Question 517775
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Let *[tex \Large r] represent the usual speed.  Let *[tex \Large t] represent the usual amount of time.  Then *[tex \Large r\ +\ 9] represents the increased speed and *[tex \Large t\ -\ 0.5] represents the reduced time.


Time is equal to distance divided by rate so for the normal trip:


*[tex \LARGE \ \ \ \ \ \ \ \ \ \ t\ =\ \frac{170}{r}]


And for the 'catch-up' trip:


*[tex \LARGE \ \ \ \ \ \ \ \ \ \ t\ -\ 0.5\ =\ \frac{170}{r\ +\ 9}]


Which is to say:


*[tex \LARGE \ \ \ \ \ \ \ \ \ \ t\ =\ \frac{170}{r\ +\ 9}\ +\ 0.5]


A little Algebra Music, Maestro:


*[tex \LARGE \ \ \ \ \ \ \ \ \ \ t\ =\ \frac{170\ +\ 0.5(r\ +\ 9)}{r\ +\ 9}]


*[tex \LARGE \ \ \ \ \ \ \ \ \ \ t\ =\ \frac{174.5\ +\ 0.5r}{r\ +\ 9}]  


Now that we have two expression that are each equal to *[tex \Large t], set them equal:


*[tex \LARGE \ \ \ \ \ \ \ \ \ \ \frac{170}{r}\ =\ \frac{174.5\ +\ 0.5r}{r\ +\ 9}]


All that is left is to cross multiply, collect terms, and solve for *[tex \Large r]


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