Question 426006
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There are two ways to go about this.  The direct way goes right at the quantity we want to determine, but requires more and more difficult calculations.  We'll look at this way first.


Let *[tex \Large d] represent the distance between home and the breakdown point.  Let *[tex \Large t] represent the amount of time spent driving.  And then *[tex \Large 2.25\ -\ t] must represent the amount of time spent walking.


We know that distance equals rate times time.  This implies that time is equal to distance divided by rate.


For the outbound, or driving, part of the trip, we can describe the situation thus:


*[tex \LARGE \ \ \ \ \ \ \ \ \ \ t\ =\ \frac{d}{40}]


And the return, or walking, part of the trip, the situation is:


*[tex \LARGE \ \ \ \ \ \ \ \ \ \ 2.25\ -\ t\ =\ \frac{d}{5}]


Manipulating the second equation:


*[tex \LARGE \ \ \ \ \ \ \ \ \ \ t\ =\ 2.25\ -\ \frac{d}{5}]


*[tex \LARGE \ \ \ \ \ \ \ \ \ \ t\ =\ \frac{11.25\ -\ d}{5}]


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


*[tex \LARGE \ \ \ \ \ \ \ \ \ \ \frac{d}{40}\ =\ \frac{11.25\ -\ d}{5}]


Now all you have to do is cross-multiply, collect like terms, and solve for *[tex \Large d], the requested quantity.


The indirect method is a little simpler on the arithmetic:


Use distance equals rate times time to create two equations:


*[tex \LARGE \ \ \ \ \ \ \ \ \ \ d\ =\ 40t]


*[tex \LARGE \ \ \ \ \ \ \ \ \ \ d\ =\ 5(2.25\ -\ t)\ =\ 11.25\ -\ 5t]


Since *[tex \Large d\ =\ d], set the two RHSs equal:


*[tex \LARGE \ \ \ \ \ \ \ \ \ \ 40t\ =\ 11.25\ -\ 5t]


Solve for *[tex \Large t].  Once you have *[tex \Large t], multiply times 40 to determine *[tex \Large d]



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