Question 60115
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?
:
Since we are dealing in mph, change 20 min to hrs: 20/60 = 1/3 hr
:
Let s = her speed in still water
Then (s+4) = her speed down-stream
And (s-4) = her speed up-stream
:
Time = dist/speed
:
Time to go upstream - (1/3) hr = time to go down stream 
{{{5/(s-4) - 1/3 = 5/(s+4)}}}
:
Mult equation with the common denominator; 3(x-4)(s+4), resulting in:
3(s+4)(5) - 1(s+4)(s-4) = 3(s-4)(5)
:
3(5s + 20) - 1(s^2 - 16) = 3(5s - 20)
:
15s + 60 - s^2 + 16 = 15s - 60
:
-s^2 + 15s - 15s = -60 - 60 - 16
:
-s^2 = - 136
:
s^2 = 136
:
s = SqRt(136)
:
s = 11.66 mph in still water
:
:
Check using time equation: 
5/7.66 - .33   =  1/15.66
         .65 - .33    =  .32