Addition and Subtraction Strategies Within 100

Adding and Subtracting Multiples of 10

Today we will learn how to add and subtract multiples of 10.

A multiple of 10 is a number like 10, 20, 30, 40, 50, and so on. These numbers have 0 ones.

When we add or subtract a multiple of 10, we are really adding or subtracting tens. The ones digit stays the same.

For example, in the number \(34\), there are 3 tens and 4 ones.

If we add 10, we add 1 more ten. If we subtract 10, we take away 1 ten.

That means:

Do you see what happened? The tens digit changed, but the ones digit stayed 4.

Big idea: When you add or subtract tens, the ones stay the same.

Let’s look at how this works.

• Add 10 means add 1 ten.
• Add 20 means add 2 tens.
• Add 30 means add 3 tens.
• Subtract 10 means take away 1 ten.
• Subtract 20 means take away 2 tens.
• Subtract 30 means take away 3 tens.

You can think about the tens digit moving up or down, while the ones digit does not change.

Here is a helpful pattern:

Each time we add 10, the ones digit stays 7.

Here is another pattern:

Each time we subtract 10, the ones digit stays 5.

Worked Example 1

Solve: \(23 + 10\)

The number \(23\) has 2 tens and 3 ones.

Adding \(10\) means adding 1 more ten.

So now there are 3 tens and still 3 ones.

Worked Example 2

Solve: \(46 + 20\)

The number \(46\) has 4 tens and 6 ones.

Adding \(20\) means adding 2 tens.

Now there are 6 tens and still 6 ones.

The ones digit stayed 6.

Worked Example 3

Solve: \(78 - 10\)

The number \(78\) has 7 tens and 8 ones.

Subtracting \(10\) means taking away 1 ten.

Now there are 6 tens and still 8 ones.

Worked Example 4

Solve: \(91 - 30\)

The number \(91\) has 9 tens and 1 one.

Subtracting \(30\) means taking away 3 tens.

Now there are 6 tens and still 1 one.

How to solve these problems

1. Look at the two-digit number.
2. Find the ones digit. It will stay the same.
3. Look at the multiple of 10.
4. Add or subtract that many tens.
5. Write the new number.

Let’s try a few quick thoughts:

• \(52 + 10\): the ones digit stays \(2\), so the answer is \(62\).
• \(35 + 40\): the ones digit stays \(5\), so the answer is \(75\).
• \(88 - 20\): the ones digit stays \(8\), so the answer is \(68\).
• \(64 - 30\): the ones digit stays \(4\), so the answer is \(34\).

Watch out!

• Do not change the ones digit when adding or subtracting tens.
• Remember that \(20\) means 2 tens, not 2 ones.
• Remember that \(30\) means 3 tens, not 3 ones.

For example, \(47 + 20\) is not \(49\).

Why not? Because \(20\) means 2 tens.

So we add 2 to the tens digit:

The ones digit stays 7.

You can also use a number line in your mind. If you start at \(32\) and add \(10\), you jump to \(42\). Add another \(10\), and you jump to \(52\).

That is because adding \(20\) is the same as adding 2 jumps of 10.

Let’s remember the rule:

• Adding multiples of 10 changes the tens digit.
• Subtracting multiples of 10 changes the tens digit.
• The ones digit stays the same.

Summary

When you add or subtract a multiple of 10, you are adding or subtracting tens. The ones digit does not change.

If you know how many tens to add or take away, you can solve these problems quickly and correctly.

Adding a Two-Digit and One-Digit Number

Today we will learn how to add a two-digit number and a one-digit number.

A two-digit number has tens and ones. For example, in \(27\), the \(2\) means 2 tens, and the \(7\) means 7 ones.

When we add a one-digit number, we usually add it to the ones place. Sometimes the ones make a new ten. This is called making a ten.

Making a ten helps us add more easily, especially when we cross from one ten to the next, like going from \(29\) to \(30\).

Main Idea

• Start with the two-digit number.
• Add the one-digit number to the ones.
• If the ones become 10 or more, make a new ten.
• Then find the new number.

Let’s look at how this works.

Example 1: No new ten

Solve \(24 + 3\).

The number \(24\) has 2 tens and 4 ones.

Add 3 more ones to 4 ones:

The tens stay the same. So now we have 2 tens and 7 ones.

Example 2: Make a ten

Solve \(28 + 4\).

The number \(28\) has 2 tens and 8 ones.

We add 4 ones to 8 ones:

12 ones is the same as 1 ten and 2 ones.

Now add that new ten to the 2 tens we already had:

So we have 3 tens and 2 ones.

Another way to think about it:

Start at \(28\). Add 2 to get to \(30\). There are still 2 more to add. Then \(30 + 2 = 32\).

This is making a ten.

Example 3: Cross the decade boundary

Solve \(36 + 7\).

The number \(36\) has 3 tens and 6 ones.

Add 7 ones to 6 ones:

13 ones is 1 ten and 3 ones.

Add the new ten to the 3 tens:

Now we have 4 tens and 3 ones.

Make a ten strategy:

From \(36\), add 4 to get to \(40\). We used 4 out of the 7. There are 3 left.

So \(36 + 7 = 43\).

Example 4: Another making a ten problem

Solve \(49 + 5\).

Start with \(49\). It needs 1 more to make \(50\).

Take 1 from the 5:

Add the 1 first:

Now add the 4 that is left:

So:

What to Remember

• Look at the ones place first.
• Add the one-digit number to the ones.
• If the ones make 10 or more, trade 10 ones for 1 ten.
• Making a ten can help you solve the problem faster.

