Lesson video

Hello, my name is Mr. Tazzyman, and today I'm gonna be teaching you a lesson from a unit that is all about solving problems featuring two unknowns.

You might have come across these kinds of problems previously, but in this unit, we're really going to look at comparing their structures and trying to solve them using things like trial and improvement, and maybe bar models where we can.

So make sure you're ready to learn, and let's go for it.

Here's the outcome for the lesson today then.

By the end, we want you to be able to say, "I can compare the structure of problems "with one or two unknowns." The key word today is difference.

I want you to say it back to me.

Well done.

The difference is the result of subtracting one number from another, how much one number differs from another.

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

We're gonna start by finding one or two unknowns.

Then, we're gonna think about the relationships between unknowns.

In this lesson, you're gonna meet Laura and Lucas, who will be learning with us.

They're gonna be responding to some of the prompts, trying to solve some of the problems, and they'll even be giving us some hints and tips along the way.

Hi, Laura.

Hi, Lucas.

Are you ready? Are you ready? Okay, let's start the lesson properly.

Laura and Lucas are playing with number rods.

You can see them on screen there.

They've all got different colours, and they're all different lengths.

Two rods are equivalent to the blue rod.

One of them is the red rod.

What is the other rod? Hmm, that's a really good problem.

Laura says "So we know the blue rod is equal "to the red rod and one more." There's the blue rod.

There's the red rod.

Here's our unknown.

We don't know which rod needs to go into that position.

"There can only be one solution.

"I think the missing rod might be the brown rod," says Lucas, so he's made a suggestion.

Hmm, that's the brown rod, but I'm not sure that the red rod and the brown rod are equivalent to the blue rod here.

Too long.

Lucas is right.

"I think it's the black rod," says Laura.

Aha! Success.

"Perfect!" says Laura.

"We can represent this as an equation too." B is equal to r + blk.

Blue is equal to red plus black.

Ah, that makes sense now.

So we've got the letters or groups of letters representing the different rods.

We used initials to represent each rod.

Thanks, Lucas.

That's cleared that up.

Two rods are equivalent to the blue rod.

What could the two rods be? Isn't this similar to the last question? Oh, it is, but actually the last question told us one of the rods already, which was red.

I wonder if we can start to think of our own combinations.

Well, we knew it could be red and black.

Well done, Laura.

She's remembered that from the previous question.

There they are, red and black.

"Yes, but I think there could be other solutions too," says Lucas.

If one of the parts was pink, "I think the other part would be yellow," says Laura.

There we go.

You can see that pink and yellow are the same length as the blue rod.

They are an equivalent length.

B is equal to p + y.

Blue is equal to pink and yellow.

Ah, that makes sense now.

I can see how each of those rods has been represented by a letter in the equation and initial.

"We have two different solutions so far," says Laura.

You can see them there.

We've got B is equal to r + blk, so the blue rod is equivalent to the red plus the black.

We also have B is equal to p + y, so the blue rod is also equivalent to pink plus yellow.

"I wonder how many solutions there are altogether," says Lucas.

Okay, it's time for your turn then.

We're gonna check your understanding so far.

Two rods are equivalent to the blue rod.

What could the two rods be? Pause the video and investigate.

Welcome back.

Well, we could have had white and brown.

B is equal to w + br for brown.

We could have had light green and dark green.

B is equal to light green plus dark green, or B is equal to lg + dg.

These are the last two solutions.

We already had the solutions previously in which we looked at the fact that it was red and black and pink and yellow.

Here's another chance to check your understanding.

Two rods are equivalent to the black rod.

One of them is red.

What is the missing rod? Pause the video here and have a go at finding the missing rod.

Welcome back.

You might have used the black rod to begin with, lined up the red rod alongside it, and thought about the missing rod here, and the missing rod is yellow.

There we go.

Okay, it's time for your first practise task then.

Two rods are equivalent to brown.

One of them is yellow.

What is the missing rod? Write an equation to represent this.

Remember, when you're using the equation, it's good to use initials, but be careful because the letter B crops up for quite a few of the colours.

You might want to use B to represent blue, and then further initials for brown and black.

For number two, two rods are equivalent to brown.

One of them is dark green.

