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Hi everyone, my name is Ms. Koo 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 checking and securing understanding of the distributive law.

And by the end of the lesson you'll be able to state the distributive law and use it to calculate efficiently.

So let's have a look at some keywords starting with the distributive law.

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

So let's have a look at example.

We know 12 multiplied by the four add the three in brackets, is the same as 12 multiplied by seven, which is 84.

But now what I'm gonna do is show the distributive law.

The distributive law states if you're multiplying 12 by the four add the three, it's the same as 12 multiplied by the four add the 12 multiplied by the three, which is 84.

Notice how both of our answers are the same.

Another example would be 12 multiplied by four, subtract three, well we know this is the same as 12 multiplied by four, add the negative three and we know this is the same as 12 multiplied by one, which is 12, but I'm gonna use the distributive law now.

12 multiplied by our four is 48.

Add the 12 multiplied by the negative three is negative 36.

Summing them together still makes our 12.

These are two nice examples of the distributive law.

We'll also be looking at another key word, common factor.

Just to remind you, a common factor is a number which is a factor of two or more numbers.

For example, if we were asked to identify common factors of 12, 16 and 32.

Two is a common factor because two times six is 12, two times eight is 16 and two times 16 is 32.

Four is also a common factor because four times three is 12, four times four is 16 and four times eight is 32.

So you can see how we've formed our common factors.

We'll be looking at these keywords throughout our lesson.

Our lesson today will consist of two parts.

The first part, we'll be looking at the distributive law and common factors, and the second part we'll be looking at efficiently using the distributive law.

So let's have a look at the distributive law and common factors.

Well, remember the distributive law says that multiplying a sum is the same as multiplying each addend and then summing the result.

And the distributive law allows us to write an equation in lots of different ways, but still equating to the same answer.

The great thing is some of these equations are easier to calculate than others.

So let's inspect using arrays.

Here I'm going to be using a two by eight array.

So you can see we have 16 dots.

I also want to show you the same 16 dots, but how I've split it into a two by three, add a two by five, still making our 16 dots.

Here we have a two by three and we're summing up the two by five.

Using the distributive law, it's the same as two multiplied by three add five.

In other words, we have two rows of three add five.

This is a nice way to visually see the distributive law because it shows those two rows of three times five is exactly the same as a two by three, adds that two by five.

So let's have a look at another array, and what I want you to do is see if you can identify another example using the distributive law of the four by three array.

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

Great work.

Just to let you know, there's lots of examples out there.

And here is one.

And you can check by ensuring the number of dots is the same.

So I'm going to use the four by three array and I'm going to make a four by two and a four by one.

And you can still see we still have those 12 dots and this is exactly the same as four, multiplied by two, add one.

In other words, we have four rows of the two add one.

Really nice examples showing the distributive law that four multiplied by two, add one four rows of our two add one, is the same as a four by two, add a four by one.

So now what we're going to do is we're going to use numbers and an area model to show you how the distributive law works.

Looking at a seven by 13.

So you can see I've illustrated it here.

Now this is the same as seven multiplied by 10 add three.

So you can see how I've split that 13 into 10 and three.

Well we know this is the same as seven multiplied by 10 add three, whereby we're multiplying our seven by our ten first and then summing the seven by the three, thus giving us 70 add our 21, which is our final answer of 91.

Another example could be three times two, add three times nine.

Well I'm going to show you with this area model, we know three times two is six and we're gonna add it to three times nine, which is 27.

And we can see we have a common factor of three.

So that means what I can do is I can combine both of those areas together to make three times the two add the nine, which then gives us three times 11, which is 33.

These are really nice ways to demonstrate how the distributive law works.

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

You're asked to fill in the blanks using your knowledge on the distributive law.

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

Great work.

So let's see how got on.

13 multiplied by the seven add the nine, subtract two is the same as 13 multiplied by seven add or 13 multiplied by nine add or 13 multiplied by that negative two.

The next stage of working out is 91, add 117.

Subtract our 26.

Gives me a final answer of 182.

Well done.

Have you got that one right? For b, five times something add 3.

2 times three is the same as 3.

2 times five, add something.

Well hopefully you've spotted, it had to be 3.

2 because we have a common factor of 3.

2 in this calculation.

So that means 3.

2 times the five, add the three.

