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 are going to use our knowledge of the order of operations to write equations and to complete equations with missing numbers or operations.

So we're going to be solving lots of problems involving the order of operations.

We've got two key phrases in this lesson, we've got the distributive law, and the order of operations.

So let's just practise saying them.

I'll take my turn, then it'll be your turn.

So my turn, distributive law.

Your turn.

My turn, order of operations.

Your turn.

Well done.

I'm sure there are phrases that you're familiar with.

You may have been using them quite a bit recently in your learning in maths.

So anyway, let's just remind ourselves of what they mean, because they are going to be important phrases to use 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, 4 X 3 = 2 X 3 + 2 X 3.

And the order of operations is a set of rules that tell you which operations have priority over others in an equation.

And they help us to make sure that when there's no context or no problem to be solved, an equation will have the same result, whoever solves it.

There are two parts to our lesson.

In part one, we're going to be using equations to show how we've solved a problem.

And in part two, we're going to use the order of operations to complete equations.

So let's get going with part one.

And we've got Jun, Aisha, and Sam helping us in our lesson today.

Jun and Sam have been working hard to understand the order of operations and the distributive law.

Jun says, "We know that in the order of operations when there are no brackets, multiplication and division have priority over addition and subtraction." Sam says, "We know that when there is a common factor or a common divisor, we can use the distributive law to work more efficiently." They've asked Aisha to give them some problems to solve.

They could be multiplication or division, combined with addition or subtraction.

Aisha says, "I have lots of problems I can give you to solve.

For each one, think carefully about whether you need to multiply or divide, and add and subtract.

Remember to choose an efficient strategy." So here's Aisha's first problem.

Are you ready? A pack of biscuits weighs 150 grammes, and a pack of cakes weighs half this amount.

What is the total mass of this shopping? We've got two cakes, and three packs of biscuits.

Jun says, "It says we need to find the total mass, so this will involve addition." Sam says, "There are groups of cakes and groups of biscuits, so I know this will involve multiplication." So it looks as though we may have multiplication and addition in this problem.

There are three packs of biscuits, and two packs of cakes.

3 X 150, 'cause that was the mass of a pack of biscuits, 3 X 150 + 2 X, what? "Oh," says Sam, "A pack of cakes weighs half the amount of a pack of biscuits.

Half of 150 is equal to 75." So we need 3 X 150 + 2 X 75.

So we need to do the multiplication first here.

3 X 150 = 450.

And 2 X 75 = 150.

And if we add those together, we get 600.

So the total mass of the shopping is 600 grammes.

Aisha changes the shopping in this problem.

We've still got the same facts, that the biscuits weigh 150 grammes and a pack of cakes weighs half this amount.

What's the total mass of this shopping though? Oh, Jun notice, now there are four packs of biscuits and four packs of cakes.

So we've got 4 X 150 + 4 X 75.

Have you spotted something? Yes, Sam says that means there is a common factor of four.

So we can use the distributive law to help us this time.

So we can add our uncommon factors.

They are 150 and 75.

So we can add those together, because we know that our shopping is four times the mass of cakes and biscuits added together.

So that's 225 X 4, which is equal to 900.

The total mass of the shopping this time is 900 grammes.

Time to check your understanding.

Which shopping problem involves common factors, meaning you would be able to solve easily using the distributive law? So we know that a tin of carrots weighs 400 grammes, and a tin of sweet corn weighs half this amount.

We're asking to find what the total mass of the shopping is.

So could you use the distributive law to solve the problem for A or B? Pause the video, have a think, and when you're ready for some feedback, press play.

So for the distributive law, we are looking for a common factor.

So where can we see a common factor? In A, can't we? In A, there are three tins of carrots, and three tins of sweet corn.

So we've got a common factor of three.

So we can add our tin of carrots, which is 400, and our tin of sweet corn, which is half that which is 200, and then multiply the sum of those two by three.

In B, we don't have the same number of tins so there's no common factor.

Aisha has another problem for Jun and Sam.

Class one and class two have made some badges to sell at the school fair.

Class one have made 16 badges, and class two have made 20 badges.

The badges are shared equally between four baskets to be sold.

How many badges are in each basket? Can you picture class one and class two putting their badges together to sell perhaps, and then dividing them between four baskets? Yes, Jun says, "It says the badges are shared, so this will involve division." "Class one and class two both sell their badges, so I know this will also involve addition," says Sam.

