Solve the following system of equations:

$$
\begin{gathered}
2^{\frac{x-y}{2}}-2^{\frac{x-y}{4}}=2 \\
3^{\lg (2 y-x)}=1 .
\end{gathered}
$$
(1) can be written as

$$
\left(2^{\frac{x-y}{4}}\right)^{2}-2^{\frac{x-y}{4}}-2=0
$$

This is a quadratic equation in terms of $2^{\frac{x-y}{4}}$, from which

$$
2^{\frac{x-y}{4}}=\frac{1 \pm \sqrt{1+8}}{2}=2, \quad \text { or } \quad-1
$$

Since any real power of 2 is positive, the negative root cannot be considered. Therefore,

$$
2^{\frac{x-y}{4}}=2
$$

from which

$$
\frac{x-y}{4}=1, \quad \text { or }
$$

from (2)

$$
\lg (2 y-x)=0
$$

which means

$$
2 y-x=10^{0}=1
$$

from (3) and (4)

$$
x=9 \quad \text { and } \quad y=5 \text {. }
$$

Kolonits Ferenc (Bp. VIII., Piarist g. I. o. t.)