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Hi everyone! My name is Ms. Coe.

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.

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 * (4 + 3) is the same as 12 multiply by 7, 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 (4+3), it's the same as 12 multiplied by the 4 add the 12 multiply by 3, which is 84.

Notice how both of our answers are the same.

Another example would be 12 multiply by (4-3).

Well, we know this is the same as 12 * (4 +(-3)) and we know this is the same as 12 multiply by 1, which is 12, but I'm gonna use the distributive law now.

12 multiplied by 4 is 48, add the 12 multiplied by (-3) is -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, 2 is a common factor because 2 times 6 is 12, 2 times 8 is 16 and 2 times 16 is 32.

4 is also a common factor because 4 times 3 is 12, 4 times 4 is 16 and 4 times 8 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 be looking at the distributive law and common factors, and the second part will 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 2 by 8 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 2 by 3 add a 2 by 5, still making our 16 dots.

Here we have a 2 by 3 and we're summing up the 2 by 5.

Using the distributive law, it's the same as 2*(3+5).

In other words, we have two rows of (3+5).

This is a nice way to visually see the distributive law because it shows those two rows of 3*5 is exactly the same as a 2 by 3 add that 2 by 5.

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 4 by 3 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 4 by 3 array and I'm going to make a 4 by 2 and a 4 by 1.

And you can still see we still have those 12 dots.

And this is exactly the same as 4 multiply by 2 by 1.

In other words, we have 4 rows of the (2+1).

Really nice examples showing the distributive law that 4 multiply by 2+1, 4 rows of our 2+1 is the same as a 4 by 2 add a 4 by 1.

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 7 by 13.

So you can see I've illustrated it here.

Now this is the same as 7 multiply by 10+3.

So you can see how I've split that 13 in two, 10 and 3.

Well we know this is the same as 7 *(10+3) whereby we're multiplying our 7 by our 10 first and then summing the 7 by the 3, thus giving us 70+21, which is our final answer of 91.

Another example could be 3 * 2 + 3 * 9.

Well I'm going to show you with this area model.

We know 3 * 2 is 6 and we're gonna add it to 3 * 9 which is 27.

And we can see we have a common factor of 3.

So that means what I can do is I can combine both of those areas together to make 3*(2+9) which then gives us 3*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 bunks 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 you've got on.

13 *(7+9-2) is the same as 13 multiply by 7, add our 13 multiply by 9, add our 13 multiply by that -2.

The next stage of working out is 91 add 117, subtract our 26 gives me a final answer of 182.

Well done if you got that one right.

For B, 5 * something add 3.

2 * 3 is the same as 3.

2 * 5 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 5 add the three.

I can work out the 5 add the 3 to be the 8, 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 2.

Question 2 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 1, we had to fill in the gaps.

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

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

2 * (11 + 9) is using the distributive law.

So the first line must be 11 and a 9.

Summing those 11 and 9 makes 20.

So 1.

2 * 20 gives us 24.

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

2 + 9 * 1.

For B, we have 9 *( 8 - 4 - 3).

Well you can spot we have the common multiplier of 9.

So that means using the distributor law, 9*8 - 9*4 - 9*3.

This gives us 9 * 8 is 72, 9 * 4 is 36, 9 * 3 is 27.

For C, 9.

9 multiply by 1.

1 add 0.

9 multiply by 9.

We have a common multiplier of 9.

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

9 multiplied by the sum of 1.

1 and 0.

Much easier to calculate.

9 times 2 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 2.

Question 2 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 multiply 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 1 is the same as 1.

3 times 3.

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 3 multiplied by the 0.

9 add the 3 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 the group together those three lots of 1.

3 subtract 0.

5 times 1.

3 gives us 2.

5 multiply by 1.

3, then multiply by the 4.

That gives us our 10 multiply 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, 5 multiply by 0.

75 add five multiply by 0.

25.

