Question 991189
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Let one of the numbers be *[tex \Large x] and another be *[tex \Large y].  The average of these two numbers is *[tex \Large \frac{x\ +\ y}{2}], so twice the average of these two numbers is *[tex \Large x\ +\ y].  So the sum of the three numbers is:


*[tex \LARGE \ \ \ \ \ \ \ \ \ \  2x\ +\ 2y\ =\ 64]


and


*[tex \LARGE \ \ \ \ \ \ \ \ \ \  x\ -\ y\ =\ 6]


is also true since the choice of *[tex \Large x] or *[tex \Large y] for the larger of the two is arbitrary.


Solve the 2X2 system for *[tex \Large x] and *[tex \Large y], then calculate *[tex \Large x\ +\ y].


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

*[tex \Large \ \
*[tex \LARGE \ \ \ \ \ \ \ \ \ \