David walked 19 kilometers. The first part of the walk was at 5kph and the rest at 3kph. he would have covered 2 kilometers less if he had reversed the rates, that is, if he had walked at 3kph and at 5kph for the same times that he actually walked at 5kph and 3kph, respectively. How long did it take him to walk the 19 kilometers?
algebra - drwls, Sunday, December 7, 2008 at 1:16am
Let T1 (in hours) be the time of the first part of the walk and T2 be the time of the second part.
5 T1 + 3 T2 = 19 (km)
3 T1 + 5 T2 = 17 (km)
Solve those two equations for T1 and T2.
15 T1 + 9 T2 = 57
15 T1 + 25 T2 = 85
16 T2 = 28 etc. That tells you T1. Then solve for T2
T1 + T2 is the total time that it took.
algebra (correction) - drwls, Sunday, December 7, 2008 at 5:54am
The 16 T2 = 28 equation tells you that T2 = 1.75 hours
Therefore 5T1 + 5.25 = 19
T1 = 2.75 hours
Total time = 4.5 hours