Question 206828
the trip to the wedding takes 3 hrs the wedding and reception takes 2 hrs and as
 a result of the celebration the trip home takes 4 hrs at a rate of 15 mph less
 than the going rate. what was the couples rate of travel on the way home. 
:
let r = rate of travel home
then
(r+15) = rate of travel to the wedding (15 mph faster)
:
The time of the wedding and reception are not relevant to the problem
:
The trip there and trip back are equal distance
:
Write a distance equation: Dist = time * rate
:
4r = 3(r+15)
:
4r = 3r + 45
:
4r - 3r = 45
:
r = 45 mph hr on the return trip
:
:
Check solution by finding the distances (should be equal)
outbound trip: 45 + 15 = 60 mph
3(60) = 180 mi
4(45) = 180 mi
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Not that hard, right?