Helpful Steps

1. Read the problem.
2. Find how many ones are in the two-digit number.
3. Add the one-digit number.
4. If you make a new ten, add it to the tens place.
5. Write the answer.

Try thinking like this:

• \(27 + 5\): 7 ones plus 5 ones is 12 ones, so the answer is \(32\).
• \(58 + 3\): 8 ones plus 3 ones is 11 ones, so the answer is \(61\).
• \(41 + 6\): 1 one plus 6 ones is 7 ones, so the answer is \(47\).

Brief Summary

When you add a two-digit number and a one-digit number, add to the ones place first. If the ones make 10 or more, make a new ten. This helps you cross to the next ten, like from \(28\) to \(30\), and then finish adding the rest.

Base-Ten Models for Two-Digit Addition

Today we will learn how to add two-digit numbers using base-ten models.

Base-ten models help us see the numbers. A ten rod stands for 10. A unit cube stands for 1.

When we add with base-ten models, we put the tens together and the ones together. If we get 10 ones, we can trade them for 1 ten rod.

This is called regrouping. It means we exchange 10 ones for 1 ten because both are worth 10.

Let’s remember:

• 1 ten = 10 ones
• A two-digit number has tens and ones
• Add tens to tens
• Add ones to ones
• If the ones make 10 or more, trade 10 ones for 1 ten

For example, the number \(24\) has:

• 2 tens
• 4 ones

We can write that as:

$$24 = 2\text{ tens } + 4\text{ ones}$$

The number \(35\) has:

• 3 tens
• 5 ones

We can write that as:

$$35 = 3\text{ tens } + 5\text{ ones}$$

How to add using base-ten models

1. Build the first number with ten rods and unit cubes.
2. Build the second number with ten rods and unit cubes.
3. Put all the tens together.
4. Put all the ones together.
5. If there are 10 ones, trade them for 1 ten rod.
6. Count the tens and ones to find the sum.

Worked Example 1: No trade needed

Add \(23 + 14\).

First, build each number:

• \(23\) is 2 tens and 3 ones
• \(14\) is 1 ten and 4 ones

Now combine the tens and ones:

• Tens: \(2 + 1 = 3\) tens
• Ones: \(3 + 4 = 7\) ones

So the total is 3 tens and 7 ones.

$$23 + 14 = 37$$

We did not need to trade because 7 ones is less than 10.

Worked Example 2: Trade 10 ones for 1 ten

Add \(28 + 15\).

Build each number:

• \(28\) is 2 tens and 8 ones
• \(15\) is 1 ten and 5 ones

Combine the tens and ones:

• Tens: \(2 + 1 = 3\) tens
• Ones: \(8 + 5 = 13\) ones

Now we have 13 ones. We can trade 10 ones for 1 ten.

After the trade:

• 13 ones becomes 1 ten and 3 ones
• Add that extra ten to the 3 tens

So now we have:

• 4 tens
• 3 ones

$$28 + 15 = 43$$

Worked Example 3: Another regrouping problem

Add \(36 + 27\).

Build the numbers:

• \(36\) is 3 tens and 6 ones
• \(27\) is 2 tens and 7 ones

Combine them:

• Tens: \(3 + 2 = 5\) tens
• Ones: \(6 + 7 = 13\) ones

Trade 10 ones for 1 ten.

Then we have:

• 5 tens + 1 more ten = 6 tens
• 3 ones left

$$36 + 27 = 63$$

Worked Example 4: Adding bigger two-digit numbers

Add \(47 + 26\).

Build the numbers:

• \(47\) is 4 tens and 7 ones
• \(26\) is 2 tens and 6 ones

Combine the base-ten models:

• Tens: \(4 + 2 = 6\) tens
• Ones: \(7 + 6 = 13\) ones

Trade 10 ones for 1 ten.

Now we have:

• 7 tens
• 3 ones

$$47 + 26 = 73$$

What base-ten models show us

Base-ten models help us understand why the answer makes sense.

When we trade 10 ones for 1 ten, the value stays the same. We are just showing the number in a new way.

For example:

$$10\text{ ones} = 1\text{ ten}$$

That is why 13 ones can become 1 ten and 3 ones.

A helpful way to think

• Count the tens
• Count the ones
• Look for 10 ones
• Trade if needed
• Write the final number

Common mistakes to watch out for

• Do not mix up tens and ones.
• Do not forget to trade when there are 10 or more ones.
• After trading, make sure to add the new ten to the tens place.

Let’s check one more quickly

Add \(32 + 16\).

\(32\) is 3 tens and 2 ones. \(16\) is 1 ten and 6 ones.

Combine:

• Tens: 4 tens
• Ones: 8 ones

No trade is needed.

$$32 + 16 = 48$$

Summary

Base-ten models use ten rods and unit cubes to help us add two-digit numbers.

We add the tens, add the ones, and trade 10 ones for 1 ten when needed.

This helps us see place value clearly and find the correct sum.

Base-Ten Models for Two-Digit Subtraction

When we subtract two-digit numbers, base-ten models can help us see what is happening.

Base-ten models use:

• Ten rods to show tens
• Unit cubes to show ones

For example, the number \(34\) has:

• \(3\) tens
• \(4\) ones

That means \(34\) can be shown with 3 ten rods and 4 unit cubes.

Subtraction means taking away. We start with a number, then remove some tens and ones.

Sometimes subtracting is easy. But sometimes the second number has more ones than the first number. Then we need to break apart a ten rod.

This is called regrouping.

When we regroup, we trade:

• 1 ten for 10 ones

This does not change the number. It only changes how we show it.

For example:

$$1\text{ ten} = 10\text{ ones}$$

So if we have \(52\), we can show it in two ways:

• \(5\) tens and \(2\) ones
• \(4\) tens and \(12\) ones

Both models still mean \(52\).

