Hello, janvdl!

You should get a number for the value of x.

A Passenger train is *x* times faster than a Goods train.

If the trains move in the same directrion, it takes them *x* times longer to pass each other

than when they're moving in opposite directions. .What is *x*?

Let v = speed of the Goods train.

Let xv = speed of the Passenger train.

Going in __opposite__ directions, their relative speed is: .xv + v mph.

It is as if the G-train is stopped,

. . and the P-train is approaching at xv + v mph. Code:

* - - - - - * * - - - - - *
| Passenger | → | Passenger |
* - - - - - * * - - - - - *
* - - - *
| Goods |
* - - - *

To pass the G-train, the P-train must move the length of the G-train (LG)

. . plus the length of the P-train (LP).

Suppose it takes *t* hours for this to happen.

. . Then we have: .(xv + v)t .= .LG + LP .**[1]**

Going in the __same__ direction, their relative speed is: .xv - v mph

It is as if the G-train is stopped,

. . and the P-train is approaching at xv - v mph.

To pass the G-grain, the P-train must move the length of the G-train

. . plus the length of the P-train.

We are told that this takes *xt* hours.

. . Then we have: .(xv - v)xt .= .LG + LP .**[2]**

Equate [2] and [1]: .(xv - v)xt .= .(xv + v)t . . → . . v(x - 1)xt .= .v(x + 1)t

. . Divide by *vt*: .(x - 1)x .= .x + 1

. . and we have the quadratic: .x² - 2x - 1 .= .0

. . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . **_**

Using the Quadratic Formula, the positive root is: .**x .= .1 + √2**