Lesson video

Hello, Mr. Robson here.

Welcome back to maths.

What a good place to be.

Additive relationships in algebra today.

I love this bit of maths.

It's so useful in so many other parts of maths.

So, it's going to be an awesome lesson.

What are we waiting for? Our learning outcome is that we'll be able to appreciate that an additive relationship between variables can be written in a number of different ways.

Keywords that you'll hear today, equations, expressions.

An equation is used to show two expressions that are equal to each other.

You've seen a lot of equations and expressions in your learning in mathematics.

We'll see a few more today.

Two parts to our lesson, and we're going to start by rearranging additive relationships.

Let's start with something really rather familiar.

You might have heard the word fact families.

You might or definitely have seen this equation, 7 + 3 = 10.

That's an equation.

The expression on the left hand side, 7 + 3 is equal to the expression 10 on the right hand side.

If you know that 7 + 3 is equal to 10, what other addition and subtraction facts do you know that relate 7, 3 and 10? I'd like you to pause and see if you can fill in these blanks, and give me those other addition and subtraction facts.

Welcome back.

I hope you wrote down 3 + 7 = 10.

You may have heard this called commutativity.

Commutativity.

Addition is commutative because the terms 7 and 3 can change position without it changing the calculation.

If 7 + 3 = 10, then 3 + 7 = 10.

You may have seen it described like this, this additive relationship, that 7 is an addend, 3 is an addend, and 10 is what we call the sum.

If that's true, then the two addends make our sum.

When we subtract either addend from our sum, we're left with the remaining addend.

What does that mean? Well, start with the sum 10, subtract the addend 3, and you're left with the addend 7.

The other way round, we could start with the sum 10, remove the addend 7, and we're left with 3.

You might have seen this called a fact family.

We could also call it a family of rearrangements for the equation 7 + 3 = 10.

That's a bit more like the technical language we'll see in algebra.

7 + 3 = 10 is an equation, and it can be rearranged.

Bar models are a good way to visualise any family of arrangements.

Here's a bar model for 7 and 3 equals 10.

The length of 7 and 3 make the length of 10.

You can see the other way round, the length of 3 and 7 making the length of 10.

You can also see the length of the 10 bar minus the 3 bar, leaving the 7 bar.

And the length of the 10 bar minus the 7 bar, leaving the 3 bar.

Quick check that you've got this.

I'd like you to complete the family of rearrangements for the equation 65 + 35 = 100.

I'd like another addition fact, I'd like two subtraction facts, and you might choose to use that bar model to help you.

Pause and I'll see you in a moment.

Welcome back.

The first blank we can fill in is the addition fact 35 + 65 = 100, another example of commutativity.

A subtraction fact, 100 - 35 = 65, and 100 - 65 = 35.

Next, Alex and Jun are completing the family of rearrangements for the equation 81 - 27 = 54.

Notice this time, we're starting with a subtraction fact, not an addition fact.

Alex says, "I think in this family we'll see 27 - 81 = 54." Jun says, "I think we'll see 54 - 27 = 81." Laura steps in and says, "I'd recommend checking your work and using bar models." Well, they're two sensible ideas in mathematics, checking your work.

Alex says, "Of course, this can't be right." If Alex starts with 27 and subtracts 81, he'll be on a negative number, so the answer can't be 54.

And Jun says, "Nor mine," because if you start at 54 and take away 27, you can't go up to 81.

"If you're not sure, build a bar model to visualise it." There's Laura with more wonderful advice.

So what would a bar model of 81 - 27 = 54 look like? Well, we'd start with 81 and we'd lose a length of 27 to leave a length of 54.

From there, you can see 54 + 27 = 81 is the first rearrangement.

We can commute that, 27 + 54 = 81.

And from there our next subtraction fact, 81 taking away the length of 54 will leave the length of 27.

"Thanks, Laura," say Alex and Jun.

Quick check you've got that now.

I'd like you to complete the family rearrangements for the equation 5.

36 - 1.

9 = 3.

46.

I've left a blank bar model there, which you might choose to copy, fill in, and support you in finding these rearrangements.

Pause, give this a go.

