Lesson video

Hello, my name is Mr. Tazzyman and I'm gonna be teaching you today's lesson from the unit that's all about equivalence, compensation and addition.

I hope that with some of the learning that you experience today, you'll have a better understanding of some of the underlying structures for addition problems that you face in maths.

Here's the outcome for today's lesson then.

I can use a balanced equation to calculate unknown parts.

These are the key words that you might hear during the lesson.

I'm gonna say them and I want you to repeat them back to me.

So I'll say my turn, say the word, and then I'll say your turn and you can repeat it back.

Understand? Ready? My turn, equation.

Your turn.

My turn, expression.

Your turn.

My turn, whole.

Your turn.

My turn, part.

Your turn.

Here's what those words mean then.

An equation is used to show that one number calculation or expression is equal to another.

An expression contains one or more values where each value is separated by an operator.

Two or more expressions with the same value can be separated by an equal sign to create an equation.

And you can see an expression example at the bottom there, but it's also an example of an equation as well, because there's an equal sign.

The whole is all the parts or everything, the total amount.

A part is sum of the whole.

That bar model shows that relationship.

Here's the outline then.

The lesson is use a balanced equation to calculate the unknown parts, and to begin with, we're gonna find an unknown using a balanced equation.

Then we're gonna look at larger numbers and decimal fractions.

Let's start with the first part then.

In this lesson you will meet Jun and Laura and they're gonna be really helpful to us by discussing the maths, thinking about some of the different problems, and giving us some hints and tips along the way.

Hi, Jun.

Hi Laura.

Ready? Let's go.

Laura and Jun have a set of balanced scales.

I'm gonna roll up a whole pack of plasticine and put it on my side.

I'll do the same.

What do you notice? The scales are balanced.

Each side has the same massive plasticine.

Laura and Jun split their plasticine into parts, Jun replaces his plasticine parts with some weights.

The pack of plasticine was 500 grammes.

I'll divide that into two equal parts.

There are the weights.

So each part should have a mass of 250 grammes.

What do you notice? Both sides still have equal mass.

The scales are still balanced.

I must have 500 grammes of plasticine altogether, but my parts are unequal so they can't be 250 grammes each.

Laura estimates the mass of her plasticine parts and replaces it with weights.

I know it's less than 500 grammes because that's the whole.

I think the larger part is 400 grammes.

What do you notice? 400 grammes is too much.

The scales are imbalanced.

Try taking some weights away.

Okay, I'll take a 100 gramme weight away to leave 300 grammes.

What do you notice? That's better.

Now let's do the last part.

They think about the final plasticine part.

How can Laura do it? How can she work out what that would be in mass? I can calculate the last part she says.

But it doesn't have a value.

It's an unknown.

True, but I know the rest of the parts.

Your side is 250 grammes added to 250 grammes.

My side is 300 grammes plus something grammes.

I spy an equation, says Jun.

They use an equation to calculate the unknown.

An equation has an equal symbol.

Here's my side as an expression.

250 grammes plus 250 grammes.

Here's your side as an expression with the unknown part.

300 grammes plus an unknown grammes.

I can see that my first part is 50 grammes more than yours.

That means my second part must have 50 grammes less.

So the plasticine must be 200 grammes.

Let's test it.

Replace the plasticine.

It balances so it's correct.

Good reasoning.

Okay, it's time to check your understanding then.

Use the balance scales to write an equation with an unknown part.

Then calculate the mass of the plasticine.

So you've got to write an equation looking at that image to begin with.

Okay, pause the video here and have a go.

Welcome back.

Here's the equation then to begin with, an unknown number of grammes plus 250 grammes is equal to 200 grammes plus 200 grammes.

The second part has 50 grammes more.

So the unknown part must have 50 grammes less.

Here is 150 grammes.

Did you get that? I hope so.

Okay, let's move on.

There are the same number of children in Year 5, as there are in Year 6.

In Year 5, there are 28 children in Class 1 and 32 children in Class 2.

In Year 6, there are 29 children in Class 1 How many children are there in Year 6, Class 2.

We can write this as an equation.

Yes, with an unknown.

Here's an expression for the number of children in Year 5, 28 plus 32.

Here's Year 6 with the unknown as a question mark.

29 plus an unknown.

The expressions are of equal value.

Let's compare the first parts in each expression.

28 and 29.

There is one more.

So the unknown part must be 1 less so it balances.

That means it's 31 so there are 31 children in Year 6, Class 2.

