Lesson video

Hello there, how are you today? My name's Ms. Coe, and I'm really excited to be learning with you today, in this unit on using knowledge of part-whole structures to solve additive problems. If you're ready to get started, let's get going.

So this lesson is thinking about parts and wholes, and additive structures.

And by the end of this lesson, you will be able to say that you can identify the missing parts using knowledge of part-whole relationships and structures.

Let's get going.

There are two key words in this lesson today.

I'm going to say them, and I'd like you to say them back to me.

My turn, equivalent.

Your turn.

My turn, solution.

Your turn.

Great job, let's take a look at what these words mean.

So two or more things are equivalent if they have the same value.

So you might think about things being equal to something else.

And a solution is a value, or a set of values that we can put in place of an unknown, and that makes the equation true.

So if we think about an example for that, the solution for this unknown number, three plus mm, is equal to five, would be two.

The solution is the value that we can put in place of that unknown, which makes the equation true.

In our lesson today, we're going to identify missing parts using knowledge of part-whole relationships and structures.

And our lesson is split up into two cycles.

We're going to start by finding one unknown, and then we're going to look at finding two unknowns.

So if you're ready, let's get going with the first cycle of our lesson.

In this lesson today, you're going to meet Laura and Andeep, and they're going to be helping us with our thinking along the way.

Laura and Andeep are looking at number rods.

Now, you may have seen these before and you may have used them in your classroom.

As you can see, number rods have different colours and different lengths.

They start off by thinking about how many different ways they can use number rods to create an equivalent length to the orange rod.

So the orange rod is the whole, and they're thinking about how they can make an equivalent length using different number rods.

You might want to have a go at this yourself.

Laura is adding a rule though.

She says that you have to use at least two different rods.

So that's our rule for today.

Andeep reckons, no problem, he can handle this.

Let's see what he does.

There we go, has he used at least two different rods? Yes, he has.

So we can say that the orange rod is equal to the pink rod, plus the dark green rod.

And we can write an equation to show that.

In this equation the initials are used to represent the different colours.

So pink is P, for example.

Andeep makes a different way.

Has he stuck to Laura's rule? Yes, we have used two different rods to make the length equivalent to the orange rod.

And again, we can represent that as an addition equation.

Ah, Andeep is clearly on a roll because he's found a third way.

He's used two different rods and they are equivalent to the orange rod.

And we can say that the white rod, plus the blue rod is equal to the orange rod.

I wonder if he's got any more examples? Actually, he thinks this is far too easy.

He's saying there are loads of ways, and you might want to experiment for yourself.

Can you find different ways of making rods that are equivalent in length to the orange rod? "Well," says Laura, "let's spice it up a little bit.

Let's make it a little bit more challenging." I wonder what she's got in store? Now she's saying, you have to use at least three parts.

So all of Andeep's examples so far, have used two parts, like we can see here.

Now, he's got to use three parts.

Hmm, that's definitely a little bit more challenging.

"Challenge accepted," says Andeep.

I wonder what he's going to come up with? He's found a few ways, he says.

So let's look at this example.

We can see here that we have three rods that are equivalent in length to the orange rod.

And we can say that the light green rod, plus the dark green rod, plus the white rod is equal to the orange rod.

And we can write this as an addition equation.

He's also found that the black rod, the red rod, and the white rod are equal to the orange rod, and we can write that as an equation.

And he's found another way.

So in this example we have the pink rod, a light green rod, and another light green rod, which is equal in length to the orange rod, and we can write that as an addition equation also.

"Nice," says Laura, she's really impressed and she says she's extra impressed because Andeep has found an example with two rods that are the same.

So we have two light green rods in this example.

Well clearly, Andeep has not been challenged enough, so Laura's going to challenge him even more.

I wonder what she's going to do? Hmm, what is the missing rod? I wonder.

Now that feels a little bit more challenging.

So now, we have an unknown rod, we have a missing rod.

Let's think about what that equation is telling us.

The whole is the orange rod, and we have a light green rod, plus a red rod, plus something, is equal to orange.

Laura is asking, "what is the missing rod?" Andeep thinks it might be the dark green rod.

