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Here you will learn strategies on how to add and subtract, including using visual models, the number line, and algorithms.
Students will first learn about addition and subtraction in kindergarten, continue to add and subtract larger numbers in 2 nd and 3 rd grade, and learn the standard algorithm in 4 th grade. In 7 th grade, students learn to add and subtract with integers and then rational numbers.
Addition is the operation of combining two or more numbers together. Subtraction is the operation of taking away from a number.
Students as early as pre-k have a basic, informal understanding of addition and subtraction. By the end of Kindergarten, students can operate within 10, heavily relying on models like counters or 10 frames to solve.
By the end of first grade, students should be able to add and subtract fluently within 20. As students work on repeatedly adding and subtracting with single digit numbers, they deepen their number sense around the operations by solving in different ways.
This includes beginning to build number bonds.
For example,
This aids students as they go into 2 nd and 3 rd grade and learn to add and subtract multi-digit numbers (within 1,000 ). At this level, students are still relying on number relationships, properties of operations, place value ideas, models, etc. to solve.
For example,
When adding 3- digit numbers, students may combine each place value of the addends separately.
235 + 149
Hundreds \hspace{1.1cm} Tens \hspace{1.2cm} Ones
200 + 100 \hspace{1cm} 30 + 40 \hspace{1cm} 5 + 9 \hspace{1cm}
= 300 \hspace{1.5cm} = 70 \hspace{1.3cm} = 14
And then add in the partial sums:
300+70+14=384
For example,
When subtracting numbers that are larger, students may use what they know about place value to solve.
781-395
The distance from 781-395 is the same as 786-400, it is just shifted up 5 on the number line.
Now solve 786-400 by taking away 4 hundreds.
786-400=386, so 781-395=386.
Once students have developed a flexible understanding of multi-digit addition and multi-digit subtraction in 2 nd grade and 3 rd grade, they learn the standard algorithm in 4 th grade.
For example,
It is important to consider the place value of each digit and line up the corresponding digits in each number.
In 5 th grade and 6 th grade, students continue to add and subtract with whole numbers and apply these strategies to decimal operations.
See also: Adding and subtracting decimals
In 7 th grade, students learn to add and subtract integers and rational numbers.
You can add and subtract integers using visual models or a number line.
Step-by-step guide: Adding and subtracting integers
You can also add and subtract rational numbers using visual models or a number line.
Use this quiz to check your grade 2, 3, 4 and 7 studentsβ understanding of addition and subtraction. 15+ questions with answers covering a range of 2nd, 3rd, 4th and 7th grade addition and subtraction topics to identify areas of strength and support!
DOWNLOAD FREEUse this quiz to check your grade 2, 3, 4 and 7 studentsβ understanding of addition and subtraction. 15+ questions with answers covering a range of 2nd, 3rd, 4th and 7th grade addition and subtraction topics to identify areas of strength and support!
DOWNLOAD FREEHow does this apply to 2 nd grade math, 3 rd grade math, 4 th grade, and 7 th grade math?
There are a lot of ways to add and subtract. For more specific step-by-step guides, check out the individual pages linked in the βWhat are addition and subtraction?β section above or read through the examples below.
Solve 34 + 58.
You can use a model of the tens and ones.
2Solve.
Combine the tens and ones.
You can create one new group of ten from 10 ones.
This leaves 9 tens and 2 ones, which is 92.
So, 34+58=92.
Solve 340-291.
Choose a strategy.
You can use a number line to count up to 340 from 291.
Solve.
The difference between 291 and 340 is 49.
So, 340-291=49.
Add 4,297 + 381 with the algorithm.
Stack the numbers and line up the place values.
Add each column from right to left and regroup the numbers when necessary.
Write your final answer and check to make sure your answer is reasonable.
You can estimate to decide if your answer is reasonable.
4,297 rounds to 4,300.
381 rounds to 400.
4,300+400=4,700, so the answer should be close to 4,700. Since 4,678 is close, the answer is reasonable.
4,297+381=4,678
After the harvest, a farmer has 2,148 pears. 213 of the pears are rotten and cannot be sold. If the farmer sells the rest, how many pears will she sell?
Write an equation that represents the word problem.
2,148 - 213 = β¬
Stack the numbers and line up the place values.
Subtract each column from right to left and exchange for a \bf{10} when necessary.
Write your final answer and check to make sure your answer is reasonable.
You can estimate to decide if your answer is reasonable.
