Two motorcyclists left the settlements M and N, 50 km apart from each other, and met in 30 minutes.

Two motorcyclists left the settlements M and N, 50 km apart from each other, and met in 30 minutes. Find the speed of each motorcyclist if it is known that one of them arrived at point M 25 minutes earlier than the other from point N. P.S If not difficult, outline this problem schematically so that in the future I understand how these problems are solved.

Since the meeting took place after 30 minutes (1/2 hour), the speed of convergence was:

50: 1/2 = 100 km / h.

Let’s take the speed of motorcyclists for x and y, since they were moving towards each other, the speed of approach is the sum of their speeds:

x + y = 100.

50 / x – travel time of the first;

50 / y – second.

Second equation:

50 / x – 50 / y = 25/60.

1 / x – 1 / y = 1/120.

From the first:

x = 100 – y.

Let’s substitute in the second:

1 / (100 – y) – 1 / y = 1/120;

y – (100 – y) = 1/120 * y * (100 – y);

y ^ 2 – 20y – 1200 = 0.

y = 48;

x = 52.