SOLUTION: Before I begin I thank you so much for your time!! it means a lot to me and you are helping me out of a jam. The problem: Tonya's outboard can drive her boat at 7mph in still wate

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Question 761700: Before I begin I thank you so much for your time!! it means a lot to me and you are helping me out of a jam. The problem: Tonya's outboard can drive her boat at 7mph in still water. it takes her 10 minutes more to reach her friends camp 4 miles up the river than to return to her camp down river. What is the speed of the current?
I set up two different equations, one for coming and one for going. t=time c=current speed
(1)going: 4=(7-c)*(t+10)
(2)returning: 4=(7+c)*t
I rearranged equation (2) to equal: {4/(7+c)}=t
I then plugged this into The first equation to create this: 4=(7-c)*{4/(7+c)+10}
This is where I start getting confused... I dont know how to get the (7+c) out of the denominator. Do I multiply everything by (7+c)? Or just the 4? OR should I not even get it out of the denominator and just multiply everything out? Thanks again! -Abby
Answer by stanbon(75887) (Show Source):
You can put this solution on YOUR website!
Tonya's outboard can drive her boat at 7mph in still water. it takes her 10 minutes more to reach her friends camp 4 miles up the river than to return to her camp down river. What is the speed of the current?
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Upstream DATA:
distance = 4 miles ; rate = 7-c mph ; time = d/r = 4/(7-c) hrs
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Downstream DATA:
distance = 4 miles ; rate = 7+c mph ; time = d/r = 4/(7+c) hrs
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Equation:
down time - up time = 1/6 hr
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4/(7+c) - 4/(7-c) = 1/6
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24(7-c) - 24(7+c) = 7^2-c^2
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-48c = 49 - c^2
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c^2 - 48c - 49 = 0
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(c-49)(c+1) = 0
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Usable answer: c = 49
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Note: that answer doesn't make much sense if the boat's speed
in still water is only 7 mph.
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Cheers,
Stan H.
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