Additive Reasoning and Fluency

Part-Part-Whole Relationships help us understand how numbers go together and come apart.

In math, a whole is the total amount. The parts are the smaller amounts that make the whole.

When we add, we put parts together to make a whole. When we know the whole and one part, we can find the missing part.

This idea helps us solve addition problems and think clearly about numbers.

Think of it like this:

• 2 parts can make 1 whole.
• Sometimes 3 or more parts can make 1 whole.
• The whole is always all the parts together.

We can show part-part-whole relationships with number bonds and tape diagrams.

A number bond shows how numbers are connected. The whole is one number, and the parts are the numbers that make it.

For example, if the parts are 3 and 5, the whole is:

$$3 + 5 = 8$$

So in this number bond:

• Part: 3
• Part: 5
• Whole: 8

A tape diagram is a drawing made of bars. Each bar or section stands for a number. Smaller sections show the parts, and the full bar shows the whole.

If one part is 4 and another part is 6, the whole bar shows:

$$4 + 6 = 10$$

You can imagine one long bar split into 2 pieces: 4 and 6. Together, they make 10.

Main idea: If you know the parts, add to find the whole.

$$\text{part} + \text{part} = \text{whole}$$

Another main idea: If you know the whole and one part, subtract to find the missing part.

$$\text{whole} - \text{part} = \text{missing part}$$

This is useful because addition and subtraction work together.

Worked Example 1: Find the whole

There are 7 red apples and 2 green apples. How many apples are there in all?

The parts are 7 and 2.

Add the parts:

$$7 + 2 = 9$$

Answer: The whole is 9 apples.

You can think of a number bond:

• Part: 7
• Part: 2
• Whole: 9

Worked Example 2: Find the missing part

A class has 12 students. 8 are wearing sneakers. The rest are wearing boots. How many students are wearing boots?

The whole is 12 students.

One part is 8 students.

The missing part is:

$$12 - 8 = 4$$

Answer: 4 students are wearing boots.

You can check with addition:

$$8 + 4 = 12$$

Worked Example 3: Use a tape diagram

A ribbon is 15 inches long. One piece is 6 inches long. The other piece is how many inches long?

Draw a tape diagram in your mind:

• The full bar is 15.
• One part is 6.
• The other part is missing.

Find the missing part:

$$15 - 6 = 9$$

Answer: The other piece is 9 inches long.

Check:

$$6 + 9 = 15$$

Worked Example 4: More than two parts

Mia has 3 blue stickers, 4 yellow stickers, and 5 pink stickers. How many stickers does she have altogether?

The whole is made from 3 parts: 3, 4, and 5.

Add the parts:

$$3 + 4 + 5 = 12$$

Answer: Mia has 12 stickers.

You can add in steps:

$$3 + 4 = 7$$

$$7 + 5 = 12$$

How to solve part-part-whole problems

1. Read the problem carefully.
2. Ask: What is the whole?
3. Ask: What are the parts?
4. If the parts are known, add.
5. If the whole and one part are known, subtract.
6. Check if your answer makes sense.

Helpful clues in word problems

• In all, altogether, or total often mean find the whole.
• How many more or how many are left for the other part can mean find a missing part.

Important things to remember

• Parts are smaller amounts.
• The whole is the total amount.
• Adding parts gives the whole.
• Subtracting a part from the whole gives the missing part.
• Number bonds and tape diagrams help you see the math clearly.

Let’s look at one more quick way to think about it:

If you know:

$$5 + 3 = 8$$

Then you also know:

$$8 - 5 = 3$$

and

$$8 - 3 = 5$$

These number sentences all belong to the same part-part-whole relationship.

Summary

Part-part-whole relationships show how numbers fit together. The parts join to make the whole. You can use addition to find the whole and subtraction to find a missing part. Number bonds and tape diagrams are helpful tools for seeing these relationships.

Commutative and Identity Properties help us add numbers in smart, easy ways. These properties are special rules about addition.

When we know these rules, we can solve math problems faster and with more confidence. We can also check our work and make mental math easier.

In this lesson, you will learn two important addition rules:

• Commutative Property of Addition: changing the order of the addends does not change the sum.
• Identity Property of Addition: adding zero does not change the number.

Let’s learn what each one means.

1. Commutative Property of Addition

The word commutative means that numbers can switch places when we add, and the answer stays the same.

For example:

$$3 + 5 = 5 + 3$$

Both sides equal 8.

This means the order of the addends does not matter in addition.

You can think of it like this: if you have 3 red apples and 5 green apples, you still have 8 apples total. It does not matter which group you count first.

Here is the rule in a general way:

$$a + b = b + a$$

The letters just stand for numbers. No matter what numbers you use, switching their order keeps the same sum.

2. Identity Property of Addition

The identity property tells us that when we add zero to a number, the number stays the same.

For example:

$$7 + 0 = 7$$

and

$$0 + 7 = 7$$

Zero means nothing is being added. So the number does not change.

Here is the rule in a general way:

$$a + 0 = a$$

If you have 9 blocks and add 0 more blocks, you still have 9 blocks.

Why these properties are useful

These properties help us add in easier ways.

• The commutative property lets us change the order to make a fact easier to solve in our heads.
• The identity property helps us quickly know what happens when zero is added.

For example, if you see:

$$2 + 9$$

You might know the fact more quickly as:

$$9 + 2 = 11$$

The answer is still 11 because the order does not change the sum.

If you see:

$$14 + 0$$

You can answer right away: 14.

Worked Example 1

Solve:

$$4 + 6$$

Using the commutative property, we can turn it around:

$$4 + 6 = 6 + 4$$

Now add:

$$6 + 4 = 10$$

So:

$$4 + 6 = 10$$

What we learned: switching the numbers did not change the answer.

Worked Example 2

Solve:

$$12 + 0$$

Use the identity property. Adding zero keeps the number the same.

$$12 + 0 = 12$$

What we learned: zero does not change the sum.

