6.205. $\left\{\begin{array}{l}\frac{3}{x^{2}+y^{2}-1}+\frac{2 y}{x}=1 \\ x^{2}+y^{2}+\frac{4 x}{y}=22\end{array}\right.$
## Solution.

Domain of definition: $\left\{\begin{array}{l}x \neq 0, \\ y \neq 0\end{array}\right.$

Let $\left\{\begin{array}{l}x^{2}+y^{2}=u, \\ \frac{y}{x}=v .\end{array}\right.$ Then, with respect to $u$ and $v$, the system of equations has the form

$$
\left\{\begin{array} { l } 
{ \frac { 3 } { u - 1 } + 2 v = 1 } \\
{ u + \frac { 4 } { v } = 2 2 }
\end{array} \Leftrightarrow \left\{\begin{array}{l}
\frac{3}{u-1}+2 v=1 \\
u=22-\frac{4}{v}
\end{array}\right.\right.
$$

Substituting the value $u=22-\frac{4}{v}$ from the second equation into the first, we have

$$
\frac{3}{22-\frac{4}{v}-1}+2 v=1 \Leftrightarrow 21 v^{2}-13 v+2=0
$$

from which $v_{1}=\frac{2}{7}, v_{2}=\frac{1}{3} ; u_{1}=8, u_{2}=10$.

$$
\begin{aligned}
& \left\{\begin{array} { l } 
{ x ^ { 2 } + y ^ { 2 } = 8 , } \\
{ \frac { y } { x } = \frac { 2 } { 7 } }
\end{array} \text { or } \left\{\begin{array} { l } 
{ x ^ { 2 } + y ^ { 2 } = 1 0 , } \\
{ \frac { y } { x } = \frac { 1 } { 3 } }
\end{array} \Rightarrow \left\{\begin{array} { l } 
{ x _ { 1 } = - 1 4 \sqrt { \frac { 2 } { 5 3 } } , } \\
{ y _ { 1 } = - 4 \sqrt { \frac { 2 } { 5 3 } } ; }
\end{array} \left\{\begin{array}{l}
x_{2}=14 \sqrt{\frac{2}{53}} \\
y_{2}=4 \sqrt{\frac{2}{53}}
\end{array}\right.\right.\right.\right. \\
& \left\{\begin{array} { l } 
{ x _ { 3 } = 3 , } \\
{ y _ { 3 } = 1 ; }
\end{array} \left\{\begin{array}{l}
x_{4}=-3 \\
y_{4}=-1
\end{array}\right.\right. \\
& \text { Answer: }\left(\frac{-14 \sqrt{106}}{53} ; \frac{-4 \sqrt{106}}{53}\right),\left(\frac{14 \sqrt{106}}{53} ; \frac{4 \sqrt{106}}{53}\right),(3 ; 1),(-3 ;-1) \text {. }
\end{aligned}
$$