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 be explaining how the distributive law can help to solve multiplication problems more efficiently and write equations that show this.

Have you come across the distributive law? We'll have a think about it as we go through this lesson, but watch out for it.

Let's make a start.

So we've got the key phrase, distributive law and the word common in our key words today.

So I'll take my turn to say them and then you can have a go at saying them.

My turn, distributive law, your turn.

My turn, common, your turn.

Let's have a look at what they mean.

They're gonna be useful to us today.

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

For example, four times three is equal to two times three plus two times three, and that's equal to 12.

You may have used grid multiplication or looked at arrays and broken down arrays in different ways.

This is the same as using the distributive law.

And common can mean that something is the same as something else.

For example, when multiplication expressions have one factor that is the same, they are said to be common factors.

So here we can see seven times three is equal to five times three plus two times three.

Those threes are common factors in each of those multiplication expressions.

And we're going to look at how we can make use of common factors to make our work more efficient in this lesson.

There are three parts to our lesson.

In the first part, we're going to compare expressions with common factors.

In the second part, we're going to think about the distributive law and addition.

And in the third part we're going to think about the distributive law and subtraction.

So let's get started with part one.

And Jun and Sam are with us in the lesson today.

They're thinking about how they can compare expressions.

So we've got 50 times two plus 50 times 5, and then we've got 50 times 8 and we need to compare those expressions.

Is one greater than the other or are they equal? Jun says we could solve each expression using multiplication to find out which is greater.

We could do that Jun.

Sam says, I'm not sure we need to.

I've noticed something that might help us to compare these expressions easily.

Can you think what Sam might have noticed? Think about what we've just talked about in our keyword definitions.

She says, 'All of these expressions have a common factor of 50.

' 50 times two, 50 times five, and 50 times eight.

50 is common to all of those expressions.

Jun says, we could think about how many groups of 50 are in each expression.

So we've got two groups of 50, five groups of 50, and then on the other side of our symbol, we've got eight groups of 50.

Jun says there are seven groups of 50 in the first expression, two groups of 50 and another five groups of 50.

Sam says the second expression has eight groups of 50.

So 50 times eight is greater than 50 times two plus 50 times five.

So we can put that symbol in to show that the 50 times eight has a greater value than the two other expressions combined.

Jun says, what if I calculated the first expression in a different order? If I did two plus 51 first, the total would be greater.

He'd have 52 times 50 times five.

Can you spot something that's going wrong there? 'Ah, yes', says Sam, 'Remember the order of operations, Jun? You have to multiply before you add unless there are brackets.

' So we do 50 times two and 50 times five first and then add the two products together.

He says, 'Looking at these groups of 50 again, I'm noticing something that could be important.

' 'Me too!' says Sam, 'When one factor is common, you can add the uncommon factors to write one multiplication expression.

' So our common factor is 50.

Our uncommon factors are two and five.

There they are.

Two plus five is equal to seven.

So we can rewrite those two expressions as 50 times seven.

We had 50 times two and 50 times five, two groups of 50 and five groups of 50.

We can combine them to make seven groups of 50.

And you can see that with the counters we've used as well.

Jun and Sam look at two more expressions.

Do you want to have a little look before Jun and Sam share their thinking? You might want to pause and have a think first.

Jun says, 'I can see that there is a common factor again.

' What's the common factor this time? That's right, it's 63.

We've got 63 times seven, seven groups of 63.

We've got nine groups of 63 and then we've got 63 times four, four groups of 63.

So there are the seven groups of 63.

And there we can see nine groups of 63 with four removed.

And Sam says, 'When one factor is common, you can subtract the uncommon factors and write one expression.

' We added before, this time we're subtracting.

So nine and four are our uncommon factors.

63 is the common factor.

And Jun says, 'I know that nine subtract four is equal to five.

' So that whole side can be rewritten as five times 63.

That's what we get when we subtract four lots of 63 from nine lots of 63.

'This means' says Sam, 'That the first expression is greater because we can say it has more groups of 63.

