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:

\(\displaystyle \frac{\begin{array}[b]{r}6{\color{Red} 7}\\ -\ 3{\color{Red} 3}\end{array}}{\ \ \ \ \ {\color{Red} 4} }\)

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

\(\displaystyle \frac{\begin{array}[b]{r}{\color{Red} 6}7\\ -\ {\color{Red} 3}3\end{array}}{\ \ \ \ {\color{Red} 3}4}\)

Your final answer should be \(\displaystyle 34\)

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:

\(\displaystyle \frac{\begin{array}[b]{r}5\\ -\ 9\end{array}}{\space }\)

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 \(\displaystyle 5\) since \(\displaystyle 5\) is the smaller number. In this case, we are going to look to the \(\displaystyle 3\). We only ever need to take \(\displaystyle 1\) away from the number to the left. For this problem, that will leave us with a \(\displaystyle 2\) to replace the \(\displaystyle 3\). So far, your work should look something like this:

\(\displaystyle \frac{\begin{array}[b]{r}^{2} \not35\\ -\ 19\end{array}}{\space }\)

Remember, we've borrowed \(\displaystyle 1\) from the tens place, and one ten is equal to \(\displaystyle 10\). We now add that \(\displaystyle 10\) to the digit in the ones place, \(\displaystyle 5+10=15\). Your work should look something like this:

\(\displaystyle \frac{\begin{array}[b]{r}^{2} \not3\ ^{15}\not5\\ -\ 1 \ \ \ \ \ 9\end{array}}{\space }\)

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

\(\displaystyle \frac{\begin{array}[b]{r}15\\ -\ 9\end{array}}{\ \ \ 6}\)

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

\(\displaystyle \frac{\begin{array}[b]{r}2\\ -\ 1\end{array}}{\ \ \ 1}\)

Your final answer should be \(\displaystyle 16\)

\(\displaystyle \frac{\begin{array}[b]{r}^{2} \not3\ ^{15}\not5\\ -\ 1 \ \ \ \ \ 9\end{array}}{\ \ \ 1\ \ \ \ \ 6}\)

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:

\(\displaystyle \frac{\begin{array}[b]{r}5\\ -\ 6\end{array}}{\space }\)

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 \(\displaystyle 6\) away from \(\displaystyle 5\) since \(\displaystyle 5\) is the smaller number. In this case, we are going to look to the \(\displaystyle 5\). We only ever need to take \(\displaystyle 1\) away from the number to the left. For this problem, that will leave us with a \(\displaystyle 4\) to replace the \(\displaystyle 5\). So far, your work should look something like this:

\(\displaystyle \frac{\begin{array}[b]{r}^{4} \not55\\ -\ 36\end{array}}{\space }\)

Remember, we've borrowed \(\displaystyle 1\) from the tens place, and one ten is equal to \(\displaystyle 10\). We now add that \(\displaystyle 10\) to the digit in the ones place, \(\displaystyle 5+10=15\). Your work should look something like this:

\(\displaystyle \frac{\begin{array}[b]{r}^{4} \not5\ ^{15}\not5\\ -\ 3 \ \ \ \ \ 6\end{array}}{\space }\)

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

\(\displaystyle \frac{\begin{array}[b]{r}15\\ -\ 6\end{array}}{\ \ \ 9}\)

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

\(\displaystyle \frac{\begin{array}[b]{r}4\\ -\ 3\end{array}}{\ \ \ 1}\)

Your final answer should be \(\displaystyle 19\)

\(\displaystyle \frac{\begin{array}[b]{r}^{4} \not5\ ^{15}\not5\\ -\ 3 \ \ \ \ \ 6\end{array}}{\ \ \ 1\ \ \ \ \ \ 9}\)

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:

\(\displaystyle \frac{\begin{array}[b]{r}5\\ -\ 8\end{array}}{\space }\)

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 \(\displaystyle 8\) away from \(\displaystyle 5\) since \(\displaystyle 5\) is the smaller number. In this case, we are going to look to the \(\displaystyle 8\). We only ever need to take \(\displaystyle 1\) away from the number to the left. For this problem, that will leave us with a \(\displaystyle 7\) to replace the \(\displaystyle 8\). So far, your work should look something like this:

\(\displaystyle \frac{\begin{array}[b]{r}^{7} \not85\\ -\ 18\end{array}}{\space }\)

Remember, we've borrowed \(\displaystyle 1\) from the tens place, and one ten is equal to \(\displaystyle 10\). We now add that \(\displaystyle 10\) to the digit in the ones place, \(\displaystyle 5+10=15\). Your work should look something like this:

\(\displaystyle \frac{\begin{array}[b]{r}^{7} \not8\ ^{15}\not5\\ -\ 1 \ \ \ \ \ 8\end{array}}{\space }\)

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

\(\displaystyle \frac{\begin{array}[b]{r}15\\ -\ 8\end{array}}{\ \ \ 7}\)

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

\(\displaystyle \frac{\begin{array}[b]{r}7\\ -\ 1\end{array}}{\ \ \ 6}\)

Your final answer should be \(\displaystyle 67\)

\(\displaystyle \frac{\begin{array}[b]{r}^{7} \not8\ ^{15}\not5\\ -\ 1 \ \ \ \ \ 8\end{array}}{\ \ \ \ 6\ \ \ \ \ 7 }\)

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:

\(\displaystyle \frac{\begin{array}[b]{r}1\\ -\ 2\end{array}}{\space }\)

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 \(\displaystyle 2\) away from \(\displaystyle 1\) since \(\displaystyle 1\) 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 an \(\displaystyle 7\) to replace the \(\displaystyle 8\). So far, your work should look something like this:

