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Question 61280: To be quite honest I am not sure what the teacher is wanting here. We were on Algebra Linear equations and he threw these word problems in as credit. Your help will be very much appreciated. We are supposed to solve the problem. I don't even know where to start. Please help.
A river boat can head 340 miles up-river in 19 hours, but the return trip takes only 14 hours. Find the current of the river and find the speed of the ship in still water to the nearest tenth of a mph.
Thank You very much for your time and efforts. I really appreciate it!
Answer by ankor@dixie-net.com(22740)   (Show Source):
You can put this solution on YOUR website! A river boat can head 340 miles up-river in 19 hours, but the return trip takes only 14 hours. Find the current of the river and find the speed of the ship in still water to the nearest tenth of a mph.
:
one method is to make the problem into two equations with two unknowns and solve by elimination:
:
Let x = boat speed in still water, let y = speed of the current
:
Upriver equation:
Speed up the river = (x-y)
Remember, dist = time * speed
:
19(x-y) = 340
19x - 19y = 340
:
:
Down river equation:
Speed down river = (x+y)
:
14(x+y) = 340
14x + 14y = 340
:
:
Mult "Up" eq by 14 and the "Down" equation by 19 and you have:
266x - 266y = 4760
266x + 266y = 6460
--------------------- add
532x + 0 = 11220
x = 11220/532
x = 21.1 mph speed of the boat
:
Find current (use the "down" equation:
14(21.1)+ 14y = 340
:
294.4 + 14y = 340
14y = 340 - 295.4
y = 44.6/14
y = 3.2 mph speed of the current
:
:
Check our solutions in the "up" equation
19(21.1) - 19(3.2) =
400.9 - 60.8 = 340.1 close enough
:
Make sense to you? Any questions?
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