There is a common divisor of four, because we are dividing all the badges between four baskets, so we can use the distributive law.

We can add together the total number of badges, and then divide that total by four.

Oh, but Sam's reminding us, "Don't forget to use brackets to show that the dividends from each expression are added before the division takes place." If we leave it like that, we'll calculate our division of 20 ÷ 4 first, and then add 16, and that's not going to solve the problem.

This reflects how we're going to solve the problem.

The classes are going to put their badges together, and then divide them between the four baskets.

16 + 20 = 36, and 36 ÷ 4, or shared between four baskets will mean there's nine badges in each basket, as Jun says.

Aisha's got another problem for Jun and Sam.

She wasn't wrong when she said she had lots of problems for them to solve.

She says, "I have a rectangular garden pond which is seven metres by two metres." So one side of the pond is seven metres long, and the other side is two metres, and we say that as seven by two metres.

"My school," she says, "has a square pond with sides of four metres.

Which pond has the largest area and by how much? June says, "We have to find the area, so I know that this problem will involve multiplication." Can you think what else it's going to involve? Sam says, "We are finding the difference between the two areas, so this problem will also involve subtraction." Finding the difference is all about subtraction.

And Jun says, "There are no common factors.

The rectangular pond is seven by two metres, and the square pond is four by four metres." So no, there are no common factors there.

So we will multiply first to find the area of each pond.

Aisha's pond is seven times two metres, because we multiply the length by the width to work out the area.

7 X 2 = 14.

And the school pond is a square, 4 X 4 = 16.

Sam says, "I wonder what this would be like if we recorded it as one equation." So remember we are finding the difference, so we're subtracting.

We know now that the school pond has the greater area, so we can start with that expression.

4 X 4 - 7 X 2 = 2.

This time we need to do the multiplication first.

16 - 14 = 2.

So the school pond has a larger area by two metres squared.

Time to check your understanding.

Can you match the problems to the equations? Read the problems carefully, and see which equation matches with each problem.

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

How did you get on? So the first problem said, I bought five robot toys that cost three pounds each, and two books that cost four pounds each.

How much did I spend altogether? So we've got 5 X 3, and we are adding on 2 X 4.

So that matches with the bottom equation, even though the books are the first part of the equation, 2 X 4.

What about the bottom one? I bought five robot toys and two books, which item did I spend the least on, and by how much? So we've got 5 X 3, which is 15, and 4 X 2 = 8.

So the 5 X 3 is the greater value expression.

So 5 X 3, we're finding the difference this time, so we're subtracting.

So the top problem matches the bottom equation, and the bottom problem matches the top equation.

The top one talked about how much did I spend altogether, which links with addition.

And that, by how much, is often indicating a difference, so that indicates that we are looking at subtraction this time.

Aisha says, "Now you've heard lots of different problems from me.

Let's see if you can create some problems of your own to match my equations." Ooh, that's a challenge, Aisha.

Here is Aisha's equation.

You might want to have a look at it before Jun and Sam share their thoughts.

Jun says, "4 + 20 has been calculated first, and then the total has been divided by four." The 4 + 20 is in brackets, isn't it, so it must be dealt with first.

It takes priority over the division.

Sam says, "We need to think of a problem where two groups are combined, and then shared." Jun and Sam start thinking of a problem that could match this equation.

You might want to have a think as well, before they share their problem.

Sam says, "Let's use flowers.

Aisha has four roses and 20 tulips." There's our 4 + 2.

She shares them equally between four vases.

So there will be six flowers in each vase.

4 + 20 = 24, and 24 ÷ 4, or shared between four vases, is six flowers in each vase.

So there's a problem where we combine the total and then divide.

Aisha is impressed with Jun and Sam's work so far.

She gives them a slightly different equation.

Oh, can you see she's taken the brackets away this time.

So this time we're going to do the division first.

20 ÷ 4 = 5, and 4 + 5 = 9.

What's gonna have to be different about this problem? Well, Jun says, "There are no brackets this time, that means we divide first." So we're sticking with the flowers.

Sam says, "Aisha has 20 tulips, she divides them equally between four vases." So there's our 20 ÷ 4.

Oh, and Jun says, "She also puts four roses into the first vase.

