Question 155108
Distance(d) equals Rate(r) times Time(t) or d=rt;  r=d/t and t=d/r
Let t=amount of time that passes before the second cyclist catches up with the first from the time the second cyclist starts
distance first cyclist travels=6*3+6*t(3 hour head start)
distance second cyclist travels=10*t 
Now when the above two distances are equal, the second cyclist will have caught up with the first cyclist, so our equation to solve is:
10t=18+6t  subtract 6t from each side
10t-6t=18-6t
4t=18
t=4.5 hours 

CK
3*6+6*4.5=10*4.5
18+27=45
45=45

Hope this helps---ptaylor