6.099. $\left\{\begin{array}{l}x^{2}+x y+y^{2}=91 \\ x+\sqrt{x y}+y=13\end{array}\right.$

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6.099. $\left\{\begin{array}{l}x^{2}+x y+y^{2}=91 \\ x+\sqrt{x y}+y=13\end{array}\right.$
Solution.

Domain of definition: $x y \geq 0$.

Let $\left\{\begin{array}{l}\sqrt{x}=u \geq 0, \\ \sqrt{y}=v \geq 0,\end{array}\right.$ then $\left\{\begin{array}{l}x=u^{2}, \\ y=v^{2},\end{array}\left\{\begin{array}{l}x^{2}=u^{4}, \\ y^{2}=v^{4} .\end{array}\right.\right.$

With respect to $u$ and $v$, the system of equations takes the form

$$
\left\{\begin{array} { l } 
{ u ^ { 4 } + u ^ { 2 } v ^ { 2 } + v ^ { 4 } = 9 1 , } \\
{ u ^ { 2 } + u v + v ^ { 2 } = 1 3 }
\end{array} \Leftrightarrow \left\{\begin{array}{l}
\left(u^{2}+v^{2}\right)^{2}-2 u^{2} v^{2}+u^{2} v^{2}=91 \\
u^{2}+v^{2}=13-u v
\end{array}\right.\right.
$$

From the first equation of the obtained system, we have

$$
(13-u v)^{2}-u^{2} v^{2}=91,169-26 u v+u^{2} v^{2}-u^{2} v^{2}=91, u v=3 .
$$

From the second equation of the system, we get

$$
u^{2}+v^{2}=13-3, \quad(u+v)^{2}-2 u v=10, \quad(u+v)^{2}=10+2 u v=16
$$

from which

$$
u+v=-4
$$

or

$$
u+v=4 \text {. }
$$

We have obtained two systems of equations:

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

from which, by Vieta's theorem, we find

$$
\begin{aligned}
& \left\{\begin{array} { l } 
{ u _ { 1 } = - 1 , } \\
{ v _ { 1 } = - 3 , }
\end{array} \left\{\begin{array}{l}
u_{2}=-3, \\
v_{2}=-1,
\end{array},\left\{\begin{array} { l } 
{ u _ { 3 } = 1 , } \\
{ v _ { 3 } = 3 , }
\end{array} \left\{\begin{array}{l}
u_{4}=3, \\
v_{4}=1 .
\end{array}\right.\right.\right.\right. \\
& \text { Then for }\left\{\begin{array} { l } 
{ \sqrt { x } = - 1 , } \\
{ \sqrt { y } = - 3 }
\end{array} - \varnothing , \text { for } \left\{\begin{array}{l}
\sqrt{x}=-3, \\
\sqrt{y}=-1
\end{array}-\varnothing,\right.\right. \\
& \left\{\begin{array} { l } 
{ \sqrt { x } = 1 , } \\
{ \sqrt { y } = 3 , }
\end{array} \text { from which } \left\{\begin{array} { l } 
{ x _ { 1 } = 1 , } \\
{ y _ { 1 } = 9 , }
\end{array} \left\{\begin{array} { l } 
{ \sqrt { x } = 3 , } \\
{ \sqrt { y } = 1 , }
\end{array} , \text { from which } \left\{\begin{array}{l}
x_{2}=9, \\
y_{2}=1 .
\end{array}\right.\right.\right.\right.
\end{aligned}
$$

Answer: $(1 ; 9)(9 ; 1)$.