The sum of a two-digit number and the number obtained by reversing its digits is 132. If the digits differ by 2, find the number.

Step 1: Set up expressions for the numbers

Let the tens digit be $x$ and the units digit be $y$. The original two-digit number is $10x + y$. When the digits are reversed, the new number formed is $10y + x$.

Step 2: Form the first linear equation

We are given that the sum of the original number and the reversed number is 132. Combining like terms gives $11x + 11y = 132$. Dividing both sides by 11 yields $x + y = 12$.

Step 3: Account for both cases of the digit difference

Since the problem states that the digits differ by 2 without specifying which digit is greater, we must consider two cases: $x - y = 2$ or $y - x = 2$.

Step 4: Solve Case 1 where $x - y = 2$

Adding $x + y = 12$ and $x - y = 2$ eliminates $y$, giving $2x = 14$, so $x = 7$. Substituting $x = 7$ into $x + y = 12$ gives $y = 5$. Thus, the original number is 75.

Step 5: Solve Case 2 where $y - x = 2$

Adding $x + y = 12$ and $y - x = 2$ eliminates $x$, giving $2y = 14$, so $y = 7$. Substituting $y = 7$ into $x + y = 12$ gives $x = 5$. Thus, the original number is 57.

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