I'll see you in a moment for the answers.

Welcome back.

If you chose to use the bar model to help you, then you would've filled it in like this, I hope.

5.

36 subtracting a length of 1.

9 to leave a length of 3.

46.

So we can see 3.

46 + 1.

9 must equal 5.

36.

Commute that.

1.

9 + 3.

46 must be the same thing, 5.

36.

And you can also see the second subtraction fact, 5.

36 - 3.

46 leaving 1.

9.

In mathematics, generalising is a really powerful tool.

In this lesson so far, we've seen these families of rearrangements for those three equations.

Can you write a generalisation that works for any additive relationship? Well, that sounds like a curious question.

What do I mean by a generalisation? If I started you with a mathematical truth that a + b, whatever they are, make c, then what other three mathematical truths can you tell me? This will be a really powerful generalisation.

Pause, tell the person next to you what are the facts do you think are true if a + b = c.

See you in a moment.

Welcome back.

A generalisation for an additive relationship.

If a + b = c, then b + a must also equal c.

If c is the sum, then c subtract b must leave us with A, and c subtract a must leave us with b.

In mathematics, generalising is a really powerful tool.

a, b and c can take any values or be any terms, but so long as a + b = c, all of these rearrangements will be true.

So far, we've seen a, b and c taking on numerical values, but in mathematics it's really useful to know that they could also be any terms. Instead of a numerical relationship, if I introduce the idea of an unknown here and give you an algebraic equation, x + 7 = 31, this generalisation is still true.

We could make a family of facts for this equation, x + 7 = 31.

What will this family of rearrangements look like? Pause, see if you can write down the other additive facts and the two subtraction facts that will fit into that family.

See you in a moment.

Welcome back.

How did we do? Let's find out.

Did you write a commutative example? If x + 7 = 31, 7 + x = 31.

Did you identify that 31 is the sum, therefore 31 - 7 = x and 31 - x = 7.

Well done.

Why is this useful? Oh, it's really useful because this rearrangement hands us the solution to this equation.

Look at that rearrangement where X is isolated.

Rearranging x + 7 = 31 to read 31 - 7 = x, that gives us the immediate solution to the value of x.

How nice is that? It doesn't matter how complex looking the terms are, we can still rearrange the equation.

This might look a little bit mean when you first read it.

63 - 2y squared = 13, but it's no different to the numerical example we saw earlier.

These terms can be rearranged to give us a family of facts.

I like to use a bar model just to visualise what's going on.

It helps give a bit of certainty to my thinking.

I start with the sum of 63 and I lose 2y squared, I'll be left with 13.

That's what that equation is telling me.

When I write out as a bar model, I can now see all the facts, and I can see those facts.

I can see the length 13 and the length to 2y squared giving us the length 63.

I can see those two terms commuted, 2y squared + 13 = 63.

And I can clearly see the sum is 63, therefore 63 - 13 must leave me with 2y squared.

Quick check you've got that now.

Which of these are accurate rearrangements of this equation? The equation 5x + 7y = 38.

Which are accurate rearrangements and which are not? Pause, I'll see you in a moment for the answers.

Welcome back.

Here's my bar model for 5x + 7y = 38.

I can clearly see that 7y + 5x = 38.

A is definitely true.

B is also true and I can also see that C is true.

They were all valid rearrangements of the equation 5x + 7y = 38.

Let's check this one now.

Which of these are accurate rearrangements of this equation? 2e - 5f cubed = g.

See you in a moment.

Welcome back.

A is true, g + 5f cubed gives us the sum 2e.

B is not true.

I can see from my model that 5f cubed - 2e will not give me g.

C is true.

The sum 2e - g leaves me with 5f cubed.

D is not true.

Practise time now.

For question one, I'd like you to write the family of rearrangements for the below equations.

Pause, fill in the blank spaces, and give me the family of rearrangements for those four equations.

For question two, Lucas is rearranging equations.

Lucas says, "I see a lot of it as swapping positions in the equation." For question A, Lucas says, "If 3v + 5w = 52, then 5w + 3v = 52 is also true." For B, Lucas says, "If 73 - 8x = 49, then 73 - 49 = 8x." And you see the same thing in question C.