How many flowers are in the first vase now?" So there's an equal share of the tulips, plus four roses, and we do the division first.

There are nine flowers in the first vase.

20 ÷ 4 = 5 for the tulips, plus the four extra roses.

Time to check your understanding.

Whose problem matches this equation? So have a look, is it Jen's problem or Sam's problem? Pause the video, have a read of the problems, and when you've decided, press play for some feedback.

What did you think? So we had 5 + 10 X 4.

There are no brackets this time, so we are looking for a problem where we do the multiplication first.

Oh, it's Sam's problem.

Let's have a read.

There are four flowers in a bunch.

Aisha buys 10 bunches of tulips.

She also buys five single roses that are not in a bunch.

How many flowers did she buy altogether? So she bought 10 bunches of four, so that's her 10 X 4, which is 40.

And then she bought five more, which is 45.

There are no brackets, so the multiplication has to be completed before the addition, which matches Sam's problem.

The equation for Jun's problem would need to include brackets.

Let's just check.

There are four flowers in a bunch.

Aisha buys five bunches of roses, and 10 bunches of tulips.

So in Jun's problem we'd have combined the five and the ten first, we'd have needed brackets around there.

5 + 10 = 15, and 15 X 4 = 60.

So we would've had a different result to our equation as well.

Time for you to do some practise now.

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

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

Remember to check to see if you can use the distributive law to work more efficiently.

So we're looking for those common multiples, or common divisors.

And for question two, you're going to create your own problem to match each equation.

How many different contexts can you think of? And Jen saying, "Remember to think about the order of operations.

Which part of the equation will need to be solved first?" Pause the video, have a go at questions one and two, and when you're ready for some feedback, press play.

How did you get on? Let's look at those problems in question one.

So in A, tennis balls are stored in packs of three, Jun has 12 balls to put away, and Sam has 21 balls to put away.

How many packs are put away in total? So this problem has a common divisor of three, so you can use the distributive law.

So altogether we've got 12 + 21 balls, so we need to add that together first, so we put that in brackets.

12 + 21 = 33.

And then we can divide the total number of tennis balls by three to put them into their packs of three.

And 33 ÷ 3 = 11.

So 11 packs of tennis balls have been put away in total.

For two, Sam has six books with 30 pages in each book, Jun has four books with 42 pages in each book.

How many more pages does Sam have altogether than Jun? So this time we've got no common factors, but we know that Sam has more pages.

So 6 X 30, and we're going to subtract 4 X 42.

So we're gonna have to do our multiplication first.

180 - 168 = 12.

So Sam's books have a total of 12 more pages than Jun's books have.

Sam buys five chocolate eggs that cost £3 each, one week later the same eggs are reduced to £2.

50 each.

How much would Sam have saved if she'd bought the eggs a week later? We've got a common factor of five here, haven't we? So we can use the distributive law.

The difference in the cost of the eggs is £3 - £2.

50.

So once we work out the difference, we can multiply that by the number of eggs that Sam bought.

£3 - £2.

50 = 50p, and five lots of 50p, is equal to £2.

50.

We need the brackets there to show that we need to do the subtraction first.

So Sam would've saved £2.

50 if she'd waited a week.

D, this table shows how long is spent in maths lessons every day.

How many minutes are spent in maths lessons over the week in total? Well, we can see here that we've got three lots of 40 minutes and two lots of 55 minutes.

So we can group those together, and do the multiplication first.

3 X 40 + 2 X 55 = 230.

So 230 minutes are spent in maths lessons over the week.

And in E, Jun needs to make six party bags.

He buys six pencils costing 30p each, six erases costing 40p each, and six notebooks costing 80p each.

How much has he spent altogether? We've got a common factor here of six, so we can use the distributive law.

We can add up the cost of the individual items, and multiply it all by six.

So we've got six multiplied by 30p + 40p + 80p.

So that's six lots of £1.

50, which is equal to £9.

Jun spent £9 altogether.

And for two you were creating your own problem to match the equations.

There were many different problems you could create to match each equation, so here are some examples.

So for A, we had 6 X 10 + 4 X 4, and we could have said, a restaurant has six large tables with 10 chairs each, and four small tables with four chairs each.

How many chairs are in the restaurant in total? For B, 6 + 10 X 4, we could have said a restaurant has 10 small tables with four chairs each, and six spare chairs.

