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.

I can solve problems with two unknowns in a range of contexts.

That's what you need to be able to say by the end of the lesson.

The key word today is represent, repeat that back to me.

Represent.

Okay, but what does it mean? To represent something, means to show something in a different way.

Here's the outline then for today's lesson, we're gonna start with problems with two unknowns.

Then we're gonna move on to problems with more than two unknowns.

Lucas and Jacob are gonna be joining us today.

They'll discuss a lot of the mass that you encounter during this lesson.

Hi, Lucas.

Hi, Jacob.

Are you ready? Are you ready? Okay, let's get learning.

One eraser and five pencils cost three pounds 35 pence, five erasers, and five pencils cost four pounds 75 pence.

How much does an eraser cost? How much does a pencil cost? "Shall we record each example as an equation first?" "Let's use a to represent erasers and b to represent pencils." So one eraser and five pencils is equal to three pounds 35, a plus 5b is equal to three pounds 35 and five erasers, and five pencils is equal to four pounds 75.

5a plus 5b is equal to four pounds 75.

"Now, let's represent the problem as a bar model." Here is the bar for one eraser and five pencils.

So you can see there are six parts in this bar.

One of the parts is labelled a, and the other five are labelled b.

The total of three pound 35, is written alongside, and here is the bar for five erasers and five pencils.

It's much longer.

There are more parts.

In fact, in total you can see that there are 10 parts.

That's because we have five parts which are labelled a and five parts labelled b.

The total of four pounds 75 is written alongside.

What do you notice? Hmm.

Have a look at that bar model and look at that pair of equations, what can you see? Lucas says, "I can see that both bars have the same number of pencils." "And I can see that the second bar has four more erasers than the first bar." That means the difference in cost is equivalent to four erasers.

We can calculate the difference by subtracting the total costs from one another.

Four pounds 75 subtract three pounds 35 is equal to one pound 40.

So the difference is one pound 40.

We can divide one pound 40 into four parts to find the value of each eraser.

One pound 40 divided by four is equal to 35 pence.

And they've relabeled all of the parts that were labelled a as 35 pence because that's their value that we've just calculated.

Each eraser has a value of 35 p.

Now we can find the cost of five pencils by subtracting the cost of one eraser from the first bar.

Three pounds 35 subtract 35 pence is equal to three pounds.

So we know that those five parts labelled b which are all the same size and represent the pencils, are worth three pounds.

And finally, if five pencils cost three pounds, then we can divide that cost by five to find the cost of one pencil.

Time to use our division skills.

Three pounds divided by five is equal to 60 pence.

Each pencil has a value of 60 pence.

Time to check.

Lucas says, "We have better check to see if we've got it right." 35 pence plus five lots of 60 pence is equal to three pounds 35.

Five multiplied by 35 plus five multiplied by 60 is equal to four pounds 75 pence.

"We did it." Certainly did.

Well done Lucas.

Well done Jacob.

Some excellent reasoning there.

These kinds of problems take time, but if you work through them carefully, then you can see a path.

An adult ticket for the zoo costs two pounds more than a child ticket.

Luke's family spend 33 pounds buying three adult and two child tickets.

How much does each type of ticket cost? "I think we can write two equations to represent this new problem." "I agree.

Let's use a to represent the cost of an adult ticket and b, to represent the cost of a child's ticket." "The difference in cost between an adult and child's ticket is two pounds." And there's the equation they've written for that, a subtract b is equal to two pounds.

"And the cost of three adult tickets and two child tickets is equal to 33 pounds.

3a plus 2b is equal to 33 pounds.

"Because we have two separate equations with the same unknowns, we can record this as a bar model." "Three adult tickets and two child tickets are equal to 33 pounds." There's the bar model for it.

You can see they've got three parts labelled a and two parts labelled b, and the total is 33 pounds.

They've written the equation above it just to make sure that everything is clear.

"To represent the second equation, we can show that the cost of each adult ticket is equal to the cost of a child's ticket and an extra two pounds." There's the second row of their bar model, and they've done exactly what Jacob has just said.

They've replaced each part, which was labelled a with b and a two.

Lucas says, "We can rearrange our bar model to make it easier to see." So now they've got five parts labelled b and three parts labelled two.

All of them still have the same total together though that equals 33 pounds.

They've replaced the three lots of two with six, and Jacob says, "Now we can see that five lots of child tickets and additional six pounds is equal to 33 pounds." Okay, it's time for you to check your understanding then.

Here's the bar model that we've just developed.

Can you find the value of b? Pause the video and give it a go.

Welcome back.

Well, if we subtract six, we end up with 27 pounds.

So now we know that five lots of b is equal to 27 pounds, and we can write that as an equation.

27 divided by five is equal to five pounds 40, b is equal to five pounds 40.

Okay, your turn again, this time you've got to find the value of a.

We've replaced all the b parts with five pounds 40.

Pause the video and give it a go.

Welcome back.