3x - 2 = 8y, then 8y - 2 = 3x is also true.

In each case, I'd like you to decide if Lucas is correct, and I'd like you to write a sentence to justify your decision for all three.

Pause and do that now.

Welcome back.

Question one, I asked you to write the family of rearrangements for some equations.

For the equation 58 + 76 = 134, I hope you gave me the rearrangement 76 + 58 must equal 134, the sum 134 - 76 must leave us with 58, and the sum 134 - 58 must leave us with 76.

For part B, you might have drawn a bar model that looked like that.

That's 19 subtracting h to leave you with 11.

From there, we can see these rearrangements, 11 + h = 19, h + 11 = 19, and the sum 19 - 11 leaves us with h.

There we are, that last one, taking us straight to the solution to that equation.

For part C, we could have written x + 3 = y, 3 + x = y, and the sum y - 3 = x.

For D, 100 is the sum.

Subtract 2x to leave 5y squared.

That's represented in that model.

So, 5y squared + 2x = 100, and 2x + 5y squared = 100 and the sum 100.

When you subtract 5y squared, you're left with 2x.

There's your family of rearrangements for that one.

For question two, this was interesting, the way Lucas is viewing this.

"I see a lot of it as swapping positions in this equation." So when can things swap positions and when can't they? If you understand this, you've got a really good grasp of this topic.

In the first case, absolutely true.

It's an example of the commutativity of addition.

3v + 5w is the same as 5w + 3v, so absolutely, Lucas could swap those positions.

It's just commutativity.

For B, this too was true.

8x and 49 can swap positions.

Why? It's because we can think of 73 as the sum.

When we start with 73 - 8x equaling 49, 73 is the sum, whichever one of the addends you remove, you leave the other addend.

So subtracting either addend from the sum leaves the other addend.

That means that the 8x and the 49 can swap positions on this occasion.

You've also seen this, swapping of positions, in our generalisation, a + b = c.

We had these rearrangements in our generalisation, c - b = a and c - a = b, because c's the sum, a and b can swap positions.

Note the efficiency here.

If we were asked in mathematics to solve the equation 73 - 8x = 49, the rearrangement 73 - 49 = 8x leaves us really close to the solution.

This idea of rearranging additive relationships can work for us very powerfully.

C was a tricky one, and I hope you said it's not true.

3x and 8y cannot swap positions.

Can you see why? It's because if I write the model 3x - 2y leaving 8y, it's 3x which is the sum.

So it can't be true that 8y - 2 leaves us with 3x.

3x is the sum, so it would only be the 2 and the 8y that could swap positions.

The last one was tricky, and importantly not true.

Onto the second part of our lesson now, where we're going to talk about the sum of multiple terms as a term.

What an interesting sentence that is.

It might not immediately make sense.

It definitely will in a few moments time.

Let's start with something familiar.

You know that 6 + 3 + 1 = 10, and you know it's an equation.

But does this equation have a fact family? It does.

I wonder how many facts are in that family.

Pause for a minute, maybe two, maybe three, and see how many related facts you can write additive and subtraction that involve 6, 3, 1 and 10.

Pause for a few minutes and make me a list.

See you in a moment.

Welcome back.

6 + 3 + 1 = 10.

Does this equation have a fact family? You bet it does, but it's a significantly larger family than we've seen so far in this lesson.

There's your commutative examples where I've just commuted the terms on the left hand side.

And if 6 + 3 + 1 = 10 then 6 + 3 = 10 - 1.

And I can write something similar for all of my equations on the left hand side.

And if 6 + 3 = 10 - 1, then 6 = 10 - 1 - 3, and I can write that list for all of the equations in the middle.

That was a lot of work.

It's a huge fact family.

So, is there a better way to treat these cases? There is a more efficient way to consider the rearrangement of multiple terms like this.

I'm glad to hear that.

7 + 2 + 1 = 10.

If we wanted to leave 7 on its own on the left hand side, we could do that, subtract the term 1 from both sides, and then subtract the term 2 from both sides, and we've left 7 on its own on the left hand side, and we still have a valid equation.

