13.415 Two cyclists set out simultaneously from points $A$ and $B$ towards each other. Four hours after their meeting, the cyclist who had set out from $A$ arrived at $B$, and nine hours after their meeting, the cyclist who had set out from $B$ arrived at $A$. How many hours was each cyclist on the road?
Solution. Let $x$ and $y$ be the travel times of the first and second cyclists, respectively. Then $x-4=y-9$ is the time each of them traveled until they met. Since each of them traveled a distance after the meeting equal to the distance the other had traveled before the meeting, $\frac{4}{y-9}=\frac{x-4}{9}$.

$$
\text{Thus, }\left\{\begin{array} { l } 
{ x = y - 5 } \\
{ \frac { 4 } { y - 9 } = \frac { x - 4 } { 9 } }
\end{array} \Leftrightarrow \left\{\begin{array}{l}
x=10 \\
y=15
\end{array}\right.\right.
$$

Answer: 10 hours, 15 hours.