What is the missing rod? Write an equation to represent this.

For number three, two rods are equivalent to orange.

What could the rods be? How will you know if you have all of the solutions? Okay, pause the video there, and I'll be back in a little while for some feedback.

Enjoy.

Welcome back.

Here are some of the answers then.

For number one, we were trying to find a missing rod.

It was light green, but what should the equation have been then? br is equal to y + lg.

For number two, it was a red rod that was missing, and the equation might have read br is equal to r + dg.

Remember, you might have chosen to use different initials or letters to represent these colours.

For example, some of you may have chosen B for brown.

That's up to you.

What you might want to do is pause the video now, just to see what other people have put, and to double check that you fully understood the concept behind this.

Here's number three then.

Two rods are equivalent to orange.

What could the rods be, and how will you know if you have all of the solutions? Laura says "If we start with the smallest part, "we can work systematically to find all the solutions." Really good mathematical thinking.

Well done, Laura.

So we've got white and blue.

White is the smallest part, so that's why Laura started with that.

That gives us this equation, o for orange is equal to w for white plus b for blue.

Now, we've got red and brown.

What will the equation be here? o is equal to r + br.

Then, we've got light green and black.

o is equal to lg + blk.

The next one is pink and dark green.

o is equal to p + dg.

And lastly, we've got two rods of equal length that are both yellow.

o is equal to y + y.

We know that we've got all the solutions here because if we were to continue, we would end up repeating some of the different parts we've already got.

If we carried on extending the first part, we'd end up using, on the next row, dark green and pink.

Now, dark green and pink we've already done, but because of commutativity, we've swapped round the addends.

So if we carried on, we would end up finding the same solutions we've already found.

Okay, let's move on to the second part of the lesson, relationships between unknowns.

Two rods are equivalent to the blue rod.

One of them is twice as long as the other.

What rods could they be? Laura says "Let's bring back all "of the possible pairs of rods." (mouse clicking) Which pair fit the description? Remember, one of them is twice as long as the other.

So have a look at those rods, the two parts, the bottom row in each of these.

"Is the brown twice as long as the white?," asks Laura.

You can see there, it's not.

Two lots of white are not equal to brown, so brown is not twice as long as white.

Really good reasoning.

Well done, Lucas.

Is the black twice as long as the red? Two lots of red are not equal to black, so black is not twice as long as red.

Is the dark green twice as long as the light green? Yes, two lots of light green are equal to dark green, so light green is twice as long as dark green.

We can say these two rods are related multiplicatively.

Let's record this as an equation.

What do you think the equation's going to be? Here it is.

dg is equal to 2 multiplied by lg, where dg stands for dark green, and lg stands for light green.

Dark green is equal to two lots of light green.

Makes sense.

So B is equal to lg + dg.

B is also equal to lg plus two lots of lg.

Hmm.

Lucas says "I've noticed that the blue rod "would be equivalent to three lots of light green rods." You can see that in that diagram.

We could also have an equation here that would be B is equal to three multiplied by lg.

Okay, let's check your understanding so far.

Two rods are equivalent to the brown rod.

One of them is three times as long as the other.

Which rods could it be? In A, you can see we've got white and black.

In B, we've got red and dark green.

In C, we've got light green and yellow.

And in D, we've got pink and pink.

Pause the video and decide which rods it could be.

Welcome back.

The answer was B.

Dark green is three times the length of the red rod.

You can see that shown here.

Three red rods make up one dark green rod.

Two rods are equivalent to the blue rod.

There is a difference of white between the rods.

What could the rods be? Laura says, again, "Let's bring back all of the possible pairs of rods again." (mouse clicking) Which two rods have a difference of a white rod? What is the difference between the brown rod and the white rod? And we set out those rods, so that you can see the difference in a comparison model.

The difference is greater than a white rod, so it isn't these two unknowns.

Can you see there, we put a white rod in, and that doesn't make it equivalent, so that's not the difference between the brown and white rod.

What is the difference between the black and the red rod? Let's put them in a comparison model again.

There they are.

The difference is still larger than a white rod, so it can't be these two unknowns.

How does Lucas know? Well, there's a white rod, and that doesn't make the red and the white equivalent to the black.

