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Right, well done for loading this video today.
My name is Ms. Davies and I'm gonna help you as you work your way through this lesson.
Please feel free to pause bits and rewind bits to help you as you are learning this new topic.
Anything that I can do to help, I'll try to add in as we work our way through.
I really do hope there's bits of these algebra that you're really, really gonna enjoy.
Lots of chances for you to explore things and to develop your algebraic reasoning and thinking skills.
Right, let's get started then.
All right, welcome to today's lesson where we're gonna look at representing an unknown or a variable.
By the end of today's lesson, you'll be able to recognise that a letter can be used to represent a specific unknown or a variable, and hopefully be able to tell the difference between the two.
So these keywords are words that you've probably come across before, but if you're not super confident with them, you might wanna just pause the video and read through them because I'm gonna be referring to these lots today.
So there's gonna be two parts to today's lesson.
We're gonna look at identifying unknowns and variables, and then we're gonna look at solving problems with unknowns and variables.
So to start with, we're gonna look at identifying when something is an unknown and when something is a variable.
So unknowns can be represented in variety of ways.
Some of these you would've seen before.
Generally if we're given enough information, we can work them out.
They have a set value that exists.
One way that we can represent them is in a bar model.
In fact, just looking at this bar model, you might be able to tell me that the unknown value is nine.
We can also represent unknowns in an equation.
So we've got the equation here.
R plus 2 equals 7, but r is representing an unknown.
If we put this into a bar model, it's really clear to see that that unknown value is actually representing 5 in this case.
The unknown value is 5.
We can also have unknowns in scenarios.
So here's Aisha.
"I'm thinking of a number.
I multiply it by 10 then add 5." So Aisha's number in this case is an unknown.
We don't have enough information to calculate this unknown at the moment 'cause she's not told us what her value equals when she multiplies it by 10 and then adds 5.
But we can write that as an expression.
So let's think about how we would write that as an expression.
Okay, so you might have gone with 10a plus 5, multiplied her number by 10, then added 5.
Okay, I'm adding a bit more information now.
Aisha gets the answer of 75.
How can we write this as an equation? Okay, we've got 10a plus 5, and I'm now saying that's gonna equal 75.
So that a is an unknown and it is now something we could work out.
So that was talking about unknowns, but letters can also be used to represent variables in a number of ways.
So let's look at some of those ways.
So we can have variables in a bar model as well.
So in this case, representing x plus y equals 20.
And they can vary.
There's lots of different things x and y can be.
So when you add them together, they equal 20.
We can have a scenario with variables.
So Jun says, "My sister paints action figures quicker than me.
For every one I paint, she can paint two." So how could we write this using expressions.
So if we said Jun, the amount of action figures he can paint, we'd represent with the letter n.
Then the amount of action figures that his sister can paint would be represented as 2n.
Or we can write that as an equation.
We can have a equals 2n, where n is the number Jun paints and a is the number his sister paints.
N can be any positive integer value.
In this case, Jun could paint 1, or 2, or 3, or 10, or 50 or 100 okay? And then for every amount that he can paint, his sister can paint twice as many, okay? So n can take any positive integer value and in this case is then a variable that can vary depending on the situation.
We can also represent variables using algebra tiles.
If you've got your algebra tiles with you, you might want to create this expression with your algebra tiles.
Okay, what expression have you created? How could we write this.
Right, well done if you came up with 2x plus 3.
X in this case is a variable 'cause it can take on any value.
This shows 2x plus 3, but there's not a fixed number that x is representing in this case.
One of the most common uses for variables is in formulae, okay? So you've probably seen some formula before.
You might have used some as well.
Andeep is making a roast dinner.
Recipe gives instructions for how long to cook the lamb.
So to cook the lamb to medium, the recipe says 40 multiplied by the mass in kilogrammes plus 20.
In this case, what is the variable? How could we then write a formula for the cooking time? Pause the video and think about your answers to these questions.
Okay, so in this case, the variable is the mass of the lamb.
This is a formula for how to cook lamb of any mass.
So the idea being that you can input the mass of the piece of lamb that you have and work out how long it takes to cook.
We can use any letter to represent this.
I've gone for m.
So I've got t equals 40m, 'cause it's 40 times the mass plus 20.
