Lesson video

Hello, Mr. Robson here.

Welcome to maths, great choice to join me today.

We're looking at multiplicative relationships between terms. This is a wonderful bit of algebra because you're going to use it in so many places in mathematics.

I'm looking forward to this one, so let's get started.

Our learning outcome is that we'll be able to appreciate that a multiplicative relationship between variables can be written in a number of different ways.

Some important keywords we'll be using throughout this lesson.

Equation.

You're gonna see a lot of those.

A reminder that an equation is used to show two expressions that are equal to each other.

Secondly, the word product, which I know you would've heard before in your learning on number in algebra.

Product.

A product in algebra is a result of two or more terms multiplied together.

You're gonna see algebraic terms multiplied and I'm going to refer to the result as the product.

Two parts of today's lesson and we're gonna begin by looking at the structure of multiplicative relationships.

This is a visual representation of four times three equals 12.

What other facts does it show us? You might have heard this called a fact family.

Can you tell me the related facts to four times three equals 12? Pause, have a conversation with a person next to you or a good think to yourself.

See you in a moment.

Welcome back.

I hope you said, it also shows us that three times four equals 12.

This is an example of the commutative of multiplication.

Four times three and three times four are the same thing.

They can commute positions or swap places if you will.

There's also related division facts.

This image shows us that if we start with 12 squares and we divide them into three rows, there's four squares in each row.

The inverse of times three is divide three is another way you can think of that.

Four multiplied by three gets us to 12.

To get back from 12 to four, we have to do the inverse, divide by three.

The same's true for the second division fact, 12 divided by four equals three.

The inverse of times four is divide four.

You might have heard this called a fact family.

You could also call it the family of rearrangements of the multiplicative relationship, four times three equals 12.

When we're learning algebra, you are more likely to see these referred to as rearrangements.

Additionally, in algebra, you won't see division facts written like this.

You will see them written like that.

12 divided by three, written as 12 over three, using a fraction bar, and 12 divided by four, written as 12 over four.

We use this generalisation an awful lot.

A divided by b will be written as a over b.

Quick check you've got this so far.

I'd like you to complete this family of rearrangements for the multiplicative relationship you see here in that diagram.

What multiplication is represented there and what are the rearrangements you can write of that fact? Pause.

Scribble this down and I'll see you in a moment.

Welcome back.

I hope you spotted.

That's a visual representation of five times two equals 10, and you know you can commute the five times two to read two times five, which also equals 10.

The division facts, I hope you wrote using a fraction bar rather than 10 divided by two equals five, 10 over two equals five, and 10 over five equals two.

Let's consider the structure of this family.

When we multiply two numbers together, we can say factor multiplied by factor equals product.

We might call that factor one times factor two equals the product, which then gives us these rearrangements.

A product divided by factor two will leave us with factor one and a product divided by factor one will leave us with factor two.

This allows us to generalise a rule.

If I tell you a times b equals c is a mathematical truth, what three other relationships can you write? What do you think our generalised family looks like for this multiplicative relationship? Pause.

Copy those down and fill in those blanks.

See you in a moment.

Welcome back.

I hope you said we can commute the left hand side its multiplication.

a times b is the same as b times a, so b times a equals c is our first rearrangement.

c divided by b equals a.

That's the product c being divided by the second factor to leave the first factor.

Similarly, the product c divided by the first factor to leave us with the second factor, i.

e.

, c over a equals b.

A generalisation for rearranging multiplicative relationships.

That's that family of facts there, a really useful generalisation.

a, b and c can take the form of any term, be it a number or an algebraic variable.

This is useful in many contexts and whether you're aware or not, you've already seen this in many contexts, one of which is certainly in the area of a rectangle.

You know that you find the area of a rectangle by multiplying the base by the height, which we write more simply as a equals b times h.

Write a family of rearrangements for that.

If we can rearrange a times b equals c, then we can rearrange b times h equals a.

You should get four arrangements.

Pause and write those down now.

Welcome back.

Did you write, if area equals b times h, area equals h times b? Did you write a over h equals b, a over b equals h? I hope so.

So where might we use this family of rearrangements? You would've seen questions like this in your learning.

The base of a rectangle is 2.

4 centimetres, the area is 16.

8 centimetres squared.

Find the height.