Worked Example 3

Solve:

$$0 + 18$$

Even when zero is first, the number stays the same.

$$0 + 18 = 18$$

This is still the identity property of addition.

Worked Example 4

Solve:

$$7 + 15$$

You can use the commutative property to rewrite it:

$$7 + 15 = 15 + 7$$

Some students find $$15 + 7$$ easier to think about.

Now add:

$$15 + 7 = 22$$

So:

$$7 + 15 = 22$$

How to tell which property to use

• If the numbers switch places, that is the commutative property.
• If a number is being added to zero, that is the identity property.

Look at these examples:

• $$8 + 3 = 3 + 8$$ → Commutative Property
• $$25 + 0 = 25$$ → Identity Property
• $$0 + 11 = 11$$ → Identity Property

Tips for mental math

You can use these properties to make addition quicker in your head.

• Turn around an addition fact if it feels easier.
• If you see zero, keep the other number.
• Remember: adding numbers in a different order does not change the total.

For example:

• $$1 + 13$$ can be thought of as $$13 + 1 = 14$$
• $$36 + 0 = 36$$
• $$5 + 9$$ can be thought of as $$9 + 5 = 14$$

Let’s compare the two properties

• Commutative Property: numbers switch places.
  Example: $$2 + 6 = 6 + 2$$
• Identity Property: zero is added, and the number stays the same.
  Example: $$2 + 0 = 2$$

Important reminder

These properties are about addition. They help us understand how numbers work and make solving problems easier.

When you practice addition facts, these rules can help you notice patterns. Patterns make math easier to remember.

Summary

The commutative property of addition means you can change the order of the addends, and the sum stays the same.

The identity property of addition means adding zero does not change the number.

When you use these properties, you can solve addition problems more quickly and more easily.

Associative Property for Mental Math helps us add numbers in a smart way.

Sometimes we need to add three or more numbers. The associative property tells us that when we are adding, we can change the grouping of the numbers, and the sum stays the same.

That means we can put parentheses around numbers that are easy to add first.

For example:

$$ (2+8)+5 = 2+(8+5) = 15 $$

Even though the numbers are grouped in a different way, the total is still 15.

This is helpful for mental math, which means solving problems in your head.

When we group numbers in a smart way, we can make friendly numbers like 10, 20, or other easy sums first.

Main Idea: When adding, you can group the addends in different ways.

$$ (a+b)+c = a+(b+c) $$

You do not need to worry about the letters too much. They just stand for numbers. The important idea is this:

• The numbers stay the same.
• The order stays the same.
• Only the grouping changes.

Grouping helps you look for pairs that are easy to add.

How to use the associative property:

1. Look at all the numbers in the addition problem.
2. Find two numbers that make an easy sum, like 10 or 20.
3. Group those numbers together.
4. Add the grouped numbers first.
5. Add the last number.

Let’s look at some examples.

Example 1: Add \(3+7+4\)

First, look for numbers that make 10. We see that \(3+7=10\).

So we can group them like this:

$$ (3+7)+4 $$

Now add:

$$ 10+4=14 $$

So, \(3+7+4=14\).

Example 2: Add \(6+5+4\)

If we add from left to right, we get \(6+5=11\), then \(11+4=15\).

But mental math is easier if we make 10 first. We notice that \(6+4=10\).

So we regroup:

$$ (6+4)+5 = 10+5 = 15 $$

The answer is still 15, but it is easier to do in your head.

Example 3: Add \(9+1+8\)

We notice that \(9+1=10\).

Group those first:

$$ (9+1)+8 = 10+8 = 18 $$

So, \(9+1+8=18\).

Example 4: Add \(12+8+5\)

Look for a friendly pair. We see that \(12+8=20\).

Group those numbers first:

$$ (12+8)+5 = 20+5 = 25 $$

So, \(12+8+5=25\).

Why this works:

When you add numbers, changing the grouping does not change the total.

For example:

$$ (4+6)+3 = 10+3 = 13 $$

and

$$ 4+(6+3) = 4+9 = 13 $$

Both ways give the same answer, 13.

What to look for when using mental math:

• Two numbers that make 10
• Two numbers that make 20
• Any pair that is quick and easy to add

Be careful:

The associative property is about grouping numbers when adding.

You are not changing the numbers. You are not changing the total. You are just choosing which numbers to add first.

For example, in \(5+2+8\), you can group it as:

$$ 5+(2+8) = 5+10 = 15 $$

That makes the problem easier.

Try thinking like this:

• "Do any two numbers make 10?"
• "Do any two numbers make 20?"
• "Which pair is easiest to add first?"

Summary:

The associative property of addition means you can group addends in different ways without changing the sum.

This helps with mental math because you can make friendly numbers first.

When you see three or more numbers being added, look for an easy pair, group them, and add in a smarter way.

Compensation Strategy is a smart way to make addition and subtraction easier in your head.

Sometimes a number is close to a friendly number like 10, 20, 50, or 100. A friendly number is easy to work with. With compensation, we change a number a little to make the math easier, and then we fix the answer by changing it back.

For example, adding 9 can be easier if you think, “Add 10, then subtract 1.” That is compensation.

Why does this work? It works because you are keeping the total balanced. If you add too much, you subtract the extra. If you subtract too much, you add back what you took away.

Let’s learn the two main ways to use compensation.

• For addition: Change a number to a nearby friendly number, do the easier math, then fix the answer.
• For subtraction: Change a number to a nearby friendly number, do the easier math, then fix the answer.

Compensation in addition

When you add, one addend might be close to a friendly number.

Examples:

• 9 is close to 10
• 19 is close to 20
• 29 is close to 30
• 49 is close to 50

If you change 9 to 10, you added 1 too much. So after you add 10, you must subtract 1.

Here is the pattern:

$$a + 9 = a + 10 - 1$$

And another one:

$$a + 19 = a + 20 - 1$$

Worked Example 1

Find \(27 + 9\).

1. Notice that 9 is close to 10.
2. Add 10 instead: \(27 + 10 = 37\)
3. But 10 is 1 more than 9, so subtract 1: \(37 - 1 = 36\)

So, $$27 + 9 = 36$$

This is called compensation because we changed 9 to 10, then fixed the answer.

