7.289. $\left\{\begin{array}{l}10^{\lg 0.5\left(x^{2}+y^{2}\right)+1.5}=100 \sqrt{10}, \\ \frac{\sqrt{x^{2}+10 y}}{3}=\frac{6}{2 \sqrt{x^{2}+10 y}-9}\end{array}\right.$
## Solution.

Domain of definition: $\left\{\begin{array}{l}x \neq 0, \\ y \neq 0, \\ 2 \sqrt{x^{2}+10 y}-9 \neq 0 .\end{array}\right.$

Rewrite the first equation of the system as

$$
\begin{aligned}
& 10^{\lg 0.5\left(x^{2}+y^{2}\right)+1.5}=10^{2.5} \Leftrightarrow \lg 0.5\left(x^{2}+y^{2}\right)+1.5=2.5 \Leftrightarrow \\
& \Leftrightarrow \lg 0.5\left(x^{2}+y^{2}\right)=1 \Leftrightarrow 0.5\left(x^{2}+y^{2}\right)=10 \Leftrightarrow x^{2}+y^{2}=20
\end{aligned}
$$

From the second equation of the original system, we get

$$
2\left(\sqrt{x^{2}+10 y}\right)^{2}-9 \sqrt{x^{2}+10 y}-18=0
$$

Solving this equation as a quadratic equation in $\sqrt{x^{2}+10 y}$, we have $\sqrt{x^{2}+10 y}=-4$ (no solutions), or $\sqrt{x^{2}+10 y}=6, x^{2}+10 y=36$. The system becomes

$$
\left\{\begin{array} { l } 
{ x ^ { 2 } + y ^ { 2 } = 2 0 , } \\
{ x ^ { 2 } + 1 0 y = 3 6 }
\end{array} \Leftrightarrow \left\{\begin{array}{l}
x^{2}+y^{2}=20, \\
x^{2}=36-10 y
\end{array} \Rightarrow x^{2}=36-10 y, y^{2}-10 y+16=0\right.\right.
$$

from which $y_{1}=2, y_{2}=8$. Then $x^{2}=16, x_{1,2}= \pm 4$, or $x^{2}=-44$ does not fit.

Answer: $(-4 ; 2),(4 ; 2)$.