Question 342957
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You are making this too complex in your setup.  Try it this way:


Let *[tex \Large r] represent the rate of speed on the trip from Denver to SF.  Let *[tex \Large t] represent the amount of elapsed time for the Denver to SF trip.  Then you can say that the return trip was at *[tex \Large r\ +\ 40] mph and took *[tex \Large 9\ -\ t] hours.  Furthermore, since SF and Denver are, relatively speaking, stationary, if the total round trip distance is 1600 miles, it must be 800 miles each way.


Now we can describe the outbound trip (in terms of distance equals rate times time) as:


*[tex \LARGE \ \ \ \ \ \ \ \ \ \ 800 = rt]


And the return trip is:


*[tex \LARGE \ \ \ \ \ \ \ \ \ \ 800 = (r\ +\ 40)(9\ -\ t)]


Solve each of the above for *[tex \Large t] in terms of everything else:


*[tex \LARGE \ \ \ \ \ \ \ \ \ \ t\ =\ \frac{800}{r}]


and


*[tex \LARGE \ \ \ \ \ \ \ \ \ \ 9\ -\ t\ =\ \frac{800}{r\ +\ 40}]


*[tex \LARGE \ \ \ \ \ \ \ \ \ \ t\ =\ \frac{9r\ +\ 360\ -\ 800}{r\ +\ 40}]


*[tex \LARGE \ \ \ \ \ \ \ \ \ \ t\ =\ \frac{9r\ -\ 440}{r\ +\ 40}]


Now you have two expressions both equal to *[tex \Large t].  Set them equal to each other.


*[tex \LARGE \ \ \ \ \ \ \ \ \ \ \frac{800}{r}\ =\ \frac{9r\ -\ 440}{r\ +\ 40}]


Cross-multiply and collect like terms in the LHS, setting everything in the left equal to zero:


*[tex \LARGE \ \ \ \ \ \ \ \ \ \ 9r^2\ -\ 1240r\ -\ 32000\ =\ 0]


Solve the factorable quadratic for *[tex \Large r] for the average rate of speed on the trip to SF.  Obviously, you will discard any negative root.  Add 40 to get the average rate of speed on the return to Denver.


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