Worked Example 2

Find \(46 + 19\).

1. Notice that 19 is close to 20.
2. Add 20 instead: \(46 + 20 = 66\)
3. We added 1 too much, so subtract 1: \(66 - 1 = 65\)

So, $$46 + 19 = 65$$

Compensation in subtraction

You can also use compensation when subtracting.

If a number like 9 or 19 is being subtracted, you can subtract a friendly number instead, then fix the answer.

Examples:

• Subtract 9 by subtracting 10, then adding 1 back.
• Subtract 19 by subtracting 20, then adding 1 back.

Why add back? Because when you subtracted 10 instead of 9, you took away 1 too much. You need to put that 1 back.

Here is the pattern:

$$a - 9 = a - 10 + 1$$

And:

$$a - 19 = a - 20 + 1$$

Worked Example 3

Find \(54 - 9\).

1. Notice that 9 is close to 10.
2. Subtract 10 instead: \(54 - 10 = 44\)
3. But you subtracted 1 too much, so add 1 back: \(44 + 1 = 45\)

So, $$54 - 9 = 45$$

Worked Example 4

Find \(72 - 19\).

1. Notice that 19 is close to 20.
2. Subtract 20 instead: \(72 - 20 = 52\)
3. You took away 1 too much, so add 1 back: \(52 + 1 = 53\)

So, $$72 - 19 = 53$$

How to know what to do

Ask yourself these questions:

1. Is one number close to a friendly number like 10, 20, 30, 50, or 100?
2. Can I use that friendly number to make the math easier?
3. Did I use too much or too little?
4. How do I fix the answer?

Helpful thinking for addition

• If you add 1 more than needed, subtract 1 at the end.
• If you add 2 more than needed, subtract 2 at the end.

Example:

$$35 + 28 = 35 + 30 - 2$$

Since 28 is close to 30, add 30, then subtract 2.

Let’s solve it:

$$35 + 30 = 65$$

$$65 - 2 = 63$$

So, $$35 + 28 = 63$$

Helpful thinking for subtraction

• If you subtract 1 more than needed, add 1 back.
• If you subtract 2 more than needed, add 2 back.

Example:

$$81 - 29 = 81 - 30 + 1$$

Since 29 is close to 30, subtract 30, then add 1 back.

Let’s solve it:

$$81 - 30 = 51$$

$$51 + 1 = 52$$

So, $$81 - 29 = 52$$

When compensation is useful

Compensation is helpful when numbers are close to friendly numbers.

• 8, 9 are close to 10
• 18, 19 are close to 20
• 28, 29 are close to 30
• 48, 49 are close to 50

It helps you solve problems faster and more accurately in your head.

Be careful!

The most important part of compensation is the fix.

• If you added too much, subtract the extra.
• If you subtracted too much, add the extra back.

Look at this mistake:

Someone solves \(24 + 9\) like this:

$$24 + 10 = 34$$

But they stop there.

That answer is not correct, because they added 10 instead of 9. They added 1 too much.

The correct fix is:

$$34 - 1 = 33$$

So, $$24 + 9 = 33$$

Try the steps every time

1. Find a number close to a friendly number.
2. Use the friendly number to solve the easier problem.
3. Fix the answer by adding back or subtracting the extra amount.

Summary

Compensation is a mental math strategy. You change a number to a nearby friendly number to make the math easier.

Then you compensate, or fix the answer, because you changed the number a little.

• Addition: Add a friendly number, then subtract the extra.
• Subtraction: Subtract a friendly number, then add back the extra.

With practice, compensation can help you solve math problems quickly and confidently.

Regrouping Across Multiple Places means that when we add, sometimes we make a new ten and a new hundred in the same problem.

This happens when the ones add up to 10 or more, and then the tens also add up to 10 or more after we regroup.

We can solve these problems carefully by going place by place: ones, then tens, then hundreds.

Remember the place values:

• Ones are single cubes.
• Tens are groups of 10.
• Hundreds are groups of 100.

When 10 ones are made, they can be regrouped into 1 ten.

When 10 tens are made, they can be regrouped into 1 hundred.

How to Add with Regrouping Across More Than One Place

1. Add the ones. If the sum is 10 or more, write the ones digit and regroup 1 ten.
2. Add the tens. Do not forget to add the regrouped ten. If the sum is 10 or more, write the tens digit and regroup 1 hundred.
3. Add the hundreds. Include any regrouped hundred.

Tip: Line up the digits by place value. Ones go under ones, tens under tens, and hundreds under hundreds.

Worked Example 1

Add:

$$ 58 + 27 $$

Step 1: Add the ones.

\(8 + 7 = 15\). That is 1 ten and 5 ones.

Write 5 in the ones place. Regroup the 1 ten.

Step 2: Add the tens.

\(5 + 2 + 1 = 8\) tens.

So the answer is:

$$ 58 + 27 = 85 $$

In this problem, we regrouped once.

Worked Example 2

Add:

$$ 278 + 145 $$

Step 1: Add the ones.

\(8 + 5 = 13\). That is 1 ten and 3 ones.

Write 3 in the ones place. Regroup 1 ten.

Step 2: Add the tens.

\(7 + 4 + 1 = 12\) tens.

That is 1 hundred and 2 tens.

Write 2 in the tens place. Regroup 1 hundred.

Step 3: Add the hundreds.

\(2 + 1 + 1 = 4\) hundreds.

So the answer is:

$$ \begin{aligned} &\phantom{1}278\\ + &\phantom{1}145\\ \hline &\phantom{1}423 \end{aligned} $$

This is regrouping across multiple places because we made a new ten and then a new hundred.

Let’s Look Closely

In Example 2, the regrouping happened like this:

• \(8 + 5 = 13\), so 10 ones became 1 ten.
• Then the tens were \(7 + 4 + 1 = 12\), so 10 tens became 1 hundred.

