Question 646262
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Let *[tex \LARGE r] represent the air speed and let *[tex \LARGE d] represent the distance to the town.  Also note that 20 minutes is one-third of an hour and 24 minutes is two-fifths of an hour.


We know that distance equals rate times time, so distance divided by rate must be time.  When flying into the wind, ground speed will be air speed MINUS wind speed, hence the against the wind trip is described by:


*[tex \LARGE \ \ \ \ \ \ \ \ \ \ \frac{d}{r\ -\ 15}\ =\ \frac{7}{3}]


And the downwind trip is then:


*[tex \LARGE \ \ \ \ \ \ \ \ \ \ \frac{d}{r\ +\ 15}\ =\ \frac{7}{5}]


Cross-multiplying both we get the two-variable system:


*[tex \LARGE \ \ \ \ \ \ \ \ \ \ 3d\ =\ 7(r\ -\ 15)]


*[tex \LARGE \ \ \ \ \ \ \ \ \ \ 5d\ =\ 7(r\ +\ 15)]


This is easily solved by Elimination.  Distribute the 7s across the parentheses.  Multiply the first equation by -1.  Add the two equations to eliminate the *[tex \LARGE r] variable.  Solve for *[tex \LARGE d].  Once you have the value for *[tex \LARGE d] substitute back into either equation and then solve for *[tex \LARGE r].  Et Voilą!


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