The following example is from the time of the Bolyais; it is taken from Ferencz Pethe's work "Mathesis" (Béts 1812) and we publish it in its original wording and spelling:

Three Bombardiers threw the bomb from three mortars onto a city for a whole day. The first two together threw 26 more than the third; the last two 38 more than the first two; the first and last 24 more than the middle one: how many did each throw then?
I. Solution: Let $x, y, z$ represent the number of shots fired by each cannon, respectively. Then the relationships between the number of shots can be expressed by the following equations:

$$
\begin{aligned}
x+y & =z+26 \\
x+y+38 & =y+z \\
x+z & =y+24
\end{aligned}
$$

Subtracting (1) from (2),

$$
38=y-26, \quad \text { i.e., } \quad y=64
$$

Substituting this value into (3) and rearranging (2), we get

$$
\begin{aligned}
& x+z=88 \\
& x-z=-38
\end{aligned}
$$

From which

$$
x=25, \quad \text { and } \quad z=63
$$

Li Bek U (Békéscsaba, Vízműép. t. II. o. t.)

II. Solution. We can also arrive at the result through pure reasoning as follows:

The third one fired 26 fewer shots than the total of the first two, but together with the second, they fired 38 more. The number of shots fired by the second one makes up for the 26 shortage and the 38 excess, meaning the middle one fired $26+38=64$ shots.

However, according to the third condition of the problem, the first and third together fired $(64+24)$, that is, all three together fired $64+24+64=$ 152 shots. If we subtract 26 from this $(152-26=126)$, and imagine that we subtracted it from the number of shots fired by the first two, then—since according to the first part of the problem, the number of shots fired by the first two, reduced by 26, equals the number of shots fired by the third—we essentially obtained twice the number of shots fired by the third. Thus, the third one fired $\left(\frac{126}{2}=\right) 63$ shots.

Therefore, the first one necessarily fired $152-63-64=25$ shots.

Várallyay László (Mosonmagyaróvár, Kossuth g. II. o. t.)