Solve the following system of equations:

$$
x^{\lg y}+y^{\lg \sqrt{x}}=110, \quad x y=1000
$$
The expressions in the first equation only make sense for $x>0, y>0$. Any positive number $a>0$ can be written as a power of 10: $a=10^{\lg a}$. Using this, as well as the properties of exponentiation and logarithms, our system of equations takes the following form:

$$
10^{\lg x \lg y}+10^{\frac{1}{2} \lg x \lg y}=110, \quad \lg x+\lg y=3
$$

Here, the first equation is quadratic in $10^{\frac{1}{2} \lg x \lg y}$, with roots

$$
10^{\frac{1}{2} \lg x \lg y}=10 \quad \text{or} \quad 10^{\frac{1}{2} \lg x \lg y}=-11
$$

The latter is impossible, as any power of 10 is positive. Therefore, according to (1),

$$
\lg x \cdot \lg y=2
$$

We know the sum and product of $\lg x$ and $\lg y$, so $\lg x$ and $\lg y$ are the roots of the equation

$$
z^{2}-3 z+2=0
$$

Knowing this, we obtain the solutions to the original system of equations:

$$
\begin{array}{ll}
x_{1}=10, & y_{1}=100 \quad \text{and} \\
x_{2}=100, & y_{2}=10 .
\end{array}
$$

Both pairs are indeed solutions.

Peter Lenkei (Budapest, József A. Gymnasium, 3rd grade)

Note. If we already know that $\lg x \cdot \lg y=2$ is a solution to the equation $10^{\lg x \lg y}+10^{\frac{1}{2} \lg x \lg y}=110$, it can also be seen from the strictly increasing nature of the exponential function that there cannot be another solution.