Question 272119
the second cyclist travels a 10 miles per hour and 3 hours after cyclist who is 
traveling at 6 miles per hour how much time will pass before the second cyclist
 up with the first from the time the second cyclist started biking
:
Let t = travel time of the 2nd cyclist
then 
(t+3) = travel time of the 1st
:
When the 2nd cyclist catches the 1st, they will have traveled the same distance.
Write a dist equation: dist = speed * time
:
10t = 6(t+3)
10t = 6t + 18
10t - 6t = 18
4t = 18
t = {{{18/4}}}
t = 4.5 hrs, the 2nd catches the 1st
;
:
Check solution by confirming they both went the same dist:
4.5(10) = 45 mi
7.5(6) = 45 mi