Lesson video

Hi, everyone, my name is Miss Ku, 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 results.

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 three and four is the same as four and three, and two times six is the same as six times two.

Associative examples would be two add three using our brackets, add four is equal to two add three in our brackets.

Add four, and it applies to multiplication as well.

Two multiplied in our bracket, six times four, is exactly the same as four multiplied by, in bracket, six times two.

Remember, subtraction and division are not commutative or associative.

Some non-examples of commutative would be 10 subtract two is not the same as two subtract 10, or eight divided by four is not the same as four divided by eight.

Non-examples of the associative law would be 10 subtract grouping together two subtract one is not the same as 10 subtract two in bracket subtract one, and looking at division 24 divided by the eight divided by four in brackets is not the same as 24 divided by eight in brackets divided by four.

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 three is the same as 10 multiply by our 12, which is 120, and our 10 multiplied by three, which is 30.

We know this answer to be 150, but incorporating the associative law with addition, we could also rewrite this calculation, 10 multiplied by, in our brackets, 10 add three can written as 10 multiplied by 10 add two and three.

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

10 multiplied by 10 is 100, 10 multiplied by two is 20 and 10 multiplied by three 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 three using the distributive and associative 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 two and the three, or you could do 10 multiplied by 15, add zero, or you could do 10 multiplied by 17, subtract two, or you could do 10 multiplied by nine, add one, add two, add three.

So there's an infinite number of solutions here.

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

For example, eight multiplied by 14, add eight multiplied by six.

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

Then we can work this out to be eight 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 eight and break it into four multiplied by two.

So I have four multiplied by two times 14, add four multiplied by two, add six.

So you might notice how I've changed that eight into four multiplied by two.

This then becomes four multiplied by 28, add four multiplied by 12, which I can then work out to be four 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, eight multiplied by 14 and eight times six, and common factors and a combination of our laws to rewrite this calculation differently.

So 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 eight times 14, add eight times six, so I'm going to choose eight multiplied by two multiplied by seven, and eight multiplied by two multiplied by three.

Now, this then gives us 16 multiplied by seven and 16 multiplied by three.

Taking out 16 as a common factor, I have 16 multiplied by seven and three, 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.

Four multiplied by eight, add 12 multiplied by six, subtract the 10 multiplied by four.

So she's looked at the eight and split it into two multiplied by four.

She's looked at the 12 and split it into four multiplied by three, and she's left the 10 multiplied by four.

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

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

Sophia recognises 120 add 24 is the same as four times brackets 30 add six, and she works this out to be four 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 two multiplied by 14, add eight multiplied by five, add nine multiplied by eight.

What I want you to do is find the area of the shape in centimetres squared, and then I want you to write other calculations that represent the area of this 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 centimetres 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 two add the eight and then multiply 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 five multiplied the eight and the two, add the nine multiplied by the eight and the two.

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

This is another possible way where you could have done 40 multiplied by two, add 4 multiplied by three, add 14 multiplied by five, 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' calculation.

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

So 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 seven multiplied by eight, 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 nine and 25 multiplied by nine, add 30 multiplied by three.

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 five multiplied by 10 add two multiplied by 10, add five multiplied by nine, add two multiplied by nine.

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 nine, and still multiplying by the seven.

So 10 multiplied by seven, add nine multiplied by seven.

Another way could have been splitting the 19 into 10, five, and four, but still multiplying by our seven.

So 10 times seven, add five times seven, add four times seven.

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 seven times eight, add the 12 times 11.

Well, hopefully you've spotted the eight has been split into the four multiplied by two, and the 12 has been split into the three multiplied by four.

Then from here, we have something multiplied by 14.

Well, looking at the seven times four times two, the seven times the two gave the 14, so we just have the multiplication of four.

Next, something multiplied by 33 is the same as three times four times 11.

So hopefully you can spot the three times the 11 gave the 33, so we still have our four.

Now, we have a common factor of four, so we've used the distributive law to show four multiplied by 14, add 33, which gives me an answer of 188.

Great work if you got that one right.

For B, 15 multiplied by nine, add 25 multiplied by nine, add 30 times three.

So the 30 multiplied by three can be split in 10 multiplied by three multiplied by three, then we have 15 times nine, add 25 times nine, add 10 times three times three.

Can we make a common factor of nine? 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. While 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 eight, add 0.

9 times four.

The first question you need to ask yourself is can you spot any common factors that we could use here while looking at the integers 12, eight, and four? We can use the common factor of four for each of these numbers.

So breaking 12 into four times three, eight into four times two, and four into four times one, or just four.

We now have a common factor of four.

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 four, so I'm going to multiply the 4.

5 by the three to give 13.

5, still have a multiplication of four, the 1.

3 by the two to give 2.

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

9 is still multiplied by the four.

Now, I have a common factor of four, so using the distributive law, I have four multiplied by bracket 13, add five, add 2.

6, add 0.

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

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

5 times 12, add 1.

3 times eight, add 0.

9 times four.

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 is 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 the following statements are true or false, and 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 three multiplied by something, so it has to be five.

Then 10 multiplied by 3.

2 is broken into something times two times 3.

2, so it must be five, and 20 times 3.

2 has changed into something times four times 3.

2, so it has to be five.

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

Well, three multiplied by 3.

2 is 9.

6, which is multiplied by the five.

The two times the 3.

2 is our 6.

4, which is multiplied by our five.

And the four times 3.

2 is 12.

8, which again, is multiplied by our five.

And because we have this common factor of five, 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.

Five multiplied by 3.

2, add 3.

2 times three.

We have a common multiplier, 3.

2, so that means using the distributive law, 3.

2 times five add three gives me 3.

2 times eight, which is 25.

6.

On the right-hand side, I have 2.

5 times four times 3.

2, so I'm gonna break that four into times two times two to give me five times two 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 two, two times 17 multiplied by bracket 3.

2, add 1.

8.

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

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

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

Well, we have a common factor of 17, so I could do 17 multiplied by 17, take away seven, 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 five, seven times five, add five times three, add five.

This gives me an answer of 55.

If I chose a different number, let's say eight, 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 seven multiplied by something, add three multiplied by something, add one 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 seven add three, add one, 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 distributive 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.