Algebra
posted by Greg .
During the first part of a trip, a canoeist travels 98 miles at a certain speed. The canoeist travels 10 miles on the second part of the trip at a speed 5 mph slower. The total time for the trip is 3 hours. What was the speed on the FIRST part of the trip? What was the speed on the SECOND part of the trip?
(type an integer or a decimal. Round to the nearest hundredth)

98mi @ X mi/hr.
10mi @ (x5) mi/hr.
T = 98/x + 10/(x5) = 3hrs.
Multiply both sides by x(x5):
98(x5) + 10x = 3x(x5),
98x  490 + 10x = 3x^2  15x,
3x^2 + 108x + 15x  490 = 0,
3x^2 + 123x 490 = 0,
Solve using Quadratic Formula and get:
X = 36.53, and 4.47.
The required value of x is > 5.
X = 36.53 mi/hr = Speed during 1st part of trip.
X5 = 31.53 = Speed during 2nd part of trip.