Lesson video

Hello, I'm Mr. Tazzyman, and today, I'm going to teach you this lesson from the unit that's all about equivalence, compensation, and subtraction.

So sit back, be ready to listen and engage.

Let's do this.

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

By the end, we want you to be able to say, I can explain how using constant difference can make written calculations more efficient.

These are the key words that you are gonna hear as we are learning.

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

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

Ready? My turn.

Minuend.

Your turn.

My turn.

Subtrahend.

Your turn.

My turn.

Difference.

Your turn.

My turn.

Constant.

Your turn.

Brilliant.

It's all very well being able to say them, but we need to check their meanings as well.

The minuend is the number being subtracted from.

A subtrahend is a number subtracted from another.

The difference is the result after subtracting one number from another and there's an equation written at the bottom there, a subtraction equation that's been labelled, seven subtract three is equal to four.

In that equation, seven is the minuend, three is the subtrahend, and four is the value of the difference.

A constant is a quantity that has a fixed value that does not change or vary such as a number.

Okay, here's the outline then for today's lesson on explaining how constant difference can make written calculations more efficient.

To begin with, we're gonna think about transforming the calculation, and secondly, we're gonna look at situations which feature larger numbers and decimal fractions.

So we'll start on transforming the calculation, and here are two people who are gonna help us out.

We've got Izzy and Sofia.

Hi Izzy.

Hi Sofia.

They're gonna help us by discussing some of the maths problems that we see, giving us some hints and tips and possibly even revealing some answers along the way.

Let's start with this then.

We've got here four different representations.

They're rulers with millimetres marked on them.

What's the same and what's different? Look at those green blocks.

What do you notice on them? We've got some expressions written.

What do you notice about the expressions? Izzy says the difference is constant for each of these representations.

The subtrahend and minuend are decreasing by one millimetre each time.

Izzy writes them as a sequence of expressions.

There they are.

So those are the four representations but just written down as expressions individually.

What do you notice? The difference is constant for each expression.

The minuend and subtrahend are decreasing by the same amount.

If the minuend and subtrahend are changed by the same amount, the difference remains constant.

What an important generalisation that is.

Which is easier to calculate mentally? So look at all four of those and think which would you rather calculate mentally? You might be able to do all of them and that's fine.

But what we're looking for is which one would you find easiest? And of course because we are being mathematicians, we need to also explain, we need to reason.

Which of those do you find easiest and why? Well, Izzy says 50 millimetres, subtract 25 millimetres is easier to calculate because you are starting from a multiple of 10.

I think I agree with that.

Well done, Izzy.

You can then partition the subtrahend.

This would become 50, subtract 20, subtract five.

I agree, says Sofia, but I think it's easier because I've used the known fact that 25 is half of 50.

Good spot.

Hadn't thought of that.

Okay, let's check your understanding so far.

The expressions below have the same difference.

Which would be easier to solve.

Pause the video and have a chat about that.

I'll be back in a moment to reveal what we think.

Welcome back.

So we thought that 70 subtract 28 would be easiest.

All of the expressions show a difference of 42.

That means they've got a constant difference.

But the minuend in this expression is a multiple of 10, which makes calculating the difference easier using partitioning.

70 subtract 28 is equal to 70 subtract 20, subtract eight.

That's equal to 50, subtract eight, which is 42.

Now you may have used the different mental method here, but I think starting from a multiple of 10 makes many of the different range of mental methods that we have, much more efficient, much quicker, and much simpler to complete.

Okay, let's look at this then.

How can they transform the numbers to create an equivalent expression? We've got 53, subtract 39, and we've got an unknown.

We don't know what the difference is yet.

Izzy says we need to turn the minuend or subtrahend into a multiple of 10.

Let's change the minuend and subtrahend by the same amount so that the difference is the same.

Remember, we can change both the minuend and subtrahend by the same amount.

The difference will remain constant if we do that.

They've changed it by adding one to each of them.

Now they've got 54, subtract 40.

Oh, I prefer that as a calculation to complete in my head.

That makes 14.

That's equal to 14.

Okay, your turn.

Transform the calculation below by increasing the minuend and subtrahend by the same amount to create one multiple of 10.

We've got 79, subtract 26, and we are trying to find the missing number, which is the difference.

Pause the video and have a go at that, using that transformation tactic of changing the minuend and subtrahend by the same amount.

Welcome back.

