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 today's lesson then.

By the end, we want you to be able to say, "I can solve problems with two unknowns with one, several, "and infinite solutions." These are the key words that you're going to be hearing during the lesson, precise and infinite.

I'd like you to repeat them back to me.

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

Ready? My turn, precise.

Your turn.

My turn, infinite.

Your turn.

It's also really important that we understand these words as well, so let's look at what they mean.

If something is precise, it is exact, accurate, and careful about the details.

It is more precise to say that there are 67.

326529 million people in the UK than 67 million.

Infinite means without an end.

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

This is the outline of the lesson today.

We're gonna start by anticipating the number of solutions.

Then, we're gonna think about estimating solutions and solving.

Lucas and Sam are gonna join us in the lesson today to explain some of the maths that you see on screen, and to think about some of the prompts.

Hi, Lucas.

Hi, Sam.

Okay, is everyone ready? Let's get learning.

Lucas and Sam return to a problem they have solved before.

I spent 29 pounds 90 on fish and chips.

One fish cost three pounds 20, and one portion of chips cost one pound 50.

How many portions of each did I buy? Lucas says "This problem could be represented "as an equation like this." 3.

2f added to 1.

5c is equal to 29 pounds 90.

So the 3.

2f represents three pound 20 multiplied by the number of portions of fish, and the 1.

5c represents the one pound 50 multiplied by the number of portions of chips.

The 29 pounds 90 represents the total cost of the fish and chips combined.

"Let's look at the solution we found," says Lucas.

How many solutions are there? Hmm.

"We found one solution," says Lucas.

Are there any others? I don't think there are.

There is only one solution possible.

You can see that they've circled five portions of chips at a cost of seven pounds 50, and seven portions of fish at a cost of 22 pounds 40.

If you total those two costs, that gives you 29 pounds 90.

It has to be seven portions of fish and five portions of chips.

So why is there only one solution then? Well, the cost of the fish and chips is quite precise.

It has a decimal value, and the price of each portion is a set price.

It's unlikely that there are many different solutions with such a precise cost.

It can also only be whole numbers, as fish and chip shops don't tend to sell less than whole portions.

Okay, let's look at another problem.

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? We could also represent this problem as an equation, 4s + 6h = 72.

The 4s represents four sides multiplied by the number of squares.

6h represents six sides multiplied by the number of hexagons.

Using initials, makes sense.

72 is the total number of sides we have altogether.

"Let's look at the solutions we found for this," says Sam.

How many solutions are there? 12 and 60, 48 and 24, 60 and 12, 72 by itself, and 72 by itself.

There were many pairs of solution for this problem.

"Why did this problem have several solutions then?," asks Sam.

Great question, Sam.

So why are there many solutions to this problem? The total, 72, means that there can be more than one solution.

In fact, it is a number which has lots of factor pairs.

Once again, it can only have whole numbers for its solutions because the problem is looking for whole squares and hexagons, and each shape has a set number of sides, both of which are quite common factors of lots of numbers.

An unknown added to an unknown is equal to 20.

"One last problem," says June.

"How many solutions do you think this one has?" "I can think of loads!," says Sam.

We could use whole numbers.

15 added to five.

We could use decimal numbers.

15.

5 added to 4.

5.

You could also use a negative number, <v ->5 added to 25.

</v> I've got lots of solutions too.

I think this problem is an example of a problem that would have an infinite number of solutions.

"But why does this have an infinite number of solutions?," asks Sam.

Well, we know that there are lots of ways to make 20 using just whole numbers, but as you said, we can extend this to using all sorts of numbers, including decimals and negative numbers.

There is no context to this problem, which might limit the numbers we can use.

So how can you tell if a problem with two unknowns has a set number of solutions? There are no specific rules to tell, but there are some similarities you can look out for.

One solution: may have a context that limits the number of solutions to a set of numbers, may have a precise value as its sum or difference, the unknowns may be multiplied by more complex values.

Multiple solutions: may have a context that limits the number of solutions to a set of numbers, may have a less precise value as its sum or difference which could have lots of factors, the unknowns may be multiplied by factors that are common to the sum.

And infinite solutions: often not constrained by a context, can be solved using whole, decimal, or negative numbers.

What a useful table.

Okay, let's check your understanding so far.

How many solutions do you think this problem has? An unknown added to 30 is equal to an unknown added to 60.

Does it have A, one solution, B, multiple solutions, or C, infinite solutions? Pause the video and decide.

Welcome back.

Which did you think? A, B, or C? Well, it was C.

"It's not limited by context, "and you could use many types of number to solve it," says Sam.

Alright, time for your first practise task then.

Sort the following problems into how many solutions you think each one has.

So you can see there's a blank table there.

The first column is one solution, the second is multiple solutions, and the third is infinite solutions, and you've got to sort these questions into the correct columns.

For number one, we have an unknown is equal to an unknown added to 24.

For number two, we have 4.

3x added to 2.

1y is equal to 21.

3.

Don't forget, if you have a value next to a letter in that way, that's the same as multiplying.

For number three, Lucas is thinking of a number, he subtracts another number from it, and ends up with 11.

What could his starting number be? For number four, a baker is packing cakes into boxes to take to a shop.

He has 60 cakes.

A small box can hold eight cakes, and a large box can hold 12 cakes.

How can the cakes be packed? For number five, only one angle in an isosceles triangle is 32 degrees.

What are the sizes of the remaining two angles? And for number six, 3a + 4b is equal to 84.

