Lesson video

Hi there, I'm Mr. Tazzyman.

Today I'm gonna be teaching you a lesson from a unit that is all about solving problems that feature two unknowns.

You've probably come across these kinds of problems lots of times in the past, but you may not have known that that's what we would categorise them as.

We're gonna be looking at the structures that underlie some of these problems to really help you to understand what's going on mathematically.

So, make sure you're ready to learn.

Let's go for it.

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

By the end, we want you to be able to say, I can explain how to represent an equation with a bar model.

Here's the key word for today then.

It's represent.

Can you say that back to me? And here's what we mean when we say represent.

To represent something means to show something in a different way.

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

We're gonna start by looking at representing structures, then we're gonna move on to solving problems. We're gonna meet Laura and Andeep along the way.

Hi, Laura.

Hi, Andeep.

They're gonna be discussing some of the maths prompts that you see, and they'll be giving us some hints and tips in our learning, so make sure that you're listening carefully.

We're ready to start.

An apple costs 15 pence more than a lemon.

Two apples and three lemons cost 1.

30 pounds.

What is the cost of one apple? What is the cost of one lemon? Laura says, "I'm not sure how to start tackling this problem." I know what you mean, Laura.

It looks like a tough problem, doesn't it, when you consider it all at once? Andeep says, "Why don't we start by writing these as equations?" That's a good place to start.

"Then I was thinking we could try and represent it as a bar model." "We know that the cost of two apples and three lemons is equivalent to one pound 30," and Laura said that because you can see it in the question.

It's the second line in that question.

2a plus 3b is equal to one pound 30.

The 2a represents two lots of apples, and the 3b represents three lots of lemons.

We can start by saying that we know one apple costs more than one lemon, and we have three lemons and two apples.

We can represent that here at the end.

Okay, I can see that.

Altogether it costs one pound 30.

Okay, we have the sum, but this bar model doesn't show the difference between the lemons and apples.

"We can show that quite easily I think," says Andeep.

Firstly, we know that the difference in cost between one apple and one lemon is 15 pence, a subtract b is equal to 15 pence, that's the difference.

That means that we also know b plus 15 is equal to a using the inverse.

Again, the a represents one apple and the b represents one lemon.

We know that an apple is 15 pence more than a lemon from our equation.

So we could represent our apple bar as a lemon bar and a 15 pence bar.

Let's replace each apple bar in the original bar model now with a lemon and 15p bar.

There they go.

This is starting to look as if it's going to make a bit more sense.

We can now rearrange the bars to show clearly that one pound 30 is equal to five lemons and an additional 30 pence.

They've swapped around one of the lemon bars and one of the 15 parts, so now you can see five parts that are lemon and then two parts that are 15 each.

Two 15s are 30.

"I would've represented the bar models differently I think," says Laura.

"I would've represented the problem in one bar.

Three lemons and two apples is one pound 30.

If we rearrange the bars, I think we can more clearly see a way through the problem.

We can see that one pound 30 is equal to five lemons and an additional 30 pence." "I think you're right.

I prefer the way you have represented it using the bar model." That's good of you to say, Andeep.

Okay, let's check your understanding so far then.

Represent the following problem using a bar model.

A slice of cake costs one pound 20 more than a muffin.

Two slices of cake and three muffins cost eight pounds 90.

What is the cost of one slice of cake? What is the cost of one muffin? Now remember, we're not expecting you to solve this.

We're asking you to represent it using a bar model.

Pause the video and give it a go.

Welcome back.

Here's what you might have done.

We've got two parts, which are cake, and three parts, which are muffin.

Altogether, they cost eight pounds 90.

We could then replace each of the cake parts with a muffin and one pound 20 because we know that the cake costs one pound 20 more than a muffin.

If we rearrange those parts, we end up with two lots of one pound 20 and five muffins.

Okay, next, you're gonna represent the following equations using a bar model.

We've got 4c plus 3d is equal to 13, and d subtract c is equal to two, so represent these using a bar model.

Pause the video and give it a go.

