Add and Subtract Within 1000: CCSS.Math.Content.2.NBT.B.7 - MAP 2nd Grade Math

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Question

Solve the following:

$\frac{\begin{array}[b]{r}169\ +\ 323\end{array}}{\space }$

Answer

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:

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

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

Add the numbers in the tens place, including the that was carried over:

Add the numbers in the hundreds place:

Your final answer should be

$\frac{\begin{array}[b]{r}169\ +\ 323\end{array}}{\ \ \ 492 }$

Compare your answer with the correct one above

Question

Solve the following:

$\frac{\begin{array}[b]{r}920\ -\ 290\end{array}}{\space }$

Answer

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:

$\frac{\begin{array}[b]{r}0\ -\ 0\end{array}}{\ \ \ 0 }$

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

$\frac{\begin{array}[b]{r}2\ -\ 9\end{array}}{\ \ \ }$

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

$\frac{\begin{array}[b]{r}^{8} ot920\ -\ 290\end{array}}{\space }$

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

$\frac{\begin{array}[b]{r}^{8} ot9\ ^{12} ot2\ \ \ \ \ 0\ -\ 2 \ \ \ \ \ 9\ \ \ \ \ 0\end{array}}{\space }$

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

$\frac{\begin{array}[b]{r}12\ -\ 9\end{array}}{\ \ \ 3}$

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

$\frac{\begin{array}[b]{r}8\ -\ 2\end{array}}{\ \ \ 6}$

Your final answer should be

$\frac{\begin{array}[b]{r}^{8} ot9\ ^{12} ot2\ \ \ \ \ 0\ -\ 2 \ \ \ \ \ 9\ \ \ \ \ 0\end{array}}{\ \ \ 6 \ \ \ \ \ 3\ \ \ \ \ 0}$

Compare your answer with the correct one above

Question

Solve the following:

$\frac{\begin{array}[b]{r}169\ +\ 323\end{array}}{\space }$

Answer

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:

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

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

Add the numbers in the tens place, including the that was carried over:

Add the numbers in the hundreds place:

Your final answer should be

$\frac{\begin{array}[b]{r}169\ +\ 323\end{array}}{\ \ \ 492 }$

Compare your answer with the correct one above

Question

Solve the following:

$\frac{\begin{array}[b]{r}920\ -\ 290\end{array}}{\space }$

Answer

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:

$\frac{\begin{array}[b]{r}0\ -\ 0\end{array}}{\ \ \ 0 }$

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

$\frac{\begin{array}[b]{r}2\ -\ 9\end{array}}{\ \ \ }$

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

$\frac{\begin{array}[b]{r}^{8} ot920\ -\ 290\end{array}}{\space }$

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

$\frac{\begin{array}[b]{r}^{8} ot9\ ^{12} ot2\ \ \ \ \ 0\ -\ 2 \ \ \ \ \ 9\ \ \ \ \ 0\end{array}}{\space }$

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

$\frac{\begin{array}[b]{r}12\ -\ 9\end{array}}{\ \ \ 3}$

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

$\frac{\begin{array}[b]{r}8\ -\ 2\end{array}}{\ \ \ 6}$

Your final answer should be

$\frac{\begin{array}[b]{r}^{8} ot9\ ^{12} ot2\ \ \ \ \ 0\ -\ 2 \ \ \ \ \ 9\ \ \ \ \ 0\end{array}}{\ \ \ 6 \ \ \ \ \ 3\ \ \ \ \ 0}$

Compare your answer with the correct one above

Question

Solve the following:

$\frac{\begin{array}[b]{r}169\ +\ 323\end{array}}{\space }$

Answer

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:

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

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

Add the numbers in the tens place, including the that was carried over:

Add the numbers in the hundreds place:

Your final answer should be

$\frac{\begin{array}[b]{r}169\ +\ 323\end{array}}{\ \ \ 492 }$

Compare your answer with the correct one above

Question

Solve the following:

$\frac{\begin{array}[b]{r}920\ -\ 290\end{array}}{\space }$

Answer

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:

$\frac{\begin{array}[b]{r}0\ -\ 0\end{array}}{\ \ \ 0 }$

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

$\frac{\begin{array}[b]{r}2\ -\ 9\end{array}}{\ \ \ }$

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

$\frac{\begin{array}[b]{r}^{8} ot920\ -\ 290\end{array}}{\space }$

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

$\frac{\begin{array}[b]{r}^{8} ot9\ ^{12} ot2\ \ \ \ \ 0\ -\ 2 \ \ \ \ \ 9\ \ \ \ \ 0\end{array}}{\space }$

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

$\frac{\begin{array}[b]{r}12\ -\ 9\end{array}}{\ \ \ 3}$

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

$\frac{\begin{array}[b]{r}8\ -\ 2\end{array}}{\ \ \ 6}$

Your final answer should be

$\frac{\begin{array}[b]{r}^{8} ot9\ ^{12} ot2\ \ \ \ \ 0\ -\ 2 \ \ \ \ \ 9\ \ \ \ \ 0\end{array}}{\ \ \ 6 \ \ \ \ \ 3\ \ \ \ \ 0}$

Compare your answer with the correct one above

Question

Solve the following:

$\frac{\begin{array}[b]{r}169\ +\ 323\end{array}}{\space }$

Answer

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:

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

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

Add the numbers in the tens place, including the that was carried over:

Add the numbers in the hundreds place:

Your final answer should be

$\frac{\begin{array}[b]{r}169\ +\ 323\end{array}}{\ \ \ 492 }$

Compare your answer with the correct one above

Question

Solve the following:

$\frac{\begin{array}[b]{r}920\ -\ 290\end{array}}{\space }$

Answer

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:

$\frac{\begin{array}[b]{r}0\ -\ 0\end{array}}{\ \ \ 0 }$

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

$\frac{\begin{array}[b]{r}2\ -\ 9\end{array}}{\ \ \ }$

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

$\frac{\begin{array}[b]{r}^{8} ot920\ -\ 290\end{array}}{\space }$

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

$\frac{\begin{array}[b]{r}^{8} ot9\ ^{12} ot2\ \ \ \ \ 0\ -\ 2 \ \ \ \ \ 9\ \ \ \ \ 0\end{array}}{\space }$

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

$\frac{\begin{array}[b]{r}12\ -\ 9\end{array}}{\ \ \ 3}$

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

$\frac{\begin{array}[b]{r}8\ -\ 2\end{array}}{\ \ \ 6}$

Your final answer should be

$\frac{\begin{array}[b]{r}^{8} ot9\ ^{12} ot2\ \ \ \ \ 0\ -\ 2 \ \ \ \ \ 9\ \ \ \ \ 0\end{array}}{\ \ \ 6 \ \ \ \ \ 3\ \ \ \ \ 0}$

Compare your answer with the correct one above

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