I can work out the five, add the three to be the eight, thus giving me a final answer of 25.

This was a really good question to show the distributive law.

Your task; here, you need to fill in the blanks with your knowledge of the distributive law.

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

So let's move on to question two.

Question two says there are ten 1.

3s in each of these calculations.

In other words, the answer to each calculation is 13.

You've got to use the information to find the 1.

3s and explain why the answer is 13.

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

Great work.

So let's move on to our answers.

So for question one, we had to fill in the gaps.

You great thing is you can spot, we have a common multiplier of 1.

So that means moving onto the second line, you can see 1.

2 times 11 and nine is using the distributive law.

So the first line must be 11 and or nine.

Summing those 11 and nine makes 20.

So 1.

2 times 20 gives us 24.

This calculation is much easier to calculate than 11 times 1.

2 add nine times 1.

For b, we have nine multiplied by eight, subtract four, subtract three, well you can spot we have the common multiplier of nine.

So that means using the distributor flow, nine times eight, subtract nine times four, subtract nine times three.

This gives us nine times eight is 72, nine times four is 36, nine times three is 27.

For c, 9.

9 multiplied by 1.

1 add 0.

9 multiplied by 9.

We have a common multiplier of 9.

So that means using the distributive law, it's 9.

9 multiplied by the sum of 1.

1 and 0.

9, much easier to calculate.

9 times two is 19.

And for d, this was a tough one.

You had to really had a look at the whole calculation to identify those missing numbers.

Really well done if you've got D.

Now let's have a look at question two.

Question two states we know the answer is 13.

So can you find out how we use the distributive law? Well hopefully you've spotted for a, we have a common multiplier of 1.

3, so it's the same as 1.

3 times the summation of 5.

2 add 4.

So that's the same as 1.

3 times 10, which is 13.

For b, this was a great question.

It's spotting that 3.

9 times one is the same as 1.

3 times three.

Then we have another 1.

3 and another 1.

3, thus making our ten 1.

3s, which is 13.

For c, this was a tricky one again.

You had to spot that three multiplied by the 0.

9 and the three multiplied by the 0.

4, is the same as three lots of that 0.

9 add the 0.

So it's three lots of the 1.

Then you can see group together those three lots of 1.

3 subtract 0.

5 times 1.

3 gives us 2.

5 multiplied by 1.

3, then multiplied by the four.

That gives us our 10 multiplied by 1.

3, which is 13.

C was a really tricky one, so well done if you got that one right.

Great work so far everybody.

Now we'll be moving on to efficiently using the distributive law.

Now being able to solve problems in many different ways helps us develop the skill of making sensible choices based on the numbers involved and the relationships between them.

We'll be looking where the distributive law can make calculations more efficient.

For example, five multiplied by 0.

75, add five multiplied by 0.

25.

Alex states that the answer to the calculation is five as we are secretly multiplying by one.

What do you think Alex has noticed? Well, hopefully you've spotted we have a common factor of five in our calculation.

So that means summing up the 0.

75 and the 0.

25 gives us one.

And this is an efficient method where there's no need to calculate anything and you can spot, it's simply five multiplied by one.

If we were to show this using the distributive law, you can see it's five multiplied by 0.

75.

Add on 0.

25, which is five times one.

Now using this, let's see if you can fill the banks so you secretly are multiplying by one.

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

Great work.

So let's see how you got on.

14 times 0.

3 add 14 times what? Is the same as 14 times one? Well it had to be 0.

For b, 34.

5 times 0.

2 at 34.

5 times what? At 34.

5 times 0.

2 is the same as 34.

5 times one.

What's 0.

6? Next we have 462 times 0.

12 add 0.

33 add what, is the same as 462 times one.

Well that's 0.

55.

Great work if you got that one right.

So now what I'm going to do is show you how we can write a calculation in lots of different ways, and then we can identify which calculation would we prefer to calculate.

So here you can see how I've used the distributive law to show the calculation 13 times 99.

Now there are some which are correct and some which are not correct.

See if you can identify which is correct and outta the correct ones, which one would you prefer to calculate and why? See if you can give it a go and press pause if you need.

Great work.

So hopefully you've spotted 13 times 99 is the same as 13 times a hundred, subtract 13 times one.

