Solve the following system of equations:

$$
\begin{gathered}
5^{x}+0.1^{y}=9 \\
0.2^{x}+10^{y}=0.5
\end{gathered}
$$
Notice that both the bases under the exponents $x$ and $y$ are reciprocals of each other, so the corresponding powers are also reciprocals. By introducing the unknowns $5^{x}=u, 0.1^{y}=v$ for the two terms in the first equation, our system of equations can be written, after some rearrangement, as follows:

\[
\begin{aligned}
u+v & =9 \\
\frac{1}{u}+\frac{1}{v}=\frac{v+u}{u v}=\frac{9}{u v} & =0.5, \quad u v=18
\end{aligned}
\]

Thus, we know the sum and product of $u$ and $v$. From these, $u$ and $v$ are the roots of the equation

\[
z^{2}-9 z+18=0
\]

with roots $z_{1}=6, z_{2}=3$.

Let $5^{x}=u=z_{1}=6$ and $0.1^{y}=v=z_{2}=3$. Taking the base 10 logarithm,

\[
\begin{aligned}
& x_{1}=\lg 6: \lg 5 \approx 0.7782: 0.6990 \approx 1.113 \text{ and } \\
& y_{1}=\lg 3: \lg 0.1 \approx 0.4771:(-1)=-0.4771 .
\end{aligned}
\]

If, however, $5^{x}=u=z_{2}=3$ and $0.1^{y}=v=z_{1}=6$, then

\[
\begin{aligned}
& x_{2}=\lg 3: \lg 5 \approx 0.4771: 0.6990 \approx 0.6826, \\
& y_{2}=\lg 6: \lg 0.1=-\lg 6 \approx-0.7782
\end{aligned}
\]

Bán Tamás (Budapest, József A. Gymn., III. o. t.)