Question 513707
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Time equals distance divided by rate.  The downstream rate is *[tex \Large 6\ +\ r_c] where *[tex \Large r_c] is the speed of the current.  The upstream rate is *[tex \Large 6\ -\ r_c].  If we let *[tex \Large t] represent the amount of time to go downstream, then *[tex \Large 9\ -\ t] must represent the amount of time to go upstream.


Downstream trip:


*[tex \LARGE \ \ \ \ \ \ \ \ \ \ \frac{15}{6\ +\ r_c}\ =\ t]


Upstream trip:


*[tex \LARGE \ \ \ \ \ \ \ \ \ \ \frac{15}{6\ -\ r_c}\ =\ 9\ -\ t]


Substituting:


*[tex \LARGE \ \ \ \ \ \ \ \ \ \ \frac{15}{6\ -\ r_c}\ =\ 9\ -\ \frac{15}{6\ +\ r_c}] 


Then:


*[tex \LARGE \ \ \ \ \ \ \ \ \ \ \frac{15}{6\ -\ r_c}\ +\ \frac{15}{6\ +\ r_c}\ =\ 9]


The lowest common denominator is the product of the two conjugate denominators, hence the difference of two squares:


*[tex \LARGE \ \ \ \ \ \ \ \ \ \ 15(6\ +\ r_c)\ +\ 15(6\ -\ r_c)\ =\ 9(36\ -\ {r_c}^2)]


A little Algebra Music, Maestro (verification left as an exercise for the student):


*[tex \LARGE \ \ \ \ \ \ \ \ \ \ 9{r_c}^2\ =\ 144]


Solve for *[tex \Large r_c].  Discard the negative root.


John
*[tex \LARGE e^{i\pi} + 1 = 0]
My calculator said it, I believe it, that settles it
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