Let's call the two-digit number "AB," where A represents the tens digit, and B represents the ones digit.
According to the problem:
1. The sum of the digits is 6, so we have the equation:
A + B = 6
2. If the digits are reversed, the new number is 36 greater than the original number. This can be expressed as:
10B + A = 10A + B + 36
Now, we can solve this system of equations. First, let's simplify the second equation:
10B + A = 10A + B + 36
Now, let's isolate one of the variables from the first equation. We'll solve for A in terms of B:
A = 6 - B
Now, substitute this expression for A into the second equation:
10B + (6 - B) = 10(6 - B) + B + 36
Now, simplify and solve for B:
10B + 6 - B = 60 - 10B + B + 36
Combine like terms:
9B + 6 = 60 - 36
9B + 6 = 24
Subtract 6 from both sides:
9B = 24 - 6
9B = 18
Now, divide by 9:
B = 18 / 9
B = 2
Now that we've found B, we can find A using the first equation:
A + 2 = 6
Subtract 2 from both sides:
A = 6 - 2
A = 4
So, the tens digit (A) is 4, and the ones digit (B) is 2. Therefore, the two-digit number is 42.