Lesson video

Hello, my name's Mrs. Hopper and I'm really looking forward to working with you in this lesson on how to apply the distributive law to multiplication.

I wonder if you've come across the distributive law before.

Well, if not, we're gonna learn all about it and how it can really help us to be efficient when we're thinking about multiplications.

So if you're ready, let's make a start.

So in this lesson, we're going to be using knowledge of the distributive law to solve two-part subtraction problems. You may have come across the distributive law before.

If not, we're going to be learning all about it together.

So let's make a start.

We've got two key phrases in this lesson.

We've got common factor and distributive law.

So I'll take my turn and then it'll be your turn to have a practise saying them.

Are you ready? My turn.

Common factor.

Your turn.

My turn.

Distributive law.

Your turn.

Well done, let's just check we really understand what those words mean or those phrases mean.

They're going to be really useful to us in our lesson today.

So when we compare lists of factors of two or more numbers, any factors that are the same are common factors.

That's one way of thinking about it.

We're going to be thinking about multiplication expressions and finding factors in our multiplication expressions that are common to both, and we'll look at those.

We've got one on the screen below to have a look at.

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, we've got four times three here and we've sort of partitioned our four, our factor of four into two and two, and we can say that four times three is equal to two times three plus another two times three.

And you can see that the two and two are common factors in those expressions as are the three and three in those ones.

So look out for common factors as we go through our lesson today.

There are two parts to our lesson.

In part one, we're going to be looking at two-part problems with the common factor, and in part two, we're going to be looking at two-part problems in context.

So let's make a start on part one, And in this lesson, you're going to meet Sofia and Laura.

Laura and Sofia are looking at an equation.

Can you spot something? Laura says, "This is a two-part subtraction problem because there are two multiplication expressions and they're being subtracted." "Yes," says Sofia, "Nine groups of three subtracted from nine groups of six.

There's a common factor here," she says.

What's the common factor? Can you see it? The factor of nine is common to nine times three and nine times six, so we can say that it is a common factor in both of those multiplication expressions.

"So the six and the three are not common factors in the multiplication expressions." Let's just check that understanding of common factors.

Have a look and find out what is the common factor of the two multiplication expressions in the equation below.

Pause the video, have a go.

When you're ready for some feedback, press play.

What did you spot? That's right, the factor of seven is common to seven times five and seven times two.

And we're going to learn how to use that information to help us to solve equations like this efficiently.

So let's go back to our original equation.

And Sofia says, "Let's use an array to represent the equation." So here's nine times six and here's nine times three.

We got to think carefully about what we do with those arrays because this is a subtraction remember.

Laura says, "Both arrays have nine rows because nine is the common factor." That's absolutely right.

She says, "I'm going to calculate each of these arrays and then find the difference between them." Good thinking, Laura, off you go.

Nine times six is equal to 54 and nine times three is equal to 27 and 54 minus 27 is equal to 27.

So she says that the value of those two expressions subtracted from each other is 27.

Sofia says, "I would've calculated it differently." Go on then, Sofia, show us how you'd do it.

She says, "I would've overlaid the arrays." Ah, so she's going to overlay the arrays and then subtract those counters.

So she's subtracted nine groups of three from nine groups of six, and those are the remaining counters.

She says, "I can think of it as six groups of nine subtract three groups of nine." So she had six groups of nine altogether and she subtracted three groups of nine.

"This leaves me with an array showing nine times three, which I know because of my three times table." Nine times three or three groups of nine is equal to 27.

Same answer, of course, as the one that Laura got.

So what's the same and what's different? Laura says, "I worked out both multiplications and then subtracted," and there's nothing wrong with that, Laura.

You got the right answer and you used your times table knowledge.

But she says, "You subtracted the factors that are not common, and then calculated just one multiplication." So we can think of it as six groups of nine subtract three groups of nine, and if we take three groups away from six groups, we'll have three groups left.

So we've got three groups of nine remaining, which is equal to 27.

