Question 1117309
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Let *[tex \Large x] be the number of years ago the brother was twice as old as the sister.  Since the brother is 36 now, he was *[tex \Large 36\ -\ x] then.  Likewise, the sister was *[tex \Large 29\ -\ x] then.  Also, *[tex \Large x] years ago two times her age was his age:


*[tex \LARGE \ \ \ \ \ \ \ \ \ \ 2(29\ -\ x)\ =\ 36\ -\ x]


Solve for *[tex \Large x]


John
*[tex \LARGE e^{i\pi}\ +\ 1\ =\ 0]
My calculator said it, I believe it, that settles it
<img src="http://c0rk.blogs.com/gr0undzer0/darwin-fish.jpg">
*[tex \Large \ \
*[tex \LARGE \ \ \ \ \ \ \ \ \ \  

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