The sum of the digits of a two digit number is 9. If the digits are reversed,the number is 63 more than the original. Find the number.

Let the units digit be x.
Then the tens digit is 9 - x.
Therefore the original number is 10(9 - x) + x = 90 - 10x + x = 90 - 9x.
On reversing the order of the digits the number obtained is 10x + 9 - x = 9x + 9.
By the given condition, 9x + 9 = 63 + (90 - 9x)
(9x + 9) - (90 - 9x) = 63
9x + 9 - 90 + 9x = 63
18x = 144 (Transposing 9 and -90 to the R.H.S.)
x = 8
Therefore the original number is 90 - 9× 8 = 90 - 72 = 18.