Sofia says, "Yours needed more calculations than mine, so my method was more efficient." She says, "I used the distributive law." Knowing that we could think about the uncommon factors and deal with those and then multiply what was left by the common factor of nine.

Time to check your understanding of what using the distributive law is all about.

Can you spot the mistake in the jottings below? Pause the video, have a look, and when you're ready for some feedback, press play.

What did you spot? Ah, did you spot that those uncommon factors had been multiplied rather than subtracted? So here's what the jottings should have looked like.

We're subtracting three groups of five from seven groups of five.

So we're going to have four groups of five remaining.

Four times five is equal to 20.

And if we'd done seven times five is 35, subtract 15, we'd also have had an answer of 20.

So what's the same and what's different here in the way we've written the equations? Well, Laura says, "The factors within each multiplication expression have swapped position." So we have nine times six minus nine times three, and now we've got six times nine minus three times nine, which is actually an interesting way of thinking about it.

Six groups of nine subtract three groups of nine and that helps us to think about what we're doing.

We've got six nines and we're subtracting three nines, so we'll have three nines left.

The product is still the same because.

The products are still the same and therefore our answer is the same.

And that's because multiplication is commutative.

We can change the order of the factors and the product remains the same.

The common factor of nine has swapped position, but we can still use the distributive law because of that commutativity.

But we can still think of six groups of nine subtract three groups of nine being the same as three groups of nine.

What about this time? "Well, this time, the common factor of nine appears in different places within the expressions in the different equations." "But again, the product is the same and the answer is the same because multiplication is commutative." It doesn't matter where that nine appears.

If it's multiplied by a six and multiplied by a three, then our answer will be the same.

And we can still use the distributive law to identify the common factor and the uncommon factors.

Time for you to do some practise.

Can you match the expressions that have the same value from the first column to the second column? There are fewer expressions in the second column, so they will match to multiple expressions on the other side of the screen.

So you're going to match up the expressions in the column on the left with those on the right.

For question two, you're going to solve each equation using the distributive law and think about what you notice.

And for question three, you're going to fill in the missing numbers in the jottings below.

So look out for what are going to be our common factors and what are going to be our uncommon factors.

And you could always check because we've got the answer to the equation there as well.

Pause the video, have a go at those questions.

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

How did you get on? So first, we had some matching to do.

So did you see the common factors each time? So we had seven groups of three subtract four groups of three, so we must have had three groups of three remaining.

For the next one, we had seven times 10 and we were subtracting seven times five.

So our common factor is seven.

And we had 10 groups of seven subtract five groups of seven, 10 subtract five is equal to five, so we must have seven times five remaining, five groups of seven.

Three times nine subtract three times six.

So we've got nine groups of three and six groups of three, and we're subtracting the six groups of three, so we must have three groups of three remaining.

For the next one down, our common factor is seven.

And we can do our six subtract two, which is four, so four groups of seven.

Then we've got five groups of seven subtract seven one times.

So five seven subtract one seven, which again is equal to four-sevens.

Seven is our common factor.

And then seven times eight minus three times seven.

So our common factor is seven there and we've got eight groups and three groups and we're subtracting them, so we must have five times seven.

Well done for spotting the common factors and working with those uncommon factors to simplify those expressions.

So this time, we are going to solve each of these using the distributive law.

So we've got nine times four minus nine times two, so that's four lots of nine, subtract two lots of nine.

So that must leave us with two lots of nine, which is equal to 18.

Then we've got eight times six subtract eight times four.

So we've got eight six times subtract eight four times, so we must be left with eight two times, eight times two is equal to 16.

And for C, we've got eight times seven minus seven times six.

Well, our common factor is seven there.

So we've got eight sevens, subtract six sevens, which leaves us with two sevens or seven times two, which is equal to 14.

And then we've got six times 10 subtract six times eight.

So we've got 10 lots of six subtract eight lots of six, which leaves us with two lots of six, six times two is equal to 12.