We didn't have to calculate the whole to find it out.

We compared parts and used our understanding of a balanced equation.

For a party lunch, Andeep needs 28 purple Juice Blasts and 15 orange Juice Blasts.

When he gets to the shop they only have 10 orange Juice Blasts.

To make sure he has enough Juice Blasts all together, how many purple Juice Blasts should he buy? I can write this as an equation again.

I'll write both expressions, one with an unknown part.

So Laura's written 28 plus 15, which is what Andeep wanted in the first place, but when he gets to the shop, he found that they only had 10 orange Juice Blasts, so he had to buy an unknown number of purple ones.

Now to compare the known parts.

15 compared to 10, or 10 is 5 less than 15, 5 less so the other part needs 5 more to balance.

28 plus 5 is 33.

He needs to buy 33 purple Juice Blasts.

Okay, let's check your understanding.

Can you spot the mistake in the calculation below? Pause the video and have a go.

Welcome back.

The final expressions are unequal.

This isn't a balanced equation.

They have subtracted twice instead of using addition as the inverse.

So if you have a look there, you can see that both of those arrows are labelled with subtract 5.

Now, for the equation to balance, one of those needs to be add 5 instead, and then the part needs to be adjusted accordingly.

There it goes.

There are the same number of children in Year 5 and 6.

On a school trip to the theatre the children get to choose a snack for the interval.

What would you choose? Yum.

The table below shows the snack orders, but there has been an ink splodge on one part.

What number was written there? So you can see the table there.

We've got Year 5 and Year 6, and then we've got an apple or raisins.

The year groups have the same number of children.

That means I can write an equation.

Here is the expression for Year 5 snacks.

27 plus 33 because Year 5 ordered 27 apples, and 33 packs of raisins.

Here is the expression for Year 6 with an unknown part.

We know they ordered 31 apples, but then we need to add on an unknown.

We don't know how many raisins they've ordered.

I'll compare some parts.

27 and 31.

Well, 27 plus 4 is 31.

I need to subtract 4 to balance the equation.

33 subtract 4 is 29, so they need 29 packs of raisins.

Laura and Jun compare some worked examples.

What do you notice? Have a look at those two different examples.

What can you see? This is almost the same calculation twice, says Jun.

The parts have been swapped but the redistribution is the same.

It works because addition is commutative.

Laura and Jun compare some more worked examples.

What do you notice this time? These are almost identical.

The expressions have been swapped.

You can swap expressions around in an equation.

Laura and Jun can compare some more worked examples.

What do you notice this time? In the top equation they have compared 27 and 30.

In the bottom equation they've compared 15 and 30.

Both work though.

Okay, here's your first task then.

For number one, you need to calculate the unknown plasticine part for each of these balance scales.

Try to do it without calculating the whole.

For number two, there are the same number of children in Year 3 as there are in Year 4.

In Year 3, there are 27 children in Class 1 and 31 children in Class 2.

In Year 4, there are 29 children in Class 1.

How many children are there in Year 4, Class 2.

Again, try to calculate this without working out the totals.

For number three, there are the same number of children in Year 3 and 4.

They enjoy a special film afternoon in the hall and they have to vote for a film.

That table below shows the vote count, but they haven't finished yet.

What number will go in the box? Year 3, Year 4, "Cuddle Bears 2" and "Outside In." You can see the numbers there and use those to calculate the missing number.

Okay, good luck.

Pause the video here and I'll be back in a little while with some feedback.

Welcome back.

Let's look at number one to begin with then.

You might have written some jottings like these.

350 grammes plus 250 grammes is equal to an unknown amount plus 50 grammes.

Then you might have compared the parts that you did know.

You can see that the second part in the first expression is 200 grammes greater than the second part in the second expression.

That means that we need to add 200 grammes on to find out our unknown.

It's 550 grammes.

For number two then.

We might have started with 27 added to 31 equals 29 plus the unknown amount.

Then compared 27 and 29, there was an increase of 2, so for the second part we needed a decrease of 2 from 31 giving 29.

There were 29 children in Year 4, Class 2.

and for our film afternoon we had 13 plus 45 is equal to 41 plus an unknown.

If we compare 45 and 41, we can see that there's a decrease of 4, so we need to increase the other part by 4 giving 17.

so 17 children, Year 4 wanted to watch "Outside In." Okay, let's move on to the second part of today's lesson.

Larger numbers and decimal fractions.

Laura often keeps score for her mum's cricket team, the Tigers.