What do you think the t represents in this formula? Pause the video and come up with your answer.
Right, t is the cooking time in minutes.
And again, sometimes it's really important to add in what those letters are representing, particularly things like units of measure.
So we can use a table to see how the mass of the lamb affects the cooking time.
So we're gonna use this formula and we're gonna put that into a table to see what happens, the heavier the piece of lamb is.
So if the piece of lamb was one kilogramme, it's 40 times one plus 20.
If it was two kilogrammes, 40 times two plus 20, and so on.
Calculating those gives you 60 minutes, 100 minutes, 140 minutes, and 180 minutes.
Right, I'd like you to think about these questions then.
So how much longer does it take per additional one kilogramme of mass? How long would it take to cook if the lamb had a mass of 1.
5 kilogrammes? And then Andeep cooked the lamb for 200 minutes, what could its mass have been? Okay, pause the video and have a go at those questions and then come back and we'll see if we agree.
Right, well done.
There's lots of thinking required with those questions.
So we'll talk through it all together.
So what you might have noticed is there's a pattern in our cooking time.
Each time I add one to the mass, the cooking time goes up by 40 minutes.
So you might just said something like, "It takes 40 minutes longer per one kilogramme." And if we look at where that is in our formula, remember the formula said 40 times the mass in kilogrammes.
Okay, so for every time I add on an extra kilogramme, I'm gonna have to add on an extra 40.
Then we're looking at 1.
5 kilogrammes.
It's not in my table 'cause I chose to go up by one kilogramme each time.
But there's no reason why you can't cook a piece of lamb for 1.
5 kilogrammes.
1.
5 kilogrammes would be 80 minutes, halfway between that 1 and 2.
Well done if you managed to do this one on your own.
So a 5 kilogramme piece of lamb would take 220, just follow that pattern if you like of adding on 40, or you can do 40 times 5 and add on 20.
If 5 kilogrammes is 220, then for 200 minutes that must be between the 4 and the 5, so 4.
5 kilogrammes for that one.
Right, well done.
Let's have a go yourself then.
So to cook lamb well done this time, the recipe says the length of time in minutes is 50 times the mass in kilogrammes plus 20.
Read those statements and select the statements that are true.
Right, well done.
So let's look at this first one.
The cooking time increases by 20 minutes for every additional kilogramme.
That's not true.
You could have laid it out in a table like we did before to see what's happening, but it's actually gonna increase by 50 each time.
'Cause it's 50 times the mass in kilogramme, so 50 per kilogramme.
So every time you add on an extra kilogramme it's gonna go up by 50 minutes.
The mass of the lamb is the variable.
That is true.
That can change.
We can have any mass of lamb that we like.
And a 3 kilogramme piece of lamb would take 170 minutes.
That is true.
50 times three is 150 plus 20, 170.
Brilliant time for you to give this a go yourself then.
So for each person's statement, I want you to decide whether they're describing a variable or an unknown, okay.
Notice that some of them have extra pictures to go with them.
So Sofia is talking about that bar model that's got the n, the 7, and the 15.
And then Alex is talking about that bar model by him, which has got the b, the b, and the 3.
So read the statements, okay, and decide whether you think they're describing a variable or an unknown.
There might be some that you think you are unsure about, that could be either.
So put your ideas down in words and then we'll discuss them together in a moment.
Brilliant, we're gonna look at these one at a time.
So Sofia, "I can use this bar model to calculate n." Right, so Sofia is talking about an unknown 'cause n does have a set value.
She can calculate it.
In fact it's 8.
Laura, "I am thinking of a number.
When I multiply by 3, then add 2, I get the answer 11." Right, Laura is also talking about an unknown.
She has a number in her head that she's thinking of and we can calculate that if we want.
It's 3 if you decide to calculate that yourself.
Right, Sam, "There are 60 minutes in an hour.
To work out how many minutes there are in a certain number of hours I can multiply the hours by 60." Right, well done if you spotted this is a variable.
They can use this rule to work out the number of minutes for any number of hours.
Sam's not trying to work out a certain number of hours.
They're using that to work out any number of hours.
And then Alex, "The bar model shows my age compared to my sister's age." Right, again, this is a variable because it's representing both their ages and that's gonna change every year.