7 is 10 - 1 - 2.

Yes, that's equal.

We have an equation.

But there's a better way to do this.

To be more efficient, we could group the terms 2 and 1, treat them as one term and subtract them together.

You know that we can use brackets to group, so I put brackets around the 2 + 1.

I'm now gonna treat that as one term, and I'm gonna subtract it.

If I subtract brackets 2 + 1 from both sides, I'm immediately left with 7 on the left hand side.

I now have 10 - brackets 2 + 1 on the right hand side.

That's still true, isn't it? Yes, 2 + 1 is 3, 10 - 3 is 7.

We still have a mathematical truth.

But by making the sum of multiple terms one term, i.

e.

by grouping the 2 and 1, we made the process more efficient, and you know we mathematicians love efficiency.

This also works for algebraic terms. For example in the equation a + b + 5 = 10, I could group the b + 5, treat that as one term, and subtract it from both sides.

That'll leave A isolated alone on the left hand side, and give me 10 - bracket b + 5 on the right hand side.

And we still have a mathematical truth.

Even something as complicated looking as x + 3y squared + 4 = 7.

I could group those terms, 3y squared + 4, subtract them from both sides, and I still have a true equation.

Quick check you've got that now.

Which of these are correct rearrangements of this equation? 7x squared + 8y + 4z = 16.

Three to choose from, which are correct and which are not? Pause, have a conversation with the person next to you, or a good think to yourself.

I'll see you in a moment for the answers.

Welcome back.

Did you spot the first one was true? Treating brackets 8y + 4z as one term subtracting from both sides, that'll leave you with 7x squared on the left hand side, and you'll have 16 - bracket 8y + 4z on the right hand side.

B was not true.

The brackets 8y + 4z should be subtracted from both sides.

It was grouped and subtracted from the left hand side, but then added to the right hand side.

That means we no longer have an equation.

We'd have to subtract that term from both sides, not add it.

C was true.

You could commute our equation to read 8y + 7x squared + 4z = 16, and then group the 7 x squared and 4z, and then subtract from both sides to get C.

Alex is practising this skill in class.

a = 10 - bracket b + 5 is the same as a = 10 - b + 5.

We don't need the brackets.

Well, they sound pretty similar, Alex, so is that true we don't need the brackets? If ever we're unsure about algebraic manipulation, we can try the same manipulation with a similar number problem to test the truth.

What Alex has done is simply erase the brackets.

Instead of 10 - bracket b + 5, it's just 10 - b + 5.

Just rub them out with an eraser.

Can we do that? If we can do that, then this equation 20 = 30 - bracket 7 + 3 must be the same as this equation, 20 = 30 - 7 + 3.

I just didn't write the brackets.

So if Alex is right, these two numerical equations are true.

Are they? Pause, you'll need just 20 to 30 seconds to figure that out.

I'll see you in a moment.

Welcome back.

Did you spot the truth or not about what's going on here? So, our first equation is an equation, 20 - 20 = 30 - bracket 7 + 3.

That's the same as 20 = 30 - 10.

That's equal to 20.

That's an equation, it's true.

But on the right hand side, that is not an equation.

30 - 7 + 3 is 26, which is not equal to 20.

So if that's not an equation, then Alex does not have the mathematical truth when he says that a = 10 - bracket b + 5 is the same as a = 10 - b + 5.

Brackets matter.

So what could we have done better? Alex realises, "Of course, the brackets matter.

I need to subtract both terms inside the bracket." 20 = 30 - bracket 7 + 3, we need to subtract both terms so that becomes 20 = 30 - 7 - 3.

When we do that, we get a mathematical truth.

We still have an equation.

Quick check you've got that now.

I'm going to do an example and then ask you to do a fairly similar one.

So, in my example I'm being asked to group the y and integer terms on the left hand side, subtract from both sides and simplify where possible.

Let's start by grouping the y and integer terms. I've grouped my 7y + 27.

I'll subtract it from both sides, and then simplify where possible.

Of course I've got two integer terms on the right hand side of my equation.

They're like terms, they can be simplified.

But I need to remove those brackets first.

I need to remember that it's 85 - bracket 7y + 27.

