\begin{array}{l}u_{1}=0.6, \\ v_{1}=0.8\end{array}\right.$ or $\left\{\begin{array}{l}u_{2}=0.8, \\ v_{2}=0.6 .\end{array}\right.$

Then $\left\{\begin{array}{l}x+0.2=0.6, \\ y+0.3=0.8\end{array}\right.$ or $\left\{\begin{array}{l}x+0.2=0.8, \\ y+0.3=0.6 ;\end{array}\left\{\begin{array}{l}x_{1}=0.4, \\ y_{1}=0.5\end{array}\right.\right.$ or $\left\{\begin{array}{l}x_{2}=0.6, \\ y_{2}=0.3 .\end{array}\right.$

Answer: $(0.4 ; 0.5),(0.6 ; 0.3)$