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 and 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 you know you've found all the possible solutions to a problem with two unknowns." These are the keywords that you might expect to hear during the lesson.

Systematically and infinite.

I want you to repeat them back to me.

So I'll say "My turn," say the word, then I'll say "Your turn" and you can repeat it back.

Ready? My turn, systematically, your turn.

My turn, infinite, your turn.

Here's what those words mean, then.

To work systematically means to work in an organised way, often according to a fixed plan or system.

Infinite means without an end.

If something is unlimited, you could say it has an infinite number of things.

This is the outline for the lesson, then.

We're gonna start by looking at two unknowns and one solution, then we're gonna look at two unknowns and many solutions.

In this lesson, you're gonna meet Jun and Aisha.

Hi Jun, hi Aisha.

These two are gonna help us by discussing some of the maths prompts that you see on screen, so make sure you're listening to what they have to say carefully because they're gonna help you to learn really efficiently.

Okay, you ready to start? Let's go.

I spent 29.

90 pounds on fish and chips.

One fish costs 3.

20 pounds and one portion of chips costs 1.

50 pounds.

How many portions of each did I buy? "How could we record this as an equation?" says Jun.

"Let's use f to represent the portions of fish and c to represent the portions of chips." 3.

2f plus 1.

5c is equal to 29.

90 pounds.

"We can multiply the number of portions by their costs.

So the 3.

2f represents 3.

20 pounds multiplied by the number of portions of fish." "And the 1.

5c represents the 1.

50 pounds multiplied by the number of portions of chips." Okay, Jun and Aisha have made a good start here because they've interpreted the problem and used an equation to represent it.

"I'm not sure how we can show this as a bar model." That's a good point, Jun.

It's another good representation that will help us to understand what we might need to do.

"I've got a different strategy to help us solve this," says Aisha.

Intriguing.

Let's see what she means.

"Let's systematically record the cost of both the fish and chips as they increase." So you can see there's a table there.

We've got number of portions in the first column, fish in the second, and chips in the third.

At the moment in that top row we can see that there's one portion of each.

So the fish costs 3.

20 pounds and the chips costs 1.

50 pounds.

Now there's two.

That means that they've doubled the cost of one portion of fish and doubled the cost of one portion of chips.

Now three, now four.

And so on until we get to 10 portions of fish and 10 portions of chips.

"Now we need to find two numbers that sum to 29.

90 pounds." Now it's your turn.

We're gonna check your understanding.

The question here is, can you find the number of portions of fish and chips that were bought? We've got the equation that Jun and Aisha worked out and we've also got the table that Aisha wrote down.

Can you use that to help answer the question? Pause the video and give it a go.

Welcome back.

Seven portions of fish were 22.

40 pounds and five portions of chips were 7.

50 pounds.

If you add those together, that's equal to 29.

90 pounds, so that means that seven portions of fish were bought and five portions of chips.

We've got ourselves a solution.

So here's the solution.

And Jun says very sensibly, "Now we just need to double check using our equation." We had seven portions of fish and five portions of chips.

So f is equal to 7 and c is equal to 5.

3.

2 and in brackets you can see 7 plus 1.

5 and in brackets 5 is equal to 29.

90 pounds.

3.

2 brackets with a 7 in is the same as saying 3.

2 multiplied by 7, and 1.

5 brackets with a 5 in is the same as saying 1.

5 multiplied by 5.

"There we go," says Aisha.

The products of each of those are equal to the costs you can see in the table, and we know that if we add those costs together, then we get 29.

90 pounds.

Those answers are correct.

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

You've got to solve the following problems. Number one, I use two different lengths of wood to make a fence which is 10.

55 metres long.

One length of wood is 60 centimetres and the other is 1 metre 15 centimetres.

How many planks of each length were used? Take care here, check those measurements, and you might need to do some converting.

Let's look at number two.

In one month, Jun spends 17.

5 hours in maths and science lessons.

Each maths lesson is 45 minutes and each science lesson is 75 minutes.

How many maths and science lessons did Jun have? We've got two unknowns again there that we might need to try to calculate.

