Question 593715
A passenger train traveled to Schannesburg and back.
 The trip there took 20 hours and the trip back took 16 hours. 
It averaged 12 mph faster on the return trip then on the outbound trip.
 What was the passenger train's speed on the outbound trip?
:
Let s = train speed outbound
then
(s+12) = train return speed
:
Write a distance equation: dist = time * speed
:
out dist = return dist
20s = 16(s+12)