Hmm, what do you spot there? Well, can you see a common factor emerging in all of those? Yes, it's the two times table going down and written using a pair of expressions subtracted.

So A represented nine times two, B represented eight times two, C represented seven times two, and D represented six times two.

The not common factors all had a difference of two, meaning that there was always going to be a factor of two.

Four subtract two, six subtract four, eight subtract six, and 10 subtract eight.

And the common factors decreased by one each time and that produced our two times table.

Nine, eight, seven, and six all multiplied by two each time.

And for question three, we are going to fill in the missing numbers in the jottings.

So we've got something times seven minus eight times four, and we've got eight times something as our simplified expression.

So our common factor I think here must be eight.

And then we've got seven lots of eight subtract four lots of eight.

So we must have a three there.

So our missing numbers were eight and three.

Eight times seven minus eight times four is equal to eight times three, which is 24.

Something times five minus seven times three is equal to seven times something.

Well, we've only got one seven there, so that other missing factor must be a seven.

Then seven times five subtract seven times three, so that's five seven subtract three sevens, which must leave us with two sevens.

Seven times two is equal to 14.

Something times nine minus four times five.

Well, we've got that four repeating there.

So four times nine minus four times five, so that's nine fours subtract five fours, which must leave us with four fours, and four fours are equal to 16.

So for D, something times six minus six times one, oh, there's our common factor of six, and it's equal to 30.

Well, we know that six times five is equal to 30, so we must have had six sixes subtract one six is equal to five sixes.

In E, something multiplied by eight minus three multiplied by eight is equal to two multiplied by, well that must be eight, that's our common factor.

And something subtract three is equal to two.

Well, that's five, so five times eight minus three times eight is equal to two times eight, which is equal to 16.

So in F, something times seven minus something times three is equal to something times four.

Well, I think eight must be the missing number there.

So eight times seven minus eight times three is equal to eight times four, which is equal to 32.

And then we've got six times something minus something times four is equal to nine times two.

Well, I think nine must be that missing common factor because six subtract four is equal to two and nine times two is equal to 18.

And finally, something times five minus something times two is equal to 12 times three.

Well, five subtract two is equal to three, so our missing factor must be the common factor of 12, 12 times five minus 12 times two is equal to 12 times three.

We can think of that as five twelves minus two twelves is equal to three twelves, which is equal to 36.

Lots of really good thinking going on there.

I hope you were able to reason your way through to fill in those missing numbers as well.

And on into part two of our lesson, we're going to be looking at two-part problems in context.

So Laura gets five pounds a week pocket money.

She wants to save up for a new board game which costs 20 pounds so she decides to only spend two pounds each week.

Will she be able to afford the board game after six weeks? Hmm.

I wonder.

Laura says, "We can use the distributive law here.

I need to work out how much I get and subtract how much I will spend over six weeks." So she needs to know how much she's going to get and how much she's going to subtract.

"I agree," says Sofia, "let's start with a bar model." "So what would the expression be for how much pocket money I will get over six weeks?" says Laura.

Well, she gets five pounds a week over six weeks.

So Sofia says, "Well, there are six weeks with five pounds gained each week.

So that's six times five, six lots of five pounds." Now what about an expression for how much I spend each week? So she's only gonna spend two pounds, but she is going to spend two pounds each week.

"So there are six weeks with two pounds spent each week, so that's six lots of two pounds," which the Sofia says she'd write like this.

So the empty box is her savings, so that needs calculating and we know that if we know the whole and one part, we can subtract the known part to find the missing part.

We need to subtract six times two pounds from six times five pounds.

So there is our equation representing our bar model, subtracting the known part to work out the unknown part.

Six is the common factor and it's the number of groups, isn't it? Six groups of five pounds and six groups of two pounds.

So we need to think carefully about how we're going to handle our five pounds and our two pounds.

So the common factor and the number of groups is six.

