Lesson video

Hi everyone, my name is Miss Coo, and I'm really happy to be learning with you today.

It's going to be a fun, interesting, and challenging lesson in parts, but don't worry, I am here to help.

You will come across some new keywords and maybe some keywords you've already come across before.

We're going to work really hard today, but I am here to help and we can learn together.

In today's lesson from the unit, Arithmetic Procedures with Integers and Decimals, we'll be looking at using the associative, distributive, and commutative laws together.

And by the end of the lesson you'll be able to use these laws to flexibly and efficiently solve problems. So let's look at some keywords first.

Remember, an operation is commutative if the values it is operating on can be written in either order without changing the calculation.

And the associative law states that a repeated application of the operation produces the same result, regardless of how pairs of values are grouped.

Remember we can group using brackets.

The distributive law says that multiplying a sum is the same as multiplying each addend and summing the result.

So this lesson will be broken into two parts.

The first will be looking at using more than one law, and the second will be looking at efficiently solving problems. So let's have a look at using more than one law.

Well remember, multiplication and addition are commutative and associative.

So commutative examples would be 3 add 4 is the same as 4 add 3, and 2 times 6 is the same as 6 times 2.

Associative examples would be 2 add 3 using our brackets, add 4 is equal to 2 add 4 add 3, in our brackets, and it applies to multiplication as well.

2 multiplied, in our brackets, 6 times 4 is exactly the same as 4 multiplied by, in brackets, 6 times 2.

Remember, subtraction and division are not commutative or associative.

Some non examples of commutativity would be 10 subtract 2 is not the same as 2 subtract 10, or 8 divide by 4 is not the same as 4 divided by 8.

Non examples of the associative law would be 10 subtract, grouping together 2 subtract 1, is not the same as 2 subtract 10 in brackets subtract 1.

And looking at division, 24 divided by the 8 divided by 4 in brackets, is not the same as 24 divided by 8 in brackets, divided by 4.

So the distributive law states that multiplying a number by a group of numbers added together is the same as doing each multiplication separately.

For example, 10 multiplied by, using our brackets, 12 add 3 is the same as 10 multiplied by our 12, which is 120, add our 10 multiplied by 3, which is 30.

We know this answer to be 150.

But incorporating the associative law with addition, we could also rewrite this calculation, 10 multiply by, in our brackets, 10 add 3 can be written as 10 multiply by 10 add 2 add 3.

So you can see how we split that 12 into 10 add 2 using our associative law, then we can work it out.

10 multiplied by 10 is 100, 10 multiplied by 2 is 20, and 10 multiply by 3 is 30, thus giving us exactly the same answer of 150.

Can you find another way to rewrite the calculation of 10 multiplied by bracket 12 add 3 using the distributive and associated law with addition? See if you can give it a go and press pause if you need.

Great work.

Now just to let you know, there's an infinite number of different ways.

From all these different calculations, there are some which are easier to work out than others.

So an example would be, well, 10 multiplied by 10, add the 2 add the 3.

Or you could do 10 multiplied by 15 add zero.

Or you could do 10 multiplied by 17 subtract 2.

Or you could do 10 multiplied by 9 add 1 add 2 add 3.

So there's an infinite number of solutions here.

Here's another example of the distributive law using common factors.

For example, 8 multiplied by 14 add 8 multiply by 6.

Using the distributive law, this is the same as 8 multiplied by 14 add 6, as we have our common factor of 8.

Then we can work this out to be 8 multiplied by 20, which is 160.

Now incorporating the associative law with multiplication, we can also rewrite the calculation.

So I'm gonna look at that factor of 8 and break it into 4 multiply by 2.

So I have 4 multiplied by 2 times 14, add 4 multiply by 2 add 6.

So you might notice how I've changed that 8 into 4 multiplied by 2.

This then becomes 4 multiplied by 28 add 4 multiplied by 12, which I can then work out to be 4 times 40, which is 160, the same answer as before.

So what I want you to do is see if you can use the same calculation; 8 multiplied by 14 add 8 times 6 and common factors and a combination of our laws to rewrite this calculation differently.

See if you can give it a go and press pause if you need.

Just like before, there's an infinite number of different ways and from all these different calculations there are some which are easier to work out than others.

So looking at our calculation of 8 times 14 add 8 times 6.

So I'm going to choose 8 multiplied by 2 multiplied by 7 add 8 multiplied by 2 multiplied by 3.

Now this then gives us 16 multiply by 7 add 16 multiplied by 3.

Taking out 16 as a common factor, I have 16 multiply by 7 add 3, which is the same as 16 times 10, which is 160, the same answer as we got before.

The distributive law helps us rewrite a calculation where some are easier to work out than others.

So let's have a look at a quick check.

Laura and Sofia complete two calculations.

I want you to have a look and see who's correct.

Press pause if you need.

Well done.

So hopefully you spotted they're both correct as there are an infinite number of ways to write a calculation using the distributive law.

Laura's approach is a really nice approach; 4 multiplied by 8 add 12 multiplied by 6, subtract the 10 multiplied by 4.

