120 (989). The road from point $A$ to point $B$ is 11.5 km long and goes first uphill, then on flat ground, and finally downhill. The pedestrian spent 2 hours and 54 minutes on the journey from $A$ to $B$, and 3 hours and 6 minutes on the return journey. The pedestrian's walking speed uphill was 3 km/h, on flat ground - 4 km/h, and downhill - 5 km/h. How many kilometers does the part of the road that goes on flat ground constitute?
Solution. Let the length of the road uphill be $x$ kilometers, on the plain be $y$ kilometers, and downhill be $z$ kilometers. According to the problem, we form a system of three linear equations with three unknowns:

88

$$
\left\{\begin{array}{l}
\frac{x}{3}+\frac{y}{4}+\frac{z}{5}=2.9 \\
\frac{z}{3}+\frac{y}{4}+\frac{x}{5}=3.1 \\
x+y+z=11.5
\end{array}\right.
$$

By adding the first two equations of the system, we get the equation

$$
\frac{8}{15}(x+z)+\frac{y}{2}=6 .
$$

Substituting the value of $x+z$ from the third equation of the system into this equation, we get the equation

$$
\frac{8}{15}\left(\frac{23}{2}-y\right)+\frac{y}{2}=6
$$

from which $y=4$.

Remark. The solution to the problem can be reduced to forming and solving a system of two linear equations with two unknowns if the length of the road on the plain is expressed as the difference between the total length of the road (11.5 km) and the lengths of the roads uphill and downhill.

Answer: 4 km.