"So let's simplify this equation using the distributive law." So first, to find the difference between the two not common factors, so we've got five pounds six times and two pounds six times.

So the difference is going to be three pounds, but we're going to have that six times.

So six times three pounds, we can do the multiplication, is 18 pounds.

She says, "After six weeks, I will have saved 18 pounds, which isn't enough." It costs 20 pounds for the board game.

Another week to go I think, Laura.

Time to check your understanding.

Sofia gets four pounds per week pocket money, but only spends two pounds of it for eight weeks.

Which of the following equations shows how much she saved? So think about what the two expressions that we subtract would be and how would that simplify using the distributive law.

Pause the video, have a go.

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

How did you get on? So the equation we needed was eight lots of four pounds for how much pocket money she got and eight lots of two pounds for how much she spent.

And eight times four minus eight times two.

Well, we can take those two uncommon factors.

Four subtract two is equal to two.

So we had four lots of eight subtract two lots of eight, which leaves us with two lots of eight.

Eight times two pounds is equal to 16 pounds.

So it was the bottom equation that represented how much she saved.

Sofia puts eight tubs of tennis balls into the PE cupboard at the start of the year, and the tubs contain four tennis balls each, so there we go, we've got a picture of our one tub of tennis balls, and she put eight of them in the cupboard.

She looks again in January and three of the tubs are missing.

How many tennis balls in total are there now? Tennis balls do have a habit of going missing, don't they? Sometimes they get stuck on the roof or they go over a fence into somebody's garden maybe.

Let's think about this equation though.

So she says, "This is eight tubs of four tennis balls." Subtract three tubs of four tennis balls.

"Yup, that's right," says Sofia, "There are four tennis balls in each tub." So let's start with an expression for the number of tennis balls to begin with.

So that was eight times four, eight groups of four tennis balls.

Next, an expression for the number of balls we need to subtract.

So they lost three tubs, didn't they? So they lost three lots of four tennis balls.

That's three tubs of four balls, so that's three times four.

So how many tennis balls are left now? Well, the common is four, which represents the group size.

So there were eight lots of four, and we've lost three lots of four.

So we can subtract eight minus three is equal to five.

And then we know that we've got five lots of four tennis balls left and five times four is equal to 20.

"So there are 20 tennis balls remaining now." Let's hope they last for the rest of the year.

So let's have a look at those last two problems, the pocket money problem and the tennis balls problem, and think about what we notice about the common factor in each case.

So the common factor in the first question was the number of groups.

It was the six weeks.

So six weeks of pocket money and six weeks of saving and spending.

What about the tennis balls? Well, the common factor in the second question Was the group size.

Four was the number of tennis balls in each group and our uncommon factor was the number of groups.

It doesn't matter which factor type is the common factor.

We can still use the distributive law because we know that multiplication is commutative.

We're finding that common factor.

And actually, we then want to think about how many groups of those we have really.

So we want it to be the group size rather than the number of groups because then we can use the uncommon factor to think about the number of groups we've got and either make a total or a difference to help us to simplify and work out the value of the equation overall.

So let's check your understanding.

Laura put seven tubs of golf balls into the PE cupboard at the beginning of the year.

And the tubs contain 10 golf balls.

She looks again in March and five tubs are missing.

Which of the following equations shows how many golf balls remain? So think about the two equations that you would write to represent the beginning of the year and what's happened in March, and then how would you simplify that in order to solve the equation overall? Pause the video, have a go.

When you're ready for some feedback, press play.

How did you get on? So at the beginning of the year, there were seven tubs of 10 golf balls.

And over the course of the year, five lots of 10 golf balls went missing.

So the equation we want is seven times 10 minus five times 10.

So that's seven groups of 10 subtract five groups of 10, which leaves us with two groups of 10, and two times 10 is equal to 20, so it was that bottom equation that gave us the simplified version of our equation to show how many golf balls remained.

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

In question one, you're going to look at the two problems and decide whether the common factor is the group size or the number of groups.