What is the difference between the light green rod and the dark green rod? The difference is still larger than a white rod, so it can't be these two unknowns, and we knew that from our multiplicative reasoning earlier.

We knew that two light green rods make up a dark green rod.

That means that the difference between a light green rod and a dark green rod must be one light green rod, and a light green rod is longer than a white rod, as you can see there.

What is the difference between the yellow rod and the pink rod? Aha! This looks more likely, doesn't it? The difference is one white rod this time, and we can show that by placing a white rod into that comparison model, and see if it covers the difference.

There it is.

y subtract p is equal to w.

In other words, yellow take away pink is equal to white, and then we've also got a second equation we can put in, y is equal to p + w, so yellow is equal to pink plus white.

Let's check your understanding then.

Two rods are equivalent to the brown rod.

There is a difference of red between the rods.

What rods could they be? Pause the video and see if you can solve this problem.

Welcome back.

We've got light green and yellow.

If we put light green and yellow into a comparison model, we can see that the difference is equivalent to a red rod, so the rods should be light green and yellow.

The difference between yellow and light green is red.

Yellow subtract light green is equal to red, and yellow equals light green added to red.

Okay, it's your second practise task now.

For number one, I'd like you to draw bar models to represent each answer.

A, two rods are equivalent to the orange rod.

One of them is four times as long as the other.

What rods could they be? B, two rods are equivalent to the orange rod.

There is a difference of pink.

What could the two rods be? C, two rods are equivalent to black and have a difference of light green.

What could the rods be? D, two rods are equivalent to the blue rod.

One rod is 3 1/2 times longer than the other.

What could the rods be? E, two rods are equivalent to the orange rod.

One is 2/3 the size of the other.

What could the rods be? For number two, you've got to sort different pairs of rods into this table.

It's a table with three rows and two columns, and the column that appears first tells us what we want to put into the second column.

The sum is equal to orange, or the sum is equal to orange and the difference is equal to red, or the difference is equal to red.

And again, Laura asks, "How will you know you have all the possible solutions?" Okay, pause the video here and enjoy those questions.

I'll be back shortly with some feedback.

Welcome back.

Let's start with 1a.

So the rods could have been red and brown.

That's because the brown rod is four times the length of the red rod.

B, the rods could have been black and light green.

That's because the difference between black and light green is the same length as the pink rod.

You can see that there.

For C, what could the two rods be? Yellow and red.

That's because yellow and red have a difference of light green.

For D, you could have had red and black.

You can see that the black rod is 3 1/2 times the length of the red rod.

If we line up the red rods underneath the black rod to compare them, we can see that there are 3 1/2 red rods for every one black rod.

3 1/2 of the red rods are equivalent to the black rod, and there's a dotted line there to help us to visualise this.

For E, you could have had pink and dark green.

We can see here that we've set up a bar model in which we've taken dark green and split it into thirds.

Each third is the same length as a red rod.

1/3 of dark green is red, so 2/3 of dark green is pink.

Here's number two then.

In the first row, you had to find all the pairs of number rods whose sum was equal to the orange rod.

You could have had white added to blue, light green added to black, red added to brown, yellow added to yellow, and dark green added to pink.

I'm gonna leave the middle row just for a moment and go down to the bottom row, in which you had to find the difference being equal to red.

These were the pairs that you could have chosen for this.

Orange subtract brown, brown subtract dark green, yellow subtract light green, light green subtract white, blue subtract black, black subtract yellow, pink subtract red, and dark green subtract pink.

Again, they've all been written as expressions using initials of each of the colours.

Now, we had the middle row, and it was best to work out the middle row last.

Because when you know the sum and the difference, there's only one possible solution.

It's dark green added to pink.

That's the pair which will give you the sum that's equal to orange and difference that's equal to red.

Pause the video here if you need extra time to mark those carefully.

Let's summarise the learning today then.

Unknowns can be connected additively where the unknown rods have a difference in size.

Unknowns can be connected multiplicatively where one unknown rod is something times the size of the other unknown rod.

Representations can show if unknowns are related additively or multiplicatively.

My name's Mr. Tazzyman.

Really enjoyed that lesson.

I hope you did too.

I hope you enjoyed using number rods.

Bye for now.