Lesson video

Hi, I'm Mrs. Warehouse, and welcome to today's lesson on checking and securing understanding of simplifying.

This lesson is part of our unit on algebraic manipulation.

Algebra is so useful, so let's get started.

By the end of today's lesson, you'll be able to simplify expressions by collecting like terms. Now these are some key words or phrases that we're going to be using in our lesson today.

Now you may have heard of like terms and simplifying before, and it's excellent if you have and you remember and feel confident about these definitions.

But if you're a little uncertain, feel free to pause the video now and have a read through.

Welcome back.

Glad to see that you took the time to have a read through.

Our lesson today comprises of two parts, and in the first part of today's lesson, we'll be reviewing collecting like terms. Like terms are terms which have the same set of variables and corresponding exponents.

And it's important that it's both.

So for example, 5y and 0.

8y are like terms. Both of those terms have the same variable, that's y, and both Y's have the same exponent, which is one.

X squared and negative 3x squared are also like terms, same variable, x in both, and same exponent, both have a two because they're both being squared.

Now these ones, 2x cubed y squared and 12x cubed y squared are also like terms. They have the same variables.

Each term has an x and a y in it, and you'll notice the exponents for those respective variables are the same.

So both X's are cubed and both Y's are squared.

Which of these do you think are like terms? Pause the video and make a choice now.

Welcome back.

Did you find the like terms? So you should have seen that 6xy and 2xy are like terms. 2/3 of x and negative x are like terms. 9y squared and 2y squared are like terms, and 2.

4 and seven are like terms. And that one may have caught you out, but 2.

4 and seven are both constants.

In other words, there are no variables and that makes them like terms. Now Sofia says, "I think 3x squared y and 2yx squared would be like terms." What do you think? Do you agree with Sofia? Well, let's have a little look.

She's, of course, correct.

3x squared y is equivalent to three multiplied by x squared multiplied by y.

And as you should know, multiplication is commutative, which means it can be rewritten as three multiplied by y, multiplied by x squared, or 3yx squared, and this has the same variables and corresponding exponents, so it must be like terms. Well, I agree Sofia.

She's saying it's hard to see which of these are like terms and I agree with her.

What could she do to make it easier? Ah, Jun's spot on here.

Write the variables in alphabetical order.

By writing 'em in alphabetical order, it can help to identify the like terms. So let's do that.

Ah, now it's a lot easier to see because they're in alphabetical order.

It's so much easier to compare if two terms have the same variables with the same corresponding exponents.

For example, let's look at these terms. We can see they all have a squared in them.

By writing that like this, it's then a lot easier to compare the remaining variables and exponent to see if they're the same.

This is the group that are a cubed.

Now we've grouped them by a squared and a cubed.

We can consider any other variables that are there.

For example, in the a squared group, I can see I have two subgroups I could break it into, groups with c squared in and groups with c cubed.

I can now see that I have two groups, and within a group, we have like terms. If I look at the remaining group, so that was the a cubed lot, I can see that two of the terms have b squared and one has b cubed.

And in fact these two terms are like and the other one nothing matched up with.

Time for a quick check now.

Please match up the like terms; pause the video while you do this.

Let's see how you got on.

A matches to I, B matches to G, C matches to L, D matches to H, E matches to J, and F matches to K.

Well done if you've got all these right.

When an expression contains like terms, we can simplify the expression by collecting those like terms. We can add the coefficients of the terms to write them as a single term.

So for example, 3a add 2a is equivalent to saying five lots of a, and 5x squared plus 3x squared is equivalent to 8x squared.

6pq plus one lots of pq, and that's what you have to be careful of, remember, any term can be multiplied by one and still remain the same.

And therefore we've actually got here six lots of pq plus one lots of pq, giving us 7pq altogether.

When terms are subtracted from each other, we can write this as an addition of a negative value.

You should remember this part from your number work.

So 5a subtract 3a is the same as saying 5a add negative 3a.

In other words, the result is 2a.

When there are unlike terms in expression, the like terms can be grouped together and simplified independently.

For example here, the like terms have been indicated by the colour.