Welcome back.

Here's what your bar model may have looked like.

We had four parts which were c, and three parts which were d.

4c plus 3d, and in total that equaled 13.

Then we might have decided to replace each part that was d with c and two, because d subtract c is equal to two, so d is equal to c plus two.

If we rearrange those parts, we end up with 7c and three parts, which are labelled two.

Three parts which are labelled two can be replaced with six, so we've got seven parts that are c and one part of six.

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

You've got to draw a bar model to represent each of the following problems. Number one, Andeep completed three longer runs and two shorter runs in one week.

The longer run was 1.

5 kilometres longer than his shorter run.

Altogether, he ran 14.

5 kilometres in one week.

What was the distance of his longer and shorter run? Number two, Laura put three eggs and three portions of sugar into a cake mixture.

A portion of sugar weighed 32 grammes more than one egg.

The total mass of the mixture was 354 grammes.

What was the mass of the egg? And what was the mass of one portion of sugar? Number three, draw a bar model to represent the following pairs of equations: 2a plus 4z is equal to 24.

z subtract a is equal to three.

342 is equal to 4x plus 2y.

x subtracts y is equal to 54.

Just be reminded, we're not expecting you to solve any of these yet.

We just want you to draw a bar model to represent them.

Pause the video and enjoy.

Welcome back.

Here's number one then.

Your bar model might have looked something like this initially.

We've got three parts labelled L and two parts labelled S.

The L represents longer runs and the S represents shorter runs.

Altogether, we've got a total of 14.

5 kilometres.

We might have then replaced each of the Ls with S and a part labelled 1.

5, because the shorter run was 1.

5 kilometres less than the longer run.

If you rearrange that, you ended up with three parts of 1.

5 and five parts of S, 5S.

Okay, here's number two.

You might have started with this, three parts labelled E and three parts labelled S.

The E represented eggs and the S represented sugar.

You could then replace the sugar parts with E and a part labelled 32.

That ended up giving you 6e plus three lots of 32.

And here's number three.

So, we had a total of 24, and the parts were two parts labelled a, 2a, and four parts labelled z, 4z.

We knew that z subtract a was equal to three, so using our understanding of inverse, we also knew that z was equal to a plus three, so we could replace all the z parts with a and three.

If you collect together all of those parts to like terms, then you end up with 6a plus four lots of three.

Let's look at this next one as well.

Similar sorts of thing, but this time we had a total of 342 and 4x and 2y.

If you replace each of the x with y and 54, then you get a second row that looks something like this.

Now, we know that x is equal to y plus 54 because of x subtract y equaling 54.

If you collect all the terms, you end up with four lots of 54 and six lots of y on one row.

Let's move on to the second part of the lesson then, solving problems. An apple costs 15 pence more than a lemon.

Two apples and three lemons cost one pound 30.

What is the cost of one apple? And what is the cost of one lemon? "Okay, let's see if we can solve this problem now," says Laura.

Bit more confidence there, Laura, that's good.

We know that three lemons and two apples is one pound 30, and the cost of an apple is equivalent to the cost of a lemon and an extra 15 pence.

And as we said, we can arrange the bar model to make it clearer.

This is all very similar to the practise task you've just completed.

So we can say that the cost of five lemons plus 30p is equivalent to one pound 30.

If we subtract the 30 pence from the total cost, we will see that the cost of five lemons would be equivalent to one pound.

If five lemons is equivalent to one pound, then we can divide one pound into five to find the cost of one lemon.

L is equal to one pound divided by five.

One lemon has a cost of 20 pence.

L is equal to 20 pence.

We've relabeled the bar model with 20.

Now we know the cost of one lemon is 20 pence, we can find the cost of an apple.

How do we do that? Hmm.

Well, we've relabeled all of the lemons to be 20 pence, and the cost of one apple is equivalent to the cost of one lemon and an additional 15 pence.

a is equal to L plus 15 pence.

We know that L is 20 pence, so 20 plus 15 is equal to 35 pence.

