SOLUTION: Jeff leaves his house on his bicycle at 8:30 am, and averages 5 mph. At 9:00 am his wife leaves, following the same path and averages 6 mph. At what time will joan catch up with Je

Question 170619: Jeff leaves his house on his bicycle at 8:30 am, and averages 5 mph. At 9:00 am his wife leaves, following the same path and averages 6 mph. At what time will joan catch up with Jeff?
t= time jeff travels
using the equation d=r*t, I get 5t=6(t-1/2)
expanding- 5t=6t-3
-t=-3.
So my question is, am I solving this right? And if so, what is t=3? Does that mean she will catch up to him in 3 hrs? the only other way I could imagine it is that it means she will catch him at 3 oclock but that doesn't seem right.
thank you!

Answer by stanbon(75887) (Show Source): You can put this solution on YOUR website!
Jeff leaves his house on his bicycle at 8:30 am, and averages 5 mph. At 9:00 am his wife leaves, following the same path and averages 6 mph. At what time will joan catch up with Jeff?
-----------------------
Jeff DATA:
rate = 4 mph ; time = x hours; distance = rt = 4x miles
----------------------------------------------
Wife DATA:
rate = 6 mph ; time = (x-(1/2)) hrs; distance = rt = 6(x-(1/2)) miles
==============================================
EQUATION:
distance = distance
4x = 6(x-(1/2))
4x = 6x -3
2x = 3
x = 3/2 hours
---------------
Answer: wife will catch up to Jeff at 8:30 + 1 1/2 hrs = 10:00
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Cheers,
stan H.
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