Lesson video

Hello, my name's Mrs. Hopper, and I'm looking forward to working with you in this lesson.

This lesson comes from our unit on the order of operations.

I wonder if you've come across that before.

Any ideas what it's about? Well, let's get going with this lesson and explore more about the order of operations.

In this lesson, we're going to explain how the distributive law can help to solve division problems more efficiently and write equations that show this.

There is one key phrase and two keywords in the lesson today.

Distributive law, divisor, and dividend.

So let's have a chance to practise.

I'll say them and then you can repeat them back.

Are you ready? My turn.

Distributive law.

Your turn.

My turn.

Divisor.

Your turn.

My turn.

Dividend.

Your turn.

I'm sure you're familiar with those words.

Maybe not so familiar with the distributive law, but perhaps you've had a look at it recently.

Let's just remind ourselves what they mean.

They are gonna be useful to us in today's lesson.

The distributive law says that multiplying a number by a group of numbers added together is the same as doing each multiplication separately.

So for example, four times three is equivalent to two times three plus two times three.

They're both equal to 12.

The divisor is what we are dividing by, so the number we're dividing by, and the dividend is the whole value of the amount we're dividing.

So in this example, 20 divided by four is equal to five.

20 is the dividend and the divisor is four.

So look out for those as we go through the lesson today.

And there are two parts, and you can see the distributive law is really important.

So the first part is the distributive law and addition, and the second part is the distributive law and subtraction.

But remember, we're thinking about division expressions today.

And we've got Jun and Sam in the lesson with us today.

Jun has 16 sweets and Sam has 12 sweets.

They want to share their sweets between four bags so that they can give some to their friends later.

How many sweets will there be in each bag? Jun says, "Let's share my sweets between the four bags first." So he had 16 sweets, so he's going to take one group of four, that's one in each bag; two groups of four, two in each bag; three groups of four, three in each bag; four groups of four, four in each bag.

He says, "I could write an equation to show how I shared my sweets." 16 divided between four is equal to four in each bag.

16 divided by four is equal to four.

Now says Sam, "I can share my sweets between the four bags." One group of four, that's one each.

Two groups of four, that's two each.

Three groups of four, that's three in each bag.

She says, "I could write an equation to show how I shared my sweets." 12 divided by four is equal to three.

Now Jun and Sam know how many of their sweets would be in each bag, they need to add these numbers together to find the total of sweets in each bag.

Four plus three.

Those are our two parts.

So our whole is seven.

There are seven sweets in each bag.

Jun says, "We could write an equation to show how we shared our sweets." 16 divided by four plus 12 divided by four, and that's equal to four plus three, which is equal to seven.

The 16 represents Jun's sweets.

The 12 represents Sam's sweets.

The divisor is four because they shared the sweets between four bags.

The four here represents the number of Jun's sweets in each bag.

And the three represents the number of Sam's sweets in each bag.

And the seven represents the total number of sweets in each bag.

Jun and Sam have solved this problem using a different strategy.

Jun says, "We could find the total number of sweets first," so they put their sweets together.

16 plus 12 is equal to 28.

This could be more efficient because now we will only need to divide once.

So they're going to divide into groups of four and each time they have a group of four, they're gonna put one in each bag to share them equally.

So one group of four, two groups of four, three groups of four, four groups of four, five groups of four, six groups of four, seven groups of four is equal to 28.

So there are seven sweets in each of the four bags.

28 divided by four is equal to seven.

Jun and Sam found the total number of sweets first.

This means they added before they divided.

In an equation, the order of operations means that you would need to use brackets to show that addition should be completed before division.

So this is how we can record it.

16 of Jun's sweets plus 12 of Sam's sweets, 16 plus 12 to be completed before we divide by four.

So that's equal to 28 divided by four, which is equal to seven.

The 16 represents Jun's sweets.

The 12 represents Sam's sweets.

The divisor is four because they shared the sweets between four bags.

28 represents the total number of sweets before they're divided, and seven represents the number of sweets in each bag.