The regrouping moved from the ones place to the tens place, and then from the tens place to the hundreds place.

That is why we call it across multiple places.

Worked Example 3

Add:

$$ 396 + 128 $$

Step 1: Add the ones.

\(6 + 8 = 14\). That is 1 ten and 4 ones.

Write 4 in the ones place. Regroup 1 ten.

Step 2: Add the tens.

\(9 + 2 + 1 = 12\) tens.

That is 1 hundred and 2 tens.

Write 2 in the tens place. Regroup 1 hundred.

Step 3: Add the hundreds.

\(3 + 1 + 1 = 5\) hundreds.

So the answer is:

$$ 396 + 128 = 524 $$

Worked Example 4

Add:

$$ 467 + 356 $$

Step 1: Add the ones.

\(7 + 6 = 13\). Write 3 ones and regroup 1 ten.

Step 2: Add the tens.

\(6 + 5 + 1 = 12\) tens.

Write 2 tens and regroup 1 hundred.

Step 3: Add the hundreds.

\(4 + 3 + 1 = 8\) hundreds.

So the answer is:

$$ 467 + 356 = 823 $$

What to Watch Out For

• Always start with the ones. If you start somewhere else, it is easy to miss regrouping.
• Do not forget the regrouped 1. That extra ten or hundred must be added in the next place.
• Keep digits lined up. If the numbers are not lined up by place value, the answer can be wrong.

A Helpful Way to Think About It

Imagine you have blocks.

• If you get 10 ones, trade them for 1 ten.
• If you get 10 tens, trade them for 1 hundred.

Adding is like making bigger groups when you have enough smaller ones.

Summary

Regrouping across multiple places happens when you make a new ten and then a new hundred in the same addition problem.

To solve these problems, add the ones first, then the tens, then the hundreds.

Each time you make 10 of something, regroup it into the next bigger place.

If you work carefully and remember each regrouped 1, you can solve big addition problems correctly.

Adding Multiples of 10 and 100 means adding numbers like 10, 20, 30, 100, 200, and 300.

These are called round numbers because they end in zero. We can add them quickly in our heads by using place value.

When we add multiples of 10, we are adding tens. When we add multiples of 100, we are adding hundreds.

For example, in the number \(40\), there are 4 tens. In the number \(300\), there are 3 hundreds.

Big idea: We do not need to count by ones. We can add the tens to tens and the hundreds to hundreds.

Think about place value

• \(10 = 1\) ten
• \(20 = 2\) tens
• \(50 = 5\) tens
• \(100 = 1\) hundred
• \(400 = 4\) hundreds
• \(700 = 7\) hundreds

If we know how many tens or hundreds are in each number, we can add them easily.

Adding multiples of 10

When both numbers are multiples of 10, add the tens.

Example: \(30 + 20\)

\(30\) is 3 tens. \(20\) is 2 tens.

$$3\text{ tens} + 2\text{ tens} = 5\text{ tens}$$

$$30 + 20 = 50$$

You can also think: \(3 + 2 = 5\), so \(30 + 20 = 50\).

Adding multiples of 100

When both numbers are multiples of 100, add the hundreds.

Example: \(400 + 300\)

\(400\) is 4 hundreds. \(300\) is 3 hundreds.

$$4\text{ hundreds} + 3\text{ hundreds} = 7\text{ hundreds}$$

$$400 + 300 = 700$$

You can also think: \(4 + 3 = 7\), so \(400 + 300 = 700\).

Adding a multiple of 10 to another number

Sometimes only one number is a multiple of 10. You can still use place value.

Example: \(46 + 20\)

The \(46\) has 4 tens and 6 ones. The \(20\) is 2 tens.

Add the tens: 4 tens + 2 tens = 6 tens. The ones stay 6.

$$46 + 20 = 66$$

Notice that the ones digit did not change. We only added tens.

Adding a multiple of 100 to another number

Example: \(245 + 300\)

The \(245\) has 2 hundreds, 4 tens, and 5 ones. The \(300\) is 3 hundreds.

Add the hundreds: 2 hundreds + 3 hundreds = 5 hundreds. The tens and ones stay the same.

$$245 + 300 = 545$$

Notice that the tens and ones did not change. We only added hundreds.

Worked Examples

1. Example 1: \(20 + 50\)

   \(20\) is 2 tens. \(50\) is 5 tens.

   $$2\text{ tens} + 5\text{ tens} = 7\text{ tens}$$

   $$20 + 50 = 70$$

2. Example 2: \(600 + 200\)

   \(600\) is 6 hundreds. \(200\) is 2 hundreds.

   $$6\text{ hundreds} + 2\text{ hundreds} = 8\text{ hundreds}$$

   $$600 + 200 = 800$$

3. Example 3: \(37 + 40\)

   \(37\) has 3 tens and 7 ones. \(40\) is 4 tens.

   Add the tens: \(3 + 4 = 7\) tens. Keep the 7 ones.

   $$37 + 40 = 77$$

4. Example 4: \(128 + 500\)

   \(128\) has 1 hundred, 2 tens, and 8 ones. \(500\) is 5 hundreds.

   Add the hundreds: \(1 + 5 = 6\) hundreds. Keep the 2 tens and 8 ones.

   $$128 + 500 = 628$$

Helpful ways to think

• If you add tens, the tens digit changes.
• If you add hundreds, the hundreds digit changes.
• The other digits stay the same if there is no regrouping.
• Look at what place value you are adding: tens or hundreds.

Watch out for this mistake

Do not count by ones unless you need to. For \(50 + 30\), do not count 51, 52, 53, and so on.

Instead, think: 5 tens + 3 tens = 8 tens, so \(50 + 30 = 80\).

For \(160 + 200\), think: 1 hundred + 2 hundreds = 3 hundreds, and keep the 6 tens.

$$160 + 200 = 360$$

Summary

Adding multiples of 10 and 100 is easier when you use place value. Add tens to tens and hundreds to hundreds.

If you add a multiple of 10, the tens change. If you add a multiple of 100, the hundreds change.

