Let the original two-digit number be represented as $10a + b$, where $a$ is the tens digit and $b$ is the ones digit.

Step 1: Set up the equations

1. Sum of the digits:
   The sum of the digits is given as 8, so: $a + b = 8$

2. Reversed number condition:
   When the digits are reversed, the new number is $10b + a$, and this number is 18 greater than the original number. So: $10b + a = (10a + b) + 18$ Simplifying this equation: $10b + a = 10a + b + 18$ Rearranging terms: $9b - 9a = 18$ Dividing both sides by 9: $b - a = 2$

Step 2: Solve the system of equations

We now have two equations:

1. $a + b = 8$
2. $b - a = 2$

Add these two equations together: $(a + b) + (b - a) = 8 + 2$ Simplifying: $2b = 10$ $b = 5$

Now, substitute $b = 5$ into the first equation: $a + 5 = 8$ $a = 3$

Step 3: Find the original number

The original number is $10a + b = 10(3) + 5 = 35$.

Final Answer:

The original number is $\boxed{35}$.

Do you want more details or have any questions?

Here are 5 related questions you might explore:

1. What would the original number be if the digits were reversed and the new number was 27 greater?
2. How would you solve for a number if the sum of the digits was 10 instead of 8?
3. How can you determine the general formula for such problems with different sums and differences?
4. What happens if the number is three digits instead of two, and the same conditions apply?
5. How would this change if the difference between the reversed and original number was 9?

Tip: Always express two-digit numbers as $10a + b$ where $a$ is the tens digit and $b$ is the ones digit to simplify problems like this.