"There are two ways of doing this." You could either do 33, subtract two multiplied by five pound 40, which is equal to 27 pounds, two lots of five pounds 40 is equal to 10 pounds 80, so the equation can also read 33 pounds, subtract 10 pounds 80, is equal to 22 pounds 20.

You now know that three parts labelled a are equal to 22 pounds 20.

If you divide that by three, you get seven pounds 40, so a is equal to seven pounds 40.

That's the first way of doing it.

You could also have done it in this way.

You know that b plus two is equal to seven pounds 40.

Five pound 40 plus two is equal to seven pound 40.

Here we've replaced b with its value of five pounds 40, so a is equal to seven pounds 40.

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

Here are two different designs made with the same rectangular and triangular tiles.

The area of the whole design is given, so on the first one we have 51 centimetre squared, and on the second we have 66 centimetres squared.

What is the area of one triangular tile? What is the area of one rectangular tile? Here's number two.

The masses of some groups of large and small bags of potatoes are shown.

They're even labelled.

We've got four small bags and two big bags for 11.

2 kilogrammes, and then we've got two big bags and six small bags, which is 14 kilogrammes.

What is the mass of each of the two different sized bags? Okay.

Make sure that you're using your equations and bar modelling in these steps because they're gonna really help you to find these values.

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.

Lucas says, "Let's call the area of the rectangle r and the area of the triangle t." 3r plus t is equal to 51 centimetres squared.

3r plus 2t is equal to 66 centimetres squared.

So we've got an equation to represent each different pattern.

Now, let's draw the bar models.

We've got here four parts, three of them are labelled r, and one of them is labelled t.

This represents that first pattern.

The total of 51 centimetres squared has been written alongside.

On the next bar we've got three lots of r still, but this time two lots of t and the total is 66 centimetres squared.

You can see that the difference is one part labelled t, so it's one triangular tile.

t is equal to 66 centimetres squared, subtract 51 centimetres squared.

t is equal to 15 centimetres squared.

Now we can replace all of the t parts with 15 centimetre squared, the value.

If we do 3r it's equal to 51 centimetre squared, subtract t.

Then we know that 3r is equal to 51 centimetre squared, subtract 15 centimetre squared, which is the value, so 3r is equal to 36 centimetre squared, so r is equal to 12 centimetres squared.

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

Here's number two then.

"Let's call the larger bag x and small bag y" 2x plus 4y is equal to 11.

2 kilogrammes.

2x plus 6y is equal to 14 kilogrammes.

We've now got an equation for each group.

Here's the bar model then.

Two parts are labelled x and four parts are labelled y, and the total of 11.

2 kilogrammes is written alongside.

For the second bar, we've got two parts labelled x and six parts labelled y, and the total this time is 14 kilogrammes.

We can see the difference between the two bars is two parts labelled y, 2y.

Let's work out what the value of the difference is though.

14 kilogrammes subtract 11.

2 kilogrammes.

That means that 2.

8 kilogrammes is equal to 2y because the result of the subtraction 14 subtract 11.

2 is 2.

8.

That means that 1.

4 kilogrammes is equal to y.

We've halved 2.

8 to give us the value of y.

We can then write those on the bar model, and now we know that 11.

2 kilogrammes subtract four lots of 1.

4 kilogrammes is equal to 2x.

11.

2 kilogrammes subtract 5.

6 kilogrammes, which is the result of multiplying 1.

4 by four is equal to 2x.

That means that 11.

2 kilogrammes subtract 5.

6 kilogrammes is equal to 5.

6 kilogrammes.

5.

6 kilogrammes equals 2x, so x equals 2.

8 kilogrammes.

You've solved it.

Okay, let's move on to the second part of the lesson then.

Problems with more than two unknowns.

The diagram shows the total cost of the items in each row under each column, fill in the two missing costs.

So if we look row by row, we can see we've got 11 and two pears equaling one pound 15.

We've got two oranges and a lemon equaling one pound 25, and we've got two apples and a lemon equaling 95 pence.

If you look at them as columns now, you can see in the third column we've got a pear and two lemons, we're going down now, which is equal to 95 pence.

Then we've got missing values.

Under the first and the second column, we don't know the price yet.

That's what we've got to try to work out.

"So one lemon and two pears is one pound 15," says Lucas, "One pear and two lemons is 95 pence." "Let's write an equation for each row and column." l plus 2g is equal to one pound 15 two o plus l is equal to one pound 25, 2a plus l is equal to 0.

95, g plus two l is equal to 0.

95.

2a plus l is equal to 95 pence and g plus 2l is also equal to 95 pence.

g plus o plus a is equal to an unknown.

l plus o plus a is equal to another unknown.

"I'm using g for pears in case I represent money as pence later." That makes sense now.

I did wonder what on earth we were using g for.

Now we can compare two equations.

One pound 15, subtract 95 pence is equal to 20 pence.

"There is a difference of 20 p." So we can say that a pear is 20 P more than a lemon.

"Let's now represent it as a bar model." I think this might make it a bit clearer.

There we go.

We've got l plus 2g being equal to one pound 15.

