SOLUTION: a man walks 31 mile, partly at teh rate of 2mph and the rest is 5mph. if he had walked at teh rate of 5mph for teh same times as he actually walked at 2mph and vice versa, he woul

Question 405278: a man walks 31 mile, partly at teh rate of 2mph and the rest is 5mph. if he had walked at teh rate of 5mph for teh same times as he actually walked at 2mph and vice versa, he would have covered 15 miles more. how long did it take him to walk 31 miles
Answer by ankor@dixie-net.com(22740) (Show Source): You can put this solution on YOUR website!
a man walks 31 mile, partly at the rate of 2 mph and the rest is 5 mph.
if he had walked at the rate of 5 mph for the same times as he actually walked
at 2mph and vice versa, he would have covered 15 miles more.
how long did it take him to walk 31 miles
:
let w = the original time he walked at 2 mph
let x = the original time he walked at 5 mph
:
Write a distance equation for the original trip
2w + 5x = 31
:
Write an equation for the vice-versa scenario
5w + 2x = 31 + 15
5w + 2x = 46
:
Multiply the 1st equation by 5, multiply the 2nd equation by 2, results
10w + 25x = 155
10w + 4x = 92
------------------subtraction eliminates w, find x
0 + 21x = 63
x = 63/21
x = 3 hrs at 5 mph
and
5w + 2(3) = 46
5w = 46 - 6
w =
w = 8 hrs at 2 mph
:
Total time to travel 31 mi: 3 + 8 = 11 hrs
:
:
Confirm this using the vice-versa equation
5w + 2x = 46
5(8) + 2(3) = 46, confirms our answer of 11 hrs

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