Problem:
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?

Solution:

We are given that the sum of the digits of a two-digit number is 9. Let's assume the number has a tens digit 'x' and a units digit 'y'. We can write the equation as:

x + y = 9 ...(1)

We are also given that when the digits are reversed, the number is 63 more than the original. This can be expressed as:

10y + x = 10x + y + 63 ...(2)

Simplifying Equation (2):
To simplify equation (2), we can expand it as:

10y + x = 10x + y + 63
10y - y = 10x - x + 63 (rearranging terms)
9y = 9x + 63 ...(3)

Substituting Equation (1) into Equation (3):
We can substitute equation (1) into equation (3) to eliminate one variable. From equation (1), we have:

x = 9 - y

Substituting this value of x in equation (3), we get:

9y = 9(9 - y) + 63
9y = 81 - 9y + 63
9y + 9y = 81 + 63
18y = 144
y = 144 / 18
y = 8

Substituting the value of y into Equation (1):
Now that we have the value of y, we can substitute it into equation (1) to find the value of x. From equation (1), we have:

x + y = 9
x + 8 = 9
x = 9 - 8
x = 1

So, the tens digit (x) is 1 and the units digit (y) is 8.

The two-digit number:
Therefore, the two-digit number is 18, which satisfies both conditions: the sum of its digits is 9 and when the digits are reversed, it is 63 more than the original number.

18 is the original number.