If you have some number rods, you might want to have a look and see if he might be right.

Oh, oh dear, not quite.

Why is that not correct? Well, that's right, the dark green rod means that the sum of the parts is not equivalent to the orange rod, it's too long.

It's longer than the orange rod.

So that doesn't work, I'm sorry, Andeep, it's too long.

So he says, it must be the yellow rod.

Ah, that looks better, doesn't it? So we can say that the light green, plus the red, plus the yellow is equal to the orange rod.

Andeep has found Laura's missing part, he's found the unknown value.

What about this time then, what's the missing rod? Ooh, what do you notice this time? That's right, the missing rod is now at the start.

Does it matter? No, not really.

Remember, we can add parts in any order.

So this time, we have something, plus red, plus dark green is equal to orange.

Hmm, I wonder if you can think about what the missing part might be? If you have number rods, you might want to have a look for yourself.

Andeep says, this is easier.

I wonder why he says that, hmm? He says, "I think we're missing another red rod." You know what, I think he might be right.

If I look at the gap where the unknown is, I think it's the same size as the red rod that we already have.

And Andeep is absolutely right, the missing value was a red rod, the unknown was a red rod.

So we can say that red, plus red, plus dark green is equal to orange, they are equivalent.

He solved that problem as well, well done, Andeep.

So let's have a look at both examples that Laura set.

What's the same about them? Hmm.

Well, in both examples we knew two of the parts, we had two known parts.

And because there are three parts, there is only one possible solution for each example.

So in the first example, we had one missing part, and there was only one option, only one rod that would fit in that space.

And the same for the second one.

Because we knew there were three rods that were equal to the orange rod, there was only one possible option there.

Time to check your understanding.

Which is the correct missing part? If you have number rods, you might want to experiment.

Pause the video and have a think.

Welcome back.

So we still have the whole of the orange rod, and this time, we know that the pink, plus the light green rod, plus something, one missing part, is equal to the orange.

And looking at those options there, I can see that a light green rod would fit into that space and make it equivalent to the orange rod.

So we can say that pink, plus light green, plus light green is equal to orange.

Well done if you spotted that.

Another check for your understanding.

Take a careful look at this equation.

How many possible solutions are there for this equation? Pause the video and have a think.

welcome back.

Well, we can see from the equation that we have one unknown value, there is only one missing part because we have one space in the equation.

That means, there is only one possible solution using the number rods.

Well done, if that's what you said.

In this case, the only possible solution was the yellow rod.

It was the only one that fit to make it equivalent to the orange rod.

Time for your first practise task.

For question one, I'd like you to write an equation to represent each of these images.

Remember, we have one missing part, one unknown each time.

So your equations will have a blank space or a line or a question mark or something to show that unknown value.

For question two, I'd like you to create your own missing part problem for a friend, to write an equation for, and then solve.

So for example, Laura has created this one for Andeep.

We have, the blue rod is equal to the pink rod, plus the white rod, plus one unknown value.

I wonder if Andeep could write the equation and then solve that problem? Good luck with those two tasks, and I will see you shortly for some feedback.

Welcome back, how did you get on? So for our first three examples, let's take a closer look at why these equations represent these bar models.

We have orange as the whole each time so you may have written equal to orange or orange equals, or orange is equal to, doesn't matter which way round your equation was written.

In the first example we had light green, plus white, plus something, is equal to orange.

Remember, we just have one unknown each time.

The second example, we had three parts, but still one unknown part.

So we had light green, plus light green, plus white is equal to plus.

So we had light green, plus light green, plus white, plus something, one unknown is equal to orange.

And in the final one, we had four known parts, but still one unknown part.

So we had light green, plus light green, plus white, plus white, plus something, is equal to orange.

Well done, if your equations look like that.

And remember, you might have left a space or a question mark instead of a line.

You may also know that you can use a combination of multiplication and addition where we have equal parts.

So in the second and third example, you could have written a multiplication and addition equation.

So you might have written two multiplied by LG, light green, plus W, plus something.

Remember, if we have equal parts, we can represent those as multiplication.

