Question 206290
I assume when you say 18 mph you mean 18 miles, and 7 mph mean 7 miles.
:
the current in a river moves at 2mph. a boat travels 18 miles upstream
 and 7 miles down stream in a total 7 hours. 
what the speed of the boat in still water?
;
Let s = speed of the boat in still water
then
(s+2) = speed downstream
and
(s-2) = speed upstream
;
Write a time equation: Time = {{{dist/speed}}}
:
Time downstream + time upstream = 7 hrs
{{{18/((s+2))}}} + {{{7/((s-2))}}} = 7
:
Multiply equation by (s+2)(s-2)
(s+2)(s-2)*{{{18/((s+2))}}} + (s+2)(s-2)*{{{7/((s-2))}}} = 7*(s+2)(s-2)
:
Cancel the denominators and you have:
18(s-2) + 7(s+2) = 7*(s^2 - 4)
:
18s - 36 + 7s + 14 = 7s^2 - 28
:
25s - 22 = 7s^2 - 28
:
Arrange as a quadratic equation on the right
0 = 7s^2 - 25s - 28 + 22 
:
7s^2 - 25s - 6 = 0
:
Use the quadratic formula to find s
{{{x = (-b +- sqrt( b^2-4*a*c ))/(2*a) }}}
In this problem x = s; a = 7; b = -25; c = -6
{{{s = (-(-25) +- sqrt(-25^2 - 4 * 7 * -6 ))/(2*7) }}}
:
{{{s = (25 +- sqrt(625 - (-168) ))/14 }}}
:
{{{s = (25 +- sqrt(625 + 168 ))/14 }}}
:
{{{s = (25 +- sqrt(793 ))/14 }}}
The positive solution is what we want here
{{{s = (25 + 28.16)/14 }}}
s = {{{53.16/14}}}
s ~ 3.8 mph in still water
:
:
That is awful slow, let's see if that's true
Upstream time + downstream time =
{{{7/1.8}}} + {{{18/5.8}}} =
3.89 + 3.10 = 6.99 ~ 7 hrs