' Seven groups of 63 is greater than five groups of 63.

Now Jun and Sam will try to compare the expressions without using counters.

Jun says, 'We will look for common factors.

' Sam says, 'We know that when there is a common factor we can add or subtract the uncommon factors in the expression.

' So the common factor this time is 75.

We've got four times 75 and then on the other side, we've got eight times 75 subtract three times 75.

'Eight subtract three is equal to five', says Sam.

'So I can change this expression to five times 75.

' Eight groups of 75 subtract three groups of 75 will leave us with five groups of 75.

And five groups of 75 is greater than four groups of 75.

So we can put in our inequality symbol between those two expressions.

Time to check your understanding.

Which pair of expressions have been compared correctly? Have a look carefully at A, B, C, and D and decide which ones have been compared correctly.

Pause the video, have a think and when you're ready for some feedback, press play.

Which one was it? It was D, wasn't it? That was correct.

Let's have a look at the others and work out why they were not correct.

So in this one our common factor was 16, nine plus six is equal to 15.

So 54 groups of 16 is much greater than 15 groups of 16.

So the symbol is the wrong way round.

What about in B? What's the common factor here? Well, the common factor is seven, isn't it? 23 minus seven is equal to 16.

So 16 groups of seven is less than 23 groups of seven.

And in C, what's our common factor? Our common factor here is 15.

So we were looking at the 17 groups, the 20 groups, and subtracting eight groups.

20 minus eight is equal to 12.

So 17 groups of 15 is greater than 12 groups of 15, but in D, we were correct.

31 plus 25 is equal to 56.

So 56 groups of five is indeed less than 60 groups of five.

So well done if you identify D as the one where the expressions had been compared correctly.

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

Can you use a less than greater than or equal sign to compare each pair of expressions? And you've got eight pairs of expressions there to look at.

And in question two, what could the missing digits be and how many different solutions can you find? Pause the video, have a go at questions one and two and when you're ready for the answers and some feedback, press play.

How did you get on? So in A, we had 16 times 12 subtract seven times 12 compared to 17 times 12.

So our common factor is 12.

So 16, subtract seven is equal to nine, and nine groups of 12 is less than 17 groups of 12.

So we need a less than symbol.

In B, we're comparing again to 17 groups of 12.

This time we've got 16 plus seven.

16 plus seven is equal to 23.

23 groups of 12 is greater than 17 groups of 12.

And in C, our common factor here is 17.

So we've got 28 lots of 17 plus 17 times two plus two lots.

So we've got 30 lots of 17 and that is less than 31 lots of 17.

Let's look at D.

The common factor is 17.

28 plus two is equal to 30.

Oh, and that's equal.

28 plus two is equal to 30.

30 groups of 17 is equal to 17 groups of 30 or 17 times 30.

Let's look at E.

Our common factor is 55.

So we've got 80 groups of 55 subtract seven groups of 55 on the right so that is less than 87 groups of 55.

In F, our common factor is 87.

55 groups of 87 and 50 groups plus five groups so those are equal.

And in G, our common factor is 14.

41 groups of 14.

What have we got on the other side? 41 subtract seven so that's going to be less than.

We've subtracted seven from the 41 groups.

And in H, our common factor again is 14.

We're comparing to 41 groups.

And we've got 41 groups subtract zero groups.

So those are equal.

We've got 41 groups of 14 on both sides.

And in question two, what could the missing digit be? How many different solutions can you find? So in A, something times 20 subtract three times 20 is equal to four times 20.

So something subtract three must be equal to four.

So the only number that can go there is seven.

Because these expressions are equal, there is only one possible solution.

In B, though the left hand side is greater than the right hand side.

The right hand side is 10 groups of 20.

So we've got to make sure that whatever we have on the other side is greater than 10 groups of 20.

So we've got five groups of 20.

So we need there to be at least six groups of 20 to add to it.

So six is the smallest value we can have.

So the first expression needs to be greater than 20 times 10, and the missing digit could be 6, 7, 8, or 9 if we are looking at a single digit answer.

