Age Problem: The sum of the ages of the father and his son is 99

Let the father's age be 10x+y, with x and y being the digits of his age. (Reversing digits you get 10y+x).
Let the son's age be 10a+b, with a an b being the digits of his age. (Reversing digits you get 10b+a).
The clues give us the 3 equations below:
(1) 10a+b + 10x+y=99, or son's age + 10x + y = 99,
(2) son's age + x +10y = 72, and
(3) father's age - reversed son's age = 22, or. -10a-b +10x+y = 22
Subtracting the second equation from the first, we get
9x - 9y = 27. ---> 9(x-y)=27 ---> x-y=3
Subtracting equation (3) from egquation (1), we get
11a+11b=77 ---> 11(a+b)=77 ---> a+b=7
The son's age is not a multiple of 9.
If you divide the son's age by 9,
you get a remainder of 7.
How do I know?
When you divide a number by 9,
the remainder is the sum of the digits,
or the sum of the digits of the sum of the digits,
or ... (keep adding until you get a single digit),
and if you get 9, then the number was divisible by 9.
Since the sum of the ages is a multiple of 9,
when you divide the father's age by 9, the remainder should be 9-7=2.
The sum of the digits of the father's age cannot be 2,
and it cannot be 20 either,
so it must be x+y=11.
Along with x-y=3, that tells you that
x=7 and y=4,
so the father is 74.
We can re-write equation (3) as
74 - reversed son's age = 22, so
74-22 = reversed son's age, or
reversed son's age = 52.
So, son's age = 25.

81 16?