Question 277019
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Let *[tex \Large r] represent the speed in still water.  Then let *[tex \Large t] represent the elapsed time for the upstream trip.  The upstream speed is then *[tex \Large r\ -\ 2], the downstream speed is *[tex \Large r\ +\ 2], and the downstream time is *[tex \Large 4\ -\ t]


Using the relationship of distance equals rate times time, we can describe the upstream trip:


*[tex \LARGE \ \ \ \ \ \ \ \ \ \ 15\ =\ (r\ -\ 2)t]


and the downstream trip is:


*[tex \LARGE \ \ \ \ \ \ \ \ \ \ 15\ =\ (r\ +\ 2)(4\ - \t)]


Divide both sides of both equations by the rate expression and solve for *[tex \Large t]:


*[tex \LARGE \ \ \ \ \ \ \ \ \ \ \frac{15}{r\ -\ 2}\ =\ t]


*[tex \LARGE \ \ \ \ \ \ \ \ \ \ 4\ -\ \frac{15}{r\ +\ 2}\ =\ t]


Now we have two expressions equal to *[tex \Large t], set them equal to each other:


*[tex \LARGE \ \ \ \ \ \ \ \ \ \ 4\ -\ \frac{15}{r\ +\ 2}\ =\ \frac{15}{r\ -\ 2}]


Apply LCD {*[tex \Large \left(r^2\,-\,4\right)]}:


*[tex \LARGE \ \ \ \ \ \ \ \ \ \ \frac{4\left(r^2\,-\,4\right)\ -\ 15\left(r\,+\,2\right)}{r^2\,-\,4}\ =\ \frac{15\left(r\,-\,2\right)}{r^2\,-\,4}]


*[tex \LARGE \ \ \ \ \ \ \ \ \ \ \frac{4\left(r^2\,-\,4\right)\ -\ 15\left(r\,+\,2\right)\ -\ 15\left(r\,-\,2\right)}{r^2\,-\,4}\ =\ 0]


Expand and collect like terms:


*[tex \LARGE \ \ \ \ \ \ \ \ \ \ \frac{4r^2\ -\ 16\ -\ 15r\ + 30\ -\ 15r\ -\ 30}{r^2\ -\ 4}\ =\ 0]


*[tex \LARGE \ \ \ \ \ \ \ \ \ \ \frac{4r^2\ -\ 30r\ -\ 16}{r^2\ -\ 4}\ =\ 0]


Multiply both sides by *[tex \Large \left(r^2\,-\,4\right)]


*[tex \LARGE \ \ \ \ \ \ \ \ \ \ 4r^2\ -\ 30r\ -\ 16\ =\ 0]


Finally, solve the factorable quadratic.  Exclude the negative root (she wasn't padding backwards, after all).  The positive root is the speed in still water.


John
*[tex \LARGE e^{i\pi} + 1 = 0]
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