13 times 99 is the same as 13 times 90, add 13 times nine, and 13 times 99 is the same as 15 times 99 subtract two times 99.

Now which one would you like to work out to work out our answer? Well, for me it's easier to do 13 times a hundred and then subtract 13 times one just because it's easy to multiplied by multiples of 10 and one.

So great work if you work this one out.

Now let's have a look at a check question.

What I want you to do is use the distributive law to find the missing gaps and then choose the most efficient calculation to work out the answer.

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

Great work.

So let's see how you did.

Our calculation is 21 multiplied by 998.

It's the same as 21 multiplied by 900 and 21 multiplied by 90 and 21 multiplied by eight.

It's also the same as 998 times 20 add 998 times one.

It's also the same as 21 times a thousand.

Subtract 21 times two and it's also the same as 25 times 998.

Subtract four times 998.

All these calculations equate to the same answer, but which one do you think is the most efficient? Well for me it's going to be 21 multiplied by a thousand.

Subtract 21 multiplied by two.

This is because multiplied by multiples of 10 is easy and same as multiplying by two.

So working this out, 21 times a thousand is 21,000.

21 times two is equal to 42.

Subtract gives us 20,958.

Great work if you got that one right.

Now let's move on to your practise task.

Here we have some calculations and some are secretly multiplied by one.

Which ones do you think are secretly multiplying by one? And for those that aren't multiplied by one, can you use the distributive law and change the calculation to make it secretly multiplied by one? See if you can give this a go and press pause if you need.

Great work.

So let's move on to question two.

Question two says, using the distributive law, write three different calculations and then choose the most efficient one to work out the answer.

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

Great work.

So let's have a look at question one.

Here are some calculations which are secretly multiplied by one, but which ones are they? Well, the first one, 0.

69 times two, add two, multiplied by 0.

31, using the distributive law, you can see we have a common multiplier of two.

So it's the same as two multiplied by 0.

69, add on 0.

31, which is the same as two times one.

Hopefully you've also spotted this one.

Seven multiplied by 1.

2, take away 0.

5 multiplied by seven, add seven times on 0.

Hopefully you spotted we have a common multiplier of seven, so it's the same as seven multiplied by the 1.

Subtract on 0.

5, add on 0.

3, which is seven.

Lastly, we also have six multiplied by 0.

4, add six multiplied by 0.

Our common multiplier is six.

So it's six times our 0.

4 add on 0.

2, which is six times one.

Fantastic work if you've got that one.

Now for part B, we had to use the distributive law to change those calculations which didn't secretly multiply by one.

So I've just highlighted those calculations which did not secretly multiply by one.

So let's see what we need to do to change them, so they do secretly multiply by one.

Well six multiplied by 0.

75, add six multiplied by 0.

35 did not give us a secret multiplication of one.

Using the distributive law, it's six multiplied by 0.

75, add our 0.

35, which is six multiplied by 1.

So if we adjust the 0.

7 maybe to 0.

65 or the 0.

35 to the 0.

25, this will make a multiplication of six by one.

Now let's have a look at the other calculation.

While nine multiplied by 0.

85, subtract nine multiplied by 0.

15 is the same as nine multiplied by 0.

So by simply adjusting and amending it to addition, we would've got nine multiplied by one.

Now for the last one, it's four multiplied by 0.

5, add four, multiplied by 2.

5 is the same as four multiplied by 0.

5, add on 0.

25, which is the same as four multiplied by 0.

75.

And all we needed to do was adjust that 0.

25 to 0.

5 or the 0.

5 to 0.

75.

This was a great question whereby you had to adjust the calculation to make it secretly multiply by one.

Now let's have a look at question two.

We are asked to write three different calculations and then choose the most efficient one to work out the answer to 19 times 990.

I've just chosen three examples here.

19 times 900 at 19 times 90, 990 times 10 at 990 times nine and 19 times 1000, subtract 19 times 10.

Now you can see I've highlighted this calculation as the most efficient way because we've got multiplication of multiples of 10.

19 times 1000 is 19,000, 19 times 10 is 190.

Subtracting them gives us 18,810, a huge well done if you've got that one right.

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

There are an infinite number of ways to rewrite an equation using the distributive law, but it is important to choose the calculation which aid efficiently and ease.

Great work today.

It was wonderful working with you.

Well done.

I've finished the video