You can solve these problems mentally without counting by ones.

Partial Sums Algorithm is a way to add bigger numbers by breaking them into parts.

We add the hundreds, tens, and ones separately. Then we put those parts together to get the total.

This strategy helps us see what each digit is worth. It also helps us stay organized when adding multi-digit numbers.

Let’s remember place value first:

• The digit on the right is the ones place.
• The middle digit is the tens place.
• The digit on the left is the hundreds place.

For example, in the number \(364\):

• \(3\) means \(300\)
• \(6\) means \(60\)
• \(4\) means \(4\)

So, \(364 = 300 + 60 + 4\).

How the partial sums algorithm works:

1. Write the numbers so the place values line up.
2. Add the ones.
3. Add the tens.
4. Add the hundreds.
5. Add the partial sums together.

The separate answers you get are called partial sums.

Here is a simple example of the idea:

To add \(23 + 14\), we can think:

• Ones: \(3 + 4 = 7\)
• Tens: \(20 + 10 = 30\)

Then combine the parts:

$$30 + 7 = 37$$

So, \(23 + 14 = 37\).

Now let’s try bigger numbers.

Worked Example 1

Add \(145 + 123\).

First, break each number into hundreds, tens, and ones:

• \(145 = 100 + 40 + 5\)
• \(123 = 100 + 20 + 3\)

Add each place value:

• Hundreds: \(100 + 100 = 200\)
• Tens: \(40 + 20 = 60\)
• Ones: \(5 + 3 = 8\)

Now add the partial sums:

$$200 + 60 + 8 = 268$$

So, \(145 + 123 = 268\).

Worked Example 2

Add \(256 + 132\).

Break apart the numbers:

• \(256 = 200 + 50 + 6\)
• \(132 = 100 + 30 + 2\)

Add each place:

• Hundreds: \(200 + 100 = 300\)
• Tens: \(50 + 30 = 80\)
• Ones: \(6 + 2 = 8\)

Combine the partial sums:

$$300 + 80 + 8 = 388$$

So, \(256 + 132 = 388\).

Worked Example 3

Add \(347 + 185\).

Break apart the numbers:

• \(347 = 300 + 40 + 7\)
• \(185 = 100 + 80 + 5\)

Add each place value:

• Hundreds: \(300 + 100 = 400\)
• Tens: \(40 + 80 = 120\)
• Ones: \(7 + 5 = 12\)

Now combine the partial sums:

$$400 + 120 + 12 = 532$$

So, \(347 + 185 = 532\).

Notice something important here: sometimes a partial sum can be greater than \(9\) or greater than \(90\). That is okay. We still add the parts at the end.

In this example, \(40 + 80 = 120\) and \(7 + 5 = 12\). We do not need to worry. We just add:

$$400 + 120 + 12 = 532$$

Worked Example 4

Add \(468 + 257\).

Break apart the numbers:

• \(468 = 400 + 60 + 8\)
• \(257 = 200 + 50 + 7\)

Add each place value:

• Hundreds: \(400 + 200 = 600\)
• Tens: \(60 + 50 = 110\)
• Ones: \(8 + 7 = 15\)

Add the partial sums together:

$$600 + 110 + 15 = 725$$

So, \(468 + 257 = 725\).

You can also write the work in a column.

For \(468 + 257\), one way is:

$$\begin{aligned}468 \\ +257 \\ \hline 600\text{ hundreds} \\ 110\text{ tens} \\ 15\text{ ones}\end{aligned}$$

Then add those partial sums:

$$600 + 110 + 15 = 725$$

Why partial sums is helpful

• It uses place value.
• It helps you see what each digit means.
• It makes big addition problems easier to understand.
• It helps you check your work step by step.

Tips for success

• Line up the ones, tens, and hundreds carefully.
• Say the value of each digit, not just the digit. For example, say \(50\), not just \(5\).
• Add one place at a time.
• At the end, add all the partial sums carefully.

Common mistake to watch for

A common mistake is forgetting place value.

For example, in \(243\), the \(4\) means \(40\), not \(4\). If you forget that, your answer will be wrong.

Another mistake is not lining up the digits correctly. Always make sure hundreds are with hundreds, tens are with tens, and ones are with ones.

Let’s review the steps one more time:

1. Line up the numbers by place value.
2. Add the hundreds.
3. Add the tens.
4. Add the ones.
5. Add the partial sums to find the total.

Summary

The partial sums algorithm is a way to add by parts. We add hundreds, tens, and ones separately, then combine those sums.

This strategy helps us understand place value and solve multi-digit addition problems clearly and carefully. With practice, partial sums can make big addition feel much easier.

Standard Addition Algorithm without Regrouping

Today we will learn how to add numbers using the standard addition algorithm. This is a careful way to line up numbers and add them by place value.

We will focus on problems without regrouping. That means when we add the digits in each place, the sum is 9 or less, so we do not need to carry to the next place.

For example, in the problem \(23 + 14\), the ones digits are \(3 + 4 = 7\), and the tens digits are \(2 + 1 = 3\). Since both sums are 9 or less, there is no regrouping.

What is place value?

Place value tells us what each digit means. In a 2-digit number, the digit on the right is the ones digit. The digit on the left is the tens digit.

• In \(45\), the \(4\) means 4 tens.
• In \(45\), the \(5\) means 5 ones.

When we use the standard algorithm, we must line up the digits so that ones are under ones and tens are under tens. If the numbers are not lined up correctly, the answer will not be correct.

Steps for the standard addition algorithm

1. Write the numbers one above the other.
2. Line up the digits by place value.
3. Start with the ones column.
4. Add the digits in each column.
5. Then add the tens column.
6. Write the sum.

Since we are only working on problems without regrouping, each column will have a sum of 9 or less.

Worked Example 1: Add a 1-digit number to a 2-digit number

Let’s solve \(21 + 6\).

First, line up the ones digits. The 6 goes in the ones place.

Now add the ones: \(1 + 6 = 7\).

The tens digit stays the same: \(2\) tens.