How to subtract with base-ten models

1. Build the first number with tens and ones.
2. Look at the ones in the second number.
3. If you have enough ones, subtract the ones.
4. If you do not have enough ones, break 1 ten rod into 10 unit cubes.
5. Subtract the ones.
6. Subtract the tens.
7. Count what is left.

Worked Example 1: No regrouping

Let’s solve:

$$43 - 21$$

First, build \(43\):

• \(4\) tens
• \(3\) ones

Now subtract \(21\):

• Take away \(1\) one
• Take away \(2\) tens

What is left?

• \(2\) tens
• \(2\) ones

So,

$$43 - 21 = 22$$

Worked Example 2: Regrouping because there are not enough ones

Let’s solve:

$$32 - 15$$

First, build \(32\):

• \(3\) tens
• \(2\) ones

We need to take away \(5\) ones. But we only have \(2\) ones.

So we regroup. Break apart 1 ten rod into 10 ones.

Now \(32\) becomes:

• \(2\) tens
• \(12\) ones

Now subtract the ones:

• \(12 - 5 = 7\)

Now subtract the tens:

• \(2 - 1 = 1\)

What is left?

• \(1\) ten
• \(7\) ones

So,

$$32 - 15 = 17$$

Worked Example 3: Another regrouping example

Let’s solve:

$$61 - 24$$

First, build \(61\):

• \(6\) tens
• \(1\) one

We need to take away \(4\) ones. But we only have \(1\) one.

So we regroup:

• Take away \(1\) ten from the tens
• Trade it for \(10\) ones

Now \(61\) becomes:

• \(5\) tens
• \(11\) ones

Subtract the ones:

$$11 - 4 = 7$$

Subtract the tens:

$$5 - 2 = 3$$

What is left?

• \(3\) tens
• \(7\) ones

So,

$$61 - 24 = 37$$

Worked Example 4: Regrouping with more tens

Let’s solve:

$$74 - 38$$

First, build \(74\):

• \(7\) tens
• \(4\) ones

We need to take away \(8\) ones. But \(4\) ones are not enough.

Regroup by breaking apart 1 ten rod.

Now \(74\) becomes:

• \(6\) tens
• \(14\) ones

Subtract the ones:

$$14 - 8 = 6$$

Subtract the tens:

$$6 - 3 = 3$$

What is left?

• \(3\) tens
• \(6\) ones

So,

$$74 - 38 = 36$$

What to remember

• The first number is the number you build.
• Subtract the ones and tens from the second number.
• If there are not enough ones, trade 1 ten for 10 ones.
• Then subtract the ones.
• After that, subtract the tens.

A helpful way to think about it

If you have 3 dimes and 2 pennies, that is like \(32\).

If you need to take away 5 pennies, 2 pennies are not enough. So you can trade 1 dime for 10 pennies.

Then you have 2 dimes and 12 pennies. Now you can take away 5 pennies.

Base-ten models work the same way.

Try to spot when you need to regroup

You probably need to regroup when the ones in the second number are greater than the ones in the first number.

For example:

• \(42 - 16\): regroup, because \(6\) ones is more than \(2\) ones
• \(58 - 23\): no regrouping, because \(3\) ones is not more than \(8\) ones

Summary

Base-ten models help us subtract by showing tens and ones. When there are not enough ones to subtract, we break apart 1 ten rod into 10 ones. Then we subtract the ones, subtract the tens, and count what is left.

Open Number Lines for Addition

An open number line is a blank line that helps us add by making jumps to the right.

When we add, the number gets bigger, so we move right on the number line.

Open number lines are helpful because we can break numbers into tens and ones. Then we can add the tens first and the ones next.

For example, in the number \(34 + 25\), we can think of \(25\) as:

That means we can start at \(34\), jump right \(20\), and then jump right \(5\).

How to use an open number line

1. Write the first number on the number line.
2. Break the second number apart into tens and ones.
3. Jump right by the tens.
4. Jump right by the ones.
5. Read the ending number. That is the sum.

Important idea: A jump of one ten means jump \(10\). A jump of two tens means jump \(20\). A jump of three ones means jump \(3\).

You do not need to mark every number on the line. That is why it is called an open number line. You only write the numbers you need.

Worked Example 1

Let’s solve \(23 + 14\).

Break apart \(14\):

Start at \(23\).

• Jump right \(10\): \(23 \to 33\)
• Jump right \(4\): \(33 \to 37\)

So,

Why it works: We added one ten first, then four ones.

Worked Example 2

Let’s solve \(45 + 32\).

Break apart \(32\):

Start at \(45\).

• Jump right \(30\): \(45 \to 75\)
• Jump right \(2\): \(75 \to 77\)

So,

Here we added three tens, then two ones.

Worked Example 3

Let’s solve \(56 + 27\).

Break apart \(27\):

Start at \(56\).

• Jump right \(20\): \(56 \to 76\)
• Jump right \(7\): \(76 \to 83\)

So,

This time, the ones jump goes past the next ten, but that is okay. We still move right because we are adding.

Another way to show the ones jump

Sometimes a ones jump can be split into smaller jumps to make it easier.

For \(56 + 27\), after jumping to \(76\), we can split the \(7\) into \(4\) and \(3\):

• \(76 \to 80\) by jumping \(4\)
• \(80 \to 83\) by jumping \(3\)

We still get the same answer:

This can help when you want to land on a friendly number like \(80\).

Worked Example 4

Let’s solve \(68 + 15\).

Break apart \(15\):

Start at \(68\).