She keeps a running record using chalk.

The match ends up a tie.

But someone brushed past and rubbed off the 2nd inning score.

She has to work it out instead.

The match was tied so their total scores were the same.

That means I can write an equation.

She puts in the equal sign.

Here is the expression for the tigers.

270 plus 266.

Here is the expression for the other team with an unknown.

296 plus the unknown.

I'll compare some parts to help.

270 and 296.

We've added 26 on, that means we need to subtract 26 from the other part to give us our unknown.

It's 240.

Okay, your turn, the Tigers tie against another team and the same thing happens.

Can you work out the missing score? Pause the video here, have a go and I'll be back to reveal the answer shortly.

Welcome back.

You might have started with an equation that looks like this.

219 plus 160 is equal to 180 plus an unknown.

If we compare the parts, we can see that we've added 20 to get from 160 to 180, so we need to subtract 20 to get our unknown, which is 199.

So the missing number was 199 Jun sets Laura a challenge.

He calculates an equation with expressions featuring two parts by redistributing.

He covers up one of the numbers and challenges Laura to work it out.

I hope you're ready for a challenge.

I'm gonna use larger numbers.

47,500 plus 22,000 is equal to 50,500 plus an unknown.

There is an equal sign so I know both expressions have the same value.

I'll compare some parts to help.

I think the change is in the numbers of thousands.

It goes from 47,000 to 50,000.

That's an increase of 3000.

Because it's balanced, I need to subtract 3000 from the other part.

19,000.

Well done.

Can you challenge me now? Laura sets Jun a similar challenge but uses a different kind of number.

Here you go.

Try this.

47.

5 plus 22 is equal to 50.

5 plus an unknown.

These digits look familiar.

They're the same as my challenge to you.

The place value is different though.

I'll still compare parts and use that to help.

The answer was 19.

Jun compared two parts and saw that three had been added to go through one part to the other, which meant that 3 needed to be subtracted from the other part to get the unknown.

Laura and Jun compare some more work examples.

What do you notice? The decimal fractions are the same.

It is the integers that are different.

What do you notice? The first equation only has integers.

The second has the same integers combined with decimal fractions.

Yes, but the redistribution is still in integers.

So the decimal fractions are unchanged in each expression.

Okay, your turn.

Can you find the missing part using your understanding of a balanced equation? Pause the video and have a go.

Welcome back.

You might have started by comparing two of the parts to see what the change was.

Four hundredths were added on.

That meant four hundredths needed to be subtracted, so we had 102.

83.

Okay, it's time for your second task.

Use your understanding of redistribution and balanced equations to find the missing numbers.

Jun says, what do you notice? For number two, we've got another worded problem.

Below are two recipes for making jugs of fruit cocktail.

Each recipe creates the same volume of liquid.

However, the bottom of one recipe has been scribbled over by a toddler armed with a felt tip.

Can you work out how many millilitres of pineapple juice is needed? And here's number three, another worded problem.

Jun and Laura are doing a sponsored walk for charity.

The distance was the same for both of them, but they each stopped for lunch at different points.

Their journeys are shown below with one part missing.

How far did Laura walk in the second part, after lunch? Pause the video, have a go at those questions and I'll be back shortly with some feedback.

Welcome back.

Let's look at number one to begin with.

These were the missing numbers.

29 for a.

0.

29 for b.

3.

29 for c.

23.

29 for d.

2,329 for e.

The digits are very similar in each problem.

It is the place value that changes, but the structure is the same.

Here's number two then.

We had our toddler with their felt tip.

You might have started with an equation that looked like this.

1,345 plus 155 is equal to 1,323 plus an unknown.

Compare the parts and find that there was a 22 decrease.

So that means we had to have a 22 increase giving us 177.

177 millilitres of pineapple juice.

And here's number three.

You could have used arrows here to compare on the actual diagram.

Looking at 2.

54 to 3.

74, there's an increase of 1.

2 kilometres.

That meant there needed to be a decrease of 1.

2 kilometres in the second part to get Laura's unknown distance.

7.

06 kilometres.

Okay, that brings us to the end of our learning today.

I really enjoyed that.

Thank you very much.

Here's a summary of what we've learned.

Equations show two expressions which have equal value.

Equations are balanced.

This concept can be used to calculate unknown parts by comparing the redistribution between known parts across expressions.

This is also true for larger numbers and decimal fractions.

My name is Mr. Tazzyman and hopefully I'll see you again soon in another maths lesson.

Bye for now.