Question 884545
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Let *[tex \Large x] represent the 10s digit and let *[tex \Large y] represent the 1s digit.


We are given that *[tex \Large x\ +\ y\ =\ 14]


If we subtract 29 from the number, the 1s digit of the result will be one larger than the 1s digit of the original number and the 10s digit will be 3 smaller (3, not 2 because of the borrow that must occur).  Hence we have the relationship: *[tex \Large x\ -\ 3\ =\ y\ +\ 1] since the digits of the new number must be equal.  Rearranging we get:  *[tex \Large x\ -\ y\ =\ 4]


Solve the 2X2 system to find the two digits of the original number.


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