And in C, 17 times 6, so 6 groups of 17 is less than 9 groups of 17 subtract some groups of 17.

So our answer has got to be greater than six.

So the missing digit could be zero, one or two.

As soon as we subtract three groups of 17, our expressions would be equal.

And in D, 23 times 11, so our common factor is 11, and we've got 23 groups.

So that has got to be greater than 30 groups of 11 subtract some groups of 11.

So subtracting seven would make it equal, it's got to be less than, so it needs to be eight or nine if it is a one digit number.

So as long as our missing digit is an eight or a nine, then the right hand side expression will have a value of less than 23 times 11.

And on into the second part of our lesson, we are looking at the distributive law and addition.

So Jun bought four packs of trading cards and Sam bought three packs of trading cards.

How many cards did they buy altogether? Jun says we could represent this problem using counters.

Four groups of 80 plus three groups of 80.

So we could write this expression to represent the problem, four times 80 plus three times 80.

I know that four groups of 80 plus three groups of 80 is equal to seven groups of 80.

We've got a common factor again there, haven't we of 80? That means I know a different expression we could write to represent the problem.

We could write seven times 80, 4 times 80 plus three times 80.

And Sam says, 'We could move the two groups of counters together to represent your idea too.

Four groups of 80 plus three groups of 80 is equal to seven groups of 80.

And Jun says, I know that seven multiplied by 80 is equal to 560.

So we can say that we had bought 560 cards altogether.

That's a lot of trading cards.

Jun and Sam have used the distributive law to help them solve this problem.

When there is a common factor, the distributive law means that you can add the uncommon factors so that you only need to multiply once.

Here's the equation if we multiply each part separately and do not use the distributive law, four times 80 plus three times 80 is equal to 320 plus 240, which is equal to 560.

Here's the equation if we use the distributive law, four times 80 plus three times 80 is equal to seven times 80, which is equal to 560.

Which method do you think is more efficient? Sam says, 'In the first method, you have to complete two multiplication and add the two 3-digit numbers together.

In the second method, you add two 1-digit numbers together and complete one multiplication.

' So the distributive law can be very useful for solving problems where there is a common factor.

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

So it's one to look out for.

Can you find common factors? And if so, can you combine or find the difference between the uncommon factors.

When you've used the distributive law, there is a different way you can record your solution as an equation.

You can use your knowledge of the order of operations and use brackets to show that you need to add the two uncommon factors before you multiply.

'So instead of writing this equation, four times 80 plus three times 80 is equal to seven times 80, I can write this one.

' says Sam.

Four plus three in bracket to show that we do that first multiplied by 80.

Four plus three is equal to seven.

Seven times 80 is equal to 560.

Time to check your understanding.

Which expression would have the same solution as the expression in the box? Seven times 45 plus nine times 45.

Is it A, B, or C? Pause the video.

Have a think.

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

How did you get on? It was B, wasn't it? 45 is the common factor.

There are seven groups of 45 and nine groups of 45.

So you need to add seven and nine first.

That's why they're in the brackets and then multiply by 45.

Let's explore another problem where you can use the distributive law to work more efficiently.

Our running track is 400 metres long.

Sam runs eight laps and then stops for a rest.

She runs another two laps before having another rest.

She finishes by running two more laps.

What is the total distance she has run? Sam says I can see that each lap is 400 metres.

I expect you felt it too, doing all that running.

Sam, well done.

400 is the common factor.

And Jun says there are eight groups of 400, two groups of 400 and another two groups of 400 in this problem.

'So here is the equation written out without using the distributive law', says Sam.

Oh my goodness, that's a lot of digits.

Eight times 400 plus two times 400 plus two times 400 is equal to 3,200 plus 800 plus 800, which is equal to 4,800.

'If I use the distributive law, I can write this equation', she says.

Eight plus two plus two multiplied by 400 and because it's in brackets, we know to do the eight plus two plus two first.

So that's 12 times 400, which is equal to 4,800.