So, \(21 + 6 = 27\).

Worked Example 2: Add two 2-digit numbers

Let’s solve \(34 + 25\).

Line up tens under tens and ones under ones.

It is usually written like this:

Step 1: Add the ones. \(4 + 5 = 9\).

Step 2: Add the tens. \(3 + 2 = 5\).

So, \(34 + 25 = 59\).

Worked Example 3: Another 2-digit addition problem

Let’s solve \(42 + 17\).

Add the ones: \(2 + 7 = 9\).

Add the tens: \(4 + 1 = 5\).

So, \(42 + 17 = 59\).

Worked Example 4: Add two 3-digit numbers

Now let’s try a bigger problem: \(243 + 526\).

Line up the digits by place value: hundreds, tens, and ones.

Start with the ones: \(3 + 6 = 9\).

Next add the tens: \(4 + 2 = 6\).

Then add the hundreds: \(2 + 5 = 7\).

So, \(243 + 526 = 769\).

How do I know there is no regrouping?

Look at each column before or while you add.

• If the ones add to 9 or less, no regrouping is needed in the ones place.
• If the tens add to 9 or less, no regrouping is needed in the tens place.
• The same is true for hundreds.

For example, in \(243 + 526\):

• Ones: \(3 + 6 = 9\)
• Tens: \(4 + 2 = 6\)
• Hundreds: \(2 + 5 = 7\)

Every column is 9 or less, so there is no regrouping.

Common mistakes to watch out for

• Not lining up place values: Ones must go under ones. Tens must go under tens.
• Starting in the wrong place: Start with the ones column, then move left.
• Adding across instead of by columns: Do not add \(34 + 25\) by mixing digits in the wrong places.

Helpful tip

If you are not sure whether the numbers are lined up correctly, look straight down each column.

• ones column
• tens column
• hundreds column, if there is one

Each column should contain only one kind of place value.

Let’s review with one more quick check

Solve \(56 + 13\).

Line up the numbers:

Add the ones: \(6 + 3 = 9\).

Add the tens: \(5 + 1 = 6\).

So, \(56 + 13 = 69\).

Summary

The standard addition algorithm helps us add numbers in an organized way. We line up digits by place value, start with the ones, and add each column. When each column adds to 9 or less, we can solve the problem without regrouping.

Standard Addition Algorithm with Regrouping

When we add bigger numbers, sometimes the digits in one place add up to 10 or more. When that happens, we need to regroup.

Regrouping means making a new ten from 10 ones, or making a new hundred from 10 tens. In the standard addition algorithm, people sometimes call this a carry. But it is not just a little mark above the number. It shows that we made a new ten or a new hundred.

For example, if we add 8 ones and 7 ones, we get 15 ones. And 15 ones is the same as 1 ten and 5 ones.

$$15 = 1\text{ ten } + 5\text{ ones}$$

That new ten is written in the tens place. This is why regrouping is connected to place value.

Why place value matters

• Ones are single units.
• 10 ones = 1 ten
• 10 tens = 1 hundred

When a place has 10 or more, we regroup to the next place to the left.

Steps for the standard addition algorithm with regrouping

1. Write the numbers in columns so the ones, tens, and hundreds line up.
2. Start adding from the ones place.
3. If the sum is 10 or more, write the ones digit in that place.
4. Regroup the extra ten to the tens place.
5. Add the tens, including the regrouped ten if there is one.
6. If the tens sum is 10 or more, write the tens digit and regroup a hundred.
7. Add the hundreds.

Worked Example 1: Regrouping ones into a ten

Let us add \(27 + 15\).

First, line up the digits by place value.

$$ \begin{array}{r} \phantom{1}27 \\ +15 \\ \hline \end{array} $$

Now add the ones: \(7 + 5 = 12\).

12 ones means 1 ten and 2 ones. Write the 2 in the ones place. Regroup the 1 ten to the tens place.

$$ \begin{array}{r} \phantom{1}^1\!27 \\ +15 \\ \hline \phantom{1}2 \end{array} $$

Now add the tens: \(1 + 2 + 1 = 4\) tens.

$$ \begin{array}{r} \phantom{1}^1\!27 \\ +15 \\ \hline 42 \end{array} $$

So, \(27 + 15 = 42\).

What happened? We had 12 ones. We kept 2 ones and regrouped 10 ones as 1 ten.

Worked Example 2: Another 2-digit example

Now add \(46 + 38\).

$$ \begin{array}{r} \phantom{1}46 \\ +38 \\ \hline \end{array} $$

Add the ones: \(6 + 8 = 14\).

14 ones is 1 ten and 4 ones. Write 4 in the ones place and regroup 1 ten.

$$ \begin{array}{r} \phantom{1}^1\!46 \\ +38 \\ \hline \phantom{1}4 \end{array} $$

Add the tens: \(1 + 4 + 3 = 8\) tens.

$$ \begin{array}{r} \phantom{1}^1\!46 \\ +38 \\ \hline 84 \end{array} $$

So, \(46 + 38 = 84\).

Worked Example 3: Regrouping into the hundreds

Now let us add \(158 + 67\).

It helps to write \(67\) as \(067\) so the places line up.

$$ \begin{array}{r} 158 \\ +\,67 \\ \hline \end{array} $$

Add the ones: \(8 + 7 = 15\).

15 ones is 1 ten and 5 ones. Write 5 in the ones place and regroup 1 ten.

Add the tens: \(1 + 5 + 6 = 12\) tens.

12 tens is 1 hundred and 2 tens. Write 2 in the tens place and regroup 1 hundred.

Add the hundreds: \(1 + 1 = 2\) hundreds.

$$ \begin{array}{r} \phantom{1}^1\!58 \\ +\,67 \\ \hline 225 \end{array} $$

So, \(158 + 67 = 225\).

Let us see the place value clearly:

• \(8 + 7 = 15\) ones, so regroup to make 1 ten.
• Then \(5\) tens plus \(6\) tens plus the regrouped \(1\) ten makes \(12\) tens.
• \(12\) tens is the same as \(1\) hundred and \(2\) tens.