So you could have added one to both the subtrahend and minuend.

That would give you a new calculation of 80, subtract 27.

That seems easier to complete.

That's equal to 53.

To do that, you might have partitioned 27 into 20 and seven.

So you would've done 80, subtract 20, subtract seven.

80 subtract 20 is 60, 60 subtract seven is 53.

There it is.

How can they transform the numbers to create an equivalent expression this time? We've got 68, subtract 21.

We need to turn the minuend or subtrahend into a multiple of 10.

Let's change the minuend and subtrahend by the same amount so that the difference is the same.

This time, instead of adding one, they subtract one.

That gives them 67, subtract 20, and I can see that that's already easier to solve mentally.

It's 47, so the missing number was 47 because difference remains constant if you change the subtrahend and minuend by the same amount.

Okay, your turn.

Transform the calculation below by decreasing the minuend and subtrahend by the same amount to create one multiple of 10.

We've got 71, subtract 46.

Can you decrease the minuend and subtrahend to give you a multiple of 10 as one of them.

Pause the video and have a go.

Welcome back.

You might have decided to do it this way.

Subtract one for the minuend and the subtrahend.

That transforms the calculation to 70.

Subtract 45 and that seems much easier.

It's 25.

Again, you might have partitioned the subtrahend there.

You might have said 70, subtract 40, subtract five.

I know that 70 subtract 40, it's 30, and if I take five away from that, I get 25.

There it is.

What do you notice here? These are four different versions of using the generalisation that if you adjust the subtrahend and minuend by the same amount then the difference remains constant.

What do you notice is different across all four of these? If you increase or the minuend and subtrahend by the same amount, the difference will be the same.

So you can transform the calculations into something that can be mentally calculated more efficiently.

Okay, it's your turn to put that into practise.

Draw the following sequence of expressions on the number line below.

The first has been done for you.

And then Sofia asks, what do you notice? You can see that one has already been drawn on, 50 subtract 35.

So you've got to do the other four on that number line.

For number two, for each sequence of expressions, spot the odd one out that isn't equivalent to the others and then explain your reasoning.

For number three, below is a list of expressions that have not yet been transformed to make them more efficient to calculate.

There is also a list of adjustments for the subtrahend and minuend.

That means a list of changes.

Match the expressions with the adjustments and explain your reasoning.

Okay, pause the video here, have a go at those.

Enjoy and I'll be back in a little while to give some feedback.

Good luck.

Welcome back.

Let's start with the number line.

Compare this one to yours so you can mark it accurately.

There's 45, subtract 30, there's 40, subtract 25, there's 35, subtract 20, and there's 30 subtract 15.

Starting to make a really nice pattern, isn't it? Okay, what do you notice though? The difference is constant and the minuend and subtrahend decreased by five.

Here's number two then.

You had to find the odd one out, the expression that wasn't equivalent to the others.

For A, it was 97, subtract 77.

This is a difference of 20 compared to 18 in the other expressions.

For B, it was 52, subtract 43.

This is a difference of nine compared to five in the other expressions.

And for C, it was 71 subtract 19.

This is a different to 52 compared to 51.

in the other expressions.

We've got our list of expressions here that we needed to match with the adjustments, with the changes.

But 45, subtract 17 went to add three.

By adding three, the subtrahend will be a multiple of 10.

Then we had 71 subtrahend subtract 54.

That went to subtract one.

By subtracting one, the minuend will be a multiple of 10.

68, subtract 42, matched with add two.

By adding two, the minuend will be a multiple of 10.

And lastly 56, subtract 32 matched with subtract two.

By subtracting two, the subtrahend will be a multiple of 10.

Brilliant.

Let's move on to the second part of the lesson then.

We are gonna look at larger numbers and decimal fractions now.

Similar sort of thing to what we faced in the first part of the lesson.

Look at this expression and think how could we transform the numbers to create an equivalent expression? We need to turn the minuend or subtrahend into a multiple of 10, says Izzy.

Let's change the minuend of subtrahend by the same amount so the differences the same.

So far, very similar to what we've been thinking about already.

They add one to each, they get 439, subtract 390.

Sofia says, I don't think that's made it easier or more efficient.

I think we need to do something else instead.

What do you think? Do you think that was a good adjustment to make to the minuend and subtrahend? Is there a different place value that we could aim for that would make this calculation easier with the same difference? Izzy says, I can see that 389 is near the next multiple of 100 so let's try adjusting the minuend and subtrahend to create a multiple of 100.

