Question 859630
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Let *[tex \LARGE x] represent the 10s digit and *[tex \LARGE y] represent the 1s digit.  We can then represent the original number by *[tex \LARGE 10x\ +\ y] and we can represent the number with the digits reversed by *[tex \LARGE 10y\ +\ x].  We know the following facts:


*[tex \LARGE \ \ \ \ \ \ \ \ \ \ x\ +\ y\ =\ 8]


and


*[tex \LARGE \ \ \ \ \ \ \ \ \ \ 10x\ +\ y\ =\ 10y\ +\ x\ -\ 18]


Simplifying the second equation above yields:


*[tex \LARGE \ \ \ \ \ \ \ \ \ \ x\ -\ y\ =\ -2]


Solve the 2X2 system:


*[tex \LARGE \ \ \ \ \ \ \ \ \ \ x\ +\ y\ =\ 8]


*[tex \LARGE \ \ \ \ \ \ \ \ \ \ x\ -\ y\ =\ -2]


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