So one apple is equal to 35 pence.

Now we just need to check our workings.

It's a really important step.

Each lemon is 20 pence and each apple is 35 pence.

One pound 30 is equal to 20 pence, plus 20 pence, plus 20 pence, plus 35 pence, plus 35 pence.

There we go.

Three lots of 20p and two lots of 35p are equal to one pound 30.

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

Solve the following problem.

A slice of cake costs one pound 20 more than a muffin, two slices of cake and three muffins cost eight pound 90.

What is the cost of one slice of cake? What is the cost of one muffin? And you've already got there a bar model because this is the same problem that you faced earlier.

Pause the video here and give that a go.

Good luck.

Welcome back.

Your first step might have been to combine the two one pound 20s that were next to one another.

That would give you two pound 40.

5m plus two pounds 40 is equal to eight pounds 90.

So 5m is equal to eight pounds 90 subtract two pounds 40.

That's six pounds 50 then.

Six pounds 50 divided by five is equal to one pound 30, this gives us the value of m.

m is equal to one pound 30.

c is equal to m plus one pound 20.

c is equal to one pound 30 plus one pound 20 because we've replaced m with its value.

c is equal to two pounds 50.

A muffin is one pound 30.

A slice of cake is two pounds 50.

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

This time, you have got to solve the following problems, and these problems were the same as the ones that you modelled in the first practise task.

You can see for three and four as well, you've got to find the value of a and z and find the value of x and y.

Again, same equations that you practised in that first task.

Okay, pause the video here.

Enjoy.

I'll be back in a little while for some feedback.

Good luck.

Welcome back.

Here's number one then.

This is the bar model that you developed in the first part.

If we replace those three lots of 1.

5 with 4.

5, that can give us the equation 5s plus 4.

5 is equal to 14.

5, where s is a shorter run.

That means that 5s is equal to 14.

5 subtract 4.

5, which means 5s is equal to 10.

10 divided by five is equal to two, so s is equal to two kilometres.

We know that L, the longer run, is equal to S plus 1.

5.

If we replace S with its value of two, we get L is equal to two plus 1.

5, so L is equal to 3.

5 kilometres.

Pause the video here if you need some extra time to finish marking that.

Welcome back.

Here's number two then.

For this, we had this bar model already.

If you replace those three lots of 32 with 96, because that's 32 multiplied by three, then we get an equation.

The equation is 6e plus 96 is equal to 354.

So 6e is equal to 354 subtract 96.

6e is equal to 258.

Use your division skills now.

258 divided by six is equal to 43.

That means e is equal to 43 grammes.

S, which is the sugar, is equal to e plus 32.

If we replace the value of e, then we get S equals 43 plus 32, so S is equal to 75 grammes.

Again, pause the video here if you need some more time to finish marking that.

Here's number three then.

We had to find the value of a and z.

We already had this bar model from the first practise task.

If we replace those four threes with 12, then we end up with the equation 6a plus 12 is equal to 24.

That means that 6a is equal to 24 subtract to 12, so 6a is equal to 12.

12 divided by six is equal to two.

a equals two then.

z is equal to a plus three.

That means that z is equal to two plus three, so the value of z is five.

Pause video here if you need some extra time to mark.

Finally, number four.

This is the bar model that we developed in the first practise task.

We now need to multiply 54 by four, that's 216.

This gives us the equation 6y plus 216 is equal to 342.

That means that 6y is equal to 342 subtract 216.

6y is equal to 126.

126 divided by six is equal to 21, so y is equal to 21.

x equals y plus 54, so x equals 21 plus 54.

x is equal to 75.

Pause the video for some extra time to complete your marking.

We've reached the end of the lesson then.

Here's a summary of some of the things that we've learned.

You can represent sum and difference problems using a pair of equations.

These equations can be represented using a bar model.

Depending on the problem, some bar models may be more appropriate to represent the problem more clearly.

My name is Mr. Tazzyman.

I hope you enjoyed that, I certainly did.

I'll see you again soon.

Bye for now.