Most people would write down a equals b times h and then substitute 16.

8 for the area and 2.

4 for the base and then if 16.

8 is made up of 2.

4 times h, 16.

8 divided by 2.

4 must tell us what h is and we find that h equals seven, but there were an awful lot of steps to that mathematics.

Watch.

We could have made it a little more simple.

We were asked to find the height.

We have an arrangement of this multiplicative relationship where the height is isolated, so let's use that.

If the height equals a over b, then we can just substitute 16.

8 over 2.

4 and immediately we can see that the height must be seven.

We mathematicians like efficiency.

Rearrangements like this can help us with our efficiency.

Quick check you've got that.

I'd like you to complete the family of rearrangements for this multiplicative relationship, another example from geometry.

Circumference equals pi times diameter.

Can you give me the four rearrangements of that? Pause, write those down now.

Welcome back.

I hope you wrote C equals d multiplied by pi.

The commutativity of multiplication.

And then C over d equals pi.

C over pi equals d.

Notice that in both division rearrangements, C, the numerator, was the product of the original multiplicative relationship.

Sometimes we'll need to consider the product of two terms as a term when rearranging.

If I looked at the circumference of a circle in a slightly different way, I could call it two times pi times radius, because two lots of the radius makes the diameter, so we can write this formula in this way.

If I wanted to rearrange, I could divide both sides by pi to leave myself with C divided by pi equals two r.

I could further divide both sides by two and leave myself with C divided by pi divided by two equals r.

I've rearranged to isolate the r.

We'll write C divided by pi divided by two on the left hand side using a fraction bar.

C over two pi equals r.

Could we have done it quicker? Absolutely.

We could think of C equals two pi times r as C equals r times two pi, and then divide through by both terms at once.

Divide both sides by two pi and we immediately get to C divided by two pi equals r, which you know we'd write as C over two pi equals r.

Look how much more efficient we were on the right hand side.

Quick check you've got that.

Which other rearrangements belong in this family? The family C equals two pi r, for circumference of a circle.

I've just showed you that C over two pi equals r is in that family.

Which of these three will also be there? r over two pi equals C, C over two r equals pi, or two over C equals pi r? See if you can find the correct arrangement.

Pause, do that now.

Welcome back.

Let's see how we got on.

It was not option a.

It's was option b, C divided by two r leaves us with pi or C over two r equals pi.

C was not an option.

How did we get to the right answer? We started with C equals two pi r, separated the right hand side to read pi times two r, and then divided both sides by two r.

That's how we got to C over two r equals pi.

Notice again that in both correct rearrangements, C, the numerator, was the product of the original multiplicative relationship.

Practise time now.

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

Pause and do that now.

For question two, the volume of a cuboid can be calculated as length times width times height.

Fill in the missing parts of these arrangements for that formula.

Pause and do that now.

Let's see how we got on.

Question one, writing a family of rearrangements for four equations.

Part a.

I hope you started by commuting the multiplication to get seven times 12 equals 84.

Therefore, 84 over seven equals 12 and 84 over 12 equals seven.

For part b, commute 2.

7 times 9.

3 equals 0.

81.

Therefore, 0.

81, the product, over 2.

7 equals 0.

3, and 0.

81, the product, over 0.

3 will leave us with 2.

7.

Question c.

Include an algebraic variable, but nothing really changes.

It's the same structure, this multiplicative relationship.

We can commute the y and the seven so that it reads y times seven equals three.

We can use the inverse to write division facts.

The product three over y leaves seven, three over seven leaves y.

Did you notice that fourth arrangement is the solution to the equation? If you are faced with seven y equals three, you can rearrange it to three over seven equals y and you've solved it.

For part d, d equals s times t.

I bet you've seen that one in school before.

We can commute the s and the t, so d equals t times s.

If d is the product, then d over t equals s, and d over s equals t.

Question two, filling in the missing parts, these arrangements for the formula for volume of a cuboid.

V over something leaves us with lw.

That'll be V over h.

V over w leaves us with something.

That will be lh.

V over something leaves us with h, that'll be lw.

And V over wh leaves us with l.

V, the product, the original multiplicative relationship, remained the numerator in all of those division rearrangements.

Onto part two of the lesson now, where we're going to be rearranging multiplicative relationships.

Multiplicative relationships are made up of factor one times factor two equals product.

