SOLUTION: During the first part of a trip, a canoeist travels 76 miles at a certain speed. The canoeist travels 2 miles on the second part of the trip at a speed 5 mph slower. The total time

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Question 328458: During the first part of a trip, a canoeist travels 76 miles at a certain speed. The canoeist travels 2 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 each part of the trip?
(type an integer or a decimal. Round to the nearest hundredth)
Answer by checkley77(12844) (Show Source):
You can put this solution on YOUR website!
D=RT
FOR THE FIRST PART OF THE TRIP;
76=RT
T=76/R
FOR THE SECOND PART OF THE TRIP:
2=(R-6)T
T=2/(R-6)
76/R+2/(R-6)=3
(76(R-6)+2R)/R(R-6)=3
(76R-456+2R)/(R^2-6R)=3
(78R-456)/(R^2-6R)=3 CROSS MULTIPLY.
78R-456=3(R^2-6R)
78R-456=3R^2-18R
3R^2-18R-78R+456=0
3R^2-96R+456=0
3(R^2-32R+152)=0

R=(32+-SQRT(32^2-4*1*152])/2*1
R=(32+-SQRT[1,024-608])/2
R=(32+-SQRT416)/2
R=(32+-20.396)/2
R=(32+20.396)/2
R=52.396/2
R=26.198 MPH. ANS.
PROOF:
76/26.198+2/(26.198-6)=3
2.9+2/20.198=3
2.9+.1=3
3=3