SOLUTION: When a crew rows with the current, it travels 18 miles in 2 hours. Against the current, the crew rows 5/9 of the distance in 2 hours. Let x=the crew rowing rate in still water and

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Question 766469: When a crew rows with the current, it travels 18 miles in 2 hours. Against the current, the crew rows 5/9 of the distance in 2 hours. Let x=the crew rowing rate in still water and let y=the rate of the current. Find the rate of rowing in still water and the rate of the current.
Found 2 solutions by josmiceli, Alan3354:
Answer by josmiceli(19441)   (Show Source): You can put this solution on YOUR website!
Let = rate of rowing in still water in mi/hr
Let = rate of the current in mi/hr
------------------
Rowing with the current:
(1)
Rowing against the current:
(2)
--------------------------
(2)
and
(1)
-------------------
Add the equations:


Plug this result into (1) or (2)
(2)
(2)
(2)
(2)
(2)
------------
The rate of rowing in still water is 7 mi/hr
The rate of the current is 2 mi/hr
----------------------------
check:
(2)
(2)
(2)
and
(1)
(1)
(1)
OK

Answer by Alan3354(69443)   (Show Source): You can put this solution on YOUR website!
When a crew rows with the current, it travels 18 miles in 2 hours. Against the current, the crew rows 5/9 of the distance in 2 hours. Let x=the crew rowing rate in still water and let y=the rate of the current. Find the rate of rowing in still water and the rate of the current.
-----------------
Find the speed upstream and d/stream.
The crew's speed is the average of the 2.
The current is the difference.

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