When we add two digit numbers, we start by adding the numbers in the ones place.

$\displaystyle \frac{\begin{array}[b]{r}3{\color{Red} 6}\\ +\ 1{\color{Red} 1}\end{array}}{ \ \ \ \ \ \space7}$

Next, we need to add the numbers in the tens place. 

$\displaystyle \frac{\begin{array}[b]{r}{\color{Red} 3}6\\ +\ {\color{Red} 1}1\end{array}}{ \ \ \ \space47}$

The final answer is $\displaystyle 47$

When we add two digit numbers, we start by adding the numbers in the ones place.

$\displaystyle \frac{\begin{array}[b]{r}1{\color{Red} 8}\\ +\ 5{\color{Red} 1}\end{array}}{ \ \ \ \ \ \space9}$

Next, we need to add the numbers in the tens place. 

$\displaystyle \frac{\begin{array}[b]{r}{\color{Red} 1}8\\ +\ {\color{Red} 5}1\end{array}}{ \ \ \ \space69}$

The final answer is $\displaystyle 69$

When we add $\displaystyle 10$ to a two digit number, the only number that changes in our answer is the tens position, and it will always go up by $\displaystyle 1$. Mentally, we can add $\displaystyle 1$ to the number in the tens place to find our answer. 

$\displaystyle 1+1=2$

$\displaystyle \frac{\begin{array}[b]{r}15\\ +\ 10\end{array}}{ \ \ \ \ \space25}$

When we add $\displaystyle 10$ to a two digit number, the only number that changes in our answer is the tens position, and it will always go up by $\displaystyle 1$. Mentally, we can add $\displaystyle 1$ to the number in the tens place to find our answer. 

$\displaystyle 1+1=2$

$\displaystyle \frac{\begin{array}[b]{r}10\\ +\ 10\end{array}}{ \ \ \ \space20}$

When we add $\displaystyle 10$ to a two digit number, the only number that changes in our answer is the tens position, and it will always go up by $\displaystyle 1$. Mentally, we can add $\displaystyle 1$ to the number in the tens place to find our answer. 

$\displaystyle 2+1=3$

$\displaystyle \frac{\begin{array}[b]{r}20\\ +\ 10\end{array}}{ \ \ \ \space30}$

When we add $\displaystyle 10$ to a two digit number, the only number that changes in our answer is the tens position, and it will always go up by $\displaystyle 1$. Mentally, we can add $\displaystyle 1$ to the number in the tens place to find our answer. 

$\displaystyle 2+1=3$

$\displaystyle \frac{\begin{array}[b]{r}25\\ +\ 10\end{array}}{ \ \ \ \space35}$

When we add $\displaystyle 10$ to a two digit number, the only number that changes in our answer is the tens position, and it will always go up by $\displaystyle 1$. Mentally, we can add $\displaystyle 1$ to the number in the tens place to find our answer. 

$\displaystyle 3+1=4$

$\displaystyle \frac{\begin{array}[b]{r}20\\ +\ 10\end{array}}{ \ \ \ \space40}$

When we add $\displaystyle 10$ to a two digit number, the only number that changes in our answer is the tens position, and it will always go up by $\displaystyle 1$. Mentally, we can add $\displaystyle 1$ to the number in the tens place to find our answer. 

$\displaystyle 3+1=4$

$\displaystyle \frac{\begin{array}[b]{r}35\\ +\ 10\end{array}}{ \ \ \ \space45}$

When we add $\displaystyle 10$ to a two digit number, the only number that changes in our answer is the tens position, and it will always go up by $\displaystyle 1$. Mentally, we can add $\displaystyle 1$ to the number in the tens place to find our answer. 

$\displaystyle 4+1=5$

$\displaystyle \frac{\begin{array}[b]{r}40\\ +\ 10\end{array}}{ \ \ \ \space50}$

When we add $\displaystyle 10$ to a two digit number, the only number that changes in our answer is the tens position, and it will always go up by $\displaystyle 1$. Mentally, we can add $\displaystyle 1$ to the number in the tens place to find our answer. 

$\displaystyle 4+1=5$

$\displaystyle \frac{\begin{array}[b]{r}45\\ +\ 10\end{array}}{ \ \ \ \space55}$