It's really important that we understand what each value within the equation means so that we know exactly where it's come from and how that equation represents our problem.

Jun says, "I could also write a shorter equation where I don't record the sum of 16 and 12 like this." So he could say 16 plus 12 divided by four is equal to seven.

In their second solution, Jun and Sam used the distributive law to help them solve the problem.

When there is a common divisor, the distributive law means that you can add the dividends so that you only need to divide once.

So we added the two whole numbers of sweets and we could do that because both of those numbers were being divided by four.

The same divisor, a common divisor.

Jun says, "Here is the equation if we divide each part separately and do not use the distributive law." 16 divided by four plus 12 divided by four is equal to four plus three, which is equal to seven.

And here's the equation if we use the distributive law.

16 plus 12 divided by four is equal to 28 divided by four, which is equal to seven.

Which method do you think is more efficient? Well, Sam says, "In the first method, you have to complete two divisions and then add the quotients together.

In the second method, you add the two numbers together and then complete one division." The distributive law can be very useful for solving problems where there is a common divisor.

It means that you can work in a more efficient way by reducing the number of calculations needed to solve a problem.

But it's important that we understand what the dividends are in our problem and that we spot that there is a common divisor.

And it's time to check your understanding now.

Which expression has the same value as the expression in the box? We've got 24 divided by six plus 30 divided by six.

Is that represented by A, B, or C? Pause the video, have a go.

And when you're ready for some feedback, press play.

How did you get on? Did you spot that it was B.

Six is the common divisor and you're dividing 24 and 30 by six.

So you can add the dividends and then divide 54 by six.

So we can spot that common divisor of six.

So we can add the dividends, our wholes.

Jun and Sam explore another problem where they could use the distributive law to help them work more efficiently.

Wallet A has 20 pence in five-pence coins and wallet B has 40 pence in five-pence coins.

How many coins are there altogether? So to work out how many coins, we need to divide our total value by the value of each coin.

So 20 divided by five for A and 40 divided by five for B.

That will tell us how many coins there are, how many groups of five there are in each of those totals.

We've got a common divisor there of five.

Sam says the common divisor is five, so we can add the two dividends first and then divide to find the total number of coins.

So we can add our dividends, which is 20 plus 40 is equal to 60.

And then we can divide by our common divisor of five, 60 divided by five is equal to 12.

There are 12 coins altogether and we can record that using the distributive law.

This is how Sam's done it.

20 plus 40 in brackets, because we're doing that first, divided by five is equal to 12.

It's important to include the bracket here because without them, the division would be completed before the addition.

Time to check your understanding again.

Which of the following correctly shows how to use the distributive law to solve this problem? How many coins are there in total? Wallet A has 20 pence in two-pence coins and wallet B has 14 pence in two-pence coins? Can that be represented by expression A, B, or C? Pause the video.

Have a think.

And when you're ready for some feedback, press play.

What did you think? So we had 20 pence in two-pence coins and 14 pence in two-pence coins.

So our common divisor was two, so we're looking to divide by two, and we're going to combine our dividends, the 20 and the 14.

So C represents that problem.

20 plus 14 in brackets to be completed first, all divided by two to find out how many two p coins there are.

And Sam's reminding us when you use the distributive law to work more efficiently, you add the dividends first and then divide by the common divisor.

And it's time for you to do some practise.

You're going to solve each problem using the distributive law and write the equation that shows how you solved it.

So you've got A, B, C, and D to work on.

Pause the video, have a go.

And when you're ready for some feedback, press play.

How did you get on? So in A, Jun has 32 p in two-pence coins and Sam has 16 p in two-pence coins.

How many coins do they have altogether? So here we've got a common divisor of two for the two-pence coins.

So therefore, we can add together the total values, the dividends.

32 plus 16 is equal to 48 and 48 divided by two is equal to 24.

So they have 24 two-pence coins altogether.

48 pence in total divided by two to find out how many two-pence coins there are, 24 coins altogether.

In B, Sam has 10 litres of red paint, which she shares equally between five pots.

