Let the two-digit number be represented as \(10a + b\), where \(a\) is the tens digit and \(b\) is the units digit. According to the problem, we have the following relationships:
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The sum of the digits is 17: \[ a + b = 17 \]
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When the digits are reversed, the new number exceeds the original number by 9: \[ 10b + a = (10a + b) + 9 \]
Now, we can simplify the second equation: \[ 10b + a = 10a + b + 9 \]
Rearranging this gives: \[ 10b + a - b - 10a = 9 \] \[ 9b - 9a = 9 \] \[ b - a = 1 \]
Now we have a system of two equations:
- \(a + b = 17\)
- \(b - a = 1\)
We can solve this system by first expressing \(b\) in terms of \(a\) using the second equation: \[ b = a + 1 \]
Now substituting \(b\) into the first equation: \[ a + (a + 1) = 17 \] \[ 2a + 1 = 17 \] \[ 2a = 16 \] \[ a = 8 \]
Now substituting \(a\) back into the equation for \(b\): \[ b = 8 + 1 = 9 \]
Thus, the digits are \(a = 8\) and \(b = 9\). The two-digit number is: \[ 10a + b = 10(8) + 9 = 80 + 9 = 89 \]
To verify:
- The sum of the digits \(8 + 9 = 17\) (correct).
- The reversed number is \(98\) and \(98 - 89 = 9\) (also correct).
Therefore, the two-digit number is: \[ \boxed{89} \]