That's a good idea.

This is a three digit number, so using multiples of 100 will be more efficient.

This time, they decide to add 11 to both the minuend and subtrahend, so they get 449, subtract 400.

That seems much easier, it's 49.

I can see that just by using place value and partitioning 449.

That means the missing number is 49, because of course, difference remains constant if you adjust the subtrahend and minuend by the same amount.

Okay, let's do some thinking here.

We've got a true or false statement.

You can only adjust the subtrahend and minuend by one for the difference to remain constant.

Is that true or is that false? Decide now and pause the video.

Welcome back.

That was false, but why? Let's make sure that we include justification because that's what good mathematicians do.

Here's two justifications.

You've got to select which one you think is correct.

Is it A, As long as the amount is the same, the minuend and subtrahend can be adjusted by any amount.

B, you can adjust the minuend and subtrahend by different amounts and the difference will be the same.

Pause the video and have a think about which justification you think you would choose.

Welcome back.

A was the best justification here.

As long as the amount is the same, the minuend and subtrahend can be adjusted by any amount.

For B, you can adjust the minuend and subtrahend by different amounts and the difference will be the same.

We know that's not true.

The difference will only remain constant if the minuend and subtrahend are adjusted by the same amount.

Okay, here's another expression to look at and think about how we can transform it to create an equivalent expression that's easier to solve.

Let's try adjusting the minuend and subtrahend to create a multiple of 100.

So they subtract four from the minuend and subtrahend and they get 35,225, subtract 15,200.

Now we can use partitioning to subtract the hundreds and thousands separately.

So 35,225, subtract 15,000, subtract 200.

They've started using the thousands and they've got 20,225.

Subtract 200.

That equals 20,025.

Still quite a few steps there, but definitely easier to solve mentally by adjusting the minuend and subtrahend.

Okay, your turn to have a go at that.

Transform the calculation below by decreasing the minuend and subtrahend by the same amount to create a multiple of 100.

Pause the video and have a go.

Welcome back.

We decided that we would subtract three to the minuend and subtrahend because that would give us 72,510, subtract 50,400.

So you can see that that's a multiple of 100 in the subtrahend there.

That makes it easier.

Then we use partitioning.

We ended up with 22,510, subtract 400, which was equal to 22,110, so that was the difference.

Have a look at this one then.

Instead of larger numbers, we've got decimal fractions this time.

Let's try adjusting the minuend and subtrahend to create a multiple of 100.

Wait, I don't think that will be successful here.

Look at the numbers.

What do you think? Do you think it would be worthwhile trying to adjust the minuend and subtrahend so that one of them is a multiple of 100? You are right.

These are decimal fractions, so let's adjust to the nearest one.

So to begin with in this lesson we were adjusting to the nearest 10.

When we had large numbers, we went to the nearest hundred.

Now we've got decimal fractions.

We are looking at adjusting to the nearest one.

I can see that 3.

95 is nearest to the next hole.

So let's adjust that.

They've added five-hundredths.

Now they've got 19.

72, subtract 4.

00.

We could just say subtract four.

We get 15.

72.

Because difference remains constant, if you adjust the subtrahend and minuend by the same amount, that means that the missing number was also 15.

72.

Your turn, you are gonna do the same thing, but for these, 16.

45, subtract 5.

03.

Remember, you're going to be decreasing the minuend and subtrahend by the same amount to create a multiple of one, decreasing, not increasing this time.

Alright, pause the video and have a go.

Welcome back.

So we took away three-hundredths from each, from the minuend and from the subtrahend.

That gave us 16.

42, subtract 5.

00.

Again, a much easier calculation to complete mentally.

It gave us 11.

42, and because we know that we have constant difference if we adjust the minuend and subtrahend by the same amount, the missing number was also 11.

42.

Did you get that? I hope so.

We've got another expression here.

30,000, subtract 3,765.

Larger numbers, tricky to do.

Izzy says, I'm gonna use column subtraction as a written method to solve this.

Let's see how she gets on.

Oh, that took a long time, didn't it? And actually, look at the amount of regrouping there.

Wow, that's a lot of regrouping, says Sofia.

I would do it differently.

I'd use constant difference to transform the calculation first.

Watch this.

This is a really good tip.

I decreased the subtrahend and minuend to create an equivalent expression.

So Sofia takes away one from both the subtrahend and minuend.