So she's looked at the 8 and split it into 2 multiply by 4, she's looked at the 12 and split it into 4 multiplied by 3, and she's left the 10 multiply by 4.

The reason for this is because she spotted a common factor of 4.

Then removing that common factor of 4, we have 4 multiplied by, in brackets, the 28 add the 18, subtract 10 which is the same as 4 multiplied by 36, which is equal to 144.

Sofia recognises 120 add 24 is the same as 4 times, brackets, 30 add 6, and she works this out to be 4 times 36, which is 144.

So now let's have a look at another check question.

Here, it gives us an area of a shape can be found by calculating 2 multiplied by 14 add 8 multiplied by 5 add 9 multiplied by 8.

What I want you to do is find the area of the shape in centimetre squared and then I want you to write other calculations that represent the area of the same shape.

See if you can give it a go and press pause if you need.

Great work.

So let's work out the area.

Well, the area is 140 centimetre squared, but did you manage to write other calculations that represent the same area of the shape? Well, there are lots of different ways how you could split this shape to work out the area.

However, it's important to understand how they are formed.

For example, you may have chosen the 2 add the 8 and then multiplied by the 14.

That gives us an area of 140 centimetres squared.

Another way you may have chosen 10 multiplied by 14, which is 140.

Another way would be doing 5 multiplied the 8 add the 2, add the 9 multiplied by the 8 add the 2.

So you can see how it's shaded in, still giving us 140.

This is another possible way where you could've done 40 multiplied by 2 add 40 multiplied by 3 add 14 multiply by 5, still giving us 140.

Now let's move on to your task.

Lucas wants to find the area to this rectangle and he starts by writing this calculation.

See if you can complete Lucas's calculation.

Show how Lucas has split the rectangle, and see how many different ways you can split the rectangle to find the same area.

See if you can give it a go and press pause if you need more time.

Great work.

So let's move on to question two.

Question two wants us to fill in the blanks using your knowledge on the distributive law.

Here for A, we have 7 multiplied by 8 add 12 multiplied by 11.

Can you figure out what numbers goes in those blanks for the rest of the calculations? For B, we have 15 multiplied by 9 add 25 multiplied by 9 add 30 multiplied by 3.

Can you figure out those blanks to complete this working out? See if you can give it a go and press pause if you need more time.

Fantastic work, so let's move on to question three.

Question three asks, how many different ways can you split the calculation? Here, some have been suggested for you.

So we have 28 multiplied by 84.

Can you fill in those missing numbers and suggest a different way to split the calculation using the knowledge of our laws? See if you can give it a go and press pause for more time.

Fantastic work.

Let's go through our answers.

Well, for question one, remember Lucas wants to find the area to this rectangle and he starts with this calculation.

Finishing the calculation off, we would've had 5 multiplied by 10 add 2 multiplied by 10, add 5 multiplied by 9 add 2 multiplied by 9.

In other words, you can see how Lucas has split our rectangle.

But how many different ways can you split the rectangle to find the same area? There are so many different ways, but ensure to check your calculation.

I'm just gonna go through a few.

Here, I'm going to choose splitting our rectangle into 10 and 9 and still multiplying by the 7.

So 10 multiplied by 7 add 9 multiplied by 7.

Another way could have been splitting the 19 into 10, 5, and 4, but still multiplying by a 7.

So 10 times 7 add 5 times 7 add 4 times 7.

There are lots of different ways.

I've just given you two examples.

Now let's have a look at question two.

Question two wants you to use your knowledge on the distributive law to fill in the missing gaps.

So let's see how you got on.

The 7 times 8 add the 12 times 11.

Well, hopefully you've spotted the 8 has been split into the 4 multiplied by 2, and the 12 has been split into the 3 multiplied by 4.

Then from here we have something multiplied by 14.

Well looking at the 7 times 4 times 2, the 7 times the 2 gave the 14, so we just have the multiplication of 4.

Next, something multiplied by 33 is the same as 3 times 4 times 11.

So hopefully you can spot the 3 times the 11 give the 33, so we still have our 4.

Now we have a common factor of 4.

So we've used the distributive law to show 4 multiplied by 14 add 33, which gives me an answer of 188.

Great work if you look that one right.

For B, 15 multiplied by 9 add 25 multiplied by 9 add 30 times 3.

So the 30 multiplied by 3 can be split and 10 multiplied by 3 multiplied by 3.

Then we have 15 times 9 add 25 times 9 add 10 times 3 times 3.

Can we make a common factor of 9? Yes we can.

So therefore, using the distributive law, we can fill in our missing answers.

Great work if you got this one right.

So let's have a look at question three.

There are an infinite number of different ways and the last two are simply common examples.

So do check your answers with a calculator to see if the result is the same as 28 multiplied by 84.

So here are some examples that I've chosen to work these out.

Huge well done if you got this one right.

Fantastic work so far, everybody.

Now we'll move on to efficiently solving problems. Well, using a combination of the laws allows us to rewrite a calculation for efficiency and ease, especially when using decimals.

Identifying common factors can help identify an easier calculation.

So let's have a look at this question; 4.

5 times 12 add 1.

3 times 8 add 0.

9 times 4.

The first question you need to ask yourself is can you spot any common factors that we could use here.