So I wonder what the last one would look like? That's right, we could write two lots of light green, plus two lots of white, plus an unknown, is equal to orange.

So well done, if you thought about those using multiplication and addition.

For question two, remember, that you will have created your own problem.

So your problem may not have looked like this, but if you remember, Laura set Andeep this challenge.

She told Andeep that there was one unknown in this example, and he had to write an equation and solve the problem.

So this is what Andeep did.

He wrote pink, plus white, plus something, is equal to blue.

And then he used his number rods to think about the missing part, which would be a pink bar.

If we add one pink bar, one unknown, it is equivalent to the blue bar.

Well done, Andeep, and I'm sure you had fun solving your own problems. Time to move on to the second cycle of our learning today, which is finding two unknowns.

So Laura and Andeep are still looking at number rod problems, and they're still using orange as the whole.

But this time, Andeep is going to challenge Laura.

I wonder what he's going to come up with? This time, the rule is that there are two missing parts.

Hmm, I wonder how that will change the challenge or the equation that we're going to write? So this time we have, the orange rod is equal to light green, plus light green, plus something, plus something, there are two unknown rods that will make it equivalent, equal to that orange rod.

That does feel a bit more challenging doesn't it? Laura's going start by adding a white rod.

So you can see that one of the parts, she's decided is going to be the white rod.

I wonder if you can see what rod will fit in that gap to make it equivalent to the orange rod? "That means," says Laura, "that the last rod must be another light green rod." Do you think she solved the problem? Well, we've got two missing parts, two unknown values, and they're equivalent to the orange rod.

So we can say that light green, plus light green, plus white, plus light green is equal to orange.

I think she's cracked it.

She's got two missing parts, and they're equivalent to the orange rod.

Well done, Laura.

"Oh," says Andeep, "hmm, are there any other ways that you could do it?" So he is not quite satisfied that Laura's found one way.

Can you think of another way? Now, you may have come up with a different way.

I wonder what Laura's going to do? So we have the same two known parts, the two light green rods, and we're still looking for two unknown values.

But we can't use the same example that Laura just used.

"Hmm," says Laura, "this is a bit trickier." She completes it like this.

What do you think? Do you think that's correct? She could start with a light green rod, and then add the white rod.

So our equation is light green, plus light green, plus light green, plus white is equal to orange.

What do you think? Is that another way that she could solve this problem? "Hmm, not quite," says Andeep.

"You could do it that way, but actually, that's just the same as the original one.

We've just applied the commutative law." So remember, the commutative law tells us that we can add the parts in any order.

So adding light green, and then white, is the same as adding white, and light green.

They're the same equations, just in a different order.

So not quite there yet, Laura.

"Are there any other ways?" Says Andeep.

He's still really challenging Laura, I wonder what you think? Well, we've still got the same two known parts, and two unknown parts.

And remember, we can't use white and light green.

"Well," says Laura, "there's nothing else that could go with a white rod." So if white was one of the unknown parts, we can't use that because the only option would be to add light green, and we know we've done that.

So what about the next smallest rod, the red rod? Now, I like Laura's thinking because she's thinking systematically.

She started with the white rod, which is the smallest, and now, she's gone to the next smallest, which is the red rod.

Let's see.

Ah, so in this case, if we use one red rod, we have two red rods that fill the gap.

Andeep's really impressed by that, he says she's working systematically.

So starting with that smallest rod, and working up.

We can say that light green, plus light green, plus red, plus red is equal to orange.

We have two unknown values, and we found a possibility for both of them.

Well done, Laura.

What did you notice with the challenge that Andeep set Laura? Well, the problem needed two additional rods each time to make the rods equivalent.

So there were two unknown values this time, rather than one, that we saw earlier.

And there were also this time, multiple solutions to solve the problem, not just one.

Remember, we found that the light green and white were two unknowns that would work.

And we also found that two red rods would work as well.

So this time, if there are two unknown values, there might be more than one solution to the problem.

There might be more than one way to work it out.

Time to check your understanding.

This time we have two unknowns in our problem.

Tick two rods that would complete the whole.

