SOLUTION: It takes a rowing crew 1 hour and 30 minutes longer to go 12 miles up a river than to return. Find the speed of the rowing crew in still water if the speed of the current is 2 mph
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Question 761466: It takes a rowing crew 1 hour and 30 minutes longer to go 12 miles up a river than to return. Find the speed of the rowing crew in still water if the speed of the current is 2 mph.
I set the problem up ( s= speed and t= time)
Going up river: 12= (s+ 2)t
I then know that t= 12/(s+2)
Going down river: 12= (s-2)(t+1.5)
So making the substitution for t:
Going down river 12=(s-2)((12/(s+2)) + 1.5)
But this is where I run into problems. How do I solve this algebra problem?
Thanks
You can put this solution on YOUR website! You have the speed upriver as s + 2. That's not right. When you go up or against the current the net speed is the difference or s - 2.
Then we have for upriver
(1) 12 = (s - 2)*t or
(2) t = 12/(s-2)
and for down river (with the current) we get
(3) 12 = (s + 2)*(t-1.5)
Substituting (2) into (3) gives
(4) 12 = (s+2)*(12/(s-2)-1.5) or
(5) 12 = (s+2)*((12-1.5*(s-2))/(s-2)
(6) 12*(s-2) = (s+2)*(12-1.5*s+3) or
(7) 12*(s-2) = (s+2)*(15-1.5s) or
(8) 12*(s-2) = -1.5*(s-10)*(s+2) or
(9) -8*(s-2) = (s-10)*(s+2) or
(9) -8s + 16 = s^2 -8s -20 or
(10) 16 + 20 = s^2 or
(11) s^2 = 36 or
(12) s = 6
Use (12) in (2) gives
(13) t = 12/(6-2) or
(14) t = 3
Let's check this with (3).
Is (12 = (6+2)*(3-1.5))?
Is (12 = 8*1.5)?
Is (12 = 12)? Yes
Answer: The crew is rowing at 6mph.
PS Your equation is satisfied for s = 6, so it is OK, in spite of my comments above. So just multiply everything out as I did and you'll get the same answer.
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