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Next we will model adding two digit numbers.
Model the addition: $17+26.$
$17+26$ means the sum of 17 and 26.
Model the 17.  1 ten and 7 ones  
Model the 26.  2 tens and 6 ones  
Combine.  3 tens and 13 ones  
Exchange 10 ones for 1 ten.  4 tens and 3 ones
$40+3=43$ 

We have shown that $17+26=43$ 
Now that we have used models to add numbers, we can move on to adding without models. Before we do that, make sure you know all the one digit addition facts. You will need to use these number facts when you add larger numbers.
Imagine filling in [link] by adding each row number along the left side to each column number across the top. Make sure that you get each sum shown. If you have trouble, model it. It is important that you memorize any number facts you do not already know so that you can quickly and reliably use the number facts when you add larger numbers.
+  0  1  2  3  4  5  6  7  8  9 

0  0  1  2  3  4  5  6  7  8  9 
1  1  2  3  4  5  6  7  8  9  10 
2  2  3  4  5  6  7  8  9  10  11 
3  3  4  5  6  7  8  9  10  11  12 
4  4  5  6  7  8  9  10  11  12  13 
5  5  6  7  8  9  10  11  12  13  14 
6  6  7  8  9  10  11  12  13  14  15 
7  7  8  9  10  11  12  13  14  15  16 
8  8  9  10  11  12  13  14  15  16  17 
9  9  10  11  12  13  14  15  16  17  18 
Did you notice what happens when you add zero to a number? The sum of any number and zero is the number itself. We call this the Identity Property of Addition. Zero is called the additive identity.
The sum of any number $a$ and $0$ is the number.
Find each sum:
ⓐ The first addend is zero. The sum of any number and zero is the number.  $0+11=11$ 
ⓑ The second addend is zero. The sum of any number and zero is the number.  $42+0=42$ 
Look at the pairs of sums.
$2+3=5$  $3+2=5$ 
$4+7=11$  $7+4=11$ 
$8+9=17$  $9+8=17$ 
Notice that when the order of the addends is reversed, the sum does not change. This property is called the Commutative Property of Addition, which states that changing the order of the addends does not change their sum.
Changing the order of the addends $a$ and $b$ does not change their sum.
Add:
ⓐ  
Add.  $8+7$ 
$15$ 
ⓑ  
Add.  $7+8$ 
$15$ 
Did you notice that changing the order of the addends did not change their sum? We could have immediately known the sum from part ⓑ just by recognizing that the addends were the same as in part ⓑ , but in the reverse order. As a result, both sums are the same.
Add: $28+61.$
To add numbers with more than one digit, it is often easier to write the numbers vertically in columns.
Write the numbers so the ones and tens digits line up vertically.  $\begin{array}{c}\hfill 28\phantom{\rule{0.2em}{0ex}}\\ \\ \hfill \phantom{\rule{0.4em}{0ex}}\underset{\text{\_\_\_\_}}{+61}\end{array}$ 
Then add the digits in each place value.
Add the ones: $8+1=9$ Add the tens: $2+6=8$ 
$\begin{array}{c}\hfill 28\phantom{\rule{0.2em}{0ex}}\\ \\ \hfill \phantom{\rule{0.4em}{0ex}}\underset{\text{\_\_\_\_}}{+61}\\ \hfill 89\phantom{\rule{0.2em}{0ex}}\end{array}$ 
In the previous example, the sum of the ones and the sum of the tens were both less than $10.$ But what happens if the sum is $10$ or more? Let’s use our $\text{base10}$ model to find out. [link] shows the addition of $17$ and $26$ again.
When we add the ones, $7+6,$ we get $13$ ones. Because we have more than $10$ ones, we can exchange $10$ of the ones for $1$ ten. Now we have $4$ tens and $3$ ones. Without using the model, we show this as a small red $1$ above the digits in the tens place.
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