Question 476940
Let x = speed in still water.



Now let's form equations for the upstream and downstream journeys


Upstream: 


{{{d=rt}}}



{{{10=(x-5)t}}}



{{{t=10/(x-5)}}}



So the time to travel upstream is given by the equation {{{t=10/(x-5)}}} where 'x' is the speed of the boat in still water.



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Downstream: 


{{{d=rt}}}



{{{10=(x+5)t}}}



{{{t=10/(x+5)}}}



So the time to travel downstream is given by the equation {{{t=10/(x+5)}}} where 'x' is the speed of the boat in still water.



Now add the two time expressions and set them equal to 3 (since the total time was 3 hrs)



{{{10/(x-5)+10/(x+5)=3}}}



{{{10(x+5)+10(x-5)=3(x-5)(x+5)}}} ... Note: I'm multiplying EVERY term by the LCD (x-5)(x+5) to clear out the fractions.



{{{10(x+5)+10(x-5)=3(x^2-25)}}}



{{{10x+50+10x-50=3x^2-75}}}



{{{20x=3x^2-75}}}



{{{0=3x^2-20x-75}}}



Now solve for x:



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*[invoke quadratic_formula 3, -20, -75, "x"]


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Now approximate those answers (by use of a calculator) to get {{{x=9.34259}}} or {{{x=-2.675}}}. Ignore the negative answer (since a negative speed doesn't make sense)



So the speed of the boat in still water is approximately 9.34259 km/h