Two pedestrians went out to meet each other at the same time: the first from A to B, the second from B to A

Two pedestrians went out to meet each other at the same time: the first from A to B, the second from B to A. When the distance between them decreased 6 times, a cyclist left B to A. The first pedestrian met him at the moment when the second passed 4 / 9 the distance between A and B. B point A the cyclist and the first pedestrian at point B arrived at the same time. Determine the ratio of the speed of each pedestrian to the speed of the cyclist.

Let us take for x, y, and x the speeds of pedestrians and cyclists, respectively, denote by S the distance between points A and B, then:

5/6 * S / (x + y) – the time of the cyclist’s departure;

4 / 9S / y – time until the first pedestrian meets the cyclist;

5 / 9S / z + 5/6 * S / (x + y) – time until the meeting between the cyclist and the 1st pedestrian;

S / x – arrival time of the 1st in B;

S / z + 5/6 * S / (x + y) – the time of arrival of the cyclist at A;

We get the system:

5 / 9S / z + 5/6 * S / (x + y) = 4 / 9S / y

S / z + 5/6 * S / (x + y) = S / x