Lesson video

Thank you for choosing to learn using this video today.

My name is Ms. Davis and I'm gonna help you as you work your way through this lesson.

Feel free to pause the video and work through things at your own pace.

Lots of exciting things to look at, so let's get started.

Welcome to our lesson on equality in an equation.

By the end of the lesson, you will appreciate that in an equation, the two sides of the equals sign balance.

We're gonna talk lots today about equations, so just make sure that you are happy with what an equation is.

We're gonna start by using bar models to rearrange equations and then we're gonna look at balances and how we can use the balances to represent equations.

When we write equations, we can rearrange these to find other equations which are true for that relationship.

For example, if I know that eight plus two equals 10, I also know that two plus eight equals 10, 10 minus eight equals two and 10 minus two equals eight.

All those equations have come from the same relationship.

Lucas has written the equation 28 plus 49 equals 77.

Just pause the video and see if he can write down some other equations that he can write from that information.

Fantastic.

you could have had 77 minus 49 equals 28, 77 minus 28 equals 49 or 49 plus 28 equals 77.

There are potentially others that you could have come up with as well.

Luca says, I can also show this with a bar model.

Let's have a look at his bar model.

Because 28 plus 49 is equal to 77, remember the bars must be the same length.

Let's just see the equations that we had before and how they're related to our bar model.

So 49 plus 28 is the same length as 77.

We can see that 77 subtract 28 would leave us with 49.

We can see that 77 subtract 49 would leave us with 28.

We didn't have one this one before.

I wonder if you came up with it.

77, subtract 49, subtract 28 would leave us with nothing.

How could Lucas change his bar model to show that 28 plus 49 plus 13 equals 90? What do you think he could do? So if we add 13 to that top bar, so it now represents 28 plus 49 plus 13, we'd have to add 13 to the bottom bar.

77 add 13 can be written as 90.

How could Lucas use his bar model to show that 20 plus 28 plus 49 is not equal to 77 plus 23? Just pause the video, read through, what could we do? Okay, so the top bar at the moment is 28 plus 49.

We want it to represent 20 plus 28 plus 49.

So we need to add another 20 bar.

The bottom one at the moment is 77.

We want it to represent 77 plus 23.

Because a different value is added to the top and the bottom bar, they are no longer the same length, so make sure that you don't draw them equal because they're no longer equal, they are not the same length.

The same thing is true for bar models with variable terms. Lucas says this is a bar model for the equation, a plus c equals 50.

Laura says, I've written a different equation for this bar model.

What equation could Laura have written? Pause the video, see how many you can come up with.

I wonder which one she came up with.

We could have c plus a equals 50, 50 minus a equals c, 50 minus c equals a, or maybe you went with this one.

50 minus c minus a equals nothing so equals zero.

Just be aware that we don't know the value of a and c.

We don't know which value is supposed to be longer.

Just because we've chosen to draw c longer doesn't mean c has to be a bigger value.

Bar models are just a representation.

They do have limitations.

They don't give us all the information about a variable.

If I've written the equation, x minus y equals five, what other equations are true? Let's look at the bar model to help us.

So x, we want to subtract y, so let's draw y underneath and that leaves us with five.

So x minus y equals five.

Pause the video, what other equations are true this time? These are the four that I came up with.

I wonder if any of yours were the same.

Laura says in this case x has to be larger than five.

Do you agree? Pause the video and have a think.

I wonder if you can see why Laura has said this.

So because x minus y equals five and our x bar has been drawn the longest, that seems to suggest that x has to be larger than five.

However, if you just look at the original equation, x minus y equals five, I wonder if you can think of any values where that's not true.

In fact, if x was three and y was negative two, three subtract negative two would give you five, so that would fit that equation and in that case x is smaller than five.

So just be aware that there are limitations with bar models in terms of their size.

They are great however, for showing the relationship between terms in an equation and they really help us make sense of rearranging equations into different forms. Just be aware that we can't always use the size of the bars to give us information about the unknowns or variables, particularly involving negative values.

How could we draw a bar model for the equation? Four x plus five equals two x plus nine.

Why don't you give this one a go? Lovely, I've drawn it this way.

