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:

$\displaystyle 5+4=9$

Add the numbers in the tens place:

$\displaystyle 3+0=3$

Add the numbers in the hundreds place:

$\displaystyle 2+2=4$

Your final answer should be $\displaystyle 439:$

$\displaystyle \frac{\begin{array}[b]{r}235\\ +\ 204\end{array}}{\ \ \ \ 439 }$

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:

$\displaystyle 0+6=6$

Add the numbers in the tens place:

$\displaystyle 3+2=5$

Add the numbers in the hundreds place:

$\displaystyle 3+1=4$

Your final answer should be $\displaystyle 456:$

$\displaystyle \frac{\begin{array}[b]{r}330\\ +\ 126\end{array}}{\ \ \ 456 }$

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:

$\displaystyle 7+0=7$

Add the numbers in the tens place:

$\displaystyle 8+1=9$

Add the numbers in the hundreds place:

$\displaystyle 3+4=7$

Your final answer should be $\displaystyle 797:$

$\displaystyle \frac{\begin{array}[b]{r}387\\ +\ 410\end{array}}{\ \ \ 797}$

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:

$\displaystyle 8+8=16$

Because this sum is greater than $\displaystyle 9$, we write the $\displaystyle 6$ from the ones place and carry the $\displaystyle 1$ from he tens place over to the left. Your work should look something like this:

$\displaystyle \frac{\begin{array}[b]{r}2^{{\color{Red} 1}}2\ 8\\ +\ 6\ 2\ 8\end{array}}{\ \ \ \ \ \ \ \ 6 }$

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

$\displaystyle 1+2+2=5$

Add the numbers in the hundreds place:

$\displaystyle 2+6=8$

Your final answer should be $\displaystyle 856:$

$\displaystyle \frac{\begin{array}[b]{r}228\\ +\ 628\end{array}}{\ \ \ 856 }$

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:

$\displaystyle 5+9=14$

Because this sum is greater than $\displaystyle 9$, we write the $\displaystyle 4$ from the ones place and carry the $\displaystyle 1$ from he tens place over to the left. Your work should look something like this:

$\displaystyle \frac{\begin{array}[b]{r}1^{{\color{Red} 1}}9\ 5\\ +\ 5\ 1\ 9\end{array}}{\ \ \ \ \ \ \ \ 4 }$

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

$\displaystyle 1+9+1=11$

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

$\displaystyle \frac{\begin{array}[b]{r}^{{\color{Red} 1}}1^{{\color{Red} 1}}9\ 5\\ +\ 5\ 1\ 9\end{array}}{\ \ \ \ \ \ 1\ 4 }$

Add the numbers in the hundreds place, including the $\displaystyle 1$ that was carried over:

$\displaystyle 1+1+5=7$

Your final answer should be $\displaystyle 714:$

$\displaystyle \frac{\begin{array}[b]{r}195\\ +\ 519\end{array}}{\ \ \ 714}$

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:

$\displaystyle 2+6=8$

Add the numbers in the tens place:

$\displaystyle 1+9=10$

Because this sum is greater than $\displaystyle 9$, we write the $\displaystyle 0$ from the ones place and carry the $\displaystyle 1$ from the tens place over to the left. Your work should look something like this:

$\displaystyle \frac{\begin{array}[b]{r}^{{\color{Red} 1}}312\\ +\ 496\end{array}}{\ \ \ \ \ 08 }$

Add the numbers in the hundreds place, including the $\displaystyle 1$ that was carried over:

$\displaystyle 1+3+4=8$

Your final answer should be $\displaystyle 808:$

$\displaystyle \frac{\begin{array}[b]{r}^{1}312\\ +\ 496\end{array}}{\ \ \ \ 808 }$

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:

$\displaystyle 9+1=10$

Because this sum is greater than $\displaystyle 9$, we write the $\displaystyle 0$ from the ones place and carry the $\displaystyle 1$ from the tens place over to the left. Your work should look something like this:

$\displaystyle \frac{\begin{array}[b]{r}2^{{\color{Red} 1}}6\ 9\\ +\ 2\ 2\ 1\end{array}}{\ \ \ \ \ \ \ \ 0 }$

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:

$\displaystyle 2+2=4$

Your final answer should be $\displaystyle 490:$

$\displaystyle \frac{\begin{array}[b]{r}269\\ +\ 221\end{array}}{\ \ \ 490 }$

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:

$\displaystyle 0+4=4$

Add the numbers in the tens place:

$\displaystyle 0+2=2$

Add the numbers in the hundreds place:

$\displaystyle 4+2=6$

Your final answer should be $\displaystyle 624:$

$\displaystyle \frac{\begin{array}[b]{r}400\\ +\ 224\end{array}}{\ \ \ 624 }$

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:

$\displaystyle 8+8=16$

Because this sum is greater than $\displaystyle 9$, we write the $\displaystyle 6$ from the ones place and carry the $\displaystyle 1$ from the tens place over to the left. Your work should look something like this:

$\displaystyle \frac{\begin{array}[b]{r}1^{{\color{Red} 1}}2\ 8\\ +\ 5\ 4\ 8\end{array}}{\ \ \ \ \ \ \ \ 6 }$

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

$\displaystyle 1+2+4=7$

Add the numbers in the hundreds place:

$\displaystyle 1+5=6$

Your final answer should be $\displaystyle 676:$

$\displaystyle \frac{\begin{array}[b]{r}128\\ +\ 548\end{array}}{\ \ \ 676}$

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:

$\displaystyle 7+8=15$

Because this sum is greater than $\displaystyle 9$, we write the $\displaystyle 5$ from the ones place and carry the $\displaystyle 1$ from the tens place over to the left. Your work should look something like this:

$\displaystyle \frac{\begin{array}[b]{r}4^{{\color{Red} 1}}0\ 7\\ +\ 3\ 2\ 8\end{array}}{\ \ \ \ \ \ \ \ 5 }$

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

$\displaystyle 1+0+2=3$

Add the numbers in the hundreds place:

$\displaystyle 4+3=7$

Your final answer should be $\displaystyle 735:$

$\displaystyle \frac{\begin{array}[b]{r}407\\ +\ 328\end{array}}{\ \ \ 735}$