And Jun has 15 litres of blue paint, which he shares equally between five pots as well.

They give one pot of paint to Aisha.

How much paint does Aisha have? So to find out the value of Sam's paint pots, we have to divide 10 by five.

To find the value of Jun's paint pots, we have to divide 15 by five.

We have a common divisor of five, so we can combine the dividends of 10 litres and 15 litres.

10 plus 15 is equal to 25, and 25 divided by five is equal to five.

So Aisha has five litres of paint.

In C, there are eight tables in a classroom.

Sam shares out 32 workbooks equally, and Jun shares out eight textbooks equally.

How many books are there on each table? We've got a common divisor here of eight because there are eight tables in the classroom.

So we've got 32 divided by eight and eight divided by eight.

So we can combine our dividends.

That's 32 plus eight is equal to 40.

40 divided by eight is equal to five.

So there'll be five books on each table.

And in D, a recipe requires 250 grammes of flour to make one pizza base.

Sam has one kilogramme of flour and Jun has two kilogrammes of flour.

How many pizza bases can they make? So this time, our dividend is in kilogrammes.

Our divisor is 250 grammes, so we need to do a bit of conversion here.

So one kilogramme is equal to 1,000 grammes, two kilogrammes is equal to 2,000 grammes.

So we can combine our total amount of flour, which is 1,000 plus 2,000, which is 3,000 grammes, and we can divide that by 250.

Well, we know that there are four lots of 250 in 1,000, so that'll be three lots of that.

So 12 lots of 250 in 3,000.

So they can make 12 pizza bases using the flour that they have.

You could also have used your knowledge that 250 grammes is equal to one quarter of a kilogramme and used one quarter as the divisor rather than converting into grammes.

So then you'd have been thinking about three kilogrammes.

And how many quarters are there in three? And if you imagine one whole divided into four quarters and you've got three of them, there'll be 12, won't there? Same answer.

Different way of thinking about it.

And on into the second part of our lesson, this is thinking about the distributive law and subtraction.

So far, we've looked at the distributive law to solve problems that combine division and addition.

Sam wonders, "Do you think that the distributive law will still be helpful if we are solving problems that combine division and subtraction?" Well, let's explore a problem to see if the distributive law can still help us to be more efficient.

Jun has 60 p in 10-p coins.

Sam has 40 p in 10-p coins.

Who has more coins? And by how many? Jun says, "I could write the division expressions I can see in this problem." 60 divided by 10.

So that will work out how many 10 p coins Jun has, and 40 divided by 10, that will work out how many 10-p coins Sam has.

When two dividends are divided by the same divisor, we can subtract the dividends first and then divide.

So our dividends here are our whole amounts of money, 60 p and 40 p.

They're both being divided by 10.

So we can do 60 subtract 40, and then divide by 10.

And you can write this equation to show that you've used the distributive law.

We put our 60 subtract 40 in brackets because to show that it goes first, we need to give it that priority by putting brackets around it.

So 60 subtract 40 is equal to 20 and 20 divided by 10 is equal to two.

Jun has more money, he's got 60 p, but we wanted to know how many more coins he had.

And Sam says, "You have two more coins than I have." The difference in their amounts of money was 20 p, and that is equivalent to two 10-p coins.

So Jun has two more coins than Sam.

Here's a different problem.

Jun has 25 p in five-p coins and Sam has 40 p in five-p coins.

Who has more coins? And by how many? Again, Jun says he can write the division expressions.

25 divided by five to show how many five-p coins he has and 40 divided by five to show how many five-p coins Sam has.

Again, when two dividends are divided by the same divisor, so dividing by five in this case, we can subtract the dividends first and then divide.

And we can write this equation to show that we've used the distributive law.

So this time, we know that Sam has more money, but we need to know how many more coins.

So 40 subtract 25, that shows the difference in the amount of money they have, which is 15 p.

And that divided by five tells us how many more coins there are, how many more five-p coins.

15 divided by five is equal to three.

This time, Sam says, "The dividend in the second expression is greater, so 40 must be written first in the subtraction for a positive answer." So Sam this time has three more five-p coins than Jun.

