8.2. From two cities, the distance between which is 105 km, two pedestrians set out simultaneously towards each other at constant speeds and met after 7.5 hours. Determine the speed of each of them, knowing that if the first walked 1.5 times faster and the second 2 times slower, they would have met after $8 \frac{1}{13}$ hours.
Answer: 6 and 8 km per hour.

Solution: Let their speeds be $x$ and $y$ km per hour, respectively. From the condition, we get: $\frac{15}{2}(x+y)=105, \frac{105}{13}\left(\frac{3}{2} x+\frac{1}{2} y\right)=105$, from which $\quad x=6, y=8$.