We rearrange by first commuting.

If six eights a 48, eight six is a 48.

Then by doing the inverse of multiplication.

If six eights make 48, divide both sides by eight, and we get six equals 48 over eight and we can do the same for eight sixes make 48.

Divide both sides by six and we get eight equals 48 over six and of course, we'd write those division facts with a fraction bar.

Notice that the product is anchored in the numerator position.

It's not the first time we've seen that in this lesson.

48's the product of six and eight and it's anchored in the numerator position.

Did you notice that the factors change position? We can have 48 over eight being equal to six or 48 over six being equal to eight.

The six and the eight, the factors, can swap positions.

Sometimes we have to rearrange from product over factor, so rearranging 10 over two equals five.

Many ways to consider this.

10 over two equals five, we could write as 10 divided by two equals five and then do the inverse, multiply both sides by two and we'll end up with 10 equals five multiplied by two.

Alternatively, we could think of this as half of 10 is five, which we could write as half of 10 is five and then we'll multiply by the reciprocal of a half, the reciprocal of a half being two, and we'll get 10 equals five times two.

It's important that you are fluent in this and you can understand both methods there.

Once we've got 10 equals five times two, we can write the other multiplication facts.

10 must be two lots of five, and then the other division facts, 10 over five will leave us with two.

And we end up with all four rearrangements of that multiplicative relationship.

Rearranging from product over factor with algebraic variables in the equation is no different.

If we started with d over t equals s, we can consider it in the same way we consider 10 over two equals five.

d over t, we could write as d multiplied by one over t.

From there, we'll multiply by the reciprocal of one over t, which is t.

Once we've done that, the left hand side simplifies to just a d and the right hand side reads s multiplied by t.

Notice the algebra is behaving in a very similar way to the numerical relationship that we saw.

Once we've got d equals s times t, then it must also be true that d equals t times s and we can write the other division facts, d over s equals t.

Again, the product is anchored in the numerator position.

The factors can change positions.

Quick check you've got that.

Which of these is a correct manipulation of this equation? d equals m over V.

One of those three is the same thing, which is it? Pause, tell the person next to you, have a good think to yourself.

I'll see you in a few seconds for the answer.

Welcome back.

It was not option a.

If we multiplied the right hand side of option a, we'd get V over m, which is the reciprocal of m over V.

It's not the same as m over V.

B was the correct option.

If you multiplied the right hand side, m, multiplied by one over V, we'd get m over V, matching our original equation.

Option c was not correct.

m multiplied by m over V would give us m squared over V.

Again, not the same as our original equation.

Rearranging from product over factor with multiple terms grouped as one term is no different.

Again, if we started with the arrangement V over pi r squared equals h, we could think of that as V multiplied by one over pi r squared.

From there, we'll multiply by the reciprocal.

The reciprocal of one over pi r squared is pi r squared.

That'll leave us with just V on the left hand side and h multiplied by pi r squared on the right hand side, simplifying to pi r squared h, and there we are.

V equals pi r squared h, our formula for volume of a cylinder.

Quick check you've got that now.

Which of these is the first step to rearranging this equation? A tricky looking one.

Seven over x multiplied by y cubed equals five.

Which of those three options is the first step to rearranging? Pause, tell the person next to you, have a good think to yourself.

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

Welcome back.

It was option a, seven multiplied by one over x, y cubed equals five.

It was not option b, it was not option c.

Now that we know the first step, what do you think the next step is to rearranging this equation? We've separated the left hand side so that now reads seven multiplied by one over x, y cubed.

Is the next step, to divide both sides by x, y cubed, or multiply both sides by x, y cubed, or multiply x y cubed on the left hand side only? Pause, tell the person next to you, have a good think to yourself.

See you in a moment for the answer.

Welcome back.

I hope you said it's not option a.

We won't be dividing by x y cubed on both sides.

It is option b, we will be multiplying by x y cubed.

It's not option c.

Performing the operation to the left hand side only would lose equality.

We'd no longer have an equation.

We have to do the same thing to both sides.

So why was it b? It's because we're multiplying both sides by the reciprocal of one over x y cubed, that being, to multiply by x y cubed.