The total distance Sam runs is 4,800 metres.

That's 4.

8 kilometres, nearly 5K.

Jun says, 'I can really see how the distributive law is helping us to work much more efficiently.

' Time to check your understanding.

Which solution is correct? Sam bought three erasers that cost 65p and three pencils that cost 80p.

How much did she spend in total? A, B, or C? Pause the video.

Have a think.

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

How did you get on? It was A, wasn't it? The multiplication expressions you could write for this problem are 3 times 65 and 3 times 80.

The common factor is three.

So you need to add the uncommon factors of 65 and 80 together and then multiply by three and they're in brackets to show that we want to do the addition first.

Time for you to do some practise.

You are 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 there.

So pause the video, have a go.

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

How did you get on? So in question A, pens come in boxes of 24.

Mrs. Hopper, oh, I'm in a problem again.

I buy five boxes of black pens and three boxes of purple pens.

How many pens do I have altogether? So my common factor is 24.

The boxes of pens have 24 in each.

I buy five boxes of black and three boxes of purple.

So five plus three, those are the uncommon factors that I need to add and I'm gonna put them in brackets to show that I need to do that first.

Five plus three times 24 is equal to 192.

I have 192 pens altogether.

I wish that was true, I don't.

In B, Sam has played a ball game eight times.

In half of the games, she scored 110 points each time.

In the other half, she scored 150 points each time.

What was her total score? Hmm! Our common factor is a bit hidden here, isn't it? Except that we know that we are thinking about half of four each time.

So a common factor must be four.

Half of eight is four.

So some must have scored 110 points in four games and 150 points in four games as well.

So we can add the number of points and multiply the total by four, 110 plus 150 multiplied by four is 1040.

So that was Sam's total score.

In C, in a junior triathlon, competitors swim 50 metres, cycle 800 metres and run 600 metres.

10 competitors finish the race.

What distance have they travelled in total? Ah, so this time our common factor is 10 because we've got 10 lots of 50, 10 lots of 800 and 10 lots of 600.

So we can add those distances and multiply by 10.

50 plus 800 plus 600 added together and then multiplied by 10 is equal to 14,500.

So the competitors have travelled 14,500 metres in total.

That's 14 and a half kilometres.

And indeed Jun bought four suites that cost 24p each, four suites that cost 15p each and four chocolate bars that cost 56p each.

How much did he spend? Our common factor there is four.

So we need to add together the costs of each item and then multiply the whole lot by four.

24 plus 15 plus 56 multiplied by four is equal to 380.

So Jun spent three pounds and 80 pence.

And on into the final part of our lesson, the distributive law and subtraction.

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

Sam's got a question.

Do you think that the distributive law will still be helpful if we are solving problems that combine multiplication and subtraction? Let's have a look.

Let's explore a subtraction problem and see if the distributive law can still help us to be more efficient.

Jun bought four packs of trading cards.

Sam bought two packs of trading cards.

How many more cards did Jun buy? Trading cards come in packs of 80.

'We could represent this problem using counters', says Jun.

I have four groups of 80 and you have two groups of 80.

And we could write this expression to represent the problem.

Four times 80 subtract two times 80 'cause we want to find the difference.

We want to subtract one part from the whole, which is the largest value, which is Jun's cards.

'We are finding the difference between my total number of cards and your total', says Jun.

We could move the two groups of counters to show the difference more clearly.

So there's four lots of 80 and we're comparing those to two lots of 80.

And can you see the whole is four lots? We need to subtract those two lots to find the difference.

Sam says four groups of 80 subtract two groups of 80 is equal to two groups of 80.

And Jun says, 'I know the difference is two groups of 80, so I have 160 more cards than you.

' Sam's explanation helps us to see how Jun and Sam use the distributive law to help them solve the problem.

Let's look at Sam's again.

Four groups of 80 subtract two groups of 80 is equal to two groups of 80.

We know that the common factor is 80 here, so we can subtract the uncommon factors, four subtract two.

