Well, let's call the speed of the slower one ' S '. Then the speed of the faster one is ' S + 3 '.
They started out at two towns that are far apart, and they both kept pedaling toward each other. Now think about the space between them: The slower one was covering distance on her end of the space at the rate of 'S' mph, and the faster one was covering distance on her end of it at the rate of 'S+3' mph. So the space between them was shrinking at the rate of ' (S) + (S+3) ' mph, and that's ' 2S + 3 ' mph all together.
OK. The space between them starts out being 68 miles.
Speed = (distance) / (time) .
The speed at which the space shrank (that's 2S+3) =
(68 miles) / (the time they biked before meeting) .
I left it in that form for a minute, because I need to discuss a very important question with you before we go any farther:
I read your question very carefully. You said that they biked from 10pm until 1pm. That's 15 hours, and biking all night, and if I use that, and go ahead and complete the calculation, then it's going to turn out that the slower biker is traveling about 1.124 feet per second ... maybe the speed of a small snake in the grass. This doesn't make sense. I think you made a tiny mistake when you typed the question, and I'm almost sure they started out at 10am, and then met at 1 pm. That's 3 hours in the middle of the day, and seems to be a lot more reasonable, so I'm going ahead with it.
The speed at which the space shrank (that's 2S+3) =
(68 miles) / (3 hours) .
2S + 3 = 68 / 3 = 22 and 2/3 .
Subtract 3 from each side: 2S = 19 and 2/3
Divide each side by 2 : S = 9 and 5/6 miles per hour. That's the speed of the slower biker.
The faster biker was 3 mph faster: S + 3 = 12 and 5/6 mph .