• Jump right \(10\): \(68 \to 78\)
• Jump right \(5\): \(78 \to 83\)

So,

What to remember

• Start with the first number.
• Break the second number into tens and ones.
• For addition, always jump right.
• Add the tens first, then the ones.
• The number where you land is the sum.

Common mistake to watch for

Sometimes students jump the wrong amount.

For example, in \(34 + 21\), the \(21\) means:

It does not mean jump \(2\) and then \(1\). It means jump \(20\) and then \(1\).

So:

• \(34 \to 54\) by jumping \(20\)
• \(54 \to 55\) by jumping \(1\)

Then,

Try thinking like this:

• \(13\) is \(10 + 3\)
• \(46\) is \(40 + 6\)
• \(28\) is \(20 + 8\)

Breaking numbers apart helps you make the right jumps.

Quick practice ideas

• For \(31 + 12\), start at \(31\), jump \(10\), then jump \(2\).
• For \(52 + 24\), start at \(52\), jump \(20\), then jump \(4\).
• For \(47 + 30\), start at \(47\), jump \(30\), then no ones jump is needed.

Summary

An open number line helps us add by showing jumps to the right.

We start at the first number, break the second number into tens and ones, and make jumps for each part.

Adding tens first and ones next makes two-digit addition easier and helps us find the sum.

Open Number Lines for Subtraction

Sometimes subtraction can feel tricky. An open number line helps us see the numbers and the jumps we make.

An open number line is a line with numbers we choose to write. It does not show every number. We use it to make subtraction easier.

There are two good ways to use an open number line for subtraction:

• Jump backward from the bigger number.
• Jump forward from the smaller number to find the difference.

Both ways help us solve subtraction problems within 100.

What subtraction means

When we subtract, we are finding how many are left or how far apart two numbers are.

For example, in \(15 - 6\), we start with 15 and take away 6. The answer is 9.

On an open number line, we can show that take-away with jumps.

How to use an open number line: Jump backward

This way starts at the bigger number. Then we jump back to subtract.

1. Write the bigger number on the number line.
2. Look at the number you are subtracting.
3. Break that number into easy parts, like tens and ones.
4. Jump backward by those parts.
5. Where you land is the answer.

It is often easiest to subtract the tens first and then the ones.

Worked Example 1

Solve \(27 - 5\).

Start at 27. Jump back 5.

You can do one jump of 5:

\(27 \rightarrow 22\)

So, $$27 - 5 = 22$$

Worked Example 2

Solve \(46 - 23\).

Start at 46. Break 23 into 20 and 3.

Jump back 20, then jump back 3.

\(46 \rightarrow 26 \rightarrow 23\)

First jump: \(46 - 20 = 26\)

Second jump: \(26 - 3 = 23\)

So, $$46 - 23 = 23$$

How to use an open number line: Jump forward to find the difference

This way starts at the smaller number. Then we jump forward until we reach the bigger number.

This is helpful when the numbers are close together, or when you want to find how far apart the numbers are.

1. Write the smaller number on the number line.
2. Write the bigger number as the place you want to reach.
3. Make friendly jumps forward, like to the next ten.
4. Add the jumps together.
5. The total of the jumps is the difference.

Worked Example 3

Solve \(52 - 48\).

These numbers are close together. Start at 48 and jump forward to 52.

\(48 \rightarrow 50 \rightarrow 52\)

The jumps are 2 and 2.

$$2 + 2 = 4$$

So, $$52 - 48 = 4$$

Worked Example 4

Solve \(73 - 58\).

Start at 58 and jump forward to 73.

Make easy jumps:

\(58 \rightarrow 60 \rightarrow 70 \rightarrow 73\)

The jumps are:

• From 58 to 60 is 2
• From 60 to 70 is 10
• From 70 to 73 is 3

Add the jumps:

$$2 + 10 + 3 = 15$$

So, $$73 - 58 = 15$$

Tips for making easy jumps

• Jump to the nearest ten when you can.
• Break numbers into tens and ones.
• Use jumps that are easy for you to add or subtract.
• Check that your jumps match the subtraction problem.

Which way should I use?

You can choose the way that feels easiest.

• Use jump backward when taking away tens and ones feels simple.
• Use jump forward when the numbers are close or when finding the distance feels easier.

Both ways help you get the same answer.

Let’s compare

Look at \(64 - 61\).

Jumping backward:

\(64 \rightarrow 61\)

That is a jump of 3, so the answer is 3.

Jumping forward:

\(61 \rightarrow 64\)

That is also a jump of 3, so the answer is 3.

Either method works.

Common mistakes to watch for

• Do not start at the wrong number. For jumping backward, start at the bigger number.
• Do not forget to add all the forward jumps together.
• Be careful with tens and ones. A jump of 10 is bigger than a jump of 1.
• Check that you land on the right number.

Practice thinking

If you solve \(81 - 34\), you might jump backward:

\(81 \rightarrow 51 \rightarrow 47\)

That means subtract 30, then 4. So the answer is 47.

Or you might solve it by jumping forward from 34:

\(34 \rightarrow 40 \rightarrow 80 \rightarrow 81\)

The jumps are 6, 40, and 1. Together, \(6 + 40 + 1 = 47\).

Both ways show that $$81 - 34 = 47$$

Summary

An open number line is a helpful tool for subtraction. You can jump backward from the bigger number, or jump forward from the smaller number to find the difference.

Break numbers into easy parts, like tens and ones, and use friendly jumps. With practice, open number lines make subtraction within 100 easier to see and understand.

Partial Sums Decomposition is a smart way to add two-digit numbers.

It means we break apart each number into tens and ones. Then we add the tens, add the ones, and put the answers together.

This strategy helps us see what each number is made of. It also helps us stay organized when adding bigger numbers.

