Distance = rate x time
So, for train A : z = R1 * X
Train B : z = R2 * Y

Let t be the time travelled by train A, since train B starts 2 hr earlier, the time travelled by train B is (t+2). To catch up the train B, train A and B should travel the same distance,

R2 * (t+2) = R2 * (t)
z/Y * (t+2) = z/X * t
t+2/Y = t/X
t(x-y) = -2X
so, t = 2X/(Y-X)

There is a high speed train (train A) that takes x hours to cover z miles and a regular speed train (train B) that takes y hours to cover the same distance. If train B leaves 2 hours earlier, how long would it take train A to catch up assuming they started from the same point?

Relative speed in this case = (z/x - z/y)
Distance covered by B in these 2 hours = 2z/y
Thus time taken by A to cover this distance of 2z/y miles

= D/S = (2z/y) / (z/x - z/y)
=(2z/y) / [(yz-xz)/xy]
=2z/y * (xy/yz-xz)
=2/(y-z)

I am getting 2x/(y-x)