That's equal to two.

So the difference is 2 times 80, which is 160.

We could also write the equation this way to show how we use the distributive law.

We're doing that subtraction first and we put it in brackets to show that we need to do it first.

Four subtract two multiplied by 80, which is the same as two multiplied by 80, which is 160.

Let's look at another problem where you can use the distributive law to work more efficiently.

At Jun's birthday party, there are 9 bags of sweets with 26 sweets in each bag.

Five whole bags of sweets are eaten.

How many sweets are left? Sam says, 'I can see that each bag has 26 sweets.

26 is the common factor.

' 'That means', says Jun, 'We can use the distributive law and subtract the uncommon factors.

' I can write an equation like this to show how we solve the problem.

9 groups of 26 subtract 5 groups of 26 is equal to 4 groups of 26, and that's equal to 104.

I could also write this equation, which uses brackets to show how we use the distributive law.

9 bags subtract 5 bags, gives us the 4 bags remaining and then we multiply that by 26.

And the subtraction is in brackets to show that we do it first.

'There are 104 sweets left.

', says Jun.

Time to check your understanding.

Which expression would give the correct solution to this problem? Sam bought three erasers and three pencils.

How much more did she spend on pencils than erasers? Pause the video, have a think and when you're ready for some feedback, press play.

What did you think? It was C, wasn't it? The multiplication expressions you could write for this problem are 3 times 65 and 3 times 80.

The common factor is three.

So you need to subtract the uncommon factors of 80 and 65 and then multiply by three.

So C has 80, subtract 65 in bracket 'cause we do that first and then we multiply the difference by three.

You could also have thought about the difference between the cost of one pencil and one eraser because Sam bought three of each item.

So you'd need to find three times this difference.

It's represented with the same calculation, 80 subtract 65.

There's the difference in the cost multiplied by three.

And it's time for your final practise.

Solve each problem using the distributive law and write the equation that shows how you solved it.

Pause the video, have a go.

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

How did you get on? So in A, it's back to my pens again.

I buy five boxes of 24 black pens and three boxes of 24 purple pens.

How many more black pens than purple pens does Mrs. Hopper have? So this time we are subtracting those boxes.

The common factor is 24 again.

I've got five boxes of black and three boxes of purple.

So five subtract three is equal to two.

2 times 24 is equal to 48.

So there were more black pens.

I have 48 more black pens than purple pens.

In B, there are 12 bookshelves with 11 books on each.

Three of the bookshelves are taken away.

How many books are left? So this time, 12 is our common factor.

We've got 12 times 11, and then we're subtracting three times 11.

So 12 subtract 3 is equal to 9.

9 times 11 is equal to 99.

So there are 99 books left.

In D, normally cinema tickets cost 10 pounds.

However, on Mondays, the tickets are seven pounds.

Three friends buy cinema tickets on a Tuesday.

How much money could they have saved if they went on a Monday? Ah, so they paid 10 pounds for their three tickets, but they could have paid seven pounds, a three pound difference.

So we are going to calculate 10 subtract 7.

That's our uncommon factor.

The common factor is that there were three friends each time.

So the saving is 10 subtract 7, which is 3 pounds, and 3 times 3 is equal to 9.

So they could have saved nine pounds.

And finally in D, a running track is 800 metres long.

In a week, Jun runs 10 laps and Sam runs seven laps.

How much further has Jun run than Sam? So the common factor is 800 metres.

Jun has run 10 times 800, and Sam has run seven times 800.

So we can subtract the uncommon factors.

10 subtract 7 is equal to 3.

3 times 800 is equal to 2,400.

So Jun ran 2,400 metres further than Sam.

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

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

What have we been thinking about? What we've thought about the fact that the distributive law can be very useful for solving problems, where there is a common factor.

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

And you can use brackets in an equation to show that you've used the distributive law and to identify the addition or subtraction that needs to be calculated first.

Thank you for all your mathematical thinking in this lesson.

I hope I get to work with you again soon.

Bye-Bye!.