For example, in the number 34, the 3 means 3 tens and the 4 means 4 ones.

So:

$$34 = 30 + 4$$

And in the number 25:

$$25 = 20 + 5$$

When we use partial sums, we add the tens first and the ones first. Then we combine those two sums to get the total.

How to use partial sums

1. Break apart each number into tens and ones.

2. Add the tens.

3. Add the ones.

4. Add those two answers together.

Here is the pattern:

$$\left(\text{tens} + \text{ones}\right) + \left(\text{tens} + \text{ones}\right)$$

Then:

$$\text{tens sum} + \text{ones sum} = \text{total}$$

Worked Example 1

Add \(23 + 14\).

First, break apart the numbers:

$$23 = 20 + 3$$

$$14 = 10 + 4$$

Now add the tens:

$$20 + 10 = 30$$

Add the ones:

$$3 + 4 = 7$$

Now put the sums together:

$$30 + 7 = 37$$

So:

$$23 + 14 = 37$$

Worked Example 2

Add \(46 + 32\).

Break apart each number:

$$46 = 40 + 6$$

$$32 = 30 + 2$$

Add the tens:

$$40 + 30 = 70$$

Add the ones:

$$6 + 2 = 8$$

Put them together:

$$70 + 8 = 78$$

So:

$$46 + 32 = 78$$

Worked Example 3

Add \(27 + 35\).

Break apart the numbers:

$$27 = 20 + 7$$

$$35 = 30 + 5$$

Add the tens:

$$20 + 30 = 50$$

Add the ones:

$$7 + 5 = 12$$

Now combine the partial sums:

$$50 + 12 = 62$$

So:

$$27 + 35 = 62$$

In this example, the ones made a number bigger than 10. That is okay. We can still add that ones sum to the tens sum.

Worked Example 4

Add \(58 + 26\).

Break apart each number:

$$58 = 50 + 8$$

$$26 = 20 + 6$$

Add the tens:

$$50 + 20 = 70$$

Add the ones:

$$8 + 6 = 14$$

Now combine the sums:

$$70 + 14 = 84$$

So:

$$58 + 26 = 84$$

Why partial sums works

Every two-digit number has tens and ones. When we add tens to tens and ones to ones, we are using place value.

That means we keep the tens with tens and the ones with ones. This makes addition easier to understand.

Another way to write it

Some students like to line up the parts:

For \(34 + 21\):

$$34 = 30 + 4$$

$$21 = 20 + 1$$

$$30 + 20 = 50$$

$$4 + 1 = 5$$

$$50 + 5 = 55$$

This is the same partial sums strategy.

Tips to remember

• Look at the digit in the tens place.

• Look at the digit in the ones place.

• Add tens with tens.

• Add ones with ones.

• Then add those two sums together.

Watch out for these mistakes

• Do not mix tens and ones.

• Do not forget to combine the two partial sums at the end.

• If the ones sum is 10 or more, that is okay. Just add it to the tens sum.

Let’s try one together

Add \(41 + 27\).

Break apart the numbers:

$$41 = 40 + 1$$

$$27 = 20 + 7$$

Add the tens:

$$40 + 20 = 60$$

Add the ones:

$$1 + 7 = 8$$

Combine them:

$$60 + 8 = 68$$

So:

$$41 + 27 = 68$$

Summary

Partial sums decomposition means breaking apart two-digit numbers into tens and ones.

Then we add the tens, add the ones, and combine those sums.

This strategy helps us use place value to add numbers clearly and carefully.

Compensation for Friendlier Numbers is a smart way to add numbers more easily.

Sometimes one of the numbers is close to a multiple of ten, like 10, 20, 30, or 40. We can change the numbers a little to make the problem friendlier. Then we make sure the total stays the same.

This is called compensation. Compensation means we add some to one number and take the same amount from the other number. That way, the sum does not change.

For example, if we have \(28 + 35\), the number 28 is close to 30. We can give 2 from 35 to 28. Then the problem becomes \(30 + 33\). That is easier to add.

Both problems have the same answer:

$$28 + 35 = 30 + 33 = 63$$

Why does this work?

If one number gets bigger by a little bit, the other number gets smaller by the same little bit. The total stays the same.

Think of it like moving blocks from one pile to another. One pile gets 2 more blocks, and the other pile gives away 2 blocks. The number of blocks altogether does not change.

When should we use compensation?

• When a number is close to a ten
• When changing a number to a ten makes the addition easier
• When we can move a small amount from one addend to the other

Steps for using compensation

1. Look for a number that is close to a multiple of ten.
2. Decide how much it needs to get to that ten.
3. Take that same amount from the other addend.
4. Add the new, friendlier numbers.

Worked Example 1

Find \(19 + 24\).

The number 19 is close to 20. It needs 1 more.

Take 1 from 24 and give it to 19.

Now the problem is \(20 + 23\).

$$19 + 24 = 20 + 23 = 43$$

So, \(19 + 24 = 43\).

Worked Example 2

Find \(27 + 36\).

The number 27 is close to 30. It needs 3 more.

Take 3 from 36 and give it to 27.

Now the problem is \(30 + 33\).

$$27 + 36 = 30 + 33 = 63$$

So, \(27 + 36 = 63\).

Worked Example 3

Find \(48 + 25\).

The number 48 is close to 50. It needs 2 more.

Take 2 from 25 and give it to 48.

Now the problem is \(50 + 23\).

$$48 + 25 = 50 + 23 = 73$$

So, \(48 + 25 = 73\).

Worked Example 4

Find \(39 + 18\).

The number 39 is close to 40. It needs 1 more.

Take 1 from 18 and give it to 39.

