Question 229903
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Let *[tex \Large x] represent the 10s digit of the original number and let *[tex \Large y] represent the 1s digit.

The sum of the digits is 9:  *[tex \LARGE x + y = 9 \rightarrow\ x = 9 - y]

The value of the original number:  *[tex \LARGE 10x + y]

Double the original number:  *[tex \LARGE 2(10x + y)]

Less 36: *[tex \LARGE 2(10x+y)-36]

The number with the digits reversed:  *[tex \LARGE 10y + x]

So:

*[tex \LARGE \ \ \ \ \ \ \ \ \ \2(10x+y)-36\ =\ 10y+x]

Substituting:

*[tex \LARGE \ \ \ \ \ \ \ \ \ \2(10(9-y)+y)-36\ =\ 10y+(9-y)]

Solve for *[tex \LARGE y] and then solve *[tex \LARGE x + y = 9] for *[tex \LARGE x]

John
*[tex \LARGE e^{i\pi} + 1 = 0]
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