Jun and Sam are playing Battle Robots.

You need three robots to make a team.

Jun has 18 robots and Sam has 24 robots.

Who can make the most teams and by how many? Well, the common advisor here is three.

So we can subtract the robots from each other to find out what the difference is in the number of teams. So we know that Sam can make the most teams because she's got 24 robots.

So 24 is greater, so we're going to start with the 24.

24 subtract 18.

That shows us the difference in the number of robots.

And that's six.

And so then we've got to divide by three to find out how many teams that is.

Six divided by three is equal to two.

So Sam can make two more teams of robots than Jun can.

Time to check your understanding.

Which expression matches this problem? Jun and Sam have been shopping for more robots.

Jun now has 27 robots and Sam has 30 robots.

You still need three robots to make a team.

Who can make the most teams and by how many? So which of those expressions represents that problem? Pause the video, have a think.

And when you're ready for some feedback, press play.

How did you get on? Did you spot that it was C? A has subtracted the dividends so that the difference will be negative.

To make sure that you have a positive answer, you need to start the subtraction expression with a dividend that is greater.

B does not match the problem because it does not have any brackets, so the division would be calculated before the subtraction.

So C is correct.

And it's time for you to do some practise.

You're going to solve each problem using the distributive law and write the equation that shows how you solved it.

You've got A, B, C, and D.

Pause the video, have a go at those four.

And when you're ready for the answers and some feedback, press play.

How did you get on? So in a Jun has 32 p in two-pence coins and Sam has 16 p in two-pence coins.

Who has the most coins and by how many? We've got that common divisor of two for the two p coins.

So we can subtract our values.

32 p is greater than 16 p, so we do that first and use the brackets to show.

32 subtract 16 is equal to 16 and 16 divided by two is equal to eight.

So that eight represents eight two-p coins.

Jun has eight more coins than Sam.

Sam has 10 litres of red paint, which she shares equally between five pots.

Jun has 15 litres of blue paint, which he shares equally between five pots.

They each give one pot of paint to Aisha.

Who gave the most paint to Aisha? And by how much? So we're dividing into five pots both times.

So we're going to find the difference between Jun's and Sam's paint, so that's 15 subtract 10, and that is equal to five, and five divided by five is equal to one.

So that represents one more litre of paint that Jun gave Aisha than Sam did.

In C, there are eight tables in the classroom.

Sam shares out 32 workbooks equally and Jun shares out eight textbooks equally.

Who gave out the most books to each table? And by how many? Are you noticing something here? Have you seen these problems before? So we've got a common divisor here of eight.

So we want to work out who gave out the most textbooks.

Sam clearly gave out more because Sam shared out 32 workbooks and Jun shared out eight textbooks.

But who gave the most books to each table? And by how many? Well it was Sam, but by how many? So we can subtract eight from 32.

32 subtract eight is equal to 24 and 24 divided by eight is equal to three.

So Sam gave three more books to each table than Jun gave out.

And in D, the new version of Battle Robots uses teams of six.

Jun has 24 robots and Sam has 36 robots.

Who can make the most teams and by how many? Well, I think Sam can make more teams because Sam's got more robots to start with, but let's work out how many more teams Sam can make.

So we're going to find the difference in their number of robots and find out how many teams that is worth.

So 36 subtract 24 is equal to 12, and 12 divided by six is equal to two.

So Sam can make two more teams of robots than Jun can make.

And we've come to the end of our lesson.

We've been explaining how the distributive law applies to division expressions with a common divisor.

The distributive law can be very useful for solving problems where there is a common divisor.

When both dividends are divided by the same divisor, you can add or subtract the dividends before you divide.

So that means you can think about the whole value, whether you're combining two dividends to make a larger whole value, or whether you're finding the difference between those dividends in order to be able to divide that difference.

You can use brackets in an equation to show that you have used the distributive law, and an example here.

Thank you for all your hard work and your mathematical thinking in this lesson, and I hope I get to work with you again soon.

Bye-bye.