Math - A number contains two digit whose sum is 5 If we add 9 to number its digit are reversed find?

The number is 23

Wanted: A number n consisting of two digits a and b, so that
n = a*10^1 + b*10^0 = 10a + b..

Conditions given:
I) a + b = 5 ; implies a = 5 - b

II) 10a + b + 9 = 10b + a ;simplify that
9a - 9b + 9 = 0

Substitute a by (5 - b) in II):
9(5 - b) - 9b + 9 = 0
45 - 9b - 9b + 9 = 0
54 - 18b = 0
18b = 54
b = 3

Insert b = 3 in I):
a + 3 = 5
a = 2

So, 23 is your number.