\(\displaystyle \frac{\begin{array}[b]{r}^{7} \not81\\ -\ 12\end{array}}{\space }\)

Remember, we've borrowed \(\displaystyle 1\) from the tens place, and one ten is equal to \(\displaystyle 10\). We now add that \(\displaystyle 10\) to the digit in the ones place, \(\displaystyle 1+10=11\). Your work should look something like this:

\(\displaystyle \frac{\begin{array}[b]{r}^{7} \not8\ ^{11}\not1\\ -\ 1 \ \ \ \ \ 2\end{array}}{\space }\)

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

\(\displaystyle \frac{\begin{array}[b]{r}11\\ -\ 2\end{array}}{\ \ \ 9}\)

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

\(\displaystyle \frac{\begin{array}[b]{r}7\\ -\ 1\end{array}}{\ \ \ 6}\)

Your final answer should be \(\displaystyle 69\)

\(\displaystyle \frac{\begin{array}[b]{r}^{7} \not8\ ^{11}\not1\\ -\ 1 \ \ \ \ \ 2\end{array}}{\ \ \ \ 6\ \ \ \ \ 9 }\)

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:

\(\displaystyle \frac{\begin{array}[b]{r}1\\ -\ 7\end{array}}{\space }\)

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 \(\displaystyle 7\) away from \(\displaystyle 1\) since \(\displaystyle 1\) is the smaller number. In this case, we are going to look to the \(\displaystyle 2\). We only ever need to take \(\displaystyle 1\) away from the number to the left. For this problem, that will leave us with a \(\displaystyle 1\) to replace the \(\displaystyle 2\). So far, your work should look something like this:

\(\displaystyle \frac{\begin{array}[b]{r}^{1} \not21\\ -\ 17\end{array}}{\space }\)

Remember, we've borrowed \(\displaystyle 1\) from the tens place, and one ten is equal to \(\displaystyle 10\). We now add that \(\displaystyle 10\) to the digit in the ones place, \(\displaystyle 1+10=11\). Your work should look something like this:

\(\displaystyle \frac{\begin{array}[b]{r}^{1} \not2\ ^{11}\not1\\ -\ 1 \ \ \ \ \ 7\end{array}}{\space }\)

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

\(\displaystyle \frac{\begin{array}[b]{r}11\\ -\ 7\end{array}}{\ \ \ 4}\)

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

\(\displaystyle \frac{\begin{array}[b]{r}1\\ -\ 1\end{array}}{\ \ \ 0}\)

Your final answer should be \(\displaystyle 4\)

\(\displaystyle \frac{\begin{array}[b]{r}^{1} \not2\ ^{11}\not1\\ -\ 1 \ \ \ \ \ 7\end{array}}{\ \ \ \ \ \ \ \ \ \ \ 4}\)

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:

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

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 \(\displaystyle 8\) away from \(\displaystyle 6\) since \(\displaystyle 6\) is the smaller number. In this case, we are going to look to the \(\displaystyle 3\). We only ever need to take \(\displaystyle 1\) away from the number to the left. For this problem, that will leave us with a \(\displaystyle 2\) to replace the \(\displaystyle 3\). So far, your work should look something like this:

\(\displaystyle \frac{\begin{array}[b]{r}^{2} \not36\\ -\ 18\end{array}}{\space }\)

Remember, we've borrowed \(\displaystyle 1\) from the tens place, and one ten is equal to \(\displaystyle 10\). We now add that \(\displaystyle 10\) to the digit in the ones place, \(\displaystyle 6+10=16\). Your work should look something like this:

\(\displaystyle \frac{\begin{array}[b]{r}^{2} \not3\ ^{16}\not6\\ -\ 1 \ \ \ \ \ 8\end{array}}{\space }\)

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

\(\displaystyle \frac{\begin{array}[b]{r}16\\ -\ 8\end{array}}{\ \ \ 8}\)

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

\(\displaystyle \frac{\begin{array}[b]{r}2\\ -\ 1\end{array}}{\ \ \ 1}\)

Your final answer should be \(\displaystyle 18\)

\(\displaystyle \frac{\begin{array}[b]{r}^{2} \not3\ ^{16}\not6\\ -\ 1 \ \ \ \ \ 8\end{array}}{\ \ \ 1\ \ \ \ \ 8}\)

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:

\(\displaystyle \frac{\begin{array}[b]{r}1{\color{Red} 5}\\ -\ 1{\color{Red} 0}\end{array}}{\ \ \ \ \ {\color{Red} 5} }\)

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

\(\displaystyle \frac{\begin{array}[b]{r}{\color{Red} 1}5\\ -\ {\color{Red} 1}0\end{array}}{\ \ \ \ {\color{Red} 0}5}\)

Your final answer should be \(\displaystyle 5\)

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:

\(\displaystyle \frac{\begin{array}[b]{r}3{\color{Red} 7}\\ -\ 1{\color{Red} 1}\end{array}}{\ \ \ \ \ {\color{Red} 6} }\)

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

\(\displaystyle \frac{\begin{array}[b]{r}{\color{Red} 3}7\\ -\ {\color{Red} 1}1\end{array}}{\ \ \ \ {\color{Red} 2}6}\)

Your final answer should be \(\displaystyle 26\)

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:

\(\displaystyle \frac{\begin{array}[b]{r}5{\color{Red} 8}\\ -\ 3{\color{Red} 0}\end{array}}{\ \ \ \ \ {\color{Red} 8} }\)

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

\(\displaystyle \frac{\begin{array}[b]{r}{\color{Red} 5}8\\ -\ {\color{Red} 3}0\end{array}}{\ \ \ \ {\color{Red} 2}8}\)

Your final answer should be \(\displaystyle 28\)