Aha.

I immediately noticed that the minuend now features a lot of nines as its digits.

Well, that means that we're not gonna have to do any regrouping.

Fantastic.

I will still make use of columns here, but there won't be any regrouping now.

There it is.

Nice and quick.

Same answer, but no regrouping.

So much more efficient.

The answer is 26,235.

No regrouping.

That's much more efficient.

I think Izzy will probably use that tip in the future.

Okay, it's your turn.

Transform the expression below by decreasing the minuend and subtrahend to create an equivalent expression that won't need regrouping in column subtraction.

Then solve it using a column subtraction.

We've got 50,000.

Subtract 2,124.

Give it a go.

Pause the video.

Welcome back.

So you might have started by subtracting one from both the minuend and subtrahend.

That gives you 49,999, subtract 2,123.

That's something that you can use columns with much more efficiently.

You end up getting 47,876.

I hope you managed to get that.

Have a look at this one.

5.

1, subtract 2.

98.

Izzy says, I would adjust the minuend and subtrahend by subtracting one-tenth.

5.

0, subtract 2.

88.

Then I would count on and use partitioning to find the difference.

That gets 2.

12.

So the missing number is 2.

12.

Sofia says, I would do it differently, but still use constant difference.

Can you see a different way of using constant difference? Hmm.

Well, here's Sofia's method.

She starts again with the same expression, but she decides to add two-hundredths to each.

So she gets 5.

12, subtract 3.

00.

The result is the same, 2.

12, which means the missing number is also 2.

12 because of constant difference.

Okay, it's your turn to have a go now at the second task.

For each of the expressions below, use constant difference to transform the calculation to solve more efficiently.

Explain your choices of adjustment.

There's that particularly important part, explanation.

That's what makes us mathematicians.

For number two, each expression below can be transformed using constant difference to solve more efficiently in two different ways.

Fill in the blanks to show the two ways.

So for A, you can see that we've got the jottings laid out, but there's two versions of them.

You are gonna decide to add or subtract different amounts.

Same for B.

And then for number three, for each expression, choose whether or not you think constant difference would be a useful mental strategy by ticking yes or crossing no.

Explain your choices and include any adjustments if you tick them.

Okay, pause the video here and have a go at those.

I'll be back shortly with some feedback.

Good luck.

Welcome back.

We'll start with number one, A.

You might have adjusted it by subtracting four from the minuend and subtrahend, giving you a much simpler calculation.

The final answer was 37,221.

By subtracting four from the minuend and subtrahend, the subtrahend became a multiple of 1000, which was easier to subtract.

Let's look at B then.

This time we subtracted three-hundredths and that gave us an easier calculation, which resulted in 54.

47.

That was the difference that you needed to find.

What was the explanation? By subtracting 0.

03 from the minuend and subtrahend, the minuend became an integer, so we could use partitioning and counting on easily.

Here's number two then.

For A, we just started by subtracting 30.

That gave us 20,200.

Subtract 4,065.

For the second part of A, instead of subtracting 30, we decided to add five because that meant that the subtrahend was a multiple of 100.

We ended up with 20,235, subtract 4,100.

Let's look at B then.

These were decimal fractions, so we were trying to get to the nearest one.

To begin with, we took away three-tenths, giving us 16.

0, subtract 4.

66.

For the second part, we added four-hundredths.

That meant that this time the subtrahend was actually a multiple of one.

We got 16.

34, subtract 5.

00.

Okay, let's look at number three then.

For A, we ticked it.

We said yes.

Add seven because 93 is near to a multiple of 100.

For B, we crossed it.

Neither the minuend or subtrahend are close to the nearest one.

For C, that was a cross as well.

We said no, the subtrahend is already a multiple of 10,000, so no adjustment is needed.

But for D, we ticked it.

Subtract 0.

003.

That's three-thousandths, because 45.

003 is only 0.

003 from an integer.

That brings us to the end of the lesson then.

Here's a summary of all of the learning.

If the minuend and subtrahend are changed by the same amount, the difference stays the same or remains constant.

Constant difference is a useful strategy to transform calculations into something more efficient to solve.

By adjusting the minuend and subtrahend carefully, you can make one of them a power of 10, which is easier to mentally calculate with.

My name is Mr. Tazzyman.

I hope you've enjoyed the lesson today.

I know I have, and I hope that I might see you again in future Maths lessons.

Bye for now.