Now the problem is \(40 + 17\).

$$39 + 18 = 40 + 17 = 57$$

So, \(39 + 18 = 57\).

Let’s notice a pattern

• \(19\) becomes \(20\), so the other number loses 1
• \(27\) becomes \(30\), so the other number loses 3
• \(48\) becomes \(50\), so the other number loses 2

We always move the same amount.

Be careful!

Do not just change one number. If you add to one addend, you must take the same amount from the other addend.

For example, \(28 + 35\) should not become \(30 + 35\). That would make the total too big.

Instead, it should become \(30 + 33\).

Try thinking it through

For \(29 + 14\), 29 is close to 30. It needs 1 more. Take 1 from 14. Then we get \(30 + 13\), which equals 43.

For \(38 + 26\), 38 is close to 40. It needs 2 more. Take 2 from 26. Then we get \(40 + 24\), which equals 64.

Summary

Compensation helps us add by making a number into a friendlier ten.

We do this by giving a small amount from one addend to the other addend.

If we move the same amount, the total stays the same.

This strategy makes many addition problems within 100 quicker and easier.

Composing a Ten (Regrouping) helps us add two-digit numbers when the ones make a number bigger than 9.

Sometimes, when we add the ones, we get 10 or more. When that happens, we can make a new ten. This is called composing a ten.

It is also sometimes called regrouping or carrying. In 2nd grade, it is helpful to think of it as making a new group of ten.

Remember place value:

• The ones place tells how many ones.
• The tens place tells how many groups of ten.

When 10 ones come together, they can be traded for 1 ten.

That means:

$$10\text{ ones }= 1\text{ ten}$$

Why do we compose a ten?

We compose a ten because a ones place can only hold 0 through 9 ones.

If we add and get 10 ones or more, we need to move 10 of those ones into the tens place as 1 new ten.

For example, if we have 13 ones, that is the same as 1 ten and 3 ones.

$$13 = 10 + 3$$

So 13 ones becomes 1 ten and 3 ones.

How to add by composing a ten

1. Add the ones.
2. If the ones are 10 or more, make a new ten.
3. Write the extra ones in the ones place.
4. Add the tens, including the new ten you made.

Another way to think about it is:

• Add ones first.
• Trade 10 ones for 1 ten.
• Then add all the tens.

Worked Example 1

Add \(27 + 15\).

Step 1: Add the ones.

\(7 + 5 = 12\)

12 ones means 1 ten and 2 ones.

Step 2: Add the tens.

\(2\) tens from 27, plus \(1\) ten from 15, plus the new 1 ten.

$$2 + 1 + 1 = 4\text{ tens}$$

Step 3: Put the answer together.

4 tens and 2 ones is \(42\).

$$27 + 15 = 42$$

Worked Example 2

Add \(36 + 28\).

Step 1: Add the ones.

\(6 + 8 = 14\)

14 ones is 1 ten and 4 ones.

Step 2: Add the tens.

\(3\) tens plus \(2\) tens plus the new \(1\) ten equals \(6\) tens.

$$3 + 2 + 1 = 6\text{ tens}$$

Step 3: Put the answer together.

6 tens and 4 ones is \(64\).

$$36 + 28 = 64$$

Worked Example 3

Add \(48 + 27\).

Step 1: Add the ones.

\(8 + 7 = 15\)

15 ones is 1 ten and 5 ones.

Step 2: Add the tens.

\(4\) tens plus \(2\) tens plus the new \(1\) ten equals \(7\) tens.

$$4 + 2 + 1 = 7\text{ tens}$$

Step 3: Put the answer together.

7 tens and 5 ones is \(75\).

$$48 + 27 = 75$$

Worked Example 4

Add \(59 + 16\).

Step 1: Add the ones.

\(9 + 6 = 15\)

15 ones is 1 ten and 5 ones.

Step 2: Add the tens.

\(5\) tens plus \(1\) ten plus the new \(1\) ten equals \(7\) tens.

$$5 + 1 + 1 = 7\text{ tens}$$

Step 3: Put the answer together.

7 tens and 5 ones is \(75\).

$$59 + 16 = 75$$

What it looks like in vertical addition

We can also line the numbers up by place value.

For \(27 + 15\):

$$\begin{array}{r} 27 \\ + 15 \\ \hline 42 \end{array}$$

First add the ones: \(7 + 5 = 12\). Write the \(2\) in the ones place.

The 10 ones become 1 new ten. Add that new ten to the tens column.

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

So the answer is \(42\).

A place value picture in your mind

Imagine \(27\) as 2 tens and 7 ones.

Imagine \(15\) as 1 ten and 5 ones.

Now put the ones together:

$$7\text{ ones} + 5\text{ ones} = 12\text{ ones}$$

12 ones can be regrouped as:

$$12\text{ ones} = 1\text{ ten} + 2\text{ ones}$$

Now count the tens:

$$2\text{ tens} + 1\text{ ten} + 1\text{ new ten} = 4\text{ tens}$$

That gives us 4 tens and 2 ones, or \(42\).

How to know when to regroup

You need to compose a ten when the ones add to:

• 10
• 11
• 12
• 13
• 14
• 15
• 16
• 17
• 18

All of these numbers have at least 1 group of ten in them.

If the ones add to 9 or less, you do not need to compose a ten.

Common mistake to watch for

A common mistake is to add the ones and write the whole number in the ones place.

For example, in \(27 + 15\), the ones make 12. We do not write 12 in the ones place.

Instead, we write the 2 ones, and the 10 ones become 1 ten.

So we regroup:

$$12\text{ ones} = 1\text{ ten} + 2\text{ ones}$$

Try these steps every time

• Line up the tens and ones.
• Add the ones first.
• If the ones are 10 or more, make a new ten.
• Add the tens, including the new ten.
• Check that your answer makes sense.