"We now know the cost of a pear is equal to the cost of a lemon plus 20 pence." There's the second row, the bar model then.

We've got l, l, 20 p, l and 20 p.

So we can say that the cost of three lemons and two lots of 20 p is equivalent to one pound 15.

That's the final row you can see there.

They've collected together all the different terms. The two 20 ps can be replaced with 40 p because that's their total.

Now we've got three ls and 40 p equaling one pound 15.

"We can now subtract the 40 p from one pound 15, which will leave us with the cost of three lemons." One pound 15 subtract 40 p is equal to 75 pence.

"Now we can divide the 75 pence into three parts to find the cost of each lemon." 75 p divided by three is equal to 25 pence.

Each lemon costs 25 pence.

They return back to one of the rows.

"If a lemon costs 25 pence, we can now find the cost of a pear." 25 p plus 25 p plus g is equal to 95 pence.

"The cost of a pear is 45 pence." Okay, it's time to check your understanding then.

Your task is to find the cost of an orange, and Lucas reminds us, "Each lemon is 25 pence." Pause the video and give it a go.

Welcome back.

20 plus l is equal to one pound 25.

That's taken from the row that that equation is next to.

If we know that a lemon is 25 pence, we can replace l with its value.

20 plus 25 P is equal to one pound 25.

That means that 20 is equal to one pound 25, subtract 25 pence.

So 20 is equal to one pound, o is equal to 50 P.

So an orange costs 50 pence.

Okay, time for another understanding check now.

This time you're gonna find the cost of an apple.

Pause the video and give it a go.

Welcome back.

2a plus l is equal to 95 pence that's been taken from the bottom row.

We know that l is equal to 25 pence, so we can replace l with that value.

2a plus 25 pence is equal to 95 pence.

That means 2a is equal to 95 pence, subtract 25 pence, so 2a is equal to 70 pence.

That means a is equal to 35 pence and apple costs 35 pence.

Okay, you can now solve the problem.

l plus o plus a.

Well we know, by replacing those letters with their values that that totals one pounds 10 pence.

Now for the last column, the middle one, g plus o plus a, well, we can replace those letters with their values and we get one pounds 30 pence.

Okay, it's your second practise task now.

You've got to solve the following problem, a plus b plus c is equal to 800.

2a plus b plus c is equal to 950.

3a plus b is equal to 1,050.

What are the values of a, b, and c? For number two, you've gotta solve this problem.

Half of the sum of r and s is 35.

Half of the sum of t and r is 47.

r plus s plus t is equal to 120.

Calculate the values of r, s and t.

Remember equations, bar model and look for the difference.

Okay, pause the video here and good luck.

Welcome back.

Let's look at number one.

Did you represent each equation as a bar model? There they all are as bar models and they've been deliberately positioned each part because you can see there's a commonality between all three and that's b.

So if we start with the difference between the first two bars, that can help us out.

There's a difference of a, 950 subtract 800 is equal to a, so a must be 150.

We can insert those values in.

Now we know that three a plus b is equal to 1,050.

That's that bottom row.

Three multiplied by 150 plus b is equal to 1,050.

Three lots of 150 is equal to 450, so our equation could also read 450 plus b is equal to 1,050.

If we subtract 450 from that, then we would get the value of b.

b is equal to 600.

Now we can replace all the bs with 600.

Finally, 150 plus 600 plus c equals 800.

That's the top row.

That means that c is equal to 800, subtract 600, subtract 150.

c is equal to 50.

Pause the video here if you need some time to catch up with the marking.

Welcome back.

Here's number two then.

The bar models have been drawn out.

You can see at the top we've got the sum of r and s.

In the middle, we've got t, r and s, and we've been given a total for that, which is 120, and at the bottom we've got the sum of t and r.

Look at how the parts have all been matched up as well.

Now, if half of the sum of r and s is 35, that means that the full sum of r and s will be 70.

If half of the sum of t and r is 47, then the full sum will be 94.

Now we can start to use the difference here.

t is equal to 120 minus 70 because you can see that t is also equal to the difference between the first two bars.

t is equal to 50.

In 50 goes as a value.

t plus r is equal to 94.

We know that t is equal to 50, so now we know 50 plus r is equal to 94, 94 subtract 50 is equal to, 44.

So now we can replace them all with 44.

r plus s is equal to 70.

That's the same as saying 44 plus s is equal to 70.

That means s is equal to 70, subtract 44, so s is equal to 26.

Pause the video here if you need some time to catch up with that marking.

Okay, we've reached the end of the lesson.

I hope you enjoyed it.

I did.

Here's a summary of today's learning then.

Problems with two unknowns can be represented as a pair of equations.

Problems with more than two unknowns can be represented as equations and then paired to help calculate.

It can be helpful to represent these equations as bar models, which in turn can help you identify the difference.

This is helpful to find a missing value.

My name's Mr. Tazzyman.

I enjoyed that today.

I hope you did too.

Maybe I'll see you again soon.

Bye for now.