So I've put four x's plus five and I know that must be the same length as two x's plus nine.

Make sure your x's are the same size as each other.

Where we have a variable in an equation every time the variable appears it has the same value.

So do make sure that all your x's are the same size.

Sam, Sophia, Alex and Izzy have written some other equations.

I'd like you to see which of those are also correct for the same relationship.

Off you go.

Well then if you've spotted the all of them are correct.

If you're not sure, pause the video again and just read through those.

Looking at the bar model, can you see how all of those are still true? A check then so if 41 minus x equals 26, which of the following are also true? Have a look at the three suggestions and then draw yourself a bar model to check that you are correct.

So you might have been able to do this without a bar model, but hopefully you saw that the bar model is a really useful tool for just checking that we have rearranged our equations correctly.

So 41 subtract x equals 26 is represented by that bar model.

Now we can see clearly that 26 subtract x is not 41.

26 add x is 41 but x minus 26 is not 41.

So its that middle one was the only one correct in this case.

Let's have a go at this one.

If y minus x equals 17, which of the following are true? Again, try it without a bar model if you like and then draw yourself a bar model to check.

Lovely.

So there's our bar model and we can see that y subtract 17 is x, but x subtract 17 is not y and 17 subtract x is not y.

So in this case it's just a that is correct.

Well done, time to have go at a practise.

I would like you to match the equations to the bar models.

So there's three bar models along the top.

Each bar model can match with multiple equations.

Once you think you've got all matched up, write an equation which is also true for each bar model.

come back when you're ready for the answers.

Well done.

Let's see if we agree.

So for A that matches with F, I and J.

Pause the video and check your answers if you need to.

For B, that matches with E, D and K and for C that matches with G and H.

Let's have a look at some of the other equations you might have written.

There are many other possible answers.

So for A, you might have gone with three x plus 27 equals 42.

You might have gone with 42, subtract 27 is three x or you could have gone with 42 subtract 27, subtract three x gives you zero.

If you've got someone to check your work with, you might wanna check each other's answers from the bar models.

For B, you could have had three x minus 27 equals 42 or you could have gone with three x minus 27 minus 42 equals zero.

And lastly, there's a few options I went for here.

You could have had two x plus 27 equals x plus 42, two x plus 27 minus 42 equals x, two x equals x plus 42 minus 27.

Or you could have gone with two x plus 27, that's the top bar, minus x minus 42 equals zero.

Again, make sure you've spent enough time checking that you agree that they match with the bar models and if you've got someone to work with, check each other's answers and make sure you are really happy.

Lovely.

We're now gonna look at a different way of representing equations.

We're gonna use balances to represent equations.

We can start to make sense of manipulating equations using a balance model.

This is a balance scale.

It may or may not be something that you've used before.

You see them in real life to measure the massive objects, often ingredients for cooking or baking.

You might have one in your kitchen where you put masses on one side and then you weigh ingredients in the other side.

You also come across them in mythology for comparing weights of different objects.

For example, sometimes comparing the weight of a sole against a feather or things like that.

How they work is they have two scale pans, one on either side which objects can be placed on.

At the moment the scale pans are the same height because the scale is balanced.

There's nothing on either scale pan.

So the way the scale is made that will keep it balanced when there's nothing on either.

What do you think will happen when I put a mass of five grammes on the left hand side? Yeah, of course that left hand side is now gonna be lower 'cause the scales are no longer balanced.

What do you think will happen if I put a mass of five grammes on the right hand side? Yeah, you've got it, once I put it on the right hand side, my scales are balanced again.

They've got the same mass on both sides of the balance.

What can you tell me then about the mass of the orange on the left hand side and the lemons on the right hand side? Right, the orange has the same mass as those two lemons.

What will happen if I add a lemon to the left hand side? The scales are no longer balanced.

The left hand side will be lower.

We will be linking this to our equations in a moment, so just make sure you are really confident with how this representation is working.

What will happen if I add a mass of seven grammes to each side? Perfect.

The scales will remain balanced.

I've put the same mass on both sides of my scales.

Right, it's time for a bit of thinking.

What can you tell me about the mass of the orange and the lemon now? Perfect, their combined mass is 255 grammes.

