Question 357470
When the digits of a two-digit number are reversed, the new number is 9 more
than the original number, and the sum of the digits of the original number is 13. 
What is the original number?
:
x = the 10s digit, y = the units
then
10x + y = the original number
and
10y + x = the reversed number
:
Rev number = orig number + 9
10y + x = 10x + y + 9
10y - y = 10x - x + 9
9y = 9x + 9
simplify, divide by 9
y = x + 1
:
"the sum of the digits of the original number is 13. "
(kind of silly, the sum of the digits is the same original or reversed, right?"
anyway
x + y = 13
:
Rearrange the y = x + 1  and add to the above equation
-x + y = 1
+x + y = 13
------------addition eliminates x, find y
2y = 14
y = 7
:
I'll let you find x, and check it in the 1st equation