So whatever age Alex is, his sister is always going to be three years older.
So b plus 3.
So that's a variable again.
Right, Lucas, "This year my dad is exactly four times my age.
My age is currently x.
I can write his age as 4x." Right, this is an unknown.
This is quite a tricky one 'cause this actually only applies to his current age.
We don't have enough information at the moment to calculate it, but it's not gonna be true for any age he is.
It's the fact that this year his dad is exactly four times his age, okay? Therefore, he's x and his dad is 4x.
That won't be the case next year, okay? It might be fun if you think about your ages and some of your family's age.
Is there ever gonna be a time where one of them's gonna be double your age, or triple your age, or four times your age? Okay, it's quite a nice, fun idea to explore.
Lovely, so now we're gonna put all that into practise and have a go at solving some problems. So this is gonna require you to do a little bit of thinking.
If you've got other people you are learning with, you can do some discussions, otherwise jot down your ideas as you go through.
So I've got some cards here.
Feel free to write them on bits of paper, on post-it notes.
And I want you to have a go at arranging these cards in order.
There's no right or wrong answer, okay? I want you to think about how you are going to place them and then justify your order, okay? You might want to take a little bit of time on this.
You are welcome to write things down, think about things that might be the same, think about things that might be different, okay? And then we'll come together and we'll explore these in more detail.
Pause the video, off you go.
So as x is a variable, some of these statements are impossible to put in a numerical order without knowing the value of x.
I'm sure you discovered that as you were playing about with these.
Sometimes one is bigger than another one depending on what x is.
There are some really interesting features about some of these expressions though, which are quite fun to explore.
So some things we can consider.
Let's start by looking at x and 2x.
Is 2x always greater than x? Have a think about that one.
To start with, it looks like yes, 2x is always gonna be bigger than x 'cause it's 2 of x.
However, well done, if you spotted that it's true for positive values, but not true for negative values.
If x was negative 2, 2x is negative 4, negative 4 is not greater than negative 2, okay? So don't forget that when you're exploring things to try different types of numbers as well, and see if you can spot any of these patterns.
Okay, what about this one? Is x plus 3 always greater than x? What do you think? Yes, this is one that we can order.
X plus 3, no matter whether x is negative, integer, non integer, x plus 3 is always gonna be bigger than x.
If you think about putting it on a number line, okay, locate x wherever you like on a number line, add 3, okay? It is always going to be greater.
Okay, let's explore 2x and x plus 3 this time.
When is 2x greater than x plus 3? You might wanna try some things out.
Right, they're actually the same when x is 3, 'cause 2x would be 6 and 3 plus 3 would also be 6.
So that means that 2x is actually greater when x is greater than 3.
And then less when x is less than 3.
So there's that kind of turning point there when they're the same, when x has a value of 3.
What about this one? When is 2x greater than 3x subtract 2? This was quite a tricky one to explore.
Again, they are the same when x is 2, 2 times 2 is 4, and then you've got 6 minus 2, which is 4 as well.
Okay, so they're the same when x is 2.
2x is greater when x is less than 2.
Okay, so 1, or 0, or negative, or any non integer values between them, anything up to 2, 2x is gonna be greater, after that 3x is going to be, 3x minus 2 is going to be greater.
All right, what about this one? Is x squared always greater than x? This is an interesting one to have a look at.
You might wanna pause and try some things out if you haven't already.
Now, a lot of the time x squared is greater than x, but not always.
You might have worked out that when x is 0, 0 squared is 0, so x and x squared would be equal.
When x is 1, 1 squared is 1.
So x and x squared would be equal in that case.
Now, there's an interesting case that you might wanna check on your calculator.
If x is between 0 and 1, then x squared is going to be smaller.
So if you put 0.
2 is x and 0.
2 squared into your calculator, you'll see it gives you a smaller value.
Other than these two, x squared is always greater.
So if x is bigger than 1, then x squared is always going to be greater.
If x is less than zero, so any negative values, x squared is always going to be greater.
Remember, a negative multiplied by a negative gives you a positive value.
So negative 1 squared, for example, gives you positive 1.
So x squared is greater, okay? There's lots of quite interesting maths involved in that.
The point that we're trying to get across is that x can vary and actually being able to explore these variables by trying out different numbers is quite gives, brings up some quite interesting maths.
