Question

during the first part of a trip a canoist travels 34 miles at a certain speed. The canoeist travels 22 miles on the second part at a speed of 5 miles slower. the total time for the trip is 3 hrs what was the speed on each part of the trip?

Answer 1

d = 34 + 22 = 55miles
t = 3 hours
v(average) = 55/3 mph

v(average) = (v1 + (v1 - 5))/3 = 55/3

solve for v1 to get the speed on the first part of the trip,
v2 = v1-5

Answer 2

Let x be the speed on 1st part of trip, t1 and t2 denote time for each trip
distance = rate*time
Set up equations:
(i) 34 = x*t1
(ii) 22 = (x-5)*t2
(iii) t1 + t2 = 3
Goal is to solve for x, use substitution to get an equation in terms of only x
solve (iii) for t1
t1 = 3 - t2
substitute it into (i) then solve for t2
34 = x(3-t2)
34 = 3x - x*t2
x*t2 = 3x -34
t2 = (3x -34)/x
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Solve (ii) for t2
t2 = 22/(x-5)
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Now set t2 = t2 in terms of x
(3x-34)/x = 22/(x-5)
Cross-multiply to get rid of fractions
(x-5)(3x-34) = 22x
3x^2 -49x + 170 = 22x
3x^2 -71x + 170 = 0
Use quadratic formula:
x = [71 +-sqrt(71^2 - 4*3*170)]/6
x = (71 +- 54.781)/6
x = 2.7 or 20.964
Since x is the faster speed and speed must be greater than 0, then (x-5)>0
so x=2.7 is irrelevant
The speeds on each trip were 20.964 and 15.964