SOLUTION: A boat traveled 198 miles downstream and back. The trip downstream took 11 hours. The trip back took 33 hours. What is the speed of the boat in still water? I would really like to

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Question 880679: A boat traveled 198 miles downstream and back. The trip downstream took 11 hours. The trip back took 33 hours. What is the speed of the boat in still water? I would really like to know how to work this problem out.
Found 2 solutions by josgarithmetic, ankor@dixie-net.com:
Answer by josgarithmetic(39625) About Me  (Show Source):
You can put this solution on YOUR website!
RT=D rate time distance.

r = boat speed without a current
c = speed of current

__________________speed______________time_________distance
DOWN______________r+c________________11___________198
UP________________r-c________________33___________198

Travel uniform rates law allows the system:
highlight_green%28%28r%2Bc%29%2A11=198%29 and highlight_green%28%28r-c%29%2A33=198%29.
Simple linear system in variables r and c.
Solve for r and c.

Answer by ankor@dixie-net.com(22740) About Me  (Show Source):
You can put this solution on YOUR website!
A boat traveled 198 miles downstream and back.
The trip downstream took 11 hours.
The trip back took 33 hours.
What is the speed of the boat in still water?
:
Let s = boat speed in still water
Let c = rate of the current
then
(s+c) = effective speed down-stream
and
(s-c) = effective speed up-stream
:
Write a distance equation for each way; dist = time * speed
11(s+c) = 198
33(s-c) = 198
We can simplify this, divide the 1st equation by 11, the 2nd equation by 33
s + c = 18
s - c = 6
-----------addition eliminates c, find s
2s = 24
s = 24/2
s = 12 mph in still water
:
:
You can confirm this, find the current then substitute in the original equations