If I take an orange off, the scales will no longer be balanced.

The right hand side this time will be lower because the mass on the right hand side is larger than the mass on the left hand side.

Let's just check we're comfortable with this idea of using a balance.

So these scales are currently balanced.

Each square has the same mass.

What would happen if a square object was added to the right hand side? See if you can put it into words.

The right hand side would now be heavier so the scales are no longer balanced.

Or you might have said the right hand side will now be lower 'cause the right hand side is heavier.

Okay, what can you say about the mass of the triangle? Well then if you notice that the triangle has a greater mass than one square, you might have even said that the triangle has to have the same mass as three squares in order for those two sides to be balanced.

Brilliant, let's start linking that to our equations then.

We can use balance scales to make sense of manipulating equations to maintain equality.

As with any representation, there are limitations of using this balance method.

For this lesson we'll assume that the x tile has a positive mass.

Once we are confident with how to balance equations, we won't need to use the balances anymore and then we can start exploring values of the variables that are negative as well.

So this balance scale represents the equation three x plus one equals two x plus three.

If you have our algebra tiles with you, you might want to use them from this point onwards in the lesson.

If these two things are currently balanced, that means that three x plus one is the same as two x plus three, which we can see in our equation with the equal side.

What will happen if I take an x tile off the left hand side? What do you think? Right, if they're currently balanced as they are, if I take an x tile off the left hand side, then they'll no longer be balanced.

What could I do to balance the scales again? Well, I could put the x tile back on or I could take an x tile off the right hand side as well.

Adding the same value to both sides of the scales will keep them balanced just in the same way that subtracting a value to both sides of the scales keep them balanced.

Let's see what this looks like.

So I've got two x plus one equals five.

If I add two to each side and that will remain balanced.

Let's look at that with a bar model as well.

So if I have that bar model showing two x plus one equals five, if I add two to both bars, they remain equal.

And you might have explored this with an equation as well.

We have the equation two x plus one equals five.

Adding two to both sides of the equation keep them balanced.

I could rewrite my equation as two x plus three equals seven if I wished.

What equation does this balance scale represent? Pause the video and write it down.

If you've got algebra tiles, you might wanna build this as well.

Lovely, we can write this as two x plus two equals x plus four.

What have I done to the right hand side of my balance? Pause the video.

Think about your answer.

I think you could have said that we've added x plus four to the right hand side.

You might have noticed that in this case that's the same as doubling the right hand side.

What could I do to balance the scales again? Again, we have two choices.

We could either add x plus four to the left hand side as well or we could double the left hand side.

Which option did I choose? Good spot.

I doubled the left hand side two.

So although I added different amounts to both sides because those amounts were equivalent, then that's absolutely fine.

So it's the same essentially as doubling the left hand side and the right hand side.

Let's just summarise what we've learned.

Multiplying both sides by the same value will keep the scales balanced.

Lets look at another example.

So here I've got x plus one equals two.

If I multiply that by three, the left hand side becomes three x plus three.

I've got, now I've got three lots of x plus one and the right hand side becomes six.

They are still balanced.

We can see it in a bar model.

Multiplying both bars by the same amount will keep them the same length.

So if x plus one equals two, then three lots of x plus one equals three lots of two.

And we can write that in our equation as well.

Remember it's the whole of that left hand side that we're multiplying by three.

If you would like to expand your bracket and write it as three x plus three equals six, that's exactly the same thing.

Okay, so Jacob wants to write an equation for this balance scale.

What is the problem with this idea? Right, the scales are not balanced.

This means the values on the left and the right are not equal.

We can only write an equation when things are equal.

What Jacob could say about the values is that x plus 20 is greater than 50 he does know that.

He can write that with an inequality symbol if he wishes, x plus 20 is greater than 50.

However, he cannot write an equation.

Time for a check.

I would like you to complete each balance scale so the left and the right hand sides are balanced.

You'll see that the black scales on the left hand side show the original equation.

And then there in the middle I have done something to one of the scale pans.

What I would like you to do is fill in the final balance on the right hand side so that they are balanced again.

Off you go.

well done.

Let's look at each one separately.

So for a, I think there was two choices.

You could have realised that the left hand side mass has been doubled.

