SOLUTION: Two cyclists leave a city at the same​ time, one going east and the other going west. The west bound cyclist bikes at 15 mph faster than the east bound cyclist. After 2 hours

Algebra ->  Finance -> SOLUTION: Two cyclists leave a city at the same​ time, one going east and the other going west. The west bound cyclist bikes at 15 mph faster than the east bound cyclist. After 2 hours      Log On


   

Question 1070435: Two cyclists leave a city at the same​ time, one going east and the other going west. The west bound cyclist bikes at 15 mph faster than the east bound cyclist. After 2 hours they are 78 miles apart. How fast is each cyclist​ riding?
Answer by Boreal(15235) About Me  (Show Source):
You can put this solution on YOUR website!
eastbound=x mph
westbound = x+15 mph
after 2 hours the westbound is 2(x+15)
the east bound is 2x
They add to 78
4x+30=78
4x=48
x=12 mph
x+15=27 mph
In 2 hours, the east bound went 24 miles and the west bound 54 miles, and that adds to 78 miles