SOLUTION: Jan and Tariq took a canoeing trip, traveling 6 mi upstream against a 2-mi/h current. They then returned to the same point downstream. If their entire trip took 4 hr, how fast coul

Algebra ->  Customizable Word Problem Solvers  -> Travel -> SOLUTION: Jan and Tariq took a canoeing trip, traveling 6 mi upstream against a 2-mi/h current. They then returned to the same point downstream. If their entire trip took 4 hr, how fast coul      Log On

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Question 964015: Jan and Tariq took a canoeing trip, traveling 6 mi upstream against a 2-mi/h current. They then returned to the same point downstream. If their entire trip took 4 hr, how fast could they paddle in still water?
Answer by LinnW(1048) About Me  (Show Source):
You can put this solution on YOUR website!
Use the formula, rate * time = distance
Set x = speed in still water
(x-2) * t1 = 6 miles
(x+2) * t2 = 6 miles
t1, the time paddling upstream = distance 6 miles/ rate (x-2) or
6/(x-2)
t2, the time paddling downstream = distance 6 miles/ rate (x+2) or
6/(x+2)
t1 + t2 = 4 hours, so
6/(x-2) + 6/(x+2) = 4
A common denominator is (x-2)(x+2) so we want
6%2F%28x-2%29%2A%28x%2B2%29%2F%28x%2B2%29+%2B+6%2F%28x%2B2%29%2A%28x-2%29%2F%28x-2%29+=+4
expanding
%286x+%2B+12+%2B+6x+-12%29%2F%28x%5E2-4%29+=+4%2F1
simplify
%2812x%29%2F%28x%5E2-4%29+=+4%2F1
do cross products
12x=4%28x%5E2-4%29
12x=4x%5E2-16
0=4x%5E2-12x-16
divide by 4
0=x%5E2-3x-4
factoring
0+=+%28x-4%29%28x%2B1%29
So possible values for x are 4 and -1
Since we need a positive speed, the speed in
still water is 4 mph