So we've got x plus three and that's then become two x plus six.

That means we could double the right hand side too.

So you could have put four x's on that right hand side or you could have said that that's the same as adding x plus three.

So if x plus three has been added to the left, you can add x plus three to the right.

So that gives you three x's plus three on that right hand side.

If you need time, just pause the video and check you're happy with those.

For B, we've added a mass of 20 onto the left hand side, so you need to do the same on the right hand side.

You could write that as an 80 mass and a 20 mass, or you could have written it as a single a hundred mass.

For C, we have halved the mass on the right hand side, you see it's gone from four x and a hundred to two x and 50.

So both those terms have been halved.

So we also need to halve the mass on the left hand side.

So we end up with one x and then the missing mass would then be a hundred.

It's the whole of the scale plan on both the left and the right hand side that has been halved.

Fantastic, time to put that all into practise.

I would like you to choose one expression from the box to place on each scale pan.

The first scale pan is balanced.

The second one, the right hand side is heavier than the left.

And for C, the left hand side is heavier than the right.

Try and do it without using the same expression more than once.

There are several different ways to do it.

Okay, see if you can find the way that you think is your favourite.

The important bit with this is that you explain your reasoning.

Off you go, come back for the next bit.

Well done.

So for question two, these scales are currently balanced.

What might the mass of the triangle and the circle be? Is there more than one answer? For B, it's the same scale pan, but this time something has changed and the scales are still balanced.

I want you to then think about what the mass of the triangle and the circle might be, and if there's more than one answer.

And finally the mass on the right hand side has been replaced with a 20, so they no longer balanced.

What could you do to balance them? Again, think about whether there's more than one possible answer.

Off you go and come back for the answers.

Well done.

There are many possible answers for these.

I've just chosen some to show you.

So I've started with a, so you could have picked six times five and three times five times two 'cause they're both equivalent 'cause three times two is six.

They're both equivalent calculations they both equal 30.

You could have had a is the same as six times five because that would be true if a had the value of 30.

So you might have said, well if a has the value of 30, then this would be balanced.

You could have had a plus six is the same as a plus b when b is six.

So several different possibilities there and they're more than the ones that I've just come up with.

For the second one, so the top one is definitely gonna be true.

Eight plus 10 plus 12 is smaller than eight plus 10 plus 13.

So that would've been a good one to go with.

Looking at the second one, if a is a positive value, then two a will be larger.

So you could have gone with a and two a.

Remember that's only true if a is positive.

a plus six is always gonna be larger than a, so that would've been a nice one to go with on the bottom one.

So this is actually the same as B, just the other way round.

So you could have had any of those previous ones but the other way round, or I've come up with some other ones you could have used.

You could have had eight plus 10 plus 13 is larger than six times five.

You could have had that a plus b is larger than a and that's gonna be true when b is a positive value.

And you could have gone with a plus six is larger than a plus b.

That will be true if b is less than six.

So plenty of options do take your time reading through those answers.

If there's any you are not sure of, you might wanna discuss with a partner.

Let's look at the next question.

Okay, there are plenty of answers for this first one.

There are six integer possibilities.

So one and six, two and five, three and four, four and three, five and two, six and one.

We said earlier in the lesson that we were going to use positive values for our masses.

However, there are plenty of non integer answers as well.

Okay, let's have a look at what happened in b.

So you can see that a triangle has been added to the left hand side and a mass of three has been added to the right hand side and that's balanced.

That means those things have to be the same.

So one triangle must be worth a mass of three.

So that's the only option now, which is true.

For that last one, you could have added 10 to the left hand side and that would balance them again.

Or because the right hand side has doubled, you could double the left hand side and that would balance them again.

Perfect.

Lots of great thinking there today guys.

I'm hoping now that you can use bar models or balances to help you with your equation skills moving forwards.

We have seen that bar models can be used to rearrange equations with an additive relationship.

Equations can be represented with balances.

We're now experts at using that balance model.

We can only do that if the expressions on both sides are equal, and if they are, the scales will be balanced.

In order to keep a scale balanced, the same operation must be done to the expressions on both scale pans.

Well done with today's lesson and I look forward to seeing you again.