Final answer:
The original two-digit number with digits summing to 12 and increasing by 18 when reversed is 57.
Step-by-step explanation:
The question asks us to find a two-digit number, the sum of whose digits is 12, and which becomes a number 18 greater when its digits are reversed. Let's denote the tens digit as 'x' and the units digit as 'y'. Then, the original number can be written as 10x + y.
The conditions given to us are:
- The sum of the digits is 12, which means x + y = 12.
- Reversing the digits gives a number that is 18 more than the original number, so 10y + x = 10x + y + 18.
Let's solve the system of equations formed by these two conditions:
- From the first condition, x + y = 12.
- From the second condition, after simplifying, 10y + x = (10x + y) + 18 becomes 9y - 9x = 18, which simplifies to y - x = 2.
Now we have a new system of equations:
- x + y = 12
- y - x = 2
After solving the system, we get x = 5 and y = 7. Therefore, the original number is 57.