Question 366138
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Remember *[tex \Large d\ =\ rt]


Let *[tex \Large r] represent Larry's rate and *[tex \Large t] represent Larry's time.  Then *[tex \Large r\ +\ 2] represents Terrell's rate and *[tex \Large t\ -\ 1] represents Terrell's time.


Larry's trip:


*[tex \LARGE \ \ \ \ \ \ \ \ \ \ 195\ =\ rt]


which can be expressed:


*[tex \LARGE \ \ \ \ \ \ \ \ \ \ t\ =\ \frac{195}{r}]


And Terrell's trip:


*[tex \LARGE \ \ \ \ \ \ \ \ \ \ 134\ =\ (r\ +\ 2)(t\ -\ 1)]


which can be expressed:


*[tex \LARGE \ \ \ \ \ \ \ \ \ \ t\ -\ 1\ =\ \frac{134}{r\ +\ 2}]


and simplified to:


*[tex \LARGE \ \ \ \ \ \ \ \ \ \ t\ =\ \frac{136\ +\ r}{r\ +\ 2}]


Since we know *[tex \Large t\ =\ t] we can now write:


*[tex \LARGE \ \ \ \ \ \ \ \ \ \ \frac{195}{r}\ =\ \frac{136\ +\ r}{r\ +\ 2}]


Cross-multiplying and collecting terms results in:


*[tex \LARGE \ \ \ \ \ \ \ \ \ \ r^2\ -\ 59r\ -\ 390\ =\ 0]


Solve for *[tex \Large r] and exclude the negative root -- unless you think the two of them were going back in time.  *[tex \Large r] is Larry's rate.  Add 2 to get Terrell's rate.


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