Question 976564
Peter paddled a canoe 20 miles upstream, then paddled the canoe back to the starting point. the rate of the current was 3 mph and the total trip took 7 hours. At what rate would Peter paddle the canoe in still water? 
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let x=rate in still water
x+3=rate of canoe downtream
x-3=rate of canoe uptream
travel time=distance/rate
{{{20/(x+3)+20/(x-3)=7}}}
lcd: (x+3)(x-3)
20(x-3)+20(x+3)=7(x+3)(x-3)
20x-60+20x+60=7(x^2-9)
40x=7x^2-63
7x^2-40x-63=0
(x-7)(7x+9)=0
x=-9/7(reject)
x=7
At what rate would Peter paddle the canoe in still water?  7 mph