3.8 The front wheel of a moving model makes 6 more revolutions than the rear wheel over a distance of 120 m. If the circumference of the front wheel is increased by $\frac{1}{4}$ of its length, and the circumference of the rear wheel is increased by $\frac{1}{5}$ of its length, then over the same distance, the front wheel will make 4 more revolutions than the rear wheel. Find the lengths of the circumferences of the front and rear wheels.
3.8 The circumference of a wheel $C$, the number of revolutions $n$, and the distance $s$ are related by the formula $C n=s$. We will fill in the table of values of these quantities in the order indicated by the numbers (1), (2), .., (12).

| Wheel | Before change |  |  | After change |  |  |
| :---: | :---: | :---: | :---: | :---: | :---: | :---: |
|  | $\bar{C}$ | $\boldsymbol{s}$ | $\bar{n}$ | $C$ | $s$ | $\boldsymbol{n}$ |
| Front | (1) ${ }^{x}$ | $(3)^{120}$ | (5) $\frac{120}{x}$ | (7) $\frac{5 x}{4}$ | $9^{120}$ | (11) $\frac{120 \cdot 4}{5 x}$ |
| Rear | (2) ${ }^{y}$ | (4) $^{120}$ | (6) $\frac{120}{y}$ | (8) $\frac{6 y}{5}$ | (10) $^{120}$ | (12) $\frac{120 \cdot 5}{6 y}$ |

According to the condition, we obtain the system

$$
\left\{\begin{array}{l}
\frac{120}{x}-\frac{120}{y}=6 \\
\frac{96}{x}-\frac{100}{y}=4
\end{array}\right.
$$

from which we find $x=4(m), y=5(m)$.

Answer: 4 and 5 m.