two buses leave towns 604 mi apart at the same time and travel toward eachother. One bus travels 9 mi/h faster than the other one. If they meet in 4 hours, what is the rate of each bus?

s = 71 mph, and s+9 = f = 80 mph

Step-by-step explanation:

The distances the two busses travel add up to 604 mi.

Letting f be the faster speed and s the slower. Then f = s + 9 (mph).

Then (s + 9) (mph) (4 hr) + (s) (mph) (4 hr) = 604 mi. Solve this for s:

4s + 36 + 4s = 604 mi, or 8s = 568. Finally, s = 71 mph, and s+9 = f = 80 mph.