SOLUTION: I am not sure how to set this problem up. Can you help? You travel 5 miles in a boat. because of a 4 mph current it took you 20 minutes longer to go then return. How fast will yo

Question 193861: I am not sure how to set this problem up. Can you help?
You travel 5 miles in a boat. because of a 4 mph current it took you 20 minutes longer to go then return. How fast will you go without current?
Is the problems set up like-
10x+100=x^2+20x
Answer by josmiceli(19441) (Show Source): You can put this solution on YOUR website!
The equation is
distance = rate x time but with a modification:
Let = rate of the boat in still water in mi/hr
Let = rate of the current in mi/hr
Let = time going upstream in hrs
Let = time going downstream in hrs
= rate of boat against the current in mi/hr
= rate of the boat going with the current in mi/hr
-------------------
So, going upstream, I can write:
(1)
And going downstream, I can write:
(2)
-------------------
Given:
mi
hrs
mi/hr
-------------------
Now I can rewrite (1) and (2)
(1)
(2)
This is 2 equation with 2 unknowns, so it's solvable
(1)
(1)
(1)
(1)
And, from (2)
(2)
(2)
(2)
(2)
substitute this in (1)
(1)
(1)
(1)
(1)
Using quadratic equation:

hrs
I want to find
(2)
(2)
(2)
(2) mi/hr answer
check answer:
(1)
(1)
(1)
(1) (rounding error?)
and
(2)
(2)
(2)
(2)
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