Question 521603
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Let *[tex \Large s] represent Susan's age now.  Let *[tex \Large m] represent Mary's age now.


We know that


*[tex \LARGE \ \ \ \ \ \ \ \ \ \ s\ =\ m\ +\ 6]


Susan was as old as Mary is now 6 years ago, so Susan's age when she was as old as Mary is now is *[tex \Large s\ -\ 6] which we can see is the same as *[tex \Large m].  6 years ago Mary was *[tex \Large m\ -\ 6].  So Susan's age 6 years ago was twice Mary's age 6 years ago:


*[tex \LARGE \ \ \ \ \ \ \ \ \ \ m\ =\ 2(m\ -\ 6)]


Solve for *[tex \Large m], then calculate *[tex \Large m\ +\ 6]


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