Right, time for you to have a go at a question yourself.
So what I would like you to do for this one is I'd like you to choose a number from the box to be c.
I want you to choose a different number to be d.
And then we've got c plus d as our expression.
What I want to know is what is the largest value c plus d could be.
And then what is the largest even value c plus d could be? Have a play around and come back and check your answers.
Brilliant, so the largest c plus d could be would be 20 plus 15, which is 35.
Addition is commutative.
So you could have picked them either way round for that one.
The largest even value would be 13 plus 15, which gets you 28.
Okay, you can't use the 20 in this case because the 20 plus the 15 would be odd.
The 20 plus the 13 would be odd.
So the biggest you could make would be 20 plus 6, which is 26.
But actually you make a larger value by doing 13 add 15.
Well done if you got both of those ones.
Let's have a look at the next bit.
Brilliant, so we're gonna play around with some variables again now.
So this time we've got r plus t equals 30.
And for each question I'd like you to state whether it's possible to find r and t that satisfy these conditions.
So you're deciding whether they're possible or impossible.
R and t are both integers in this task as well.
So we're only looking for integer values.
If you think they're possible, see if you can give an example.
Can you come up with numbers that make it work? If you think it's impossible, see if you can explain why.
Why do you think it can't work for any integer values of r and t? Have a play around with those numbers and then we'll think about our answers together.
Lovely, okay, so both values being odd.
Can we have two odd values that add to give us 30? Yes, we can.
There's loads of possible solutions for this one.
I went with 29 add 1.
They're both odd numbers and they give us a total of 30.
Right, both values are even.
So again, that is possible.
I went with 20 plus 10.
But again, you could have come up with any two even numbers that add to 30.
Right, one value is even.
Now, this one's impossible because if one value's even then the other value must be odd.
And if you do an even number, add an odd number, you're always gonna get an odd number.
You can draw some diagrams to prove that, that's always correct.
Because 30's even, we can't have an even plus an odd to get 30.
Both values are prime.
Well done if you knew what prime was without having to look it up.
But it's never a problem to sort of look things up to make sure you've got the right definitions.
So yes, this is possible.
11 plus 19, they're both prime numbers that add to 30, or 13 plus 17 are both prime numbers that add to 30.
Well done if you've got either of those two examples.
Right, E, both values are multiples of four.
So this is impossible.
If you have a multiple of four and you add another multiple four, then your answer will be a multiple of four because 30 isn't, then that's not possible to add two multiples of four to make 30.
For F, G, and H, drawing a picture can really help with this one.
So one value is three times the other.
So what I've drawn is I've drawn one value, I've just called it r for the moment.
And then the other value has to be three times bigger.
So I've drawn r, r, r, so it's three times bigger.
And they have to all add up to 30.
Well, that can't work either because as we've just talked about before, 30 isn't divisible by 4.
So 30 divided by 4 would not give us an integer value.
Remember, we're looking for integer values that would, where one would be three times the other.
Let's look at the next one.
One value is four times the other.
So let's draw our diagram again.
So one value and then four times, okay.
That gives us 5 in total.
And that is possible 'cause 30 is divisible by 5.
So you could have had 6 and 24 for that one.
And the last one, one value is five times the other one.
So we've got one value and then I've drawn 5 more and in total we've got six boxes, haven't we? And 30 is divisible by 6.
So you could have had 5 and 25 to give you 30.
Right, fantastic.
There's lots of interesting maths and interesting number works that came up there.
Lovely, so let's have a look at what we've learned today.
So a letter can be used to represent a specific unknown in a variety of ways.
We looked at equations, we looked at bar models, we looked at scenarios.
A letter can be used to represent a variable in a variety of ways, and we looked at some different formulae.
It is possible to tell the difference between variables and unknowns.
And you did an activity where you were looking at telling the difference between a variable and an unknown.
And we can solve problems involving variables given constraints.
We did lots of playing around with numbers and the properties of numbers.
I really hope that some part of that exploration, investigation today shows you more about the number system that you weren't already confident with.
The number system is absolutely fantastic and there's loads of patterns and things to notice the more that you explore that.
Right, thank you for joining us today.
I hope you enjoyed yourself and it'd be really nice to see you again.