Again, if you have number rods, you might want to experiment.

Welcome back.

So we could have a red rod, and a dark green rod.

If we put the red rod, and the dark green rod in those spaces, in either order, remember, because it doesn't matter which way we add the parts, we can say that red, plus red, plus dark green is equal to orange.

Well done, if you spotted that combination.

Time for another check of your understanding.

This time, each A, B, C, and D has two rods.

Which combinations of those would complete the whole? So we have yellow, plus something, plus something is equal to orange.

Take a moment to have a think.

Welcome back.

So we could have the red rod, and the light green rod.

So we could have yellow, plus red, plus light green is equal to orange, and they would be equivalent.

Did you spot any other solutions? That's right, we could also have the pink and the white rod.

They would also sum to make up the orange rod.

We could have red and light green, or we could have pink and white.

So in this case there were more than one solution to this problem.

Well done, if you spotted both of those.

It's time for you to have a practise now.

For question one, I'd like you to think about how many different ways can you complete this equation? So we have two unknown values.

The white rod, plus something, plus something is equal to the blue rod.

Laura is reminding you to think about working systematically to find all of the solutions for this particular problem.

Remember, it's really useful if you've got number rods to use yourself.

For question two, find the number of solutions for each equation, and then think about when you've done these two, and the first one, what do you notice? Do you notice anything in particular? Can you predict how many solutions there are for the last example with the light green rod, before you actually find out how many solutions there are? I wonder if you notice something about those? Pause the video, and have a go at those two tasks, and I'll be back shortly for some feedback.

Welcome back.

How did you get on with those tasks? So for question one, we had the blue rod is equal to the white rod, plus two unknowns.

Now hopefully, you thought about working systematically.

Laura found four solutions altogether, and you can see from her workings out that she started with the smallest rod being the unknown.

So we could have white, plus white, which is the smallest length rod, plus black is equal to blue.

Then she moved up to the next smallest rod, which is red.

Red and dark green is equal blue.

And doing this systematically and carefully, looking for those two unknown values, she found four solutions altogether.

So hopefully, you found four as well.

Remember, those two parts could come in either order because we can use the commutative law to recognise that it doesn't matter which order we add those parts in.

But there are only four unique solutions.

So well done, if you've got all of those.

Now remember, that question one had four solutions.

That's really interesting to think about in a moment.

So question two, we had the blue rod is equal to the red rod, plus two unknown values.

How many solutions did you find for that particular one? For that one, working systematically again, we had three solutions.

So for example, the white rod, plus the dark green rod, plus that red rod is equal to the blue rod.

Working systematically, we could then have another red rod, and a yellow rod.

And then the light green rod, plus the pink rod.

There were three solutions for that particular example.

Did you predict how many solutions there would be for the third one? Did you say two solutions? I think, I would've said two solutions because I can see that in the first example we had the white rod, then we had the red rod, now we've got the light green rod.

So the rods are increasing by length.

And question one had four solutions, question two had three solutions.

So I would've predicted that question three had three.

I would've predicted that the third one had only two solutions.

Did you work it out though? What did you find? Well actually, the second one had three solutions, but so did the third one.

Working systematically, we could have white and yellow, red and pink, and two more light green rods.

So actually, there were three solutions for that equation as well, which I didn't really expect.

Can you explain why that happened? Well, Andeep thought the same as me.

He predicted the last one would have two solutions, but it had an extra pair that was two lots of the same number rod.

So that's something to think about.

Sometimes we might have two equal values if we've got two unknowns, which makes the solutions one more than we're expecting.

Well done, if you saw that pattern, and well done, if you predicted that there would be three, and then found three solutions for that last problem.

We've come to the end of our lesson and I've really enjoyed thinking about missing parts, looking at one unknown and two unknowns, and thinking about those structures with you.

Let's summarise what we've learned.

You can use bar models to represent problems with unknown parts.

If a problem has three parts, and we know two of those parts, there can only be one solution.

But if a problem has three parts and we know one of them, which means we have two unknowns, then there can be more than one solution.

Thank you so much for all your hard work today, and I hope to see you again soon.