SOLUTION: a couple attends an out of town wedding. 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

Question 206828: a couple attends an out of town wedding. 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 tate of travel on the way home.
just overwhelmed dont even know how to approach the problem dont even know how to declare and variable, thank you
Answer by ankor@dixie-net.com(22740) (Show Source): You can put this solution on YOUR website!
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
:
:
Not that hard, right?

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