Alex states that the answer to the calculation is 5 as we are secretly multiplying by 1.

What do you think Alex has noticed? Well hopefully you've spotted we have a common factor of 5 in our calculation.

So that means summing up the 0.

75 and the 0.

25 gives us 1.

And this is an efficient method where there's no need to calculate anything and you can spot, it's simply 5 multiply by 1.

If we were to show this using the distributive law, you can see it's 5 multiply by 0.

7 add on 0.

25, which is 5 times 1.

Now using this, let's see if you can fill the bank so you can secretly are multiplying by 1.

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 1? Well it had to be 0.

For B, 34.

5 times 0.

2 add 34.

5 times what add 34.

5 times 0.

2 is the same as 34.

5 times 1? Well, it's 0.

Next, we have 462 times 0.

12 add 0.

33 add what is the same as 462 times 1? 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 out of 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 100 subtract 13 times 1.

13 times 99 is the same as 13 times 90 add 13 times 9.

And 13 times 99 is the same as 15 times 99 subtract 2 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 100 and then subtract 13 times 1 just because it's easy to multiply by multiples of 10 and 1.

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 multiply by 998.

It's the same as 21 multiply by 900 add 21 multiply by 90 add 21 multiply by 8.

It's also the same as 998 times 20 add 998 times 1.

It's also the same as 21 times 1000 subtract 21 times 2.

And it's also the same as 25 times 998 subtract 4 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 multiply by 1000 subtract 21 multiply by 2.

This is because multiplying by multiples of 10 is easy and same as multiplying by 2.

So working this out, 21 times 1000 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 multiplying by 1.

Which ones do you think are secretly multiplying by 1? And for those that aren't multiplied by 1, can you use the distributive law and change the calculation to make it secretly by 1? See if you can give this a go and press pause if you need.

Great work! So let's move on to question 2.

Question 2 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 1.

Here are some calculations which are secretly multiplied by 1, but which ones are they? Well, the first one, 0.

69 times 2 add two multiply by 0.

31.

Using the distributive law, you can see we have a common multiplier of 2.

So it's the same as 2 multiply by 0.

69 add on 0.

31, which is the same as 2 times 1.

Hopefully you've also spotted this one.

7 multiply by 1.

2, take away 0.

5 multiply by 7 add 7 times on 0.

Hopefully you spotted we have a common multiplier of 7, so it's the same as 7 multiply by the 1.

2, subtract on 0.

5, add on 0.

3, which is 7.

Lastly, we also have 6 multiply by 0.

4, add 6 multiply by 0.

Our common multiplier is 6.

So it's 6 times on 0.

4, add on 0.

2, which is 6 times 1.

Fantastic work if you've got that one.

Now for part B, we had to use the distributor law to change those calculations which didn't secretly multiply by 1.

So I've just highlighted those calculations which did not secretly multiply by 1.

So let's see what we need to do to change those so they do secretly multiply by 1.

Well, 6 multiply by 0.

75, add 6 multiply by 0.

35 did not give us a secret multiplication of 1.

Using the distributive law, it's 6 multiply by 0.

75 add on 0.

35, which is 6 multiply 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 6 by 1.

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

Well 9 multiply by 0.

85, subtract 9 multiply by 0.

15 is the same as 9 multiply by 0.

So by simply adjusting and amending it to addition, we would've got 9 multiply by 1.

Now for the last one, it's 4 multiply by 0.

5 add 4 multiply by 2.

5 is the same as 4 multiply by 0.

5 add on 0.

25, which is the same as 4 multiply by 0.

75.

And all we needed to do was adjust that 0.

25 to 0.

5 or the 0.

5 to a 0.

75.

This was a great question whereby you had to adjust the calculation to make it secretly multiply by 1.

Now let's have a look at question 2.

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 add 19 times 90; 990 times 10 add 990 times 9 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 aids efficiency and ease.

Great work today! It was wonderful working with you.

Well done.

I've finished the video