This layout helps us see which digits to add together. The 7 and 5 are in the ones column, and the 2 and 4 are in the tens column.

Since \(7 + 5 = 12\), and 12 is more than 10, we need to exchange. We write the 2 in the ones place and carry the 1 ten.

The small 1 reminds us that we have an extra ten to add. This came from exchanging 10 ones for 1 ten.

Don’t forget to add the exchanged 1! We have \(1 + 2 + 4 = 7\) tens, so we write 7 in the tens place.

Our final answer is 72. We can check by estimating: 27 is about 30 and 45 is about 50, so \(30 + 50 = 80\). Our answer of 72 is close to 80, so it’s reasonable!

Notice what’s changed and what stayed the same: The first number is still 27 (no change), and the tens digit of the second number is still 4 (no change). The only thing that changed is the ones digit of the second number—it increased from 5 to 6.

This time \(7 + 6 = 13\), which is different from the 12 we got in Example 1. We still need to exchange because 13 is more than 10. We write 3 in the ones place and carry 1 ten.

The exchange process is exactly the same as before, even though our ones total is different (13 instead of 12).

Notice that the tens calculation is the same as in Example 1 because we’re still adding \(1 + 2 + 4\).

Compare the two answers: In Example 1, we got \(27 + 45 = 72\). In Example 2, we got \(27 + 46 = 73\). When the ones digit of the second number increased by 1 (from 5 to 6), our final answer also increased by 1 (from 72 to 73). Even though the ones totals were different (12 vs 13), the exchange process worked exactly the same way in both examples.