Question 992372
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If the digits of a two digit number are reversed and the result is a larger number (..., the number is increased...), then the tens digit of the original number must be smaller than the ones digit.  Furthermore, if the difference between the original number and the reversed digits number is 9, then the difference between the two digits is 1.  (The difference between the original number and the reversed digits number is always a multiple of 9, and the multiplier is the difference between the digits. E.g. 47 reversed is 74 and these two numbers differ by 27 which is 3 times 9, whereas 4 and 7 differ by 3).


Hence we begin with, assuming *[tex \Large y] represents the ones digit and *[tex \Large x] represents the tens digit:


*[tex \LARGE \ \ \ \ \ \ \ \ \ \ y\ -\ x\ =\ 1]


Which is to say:


*[tex \LARGE \ \ \ \ \ \ \ \ \ \ y\ =\ x\ +\ 1]


And then we are given that:


*[tex \LARGE \ \ \ \ \ \ \ \ \ \ 4y\ =\ 3x\ +\ 5]


Solve the 2X2 system by substitution.


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

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