Worked Example 4: Regrouping more than once

Add \(286 + 175\).

$$ \begin{array}{r} 286 \\ 175 \\ \hline \end{array} $$

Add the ones: \(6 + 5 = 11\).

11 ones is 1 ten and 1 one. Write 1 in the ones place and regroup 1 ten.

Add the tens: \(1 + 8 + 7 = 16\) tens.

16 tens is 1 hundred and 6 tens. Write 6 in the tens place and regroup 1 hundred.

Add the hundreds: \(1 + 2 + 1 = 4\) hundreds.

$$ \begin{array}{r} \phantom{1}^1\!286 \\ 175 \\ \hline 461 \end{array} $$

So, \(286 + 175 = 461\).

Important idea: Every regrouped number has a value.

• The small 1 above the tens place means 1 ten, not just 1.
• The small 1 above the hundreds place means 1 hundred, not just 1.

This is why we must always pay attention to the place.

Common mistakes to watch out for

• Not lining up place values. Ones must go under ones. Tens must go under tens.
• Forgetting to regroup. If a sum is 10 or more, move the extra ten or hundred to the next place.
• Forgetting to add the regrouped number. That small 1 must be added in the next column.
• Starting on the left. In the standard algorithm, start with the ones place.

Helpful reminders

• Start on the right.
• Add the ones first.
• If the sum is 10 or more, regroup.
• The regrouped digit means a new ten or a new hundred.
• Then add the next place.

Try thinking with place value

Suppose you add \(39 + 24\).

• Ones: \(9 + 4 = 13\) ones
• \(13\) ones = \(1\) ten and \(3\) ones
• Tens: \(3\) tens + \(2\) tens + \(1\) regrouped ten = \(6\) tens

So, \(39 + 24 = 63\).

This shows that regrouping is really about making new groups of ten.

Summary

The standard addition algorithm helps us add numbers neatly by place value. When the sum in a place is 10 or more, we regroup. That means 10 ones become 1 ten, or 10 tens become 1 hundred.

The little carry mark is important because it shows the new group you made. If you line up the digits, start with the ones, regroup when needed, and add the regrouped value in the next column, you can solve addition problems correctly.

Regrouping Across Multiple Places means that when we add, we may need to make a new ten and a new hundred in the same problem.

This happens when the ones add up to 10 or more, so we regroup into the tens place, and then the tens add up to 10 or more, so we regroup again into the hundreds place.

For example, in \(295 + 18\), the ones make a new ten, and then the tens make a new hundred. Let’s learn how to do that step by step.

First, remember place value:

• Ones are the last digits.
• Tens are the middle digits.
• Hundreds are the next digits.

In the number \(295\):

• 2 means 2 hundreds
• 9 means 9 tens
• 5 means 5 ones

In the number \(18\):

• 1 means 1 ten
• 8 means 8 ones

How regrouping works

1. Add the ones first.
2. If the ones total is 10 or more, make a new ten and regroup it to the tens place.
3. Add the tens.
4. If the tens total is 10 or more, make a new hundred and regroup it to the hundreds place.
5. Add the hundreds.

When we regroup, we are not changing the value. We are just trading:

• 10 ones for 1 ten
• 10 tens for 1 hundred

Worked Example 1

Add \(27 + 15\).

This example has regrouping in just one place, so it helps us get ready.

Add the ones: \(7 + 5 = 12\).

12 ones means 1 ten and 2 ones. Write the 2 in the ones place, and regroup the 1 ten.

Add the tens: \(2 + 1 + 1 = 4\) tens.

So, \(27 + 15 = 42\).

Worked Example 2

Add \(295 + 18\).

Step 1: Add the ones.

\(5 + 8 = 13\).

13 ones means 1 ten and 3 ones. Write 3 in the ones place, and regroup 1 ten to the tens place.

Step 2: Add the tens.

Now add the tens: \(9 + 1 + 1 = 11\) tens.

The tens are:

• 9 tens from 295
• 1 ten from 18
• 1 regrouped ten from the ones

11 tens means 1 hundred and 1 ten. Write 1 in the tens place, and regroup 1 hundred to the hundreds place.

Step 3: Add the hundreds.

\(2 + 1 = 3\) hundreds.

So, \(295 + 18 = 313\).

Why did we regroup twice?

• The ones made a new ten.
• Then the tens made a new hundred.

Worked Example 3

Add \(486 + 27\).

Step 1: Add the ones.

\(6 + 7 = 13\).

Write 3 ones. Regroup 1 ten.

Step 2: Add the tens.

\(8 + 2 + 1 = 11\) tens.

Write 1 ten. Regroup 1 hundred.

Step 3: Add the hundreds.

\(4 + 1 = 5\) hundreds.

So, \(486 + 27 = 513\).

Worked Example 4

Add \(168 + 57\).

Step 1: Add the ones.

\(8 + 7 = 15\).

Write 5 ones. Regroup 1 ten.

Step 2: Add the tens.

\(6 + 5 + 1 = 12\) tens.

Write 2 tens. Regroup 1 hundred.

Step 3: Add the hundreds.

\(1 + 1 = 2\) hundreds.

So, \(168 + 57 = 225\).

Helpful tips

• Always line up ones under ones, tens under tens, and hundreds under hundreds.
• Start adding from the ones place.
• If a place has 10 or more, regroup to the next place.
• Check if the tens also need regrouping after you add the regrouped ten.

A place value way to think about it

Look again at \(295 + 18\):

• \(295 = 2\) hundreds, \(9\) tens, \(5\) ones
• \(18 = 1\) ten, \(8\) ones

Add the ones: \(5 + 8 = 13\) ones, which is 1 ten and 3 ones.

Now the tens are \(9 + 1 + 1 = 11\) tens, which is 1 hundred and 1 ten.

Now the hundreds are \(2 + 1 = 3\) hundreds.

That gives us 3 hundreds, 1 ten, and 3 ones, or \(313\).