I need to subtract both terms inside the bracket.

So the right hand side of my equation now reads 85 - 7y - 27.

I'm subtracting both those terms. Simplify that, positive 85 negative 27 gives us positive 58.

So after simplification, I have the equation 3x = 58 - 7y.

Your turn now.

I'd like to group the e and integer terms on the left hand side, subtract from both sides and simplify where possible.

Your equation is 4d + 5e + 13 = 40.

Pause.

Give this a go.

Your working out should look a lot like mine.

See you in a moment for the answers.

Welcome back.

Let's see how we did.

Grouping the e and integer terms would look like that.

Subtracting as one term from both sides would look like that.

And then 40 - 5e - 13, we're subtracting both of the terms inside the bracket.

Simplify 40 - 13 to 27, and you're left with the equation 4d = 27 - 5e.

Alex is again practising this skill in class, and we get to another slightly trickier version.

What if there is a negative term on the left hand side? For example, 2x + 17z - 12 = 9y.

Can we still group? There's more than one method to doing this.

We could add 12 to both sides, subtract 17z from both sides, and we've isolated the 2x on the left hand side.

But we could group, group the 17z and the negative 12 and then subtract it from both sides.

If we then remove that bracket, we need to subtract 17z and subtract negative 12.

When you subtract negative 12, you end up with positive 12.

Both methods are fine here, but interestingly the one on the right hand side left the terms in order.

If we had the expression 9y negative 17z positive 12, you'd have the y term first, the z term second and the integer term last.

Quick check you've got that now.

Which of these are successful rearrangements of this equation? 7e + 2f - 3 = 9y.

Pause and decide which ones are successful rearrangements and which one are not.

See you in a moment.

Welcome back.

Let's see how we did.

I hope you said A is a successful rearrangement.

We've added positive 3 to both sides.

I hope you said B is not a successful rearrangement.

I hope you said C is a successful rearrangement, treating 2f and negative 3 as one term, and subtracting from both sides.

And then D was also a successful rearrangement.

It was in fact the correct expansion of the right hand side in the equation above in part C.

Practise time now.

Question one, I'd like you to fill in the blanks in these rearrangements.

Pause and try that now.

Question two.

Alex has done really well in this lesson, but he has made a few errors, which are perfectly permitted in mathematics.

We're all going to make them.

What I'd like you to do here is mark Alex's work on rearranging, and I'd like you to write a sentence to explain any errors you spot.

You might not spot any errors in some of his work, but where you do spot errors, can you write Alex's sentence to explain what he should do differently? Pause and give this a go now.

Feedback time, filling in the blanks in these rearrangements.

For part A, we should have got in the missing space there, k + 8, grouping the k + 8 to the left hand side and subtracting from both sides.

For part B, we have grouped and subtracted j + 8.

For part C, we have grouped and subtracted 8y + 9z.

For part D, we have grouped and subtracted 7x + 8y.

For part E, we've grouped and subtracted 7y cubed and negative 4.

Importantly, that's negative 4.

In part F, we have grouped and subtracted 4 and negative 7y cubed.

For question two, part A was absolutely fabulous, grouping 3y and z and subtracting from both sides.

Part B, there's an error, negative y squared and positive 1 need to be grouped before subtracting.

It should have looked like that, with a 1 and a negative y squared in that group.

Question two part C, I hope you said the first step is correct.

Alex has grouped 3y and 5 and subtracted from both sides.

But the next step had an error, and if there's an error there, that error will be compounded the further we go in our working out.

What was the error? Alex needed to subtract both terms inside the brackets.

So, 7 - bracket 3y + 5 should have become 7 - 3y - 5.

From there, you could simplify.

7 - 5 is 2, leaving us the equation x = 2 - 3y.

That would've been the correct version.

What do you think of Alex's work for part D? I hope you said super, delightful, wonderful.

Well done, Alex.

Sadly, that's the end of our lesson now.

But we've learnt that additive relationships between variables can be rewritten in a number of different ways.

Hope you've enjoyed this lesson as much as I have, and I should look forward to seeing you again soon for more mathematics.

Goodbye for now.