So if we do that and we multiply both sides by x y cubed, what will our final rearrangement of this equation look like? Will it be five equals seven x y cubed, seven equals five times x y cubed, or seven equals five x y cubed? Which do you think it is? Pause, tell the person next to you or tell me aloud to the screen.

Welcome back.

I hope you said it is not option a, it's not option b.

It's absolutely option c, seven equals five x y cubed.

If you'd said option b, you weren't mathematically incorrect.

It's just that option's not fully simplified.

On the right hand side there, we'd write five x y cubed instead of five multiplied by x y cubed.

So that's our final rearrangement.

Seven over x y cubed equals five, becoming seven equals five x y cubed.

When you become really confident at this, you could do it in one quick step.

Multiply both sides by x y cubed and go straight from that first arrangement to the arrangement seven equals five x y cubed.

Practise time now.

I'd like you to write the family of rearrangements for the below equations.

I'd like four facts for each equation and you'll notice I've given you in them in the form product over factor equals factor.

So rearrange those, pause and do that now.

Question two, Lucas is rearranging equations and he says I just swap positions in the equation.

For a, Lucas says 72 over nine equals eight can swap to 72 over eight equals nine.

For b, Lucas says five over h equals b can swap to five over b equals h.

And for c, Lucas says 13 over x equals two, two over x equals 13.

In each case, decide if Lucas is correct and write a sentence to justify your decision.

Pause and do that now.

Question three.

I'd like you to rearrange each equation into the form, a multiplied by b equals c, i.

e.

, I've given you a division fact.

In part a, x over eight equals y.

We could think of that as x divided by eight equals y.

Can you rearrange it, so I've got a multiplicative fact? Pause and do that now.

Feedback time now.

Question one.

I asked you to write the family rearrangements for this group of equations.

For part a, 95 over five equals 19 and that must come from the multiplicative fact, 19 fives make 95, which we could commute to five nineteens equals 95 and there's our other division facts.

95 over 19 equals five.

For part b, 0.

82 over 0.

2 equals 4.

1.

The product 0.

82 can remain in that numerator position and we can just switch around those two factors.

So our first arrangement, we could write 0.

82 over 4.

1 equals 0.

2.

The multiplication that came from must be 0.

2 times 4.

1 equals 0.

82, which of course we can commute.

For part c, and add your break variable in this equation.

If 56 over y equals seven, then 56 over seven equals y.

That must have come from the multiplicative fact.

Seven lots of y makes 56 or y lots of seven makes 56.

For part d, P over F equals A.

Same, P can remain in the product's position and the A and F can swap positions.

If P over F equals A, then P over A equals F.

That came from the multiplication, P equals F times A and P equals A times F.

Question two, Lucas is rearranging equations and I said in each case, decide if Lucas is correct and write a sentence to justify your decision.

For part a, Lucas is absolutely correct.

Both are rearrangements of eight times nine equals 72 where 72 is the product.

It's in the numerator position.

The nine and the eight can swap positions.

For part b, he was also correct.

Both the rearrangements of b multiplied by h equals five where five is the product, the product is sat in the enumerator position.

The b and h can switch positions.

Part c though, there was an error.

One is a rearrangement of two x equals 13 and the other is a rearrangement of 13 x equals two.

You may have written some advice for Lucas.

You might have said it's important to rearrange equations correctly.

It's not just about swapping terms around.

It's important to understand the structure a little further.

For question three, I asked you to rearrange each equation into the form a times b equals c.

For the first one, x over eight, we could think of x multiplied by one over eight.

The reciprocal of one over eight is eight, so we multiply both sides by eight.

That'll get us to x equals eight y.

For part b, we'll multiply by the reciprocal of one over three y.

Both sides multiplied by three y.

That gives us x on the left hand side and three y squared on the right hand side.

Do we need to write this level of method every time? Not really.

What reciprocal are we gonna multiply both sides by for part c? It'll be multiplied by three yz.

If we do that to both sides, we're left with five x on the left hand side and we get 12 yz cubed on the right hand side.

That's the end of the lesson now sadly, but in summary, multiplicative relationships between variables can be written in a number of different ways.

For example, from the equation 15 over c equals d, we can find a family of rearrangements.

15 over c equals d is born of the multiplicative fact, 15 equals c times d, which we can commute to 15 equals d times c, which could also be written as 15 over d equals c.

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

Goodbye for now.