By looking at the integers 12, 8, and 4, we can use the common factor of 4 for each of these numbers.

So breaking 12 into 4 times 3, 8 into 4 times 2, and 4 into 4 times 1 or just 4.

We now have a common factor of 4.

From here, I'm just gonna simply use the commutative law just to order the numbers a little bit more neatly.

Now I'm only interested in the multiplier of 4, so I'm going to multiply the 4.

5 by the 3 to give 13.

5, still have a multiplication of 4.

The 1.

3 by the 2 to give 2.

6, still multiplied by the 4, and the 0.

9 is still multiplied by the 4.

Now I have a common factor of 4.

So using the distributive law, I have 4 multiplied by bracket, 13 add 5 add 2.

6 at 0.

9, giving me 4 times 17, which is a nice 68.

So 4 times 17 is a much nicer calculation rather than working out 4.

5 times 12 add 1.

3 times 8 add 0.

9 times 4.

So let's have a look at a quick check question.

Laura and Andeep each are given the same question.

Here's their working out.

I want you to have a look at this and see if you can identify who's correct.

Great work, well done.

Hopefully you spotted they're both correct Andeep's method is a little bit more efficient and time saving, but both are right, and this is another fantastic quality about mathematics.

There's more than one way to get a solution.

Now let's move on to your task.

All I want you to do is fill in the blanks using your knowledge of the distributive law.

We have decimals and integers in here, so remember to show you're working out and look for those common factors.

See if you can give it a go and press pause if you need.

Great work.

So let's move on to question two.

Question two wants you to identify if the following statements are true or false.

I want you to explain how you know.

This is a great question.

Take your time and press pause when you're ready.

Great work.

So let's move on to question three.

Question three wants you to fill in the three gaps in this calculation with the same single-digit number.

Then I want you to try a different single-digit number in those gaps.

What do you notice? And more importantly, can you explain why this happens? This is a great question, see if you can give it a go.

Well done, great work.

So let's go through our answers.

For question one, you had 15 times 3.

2 add 10 times something, subtract 20 times 3.

2.

Well, from the second line of working out, we have a common multiplier, 3.

2, so therefore 10 must be multiplied by 3.

2.

The second line, 15 has been broken into 3 multiplied by something, so it has to be 5.

Then 10 multiply by 3.

2 is broken into something times 2 times 3.

2, so it must be 5.

And 20 times 3.

2 has changed into something times 4 times 3.

2, so it has to be 5.

Now you can spot we have a common factor of 5, so let's keep that common factor and multiply everything else.

Well, 3 multiply by 3.

2 is 9.

6, which is multiplied by the 5.

The 2 times the 3.

2 is our 6.

4, which is multiplied by our 5, and the 4 times 3.

2 is 12.

8, which again is multiplied by our 5.

And because we have this common factor of 5 using the distributive law, I can work out the final answer to be 16.

Great work if you got this one right.

For question two, this was a great question, and you had to identify if the following statements were true or false.

Well, let's work out the left hand side first.

5 multiplied by 3.

2 add 3.

2 times 3, we have a common multiplier 3.

2.

So that means using the distributive law, 3.

2 times 5 add 3 gives me 3.

2 times 8, which is 25.

6.

On the right hand side, I have 2.

5 times 4 times 3.

2.

So I'm gonna break that 4 into times 2 times 2 to give me 5 times 2 times 3.

2, which is 10 times 3.

2, which is 32.

So this is false as they are not equal.

So for part B, this was a tricky one.

In the brackets you might see we have a common factor of 17.

So I can take that out the brackets and multiply it by the 2, 2 times 17 multiplied by bracket 3.

2 add 1.

8.

This is the same as 34 multiplied by 5, which is 170.

Now for the right hand side, we have 17 squared subtract 7 multiplied by 17.

First of all, we know 17 squared is 17 times 17, subtract 7 times 17.

Well, we have a common factor of 17.

So I could do 17 multiplied by 17 take away 7, which is 17 times 10, which is 170.

So yes, these are equal.

This was a great question.

For question three, you could try any single digit number you wanted.

I'm gonna pick 5, 7 times 5 add 5 times 3 add 5.

This gives me an answer of 55.

If I chose a different number, let's say 8, I would've got an answer of 88.

So let's choose another number.

Well, I'm going to choose two and gives me an answer of 22.

So what do you notice? Well, we always multiply our chosen number by 11, but why are we doing this? Well, the reason why we're doing this is because we're using the commutative and associative laws here.

This is because we can rewrite the calculation as 7 multiply by something, add 3 multiplied by something, add 1 multiplied by something, as we only have a single unknown digit outside of the brackets.

Well, from here then this unknown number is actually our common factor.

So we can use the distributive law.

Our unknown number multiplied by brackets 7 add 3 add 1 is the same as our unknown number multiplied by 11.

So that's why we are always multiplying by 11.

That was a great question.

So, in summary, we can use our knowledge and skills on the commutative, associative, and distributed laws to rewrite calculations.

And rewriting calculations allows us to reflect on the most efficient approach in order to calculate the answer.

Fantastic work today, everybody, well done.

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