The sum of the digits of a two-digit number is 15 and the difference between…

For any two-digit number, if the tens digit is x and the units digit is y, the number can be written as 10x + y. When a problem gives the sum and the difference of the two digits, these translate into a pair of linear equations in x and y, solvable by the elimination (addition) method: adding the two equations eliminates y and directly gives 2x, then y follows by back-substitution.

1. Let the tens digit be x and the units digit be y, so the number is 10x + y.

2. From the digit sum: x + y = 15 …(1)

3. From the digit difference (tens digit greater): x − y = 3 …(2)

4. Add equations (1) and (2): 2x = 18 ⇒ x = 9.

5. Substitute x = 9 into equation (1): y = 15 − 9 = 6.

6. So the number is 10x + y = 10×9 + 6 = 96.

Check: 9 + 6 = 15 (digit-sum condition holds) and 9 − 6 = 3 (digit-difference condition holds), with the tens digit (9) greater than the units digit (6) as required — both conditions hold for 96.

Hence, the required two-digit number is 96.