Question 348296
<font face=”Garamond” size=”+2”>


Let *[tex \Large r] represent the rate of the faster train.  Then the speed of the slower train is *[tex \Large r\ -\ 10].  It is true that *[tex \Large r\ =\ \frac{d}{t}] but that is not a helpful relation for this particular problem.  What we will find to be more helpful is that *[tex \Large t\ =\ \frac{d}{r}].


The slower train's 210 mile trip can be described by:


*[tex \LARGE\ \ \ \ \ \ \ \ \ \ t\ =\ \frac{210}{r\ -\ 10}]


And the faster train's 260 mile trip can be described by:


*[tex \LARGE\ \ \ \ \ \ \ \ \ \ t\ =\ \frac{260}{r}]


Note that *[tex \Large t\ =\ t], so we can write:


*[tex \LARGE\ \ \ \ \ \ \ \ \ \ \frac{260}{r}\ =\ \frac{210}{r\ -\ 10}]


Now all you need to do is cross multiply and solve your linear equation in *[tex \Large r].  Then *[tex \Large r\ -\ 10] follows directly.


John
*[tex \LARGE e^{i\pi} + 1 = 0]
My calculator said it, I believe it, that settles it
<div style="text-align:center"><a href="http://outcampaign.org/" target="_blank"><img src="http://cdn.cloudfiles.mosso.com/c116811/scarlet_A.png" border="0" alt="The Out Campaign: Scarlet Letter of Atheism" width="143" height="122" /></a></div>
</font>