Watch out for these mistakes

• Forgetting to add the regrouped 1 to the tens place.
• Writing 13 in the ones place instead of writing 3 and regrouping 1 ten.
• Forgetting that the tens might also need regrouping.
• Not lining up the digits by place value.

Let’s review the big idea

Sometimes one regrouping leads to another regrouping. The ones can make a new ten, and then the tens can make a new hundred.

If you go one place at a time and remember your regrouped numbers, you can solve these problems carefully and correctly.

Result, Change, and Start Unknown Word Problems

Word problems can tell addition stories in different ways. Sometimes we know the two parts and need the result. Sometimes we know the start and the result, and we need the change. Sometimes we know the change and the result, and we need the start.

In this lesson, you will learn how to spot each kind of problem and how to solve it step by step.

There are 3 main kinds of addition word problems:

• Result Unknown: We know the start and the change. We find the total.
  Example form: \(A + B = ?\)
• Change Unknown: We know the start and the total. We find how many more were added.
  Example form: \(A + ? = C\)
• Start Unknown: We know the change and the total. We find what number we started with.
  Example form: \(? + B = C\)

Let’s look at what these words mean.

• Start means how many there were in the beginning.
• Change means how many were added.
• Result means how many there are in all at the end.

A helpful way to think about it is:

Start + Change = Result

In math, that looks like:

$$\text{Start} + \text{Change} = \text{Result}$$

Step-by-step strategy

1. Read the problem slowly.
2. Ask: What do I know?
3. Ask: What is missing?
4. Write an equation with a box or question mark for the missing number.
5. Solve.
6. Check if your answer makes sense in the story.

1. Result Unknown Problems

In a result unknown problem, we know the start and the change. We need to find the total at the end.

These problems often ask:

• How many are there now?
• How many in all?
• How many altogether?

Worked Example 1

Lena had 7 stickers. Her friend gave her 5 more stickers. How many stickers does Lena have now?

Step 1: Find the start. Lena started with 7 stickers.

Step 2: Find the change. She got 5 more stickers.

Step 3: Find the result. That is the part we do not know.

Equation:

$$7 + 5 = ?$$

Solve:

$$7 + 5 = 12$$

Answer: Lena has 12 stickers now.

This is a result unknown problem because the total was missing.

2. Change Unknown Problems

In a change unknown problem, we know the start and the result. We need to find how many were added.

These problems often ask:

• How many more?
• How many were added?
• How many did she get?

Even though this is an addition story, we can use what we know about missing numbers to solve it.

Worked Example 2

Marcus had 9 toy cars. Then he got some more toy cars. Now he has 15 toy cars. How many toy cars did he get?

Step 1: Find the start. Marcus started with 9 toy cars.

Step 2: Find the result. He ended with 15 toy cars.

Step 3: Find the missing change.

Equation:

$$9 + ? = 15$$

Think: What number goes with 9 to make 15?

We can count on:

• 10, 11, 12, 13, 14, 15

That is 6 numbers.

So:

$$9 + 6 = 15$$

Answer: Marcus got 6 toy cars.

This is a change unknown problem because the amount added was missing.

3. Start Unknown Problems

In a start unknown problem, we know the change and the result. We need to find the beginning number.

These can feel a little trickier, but the same thinking helps.

These problems often ask:

• How many were there at first?
• How many did she start with?
• What was the beginning number?

Worked Example 3

Ava had some books. She got 8 more books. Now she has 14 books. How many books did Ava have at first?

Step 1: Find the change. Ava got 8 more books.

Step 2: Find the result. Now she has 14 books.

Step 3: Find the missing start.

Equation:

$$? + 8 = 14$$

Think: What number goes with 8 to make 14?

We know:

$$6 + 8 = 14$$

Answer: Ava had 6 books at first.

This is a start unknown problem because the beginning number was missing.

Ways to solve missing-number addition problems

When the result is missing, you can add the two known numbers.

When the change or start is missing, you can:

• Count on
• Use a number line
• Think of a fact you know
• Use the related subtraction idea to check your answer

For example, in \(9 + ? = 15\), you can think, “What goes from 9 to 15?”

Or you can check with subtraction:

$$15 - 9 = 6$$

So the missing number is 6.

Worked Example 4

There were some birds in a tree. Then 7 more birds landed in the tree. Now there are 18 birds in the tree. How many birds were in the tree at first?

Let’s find the parts of the story.

• Change: 7 more birds landed.
• Result: 18 birds are in the tree now.
• Start: We do not know.

Equation:

$$? + 7 = 18$$

Think: What number plus 7 equals 18?

$$11 + 7 = 18$$

So the answer is 11.

Answer: There were 11 birds in the tree at first.

How to tell which kind of problem it is

You can ask yourself one simple question:

What is missing?

• If the total at the end is missing, it is result unknown.
• If the amount added is missing, it is change unknown.
• If the beginning amount is missing, it is start unknown.

Look at these forms again:

• Result Unknown: \(8 + 4 = ?\)
• Change Unknown: \(8 + ? = 12\)
• Start Unknown: \(? + 4 = 12\)

All 3 are addition stories, but the missing number is in a different place each time.

Helpful clues in word problems

Words like more, got, added, and joined often tell you the story is about addition.

But do not just look for one word. Read the whole story and decide:

• What happened first?
• What was added?
• What is the final amount?

Check your answer

After you solve, put your answer back into the story.

For example, if the problem is:

$$? + 8 = 14$$

and you say the answer is 6, then check:

$$6 + 8 = 14$$

It works, so the answer makes sense.

Summary

• Start + Change = Result
• Result unknown: Find the total. Example: \(A + B = ?\)
• Change unknown: Find how many were added. Example: \(A + ? = C\)
• Start unknown: Find the beginning number. Example: \(? + B = C\)
• Always ask, What is missing?
• Write an equation to help solve the problem.
• Check that your answer fits the story.

You can solve these problems by reading carefully, finding the missing part, and using addition facts you know. With practice, you will get faster at seeing whether the missing number is the result, the change, or the start.