How many chairs are in the restaurant in total? And in A and B, we would do the multiplication first, 'cause there are no brackets.

And when there are no brackets, multiplication takes priority over addition.

We do it first.

In C though, we've got to combine the six and 10.

We've got brackets there.

So a restaurant has six small chairs and 10 large chairs around each table.

How many chairs would be around four tables? So here we can combine the six and the 10, and then multiply by four.

Four would've been our common factor.

For D, this time again there are no brackets, so we need to do the division first.

Sam raises £84, which she splits exactly between seven different charities.

She also gives an extra £56 to the first charity.

How much does the first charity receive in total? So there we do the division first, and then the addition because there are no brackets.

For E, we've got brackets around the 56 + 84, so we need to add that first before dividing.

Sam raises £84 and June raises £56 for charity.

They divide their money equally between seven charities.

How much does each charity receive.

This time we've got a common divisor of seven, so we can add our dividends, our whole values of money, and then divide them all at once.

And in F we've got (84 - 56) ÷ 7.

So this time we're doing a subtraction.

First Sam raises £84 for charity, she gives £56 to her favourite wildlife charity and divides the rest of the money between seven animal charities.

How much money does each animal charity receive? So this time we need to work out how much money is left after she's given the £56 away, and then divide it by seven.

And on into the second part of our lesson, we're using the order of operations to complete equations.

Jun and Sam are going to try and solve equations that Aisha has written for them.

So here we've got 6 + 20 ÷ 4, is equal to something.

Jun says, "In the order of operations, division is completed before addition." So Sam says, "20 ÷ 4 = 5, so the solution to this equation is 11, because 6 + 5 = 11." Well done.

Aisha writes another equation for Jun and Sam to solve.

32 ÷ 4 -12 ÷ 4, is equal to something, divided by four.

Jun says, "This equation uses the distributive law." Can you see we've got a common divisor of four here? And Sam says, "When there is a common divisor, you can subtract the dividends before you divide." So 32 ÷ 4 - 12 ÷ 4 = 20 ÷ 4.

This time Aisha writes an equation with a missing operator.

Well, Jun says, "The four operations are addition, subtraction, multiplication, and division." Sam says, "The only operation that gives a solution of 27 is multiplication." And we'll do the multiplication first, because in the order of operations, multiplication is done before division.

4 X 6 = 24 + 3 = 27.

Aisha has one more equation for Jun and Sam to solve.

Oh, what do you notice here? Jun says, "I can see that three is squared.

We haven't solved any equations like this yet." We've got 5 + 3².

Sam says, "Let's have another look at the order of operations to see if that can help us." Oh, so brackets were at the very top, and do you remember that row below that had some strange looking symbols? Well included in there are powers.

So things like being squared or being cubed.

So the order of operations tells us that powers such as squaring or cubing should be completed after brackets, but before multiplication and division.

So now Sam and Jun can solve Aisha's final equation.

We need to calculate 3² first, and then add five to solve this equation.

We know that 3² means 3 X 3, and that's nine.

9 + 5 = 14.

So we can solve this equation.

Time to check your understanding.

True or false? The missing number in each equation is 11.

So look at those three equations, and decide whether 11 is the missing number in each equation.

Is that true or false, and can you explain why? Pause the video.

Have a go.

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

True or false? What did you reckon? It's false, isn't it, the way they're written.

So two of the equations have a missing number of 11.

The first two.

5 + 11, we do that first because it's in brackets, is equal to 16.

16 X 2 = 32.

And then for the bottom one, we do the two cubed first.

That's 2 x 2 X 2.

So 2 X 2 = 4, 4 X 2 = 8, 8 + 11 = 19.

What about the middle one though? If the missing number in the second equation was 11, it would need to have brackets like this.

6 + 3 X 11 = 99.

With no brackets, the missing number is 31.

3 X 31 = 93 + 6 = 99.

And it's time for your final task.

Question one you're going to solve each equation remembering about the order of operations.

And what do you notice that's the same and what's different, and can you discuss those ideas with your partner? And in question two, you're going to work out what's missing in each equation.

And in question three, you're going to write an equation that will be equal to each number below.

You should use the number four, four times in each equation.

No other numbers.

You can use your knowledge of the order of operations, and include brackets and powers, such as square, to help you.

So that's the only time you can use something that isn't a number four, is if you use a little two to show that you're squaring things.

