Two trains simultaneously left points A and B towards each other. The first train arrived at point B

Two trains simultaneously left points A and B towards each other. The first train arrived at point B at t1 = 4 hours after the meeting of the trains, and the second at point A at t2 = 9 hours after the meeting. Determine how many hours the first train took.

Let x be the speed of the first train, and
y – second
t is the time until the trains meet
Distance traveled by the first train to the meeting – xt
Distance traveled by the second train to the meeting – yt
t that passed from the meeting to the arrival at point B of the first train = S, which the second train passed to the meeting, divided by V of the first train – yt / x = 4
t, which passed from the meeting to the arrival at point A of the second train – xt / y = 9
Let us express t through the ratio of speeds from the first expression – 4x / y = t
9 = 4x ^ 2 / y ^ 2
x ^ 2 / y ^ 2 = 9/4
x / y = 3/2
Before the meeting, the trains were traveling – t = 4x / y = 4 3/2 = 6 hours
The first train was on the way:
t1 = t + 4 = 6 + 4 = 10 hours.
Answer: 10h