When we add with multi-digit numbers, we start with the numbers in the ones place, and then move to the left. 

Let's add the digits in the ones place:

\(\displaystyle 6+0+5+2=13\)

Because the sum is greater than \(\displaystyle 9\), we have to carry the \(\displaystyle 1\) to the tens place. So far, your work should look like the following: 

Next, we add the numbers in the tens position, including the \(\displaystyle 1\) that was carried over:

Let's add the digits in the tens place:

\(\displaystyle 1+3+2+2+1=9\)

Your final answer should be \(\displaystyle 93\)

When we add with multi-digit numbers, we start with the numbers in the ones place, and then move to the left. 

Let's add the digits in the ones place:

\(\displaystyle 9+0+6+1=16\)

Because the sum is greater than \(\displaystyle 9\), we have to carry the \(\displaystyle 1\) to the tens place. So far, your work should look like the following: 

Next, we add the numbers in the tens position, including the \(\displaystyle 1\) that was carried over:

Let's add the digits in the tens place:

\(\displaystyle 1+8+3+5+1=18\)

Your final answer should be \(\displaystyle 186\)

When we add with multi-digit numbers, we start with the numbers in the ones place, and then move to the left. 

Let's add the digits in the ones place:

\(\displaystyle 7+3+8+5=23\)

Because the sum is greater than \(\displaystyle 9\), we have to carry the \(\displaystyle 2\) to the tens place. So far, your work should look like the following: 

Next, we add the numbers in the tens position, including the \(\displaystyle 2\) that was carried over:

Let's add the digits in the tens place:

\(\displaystyle 2+8+1+9+1=22\)

Your final answer should be \(\displaystyle 223\)

When we add with multi-digit numbers, we start with the numbers in the ones place, and then move to the left. 

Let's add the digits in the ones place:

\(\displaystyle 1+9+2+1=13\)

Because the sum is greater than \(\displaystyle 9\), we have to carry the \(\displaystyle 1\) to the tens place. So far, your work should look like the following: 

Next, we add the numbers in the tens position, including the \(\displaystyle 1\) that was carried over:

Let's add the digits in the tens place:

\(\displaystyle 1+8+3+8+9=29\)

Your final answer should be \(\displaystyle 293\)

When we add with multi-digit numbers, we start with the numbers in the ones place, and then move to the left. 

Let's add the digits in the ones place:

\(\displaystyle 7+0+7+4=18\)

Because the sum is greater than \(\displaystyle 9\), we have to carry the \(\displaystyle 1\) to the tens place. So far, your work should look like the following: 

Next, we add the numbers in the tens position, including the \(\displaystyle 1\) that was carried over:

Let's add the digits in the tens place:

\(\displaystyle 1+8+3+9+6=27\)

Your final answer should be \(\displaystyle 278\)

When we add with multi-digit numbers, we start with the numbers in the ones place, and then move to the left. 

Let's add the digits in the ones place:

\(\displaystyle 6+4+8+4=22\)

Because the sum is greater than \(\displaystyle 9\), we have to carry the \(\displaystyle 2\) to the tens place. So far, your work should look like the following: 

Next, we add the numbers in the tens position, including the \(\displaystyle 2\) that was carried over:

Let's add the digits in the tens place:

\(\displaystyle 2+4+6+5+1=18\)

Your final answer should be \(\displaystyle 182\)

When we subtract multi-digit numbers, we start with the digits in the ones place and move to the left. 

Let's look at the numbers in the ones place:

\(\displaystyle \frac{\begin{array}[b]{r}0\\ -\ 0\end{array}}{\ \ \ 0 }\)

Next, let's look at the numbers in the tens place:

\(\displaystyle \frac{\begin{array}[b]{r}2\\ -\ 9\end{array}}{\ \ \ }\)

When the top number is smaller than the bottom number, we have to borrow from the number to the left because we can't take \(\displaystyle 9\) away from \(\displaystyle 2\) since \(\displaystyle 2\) is the smaller number. In this case, we are going to look to the \(\displaystyle 9\). We only ever need to take \(\displaystyle 1\) away from the number to the left. For this problem, that will leave us with a \(\displaystyle 8\) to replace the \(\displaystyle 9\). So far, your work should look something like this:

\(\displaystyle \frac{\begin{array}[b]{r}^{8} \not920\\ -\ 290\end{array}}{\space }\)

Remember, we've borrowed \(\displaystyle 1\) from the hundreds place, so we can put a \(\displaystyle 1\) in front of the number in the tens place. So far, your work should look something like this:

\(\displaystyle \frac{\begin{array}[b]{r}^{8} \not9\ ^{12}\not2\ \ \ \ \ 0\\ -\ 2 \ \ \ \ \ 9\ \ \ \ \ 0\end{array}}{\space }\)

Now, we can subtract the numbers in the tens place:

\(\displaystyle \frac{\begin{array}[b]{r}12\\ -\ 9\end{array}}{\ \ \ 3}\)

Next, we can subtract the numbers in the hundreds place:

\(\displaystyle \frac{\begin{array}[b]{r}8\\ -\ 2\end{array}}{\ \ \ 6}\)

Your final answer should be \(\displaystyle 630\)

\(\displaystyle \frac{\begin{array}[b]{r}^{8} \not9\ ^{12}\not2\ \ \ \ \ 0\\ -\ 2 \ \ \ \ \ 9\ \ \ \ \ 0\end{array}}{\ \ \ 6 \ \ \ \ \ 3\ \ \ \ \ 0}\)

When we add with multi-digit numbers, we start with the numbers in the ones place, and then move to the left to the tens place, followed by the hundreds place. 

Add the numbers in the ones place:

\(\displaystyle 9+3=12\)

Because this sum is greater than \(\displaystyle 9\), we write the \(\displaystyle 2\) from the ones place and carry the \(\displaystyle 1\) from the tens place over to the left. Your work should look something like this:

\(\displaystyle \frac{\begin{array}[b]{r}1^{{\color{Red} 1}}6\ 9\\ +\ 3\ 2\ 3\end{array}}{\ \ \ \ \ \ \ \ 2 }\)

Add the numbers in the tens place, including the \(\displaystyle 1\) that was carried over:

\(\displaystyle 1+6+2=9\)

Add the numbers in the hundreds place:

\(\displaystyle 1+3=4\)

Your final answer should be \(\displaystyle 492:\)

\(\displaystyle \frac{\begin{array}[b]{r}169\\ +\ 323\end{array}}{\ \ \ 492 }\)

When adding any number by \(\displaystyle 100\) the only number that will change in your answer is the hundreds place. 

\(\displaystyle 2+0=2\)

\(\displaystyle 3+0=3\)

\(\displaystyle 4+1=5\)

\(\displaystyle \frac{\begin{array}[b]{r}432\\ +\ 100\end{array}}{\ \ \ \space 532}\)

When adding any number by \(\displaystyle 100\) the only number that will change in your answer is the hundreds place. 

\(\displaystyle 9+0=9\)

\(\displaystyle 4+0=4\)

\(\displaystyle 3+1=4\)

\(\displaystyle \frac{\begin{array}[b]{r}349\\ +\ 100\end{array}}{\ \ \ \space449}\)