Question 1039197: Andy canoed downriver for 2 hours and returned in 3 hours. Andy's sped in still water is 2 mi/hr faster than the sped of the current. If W represents speed in still water and C represents speed of the current, what is Andy's still-water paddling speed and the speed of the current? I have tried to answer it but I keep getting 3.5 mi/hr as paddling speed and 1.5 mi/hr as speed of the current.
Answer by Boreal(15235)   (Show Source):
You can put this solution on YOUR website! W=speed in still water mi/hr
C=speed of current
2(W+C)=miles paddled in 2 hours
that equals 3(W-C) coming back.
Distributing,
2W+2C=3W-3C
W=5C
but C=W-2
distributing again
W=5W-10
-4W=-10
W=2.5 mph
C=0.5 mph;W=5C and W=C-2
3.0 mph together, 2 mph against.
In 2 hours, goes 6 miles downstream
In 3 hours, goes 6 miles upstream
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