Brief Summary

Composing a ten means making a new group of ten when the ones add up to 10 or more.

10 ones can be traded for 1 ten, because:

$$10\text{ ones} = 1\text{ ten}$$

When you add two-digit numbers, add the ones first. If needed, regroup 10 ones into 1 ten. Then add the tens.

This helps you add correctly and understand what regrouping really means.

Decomposing a Ten (Borrowing) helps us subtract when the top number does not have enough ones.

Sometimes in subtraction, the ones place on top is smaller than the ones place on the bottom. When that happens, we can decompose a ten. This means we take 1 ten and turn it into 10 ones.

This is also sometimes called borrowing, but a better way to think about it is unbundling a ten. We are not making a new number. We are just breaking apart a ten in a different way.

For example, the number \(34\) means 3 tens and 4 ones. If we decompose 1 ten, then \(34\) becomes 2 tens and 14 ones. It is still the same number: \(34\).

We can show that like this:

When do we decompose a ten?

• Look at the ones first.
• If the top number has fewer ones than the bottom number, decompose a ten.
• Then subtract the ones.
• After that, subtract the tens.

Steps for decomposing a ten:

1. Look at the ones.
2. If needed, take 1 ten from the tens place.
3. Change that 1 ten into 10 ones.
4. Add those 10 ones to the ones you already have.
5. Subtract the ones.
6. Subtract the tens.

Let’s learn with examples.

Example 1: \(32 - 5\)

First, think about \(32\). It is 3 tens and 2 ones.

We need to subtract \(5\) ones, but we only have \(2\) ones. That is not enough.

So we decompose 1 ten. Now \(32\) becomes 2 tens and 12 ones.

Now subtract the ones:

We still have 2 tens left, so the answer is:

Example 2: \(41 - 16\)

Look at the ones: \(1 - 6\). We cannot subtract 6 ones from 1 one.

So we decompose 1 ten from 4 tens. Now 4 tens becomes 3 tens, and 1 one becomes 11 ones.

Now subtract the ones:

Then subtract the tens:

So the answer is:

Example 3: \(53 - 27\)

Look at the ones: \(3 - 7\). We do not have enough ones.

Decompose 1 ten from 5 tens. Then \(53\) becomes 4 tens and 13 ones.

Subtract the ones:

Subtract the tens:

So:

Example 4: \(60 - 24\)

This one is a little different because there are 0 ones in 60.

The number \(60\) is 6 tens and 0 ones. We need to subtract 4 ones, but there are no ones to take away.

So we decompose 1 ten. Now \(60\) becomes 5 tens and 10 ones.

Now subtract the ones:

Then subtract the tens:

So:

Let’s remember what is happening.

When we decompose a ten, the number stays the same. We are just breaking it into tens and ones in a new way.

For example:

• \(25 = 2\text{ tens }5\text{ ones} = 1\text{ ten }15\text{ ones}\)
• \(48 = 4\text{ tens }8\text{ ones} = 3\text{ tens }18\text{ ones}\)
• \(70 = 7\text{ tens }0\text{ ones} = 6\text{ tens }10\text{ ones}\)

A helpful way to check yourself:

• Did you look at the ones first?
• If the top ones were too small, did you decompose 1 ten?
• Did the tens go down by 1?
• Did the ones go up by 10?
• Did you subtract ones first and then tens?

Common mistake:

Sometimes students forget to change both places.

If you take away 1 ten from the tens place, you must add 10 ones to the ones place.

For example, in \(52 - 8\):

• \(52\) is 5 tens and 2 ones.
• Decompose 1 ten.
• Now it becomes 4 tens and 12 ones.

Then:

and the 4 tens stay:

Summary

Decomposing a ten means taking 1 ten and changing it into 10 ones.

We do this when there are not enough ones to subtract in the ones place.

After we decompose, we subtract the ones first, then the tens.

This strategy helps us subtract two-digit numbers correctly and understand place value better.

Standard Algorithm for Two-Digit Addition

Today we will learn how to add two-digit numbers using the standard algorithm. This is a step-by-step way to add numbers by writing them in columns.

When we use the standard algorithm, we line up the digits by place value. That means the ones go under the ones, and the tens go under the tens.

We always start by adding the ones column first. Then we add the tens column. We work from right to left.

Step 1: Line up the numbers

Put one number on top of the other. Make sure:

• ones are under ones
• tens are under tens

For example, in the number \(24\), the \(2\) means 2 tens and the \(4\) means 4 ones.

Step 2: Add the ones

Look at the digits in the ones column. Add them first.

If the ones add to 9 or less, write that answer in the ones place.

If the ones add to 10 or more, write the ones digit down and carry 1 ten to the tens column.

Step 3: Add the tens

Next, add the tens column. If you carried a ten, add it too.

Then write the total for the tens.

Why does this work?

Each digit has a place. The digit on the right is the ones place. The digit on the left is the tens place.

When 10 ones are made, they become 1 ten. That is why we carry 1 to the tens column.

For example:

$$10\text{ ones } = 1\text{ ten}$$

Worked Example 1: No carrying

Add \(23 + 14\).

Line up the numbers:

$$ \begin{array}{r} 23 \\ + 14 \\ \hline \end{array} $$

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

Add the tens: \(2 + 1 = 3\).

So the answer is:

$$ \begin{array}{r} 23 \\ + 14 \\ \hline 37 \end{array} $$

Worked Example 2: Another no-carrying problem

Add \(45 + 32\).

