Question 997835
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<b>Shortcut Method</b>


If you reverse the digits of a two-digit number, the difference between the original two digit number and the new two-digit number is ALWAYS a multiple of 9.  And the difference between the two digits is the value of the multiple.  Hence, for your problem, where *[tex \Large x] is the original 10s digit and *[tex \Large y] is the original 1s digit:


*[tex \LARGE \ \ \ \ \ \ \ \ \ \ x\ +\ y\ =\ 7]


And, since the original two-digit number must be smaller than the new two-digit number, meaning *[tex \Large x\ <\ y]


*[tex \LARGE \ \ \ \ \ \ \ \ \ \ x\ +\ 3\ =\ y].


<b>Long Method</b>


Same definitions of *[tex \Large x] and *[tex \Large y]


Original two digit number:  *[tex \Large 10x\ +\ y]


New two digit number:  *[tex \Large 10y\ +\ x]


*[tex \LARGE \ \ \ \ \ \ \ \ \ \ 10x\ +\ y\ +\ 27\ =\ 10y\ +\ x]


*[tex \LARGE \ \ \ \ \ \ \ \ \ \ 9x\ -\ 9y\ =\ -27]


*[tex \LARGE \ \ \ \ \ \ \ \ \ \ x\ -\ y\ =\ -3]


And


*[tex \LARGE \ \ \ \ \ \ \ \ \ \ x\ +\ y\ =\ 7]


Either way, solve the 2X2 system for *[tex \Large x] and *[tex \Large y]


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

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