So for example, you could write, (4 - 4) X 4 X 4 in the first box.

We do the bracket first, 4- 4 = 0.

0 X 4 + 0 X 4 = 0, so that will be equal to zero.

Pause the video, have a go at those three questions, and when you're ready for the answers and some feedback, press play.

How did you get on? Let's have a look at question one.

You were solving each equation.

So in A, the multiplication are done first, 6 X 5 = 30 + 4 X 2 = 8, is 38.

In B, again the multiplication are done first.

6 X 5 = 30 - 4 X 2, which is eight.

Our answer is 22.

In C though our bracket takes priority, so we do that first.

So 5 - 4 = 1.

6 X 1 = 6, and 6 X 2 = 12.

In D, can you remember, we need to do the squaring first, then the multiplication, and then we can subtract.

So six squared is 36, 3 X 4 = 12.

36 - 12 = 24.

In E again, we need to do the squaring first, and then the multiplication.

So 6 X 6 = 36, 3 X 4 = 12.

12 + 36 this time, is equal to 48.

And for F, we need to do the squaring first, and then the multiplication, and our three squared is part of our multiplication.

So it's useful to know which order to do things in here.

Three squared is equal to nine, 9 X 4 = 36 + 6 = 42 Did you remember to calculate the three squared to get nine and then calculate 4 X 9 before adding six in F? I hope you did.

Sam noticed that the first three equations all have the same numbers written in the same order, but they all have different solutions.

Oh, yes.

And Jun says, "I noticed that some equations were solved in a different order according to the order of operations." Really important to understand that order of operations.

In equations like this, we just apply the rules.

So what was missing in each equation in question two? So in A, eight plus something, times two is equal to 14.

Well 8 + 3 x 2.

we've got a multiplication to do here.

So something plus eight must equal 14, so it must be 6 + 8.

So our multiplication is first, it must be three.

Again, in our second equation, the multiplication is done first.

So something subtract seven is equal to 18.

Well that must be 25, so it must be 5 X 5.

In C as well, we're going to do the division first.

So something divided by two plus six, is equal to 11.

Well, we know that 5 + 6 = 11, so it must be 10 ÷ 2.

Again, in D no brackets, o our multiplication must be done first.

20 subtract something is equal to five.

Well, that must be 15, so it must be 3 X 5.

20 - 3 X 5 = 5.

In E we've got some brackets.

So we've got something times two is equal to 16.

So what times two is equal to 16? Well, that's eight, so it must be 5 + 3.

In F we've got eight, 4 X 2 = 0.

Well, 4 X 2 = 8, so it must be 8 - 8 = 0.

There are two possible answers to G.

We could have 12 X 6 = 72 ÷ 2 = 36.

Or we could have (12 + 6) X 2 = 36.

12 + 6 = 18.

For the multiplication and division one, we wouldn't have needed the brackets, because neither multiplication nor division has to be done first.

They have the same level of importance.

With the multiplication and the addition, we need the brackets to show that the addition has to be done first.

And finally for H, we've got an answer of three here.

So nothing multiplied by five is going to be equal to three as a whole number.

So it must be something divided by five.

So it must be 15 ÷ 5, so we'd need to do 18 - 3 in the bracket, to be 15 ÷ 5, and that would be equal to three.

And finally, there are lots of different ways that you could have completed this task.

Here's one example for each number.

So have a look at these, see how they compare with the way that you worked out these possible solutions.

Really good fun.

This is a really interesting way to explore brackets and the order of operations.

So take a moment just to see if you got the same solutions as we did.

And Sam says, "Did you remember to check the order of operations when writing your equations?" I hope you did.

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

We've been using knowledge of the order of operations to solve equations.

What have we learned about? Well, we've learned that when solving a problem that involves more than one operation, you need to think carefully about which operations are involved.

You can use the distributive law to solve problems more efficiently when there is a common factor or a common divisor, that's really useful to look out for.

You can often simplify what looks like a very complicated equation by looking for those common factors and common divisors.

And you can use your knowledge of the order of operations to write, complete, and solve equations correctly.

In a problem, the order of operations is often obvious, but when there's no problem there, we need to remember those priorities so that we will get the same results to the equation as everybody else.

Thank you very much for your hard work and your mathematical thinking today, and I hope I get to work with you again soon.

Bye-bye.