Line up the tens and ones:

$$ \begin{array}{r} 45 \\ + 32 \\ \hline \end{array} $$

Add the ones: \(5 + 2 = 7\).

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

The sum is:

$$ \begin{array}{r} 45 \\ + 32 \\ \hline 77 \end{array} $$

Worked Example 3: Carrying a ten

Add \(28 + 15\).

First line up the numbers:

$$ \begin{array}{r} 28 \\ + 15 \\ \hline \end{array} $$

Add the ones: \(8 + 5 = 13\).

13 ones means 1 ten and 3 ones.

Write the 3 in the ones place. Carry the 1 ten to the tens column.

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

So the answer is:

$$ \begin{array}{r} ^1\!28 \\ + 15 \\ \hline 43 \end{array} $$

Worked Example 4: Carrying again

Add \(37 + 26\).

Write the numbers in columns:

$$ \begin{array}{r} 37 \\ + 26 \\ \hline \end{array} $$

Add the ones: \(7 + 6 = 13\).

Write 3 in the ones place. Carry 1 ten.

Add the tens: \(3 + 2 + 1 = 6\).

The sum is:

$$ \begin{array}{r} ^1\!37 \\ + 26 \\ \hline 63 \end{array} $$

Things to remember

• Line up the digits carefully.
• Add the ones first.
• Add the tens next.
• If the ones make 10 or more, carry 1 ten.
• Check that your answer makes sense.

Common mistakes

• Not lining up place values: ones must be under ones, tens under tens.
• Starting with tens: in this method, start with the ones.
• Forgetting to carry: if the ones make 10 or more, carry 1 ten.

Quick check idea

You can think about whether your answer is close to the numbers you added. For example, \(28 + 15\) should be more than 28, so 43 makes sense.

Summary

The standard algorithm helps us add two-digit numbers in a neat, organized way. First line up the tens and ones. Then add the ones, and after that add the tens. If the ones make 10 or more, write the ones digit and carry 1 ten.

Standard Algorithm for Two-Digit Subtraction

Today we will learn how to subtract two-digit numbers using the standard algorithm. This is a step-by-step way to line up numbers and subtract them.

When we subtract two-digit numbers, we work with ones and tens. We always start on the right side with the ones place. Then we move to the tens place.

This helps us stay organized and solve subtraction problems carefully.

Step 1: Line up the numbers

Write the bigger number on top and the smaller number under it. Make sure the ones are under the ones and the tens are under the tens.

For example:

Here, 4 and 2 are in the ones place. 3 and 1 are in the tens place.

Step 2: Subtract the ones

Start with the ones column. Ask, “Can I subtract the bottom ones from the top ones?” If yes, subtract.

Step 3: Subtract the tens

Then move to the tens column and subtract the tens.

Worked Example 1: No regrouping

Let’s solve \(34 - 12\).

First, subtract the ones: \(4 - 2 = 2\).

Next, subtract the tens: \(3 - 1 = 2\).

The answer is:

So, \(34 - 12 = 22\).

What if the top ones are smaller?

Sometimes the top number in the ones place is smaller than the bottom number. Then we cannot subtract the ones right away.

When that happens, we regroup. Regroup means we take 1 ten and turn it into 10 ones.

Remember:

How to regroup

• Look at the tens place in the top number.
• Take away 1 ten.
• Add 10 ones to the ones place.
• Then subtract the ones.
• Then subtract the tens.

Worked Example 2: With regrouping

Let’s solve \(52 - 27\).

Start with the ones: \(2 - 7\). We cannot do that, because 2 is smaller than 7.

So we regroup. The 5 tens become 4 tens. The 2 ones become 12 ones.

Now subtract the ones: \(12 - 7 = 5\).

Next subtract the tens: \(4 - 2 = 2\).

The answer is:

So, \(52 - 27 = 25\).

Let’s look closely at regrouping

The number 52 means 5 tens and 2 ones. If we take 1 ten from the 5 tens, we have 4 tens left.

That 1 ten becomes 10 ones. Then 10 ones plus 2 ones makes 12 ones.

So 52 can be thought of as 4 tens and 12 ones. The total is still 52. We just changed how it looks so we can subtract more easily.

Worked Example 3: Another regrouping problem

Let’s solve \(61 - 36\).

Start with the ones: \(1 - 6\). We cannot do that, so we regroup.

The 6 tens become 5 tens. The 1 one becomes 11 ones.

Now subtract the ones: \(11 - 6 = 5\).

Then subtract the tens: \(5 - 3 = 2\).

The answer is:

So, \(61 - 36 = 25\).

Worked Example 4: No regrouping again

Let’s solve \(78 - 24\).

Subtract the ones: \(8 - 4 = 4\).

Subtract the tens: \(7 - 2 = 5\).

The answer is:

So, \(78 - 24 = 54\).

Tips to help you

• Always line up tens with tens and ones with ones.
• Always start subtracting with the ones place.
• If the top ones are smaller, regroup 1 ten into 10 ones.
• After subtracting the ones, subtract the tens.
• Work slowly and check each step.

Common mistake to avoid

Do not subtract from left to right. In subtraction with the standard algorithm, we start from the right.

Also, do not forget to change the tens number after regrouping. If you take 1 ten, there is one less ten left.

Let’s review the steps

1. Write the numbers in a column.
2. Line up ones under ones and tens under tens.
3. Start with the ones place.
4. If needed, regroup 1 ten into 10 ones.
5. Subtract the ones.
6. Subtract the tens.

Summary

The standard algorithm is a neat way to subtract two-digit numbers. We line up the digits by place value, start with the ones, and then subtract the tens.

If the top ones are too small, we regroup 1 ten into 10 ones. With practice, this method helps us subtract carefully and correctly.