And in question two, those same problems, you're going to decide which is the correct equation to represent the problem and then solve it.

For question three, there are eight boxes of six eggs on the shelf.

Sofia clumsily knocks into the shelf and three boxes fall to the floor.

All the eggs inside them crack.

How many unbroken eggs remain on the shelf altogether? So for question four, the table below shows the pocket money and spending per week for some of the children.

Over six weeks, who saved the most money? Use the distributive law to calculate.

Pause the video, have a go at those questions.

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

How did you get on? So for question one, you had to decide whether the common factor was the group size or the number of groups in each problem.

So in A, Laura has four sheets of stickers each with six stickers on, and Laura gives two of the sheets of stickers to Sofia.

So she had four groups of six and she gave away two groups of six.

So the common factor was the group size, how many stickers there were in each group.

So in B, Jun has five bags of apples, each with six apples in, and he lets Izzy choose two apples from each bag to take with her.

How many apples does Jun have now? So he had five bags with six, so he had five lots of six, and he's still got five bags, but he's got two fewer apples in each bag, so he subtracted five lots of two.

So the common factor is five, which is the number of groups he had, but the group size has changed.

So let's work out which is the correct equation and then solve it.

So for the stickers, the correct equation was the top one, four lots of six stickers and she gives away two lots of six stickers.

So she had four lots of six, she subtracted two lots of six, so she's left with two lots of six, which is equal to 12.

And for B, the correct equation was the bottom one.

Jun had five bags of six apples and he gave away two apples from each bag, so he gave away five lots of two apples.

So he had six lots of five and he gave away two lots of five, so he is left with four lots of five.

It's actually five lots of four.

But we can think about the factors the other way round if it helps us to think about the distributive law.

So he's left with five bags with four apples in each bag, which is 20 apples altogether.

Question three, poor old Sofia.

It's horrible when you knock things over, isn't it? I do feel sorry for her.

So there were eight boxes of six eggs on the shelf and Sofia knocks three boxes to the floor and all the eggs inside them crack.

How many unbroken eggs are on the shelf? So we started off with eight boxes of six eggs and we've lost three boxes of six eggs, and eight subtract three is equal to five, so we've got five boxes of eggs left, which is 30 eggs.

So we've still got 30 unbroken eggs.

So there's still quite a lot of eggs there, Sofia.

Don't worry too much about it.

And for question four, the table shows pocket money and spending per week for some children.

So who saved the most? Let's have a look.

So Alex gets eight pounds a week and he spends four pounds a week.

So let's work out how much he saved.

So we've got six lots of eight pounds subtract six lots of four pounds.

Well, if we think about it the other way round, it might help us with our distributive law.

We've got eight sixes and we're subtracting four sixes.

So we've got four sixes left.

In this case, our four sixes we're interpreting as six weeks of four pounds, and that's 24 pounds, so Alex saves 24 pounds.

Izzy gets six lots of nine pounds and she spends six lots of five pounds.

So let's think about that the other way round.

Nine sixes subtract five sixes, that's four sixes.

So she also saves 24 pounds a week.

What about Sam? Sam gets six pounds a week, six lots of six subtract six lots of one or six one times, so one lot of six, so that's five lots of six.

So Sam saves five pounds a week over six weeks, which is 30 pounds.

So Sam saved the most money.

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

We've been using knowledge of the distributive law to solve two-part subtraction problems. Two-part subtraction problems feature two multiplication expressions that need to be subtracted.

Identical factors are a common factor.

We can use the distributive law by finding the difference between the two not common factors and multiplying that by the common factor.

And both the group size or the number of groups can be common factors, meaning this concept can be applied to many contexts.

And by using our knowledge of commutativity, that idea of being able to think of six groups of four or four groups of six can help us to understand the bits we're subtracting and the bits that we're multiplying.

Well done for all your hard work today, and I hope I get to work with you again soon.

Bye-bye.