Okay, categorise each of those questions into a column in the table.

Good luck.

Pause the video now to give it a go.

Welcome back.

Here's our blank table then, ready to put the questions into.

Number two had one solution.

Number six had multiple solutions.

Number one had an infinite number of solutions.

Number five had one solution.

Number four had multiple solutions.

And number three had infinite solutions.

Is that what you got? Pause the video to double check.

Okay, let's move on to the second part of the lesson, estimating solutions and solving.

Only one angle in an isosceles triangle is 32 degrees.

What are the sizes of the remaining two angles? "I thought this problem would have more than one solution," says Lucas.

Let's see why it only had one solution.

Firstly, we have an isosceles triangle, which has two equal angles at the bottom.

Because these two angles are equal, it means the remaining angle has to be the 32 degrees angle.

We now know that there can only be one solution because the interior angles must sum to 180 degrees, and the last two angles must be equal.

So we've got 32 degrees added to an unknown angle added to an unknown angle is equal to 180 degrees.

"Oh, I see.

"Okay, well, let's solve this then," says Lucas.

We can subtract the known angle from 180 degrees.

Now, we've got an unknown angle added to an unknown angle is equal to 180 degrees subtract 32 degrees.

If we complete that subtraction, we get 148 degrees.

And because the remaining angles have to be equal, we can divide the remaining degrees into two parts.

148 degrees divided by two is equal to 74 degrees, so the two remaining angles have to equal 74 degrees.

Andeep and Sofia both have some marbles.

If Andeep gave away 20 marbles and Sofia gave away 40 marbles, they would then have the same number of marbles.

How many marbles might Andeep and Sofia have? Sam says "It might help to write this problem "as an equation to help see "if we can identify how many solutions it might have." a subtract 20 is equal to s subtract 40.

We know that when Andeep gives away 20 marbles, a subtract 20, he has the same number as Sofia when she gives away 40 of her marbles, s subtract 40.

Initials have been used again.

The solutions must be positive whole numbers greater than or equal to the number being taken away.

You can't have less than a whole marble, so the solutions cannot be decimal numbers.

You also can't have negative number of marbles.

This shows how the context constrains the types of solutions we can have.

However, I think there is still an infinite number of solutions because you could just keep increasing the a and s value by the same amount.

Let's take a look.

There is a difference of 20 between the subtrahends, so s must always be 20 more than a.

So if a was 20, then s would need to be 40, which would leave them with zero marbles each.

20 subtract 20 is equal to 40 subtract 40.

Zero is equal to zero.

We could change a to 21 though, and change s to 41 too.

The difference is still 20.

(mouse clicking) This could go on forever.

Although, it might not be realistic for them to have a million marbles.

Okay, let's check your understanding.

Estimate how many solutions there are, and find some values to solve.

An unknown subtract 35 is equal to an unknown add 10.

Pause the video and give this a go.

Welcome back.

The difference between -35 and 10 is 45, so any pair of numbers with a difference of 45 would work.

I think there an infinite number of solutions.

50 and 5 is one example, but as Sam has already said, there are an infinite number of solutions.

It's time for your second practise task then.

You've got to estimate the number of solutions and then solve each problem.

For number one, fill in the missing digits to create two, two-digit numbers and solve the equation.

For number two, what are the possible values of whole numbers j and k? For number three, what are the possible values of a and b? And for number four, what are the possible values of a and b? Again, but a different type of equation.

Okay, pause the video here, and give those a go.

Good luck.

Welcome back.

Let's do some marking then.

Sam says, for number one, "I think it has multiple solutions." 39 is equal to 45 subtract six, or 38 is equal to 44 subtract six.

Lucas says for number two, "I think it has multiple solutions." 3j has to be less than 20, so j can have a value of 6, 5, 4, 3, 2, 1, or 0.

k has to be greater than 40 for 2k to be greater than 80.

It could be six and 44.

18 divided by three is equal to six, j is equal to six.

6 + k is equal to 50.

k is equal to 44.

Solutions can also be 5 and 45, 4 and 46, 3 and 47, 2 and 48, 1 and 49, 0 and 50.

Okay, pause the video here if you need some more time to mark those.

Here's number three then.

Sam says "I think it has one solution." If we put a and b into a bar model, they total 15.

We can also put it into a different type of bar model to show the difference, and the difference between them is five.

So a and b together equal 15, but we can replace a with b plus five.

You can see that second row, the bar model that's been done there.

That means 2b plus 5 is equal to 15.

If we take away the 5, we get 2b is equal to 10.

That means b is equal to 5.

If b is equal to 5, we can now calculate the value of a.

a is equal to 10 because 10 take away 5 is equal to 5.

That shows that it works.

Okay, for number four, Lucas says, "I think there are multiple solutions." If we rewrite that equation using boxes for the unknowns, we can start to fill in some examples.

Here's a table with all the different examples.

(mouse clicking) So we could have had 2 multiplied by 9 plus 3 multiplied by 10 is equal to 48.

2 multiplied by 12 plus 3 multiplied by 8 is equal to 48.

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

Okay, let's summarise the lesson then.

You can use your knowledge of a problem to help you anticipate how many solutions there might be.

Whilst there is no set rule, considering the context of the problem, the types of numbers that could be used, and the properties of those numbers can all help anticipate the number of solutions.

My name's Mr. Tazzyman.

I enjoyed that lesson.

Hope you did as well.

I'll see you again soon.

Bye for now.