Question 9351
OK, joggerA speed = x
so that joggerB speed = x+2.

Note also, if joggerA runs a distance d, then B must have run 21-d, since they have met somewhere.


now speed = distance/time so


for A: x = d/1.5 --> d = 1.5x --> eqn1


for B: x+2 = (21-d)/1.5
--> 1.5x+3 = 21-d


so, 1.5x + 3 = 21-(1.5x)
1.5x + 3 = 21-1.5x
3x = 18
so x = 6.


So joggerA runs at 6mph and B runs at 8mph


jon.