<PRE><B>Question 302233</B>
duringthe first part of a trip a canoeist travels 91 mi. at a certain speed. The canoeist travels 15 mi on the second part of the trip at a speed 5 mph slower. The total time for the trip is 4 hours. What was the speed on each part of the trip. 
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1st Part DATA:
distance = 91 mi ; rate = x mph ; time = d/r = 91/x hrs.
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2nd Part DATA:
distance = 15 mi ; rate = (x-5) mph ; time = d/r = 15/(x-5) hrs.
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Equation:
time + time = 4 hrs
91/x + 15/(x-5) = 4
Mltiply thru by x(x-5) to get:
91(x-5) + 15x = 4x(x-5)
91x - 455 + 15x = 4x^2 - 20x
4x^2 - 126x + 455 = 0
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Quadratic Formula:
x = [126 +- sqrt(126^2 - 4*4*455)]/(2*4)
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x = [126 +- sqrt(8596)]/8
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x = [126 +- 92.71]/8
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x = [126+92.71]/8 or x = [126-92.71]/8 
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x = 27.34 or x = 4.16
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Cheers,
Stan H.</PRE>