Let the speed of the faster bike be x mi/hr
Let the speed of the slower bike be y mi/hr

Whether the faster catches up to the slower when they are going in the same
direction, or whether they ride toward each other and meet,  the 30 mile
distance between them shinks to zero.

When traveling in the same direction the catch-up rate of the faster bike
to the slower bike is at a speed of x-y.

Since time = , the faster catches up the entire 30 miles
head start which the slower one has on the faster in  hours.
That is, the 30 miles between them shrinks to zero (by the faster catching up)
in 



hours, and we are given that that is 6 hours.  So we have the equation

 = 6
 
When traveling in opposite directions they approach each other at the speed
x+y.

Since time = , the 30 miles between them shrinks to zero (by them 
meeting each other) in  hours, and we are given that that is 1 hour. 
So we have the equation

 = 1

And so we have the system of equations:

 = 6
 = 1

Clearing the first equation of fractions, we have:

30 = 6(x-y)
30 = 6x - 6y
Divide through by 6
 5 = x - y

Clearing the second equation of fractions, we have:

30 = 1(x+y)
30 = x + y

Now the system is

 5 = x - y
30 = x + y

Adding the two equations term by term, we have:

35 = 2x

Dividing both sides by 2, we have:

17.5 = x

So the faster bike goes at 17.5 mi/hr.

Substituting 17.5 for x in

 5 = x - y
 5 = 17.5 - y
 y = 12.5

So the slower bike goes at 12.5 mi/hr.

Edwin