Okay, pause the video here and have a go at those problems. Remember to try to use representations where possible and work through them carefully and systematically, good luck.

Welcome back, let's look at number one to begin with.

You can see here that there's a table that's been written out with number of planks going from 1 to 10 and then Plank A and Plank B.

So we needed to find two measurements that in total were equal to 10.

55 metres.

60a plus 115b is equal to 1,055 centimetres.

1,055 centimetres converted to metres is 10.

55.

So that equation helps us to clarify the problem.

There are the two measurements that total 1,055 centimetres, 480 plus 575.

You can see that written down as an equation here.

60 multiplied by 8 added to 115 multiplied by 5 is equal to 1,055 centimetres.

Remember that if you use brackets in this way, that's the same as multiplying those two numbers together.

So we end up with 480 centimetres plus 575 centimetres is equal to 1,055 centimetres.

So our final solution then is that there were 5 lengths of Plank B used and 8 lengths of Plank A.

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

Here's number two, then.

We've started by converting hours to minutes so that all of the knowns and unknowns in the question are measured in the same way.

We can see that 17.

5 hours is equal to 1,050 minutes.

Now to draw out the table.

We've got the number of lessons, 1 through 10, then we've got maths and science.

The equation might read as follows.

45m where m stands for maths lessons plus 75s where s stands for science lessons is equal to 1,050 minutes.

Here are two lengths of time that total 1,050.

We can see that we've got 8 lessons of science and 10 lessons of maths there, so now let's put that information back into the equation to give us 45 multiplied by 10 plus 75 multiplied by 8 is equal to 1,050 minutes.

If we find those products and add them together, we have 450 plus 600 is equal to 1,050 minutes.

So the solution is 8 science lessons and 10 maths lessons.

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

Now we're gonna be having two unknowns but with many solutions rather than just one.

Ready? Let's go.

Laura has some squares and hexagons.

She chooses some of the shapes and when she counts the number of sides, she finds that the total is 72.

How many of each shape could she have? "Shall we record this in a table again?" says Jun.

"Let's write an equation for it first," says Aisha.

"Let s represent the number of squares and h represent the number of hexagons." That makes sense, using the initials again.

"Okay, so 4s represents four sides per square and 6h represents six sides per hexagon." So their equation reads out 4s added to 6h is equal to 72.

72 is the total number of sides we have altogether.

Now they've got their table.

We've got number of shapes, square sides and hexagonal sides.

There's the table complete, then.

We've got 1 through 10 in terms of number of shapes and the square sides goes up to 40, hexagonal sides goes up to 60.

You can see here some links with times tables.

It's your turn, then.

Let's check your understanding.

You've got to find the number of squares and hexagons that were used.

The equation's been written down and the table's there helpfully for you.

Pause the video here and have a go at solving that.

Welcome back.

36 added to 36 is equal to 72.

That would mean that there were 6 hexagons and 9 squares, so we can put that information back into our equation to check it.

We've got 4 multiplied by 9, remember the bracket notation, added to 6 multiplied by 6 is equal to 72.

4 9 to 36, 6 6 is 36.

And if you double 36, it's equal to 72.

It works, we had 6 hexagons and 9 squares.

"Hang on, though.

There are other solutions too," says Aisha.

Ah, what has she spotted? "We could also have had 6 squares and 8 hexagons." If we put those into the equation, you can see we have 4 multiplied by 6 added to 6 multiplied by 8 is equal to 72.

That's 24 added to 48 is equal to 72, it works.

"I wonder if there are more examples?" says Aisha.

Okay, let's check your understanding again.

Aisha continued the investigation.

You can see that she's created a bigger table with more information in it.

And from that, can you find any other solutions? Pause the video and give it a go.

Welcome back, did you find this? 15 squares and 2 hexagons.

Or you could have had 12 squares and 4 hexagons.

Or you could have had 3 squares and 10 hexagons.

Or you could have had 18 squares and no hexagons.

Or you could have had 12 hexagons and no squares.

"Sometimes, problems with two unknowns can have multiple solutions." You can see that with the hexagon column there are actually some that have no information in.