2,148 rounds to 2,000.
213 rounds to 200.
2,000-200=1,800, so the answer should be close to 1,800. Since 1,935 is close, the answer is reasonable.
2,148-213=1,935
Subtract: -4 -(-6)=?
Represent the problem with counters, identifying zero pairs with addition or adding zero pairs when necessary for subtraction.
-4 - (-6) starts with -4 and removes -6. Add 2 zero pairs in order to remove -6.
The answer is the leftover counters.
There are two positive counters left.
-4-(-6)=2
Solve -34 \cfrac{2}{3}+\left(-567 \cfrac{3}{4}\right) .
Make sense of the calculation – relating to positive numbers when necessary.
-34 \cfrac{2}{3}+\left(-567 \cfrac{3}{4}\right) combines two negative amounts, which do not create zero pairs.
This means the answer will be even more negative. You can use the algorithm for addition to see what the total amount is.
Use an algorithm.
\begin{aligned}& 34 \frac{2}{3}+567 \frac{3}{4} \\\\ & =34 \frac{2 \times 4}{3 \times 4}+567 \frac{3 \times 3}{4 \times 3} \\\\ & =34 \frac{8}{12}+567 \frac{9}{12} \\\\ & =601 \frac{17}{12} \text { or } 602 \frac{5}{12}\end{aligned}
Decide if the final answer is positive or negative.
Adding two negative numbers results in a negative number. For example, if you started on the number line and went back 34 \cfrac{2}{3} and then you went back 567 \cfrac{3}{4} more, you are further down the negative side of the number line.
-34 \cfrac{2}{3}+\left(-567 \cfrac{3}{4}\right)=-602 \cfrac{5}{12}
1. Solve 39 + 46.
You can use a model of the tens and ones.
Combine the tens and ones.
You can create one new group of ten from 10 ones.
This leaves 8 tens and 5 ones, which is 85.
39+46=85
2. Solve 763-594.
You can use a number line to count up to 594 from 763.
The difference between 594 and 763 is 169.
So, 763-594=169.
3. Solve 5,621 + 3,570.
To use the standard algorithm, stack the numbers and line up the place values.
You can estimate to decide if your answer is reasonable.
5,621 rounds to 6,000.
3,570 rounds to 4,000.
6,000+4,000=10,000, so the answer should be close to 10,000.
Since 9,191 is close, the answer is reasonable.
5,621+3,570=9,191
4. Jerome has 18,341 points in a game. Yasmine has 8,239 points in the same game. How many more points does Jerome have?
18,341-8,239 = β¬
To solve with the standard algorithm, stack the numbers and line up the place values.
You can estimate to decide if your answer is reasonable.
18,341 rounds to 18,000.
8,239 rounds to 8,000.
18,000-8,000=10,000, so the answer should be close to 10,000.
Since 10,102 is close, the answer is reasonable.
18,341-8,239=10,102
5. Subtract: -12-3= \, ?
-12-3 starts with -12 and removes 3 positives. Add 3 zero pairs in order to remove 3.
There are 15 negative counters left.
You can also think of this as the distance from 3 to -12.
-12-3=-15
6. Solve -84.4+391.02= \, ?.
-84.4 + 391.02 combines negative and positive values, which creates zero pairs. This means that -84.4+391.02=391.02-84.4, so you can use the standard algorithm to solve.
Adding a smaller negative number to a larger positive results in a positive answer.
-84.4+321.02=306.62
Regrouping is when you are adding and have more than 10 ones, tens, hundreds, etc. and you need to regroup to create a new group of ten, hundred, etc.
For example, if you add 27 + 18, you will have 3 tens and 15 ones. You regroup 10 of the ones to be a new group of ten, so you are left with 4 tens and 5 ones, or 45.
Exchanging is when you are subtracting and you do not have enough ones, tens, hundreds, etc. to take away and you need to exchange for a new group of ten, hundred, etc.
For example, if you subtract 27 - 18, you start with 7 ones and need to take away 8. There are not enough, so you exchange one of the tens in 27 for ones. Then you have 1 ten and 17 ones. Now you can subtract 8 ones from 17 ones, leaving 9 ones.
In 4 th grade students learn to add and subtract fractions with common denominators. In fifth grade they learn to add and subtraction fractions with uncommon denominators. Then in 7 th grade they are expected to add and subtract negative fractions.
See also: Adding and subtracting fractions
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