Question 304605
Debbie traveled by boat 5 miles upstream to fish in her favorite spot.
 Because of the 4 mph current, it took her 20 minutes longer to get there
 than to return. 
How fast will her boat go in still water?
:
Let s = boat speed in still water
then
(s-4) = speed upstream
and
(s+4) = speed down stream
:
Convert 20 min to hrs: 20/60 = {{{1/3}}}hr
:
Write a time equation: time = {{{distance/speed}}}

Upstream time = downstream time + 20 minutes (1/3 hr)
{{{5/((s-4))}}} = {{{5/((s+4))}}} + {{{1/3}}}
:
To clear out these denominators, multiply equation by 3(s+4)(s-4)
3(s+4)(s-4)*{{{5/((s-4))}}} = 3(s+4)(s-4)*{{{5/((s+4))}}} + {{{1/3}}}*3(s+4)(s-4)
Results in
:
15(s+4) = 15(s-4) + (s+4)(s-4)
:
15s + 60 = 15s - 60 + s^2 - 16
:
0 = 15s - 15s - 60 - 60 + s^2 - 16
:
0 = s^2 - 136
:
s^2 = 136
s = {{{sqrt(136)}}}
s = 11.66 mph speed in still water
:
:
Check solution on calc: (speed up = 7.66 and speed down = 15.66)
5/7.66 - 5/15.66 = .3333 which is 1/3 of an hr