That's because you know that the total number of sides was 72.

Once you get to 12 hexagons, that's already 72 sides, so we can't exceed 72.

"There can't be any more solutions as the sides must be whole numbers and the maximum number of sides is 72." Okay, let's look at this, then.

How many solutions are there for this problem? The missing values are integers.

"We have two unknowns again," says Jun.

"And both expressions need to be equal to one another." "If I put 1 in this box, what would go in the other box?" "Well, the expression on the left is equal to 51, so that means we need 31 to go in the second missing box." There's one solution, but I think we could find more.

They've got a table again.

The top row has 0 being the value of the first missing number, and the value of one side of the equation being 50, the value of the second missing number has to be 30.

"Let's try some multiples of 10," says Jun.

So he puts 10 in as the value of the missing number.

That means the value of one side of the equation is 60, 10 added to 50 is equal to 60, and that means the value of the second missing number has to be 40.

40 added to 20 is equal to 60.

Then we've got 20 as our value of our first missing number.

That gives 70 and then 50.

30 gives 80 and then 60.

40 gives 90 and then 70.

"There are so many solutions," says Jun.

"I've noticed that the second value is always 30 more than the first value." Good spot, Aisha, well done.

"It's because the value of the left expression is 30 more than the value on the right expression, so the value on the right expression needs to be 30 more than the value on the left so it balances out." Time for your second practise task, then.

We've got number one here, A through F.

You've got to find the following solutions.

We've got a missing number added to 50 is equal to a missing number added to 20.

A, give a solution where both missing values are three-digit integers.

B, give a solution where the first missing value is a single-digit integer.

C, give a solution where the second missing number is greater than 1,000.

D, could the first missing number be a decimal number? What would the other missing number be in that case? E, can you find a solution that no one else will think of? And F, how many solutions do you think there could be? Okay, pause the video here and have a go at those, good luck.

Welcome back, let's look at some of the solutions here and give you some feedback.

So for A, give a solution where both missing values are three-digit integers.

You could have had, for example, 500 and 530.

500 plus 50 is equal to 530 plus 20.

Now there are lots of other solutions as well, so you might want to pause the video here to compare with somebody else.

Let's look at B, then.

Give a solution where the first missing value is a single-digit number.

You could have had 4 and 34.

4 added to 50 is equal to 34 added to 20.

Again, lots of different solutions here, so pause the video if you need to check through.

Let's look at C, then.

You could have had 2,370 and 2,400, or 4,620 and 4,650.

Lots of solutions here.

If the second number was a four-digit number, then the first usually was, but not always.

For example, 970 and 1,000.

970 added to 50 is equal to 1,000 added to 20.

Provided that the second number was 30 more than the first number, then you know that the answer was correct.

Pause the video here if you need to do some more marking.

Okay, let's look at D.

Could the first missing number be a decimal number? What would the other missing number be in that case? Well, you could have had something like this.

3.

9 and 33.

9.

0.

25 and 30.

25.

Did you notice the second missing number also had to be a decimal with the same number of tenths and hundredths? And of course the difference between the two numbers needed to be 30 still.

Let's look at E, then.

Can you find a solution that no one else will think of? I wonder if you did.

Maybe you made use of fractions here.

7/8 and 30 7/8.

Mm, pause the video to see if you managed to get one that no one else could think of.

Okay, let's look at F, then.

How many solutions do you think there could be? Well, Aisha says, "I think there are an infinite number of solutions.

We could use whole numbers, decimals, fractions, and even negative numbers." Well put, Aisha.

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

Here's a summary.

Problems with two unknowns can have one solution, many solutions, or an infinite number of solutions.

This is often dependent on the context of the problem, where it may only accept positive whole numbers as solutions.

Other contexts may have an infinite number of solutions if the solutions include any positive whole numbers, decimals, or negative numbers.

Where problems have two unknowns and many solutions, you can work systematically to find the total number of solutions.

My name is Mr. Tazzyman.

I really enjoyed that learning today and I hope you did as well.

Maybe I'll see you again soon.

Bye for now.