6.204. $\left\{\begin{array}{l}x+y+\frac{x^{2}}{y^{2}}=7, \\ \frac{(x+y) x^{2}}{y^{2}}=12 .\end{array}\right.$
## Solution.

## Domain of definition: $y \neq 0$.

From the condition, we have

$\left\{\begin{array}{l}(x+y)+\left(\frac{x^{2}}{y^{2}}\right)=7, \\ (x+y) \cdot\left(\frac{x^{2}}{y^{2}}\right)=12 .\end{array}\right.$

Let $\left\{\begin{array}{l}x+y=u, \\ \frac{x^{2}}{y^{2}}=v .\end{array}\right.$

The system in terms of $u$ and $v$ becomes $\left\{\begin{array}{l}u+v=7, \\ u \cdot v=12,\end{array}\right.$ from which we find

![](https://cdn.mathpix.com/cropped/2024_05_21_3edb7be591bff54ce350g-0365.jpg?height=123&width=286&top_left_y=1105&top_left_x=170)

## Thus, we obtain

$$
\left\{\begin{array}{l}
x+y=3 \\
\frac{x^{2}}{y^{2}}=4
\end{array}\right.
$$

$$
\left\{\begin{array} { l } 
{ x + y = 4 , } \\
{ \frac { x ^ { 2 } } { y ^ { 2 } } = 3 . }
\end{array} \Rightarrow \left\{\begin{array} { l } 
{ x _ { 1 } = 6 , } \\
{ y _ { 1 } = - 3 ; }
\end{array} \left\{\begin{array} { l } 
{ x _ { 2 } = 2 , } \\
{ y _ { 2 } = 1 ; }
\end{array} \left\{\begin{array} { l } 
{ x _ { 3 } = \frac { 4 \sqrt { 3 } } { \sqrt { 3 } - 1 } , } \\
{ y _ { 3 } = \frac { 4 } { 1 - \sqrt { 3 } } ; }
\end{array} \left\{\begin{array}{l}
x_{4}=\frac{4 \sqrt{3}}{\sqrt{3}+1} \\
y_{4}=\frac{4}{1+\sqrt{3}} .
\end{array}\right.\right.\right.\right.
$$

Answer: $(6 ;-3),(2 ; 1),(6+2 \sqrt{3} ;-2-2 \sqrt{3}),(6-2 \sqrt{3} ;-2+2 \sqrt{3})$.