So 6x add negative 2x, the 6x and the negative 2x are like terms, and 3y and 4y are like terms. Bear in mind that addition is commutative.

I can rewrite this.

So 6x add negative 2x gives us 4x, and 3y add 4y gives us 7y.

Let's do a quick practise here.

I'll go first and then it's your turn.

Let's do a quick check here.

I'll go first and then it'll be your turn.

I'm going to simplify 7x takeaway five, takeaway x, add 12.

Well, by rewriting this I can group the like terms together.

So 7x, add negative 1x, add negative five, add 12.

Well, grouping these, I can see that I have 6x add seven.

It's now your turn.

Welcome back.

Let's see how you got on.

Well, you should have grouped the terms that have y and the constant terms. Doing that should result in 35y take away 15.

Well done if you've got that right.

Let's try another one.

I'm going to simplify 3a, subtract 2a squared, add six, subtract 7a, add 5a squared, add one.

Again, I'm going to group the terms that are like, so 3a subtract, or in this case, I'm going to add the negative.

So 3a, add negative 7a.

Then I'm gonna consider negative 2a squared, add 5a squared, and then add six, add one on the end.

By grouping these together, I'll end up with negative 4a plus 3a squared plus seven.

Now I can rewrite these terms in any order but either answer is correct, so you don't have to write your terms the same way I did.

It's now your turn.

Please simplify fully the following expression, 8xy, add 4x squared, subtract 6x squared, add five, add 2xy, subtract three.

Pause and do this now.

Welcome back.

If you follow through my working carefully, you'll see that I grouped the like terms first.

I then used the reverse, the distributive law.

We've been using this a lot, haven't we, so that I can see which coefficients go together, and then I'm correctly resolving that.

So eight lots of xy, add two lots of xy gives us 10xy.

I've then got subtract 2x squared, add two.

Again, you can write your terms in any order.

It's now time for our first task.

For question one, for each pair of terms, explain why they are not like terms. Pause the video and do this now.

Welcome back.

Question two, fully simplify each expression by collecting the like terms. Pause and do that now.

Welcome back.

Question three, fill in the missing terms to make the expressions equivalent.

So you'll notice here we are not fully simplifying, but you do need to make sure whatever you add in the missing gap allows two expressions, so either side, to be equivalent.

Pause and do this now.

Welcome back.

It's time to go through our answers.

For question one, you had to explain why they are not like terms. So for part A, you'll notice that 5x and 5y have different variables.

The fact that coefficients are the same, it is completely irrelevant.

In B, the variable has a different exponent in each term.

One is x squared and the other is just x.

For C, I have a constant term and the other term has a variable.

So they're definitely not like terms. For D, did you spot? The variables do not have the same corresponding exponents in both terms. In the first term, a is to the power of 10, but in the second term, it is b to the power of 10.

And then E, goodness me, that was a tricky one.

But if you write the variables out alphabetically, you'll notice that there is a t in the first term, but not the second.

Therefore they're not like terms. Well done if you spotted that one.

Question two, you had to fully simplify each expression by collecting the like terms. So for 2A, you should have got 11p takeaway 2q.

For 2B, you should have 32a takeaway 11b.

For 2C, it should be x minus 4x squared.

For 2D, it should be 20ab add seven, subtract 3a squared b, subtract 7ab squared.

And then lastly for 2E, you should have negative 14p to the power of five, q to the power of eight r, add 6p to the power of eight, q to the power of five r squared.

And then adding 4p to the power of five, q to the power of eight, and r squared.

Well done if you got that last one right.

Question three, you had to fill in the missing terms to make the expressions equivalent.

Well, for question 3A, you'll notice I have 6x, and I have subtract 10x.

That's a result of negative 4x.

I can see on the right hand side I have negative 5x.

So to get to negative 4x, I need to add on an x term.

So that's why I get the x.

And then for the Y's, you'll notice that on the left hand side, I have a result of just y.

So on the right hand side to get to the same result, I've got 5y takeaway 7y, negative 2y.

So I need to add 3y to get back to just 1y.

For 3B, similar reasoning will help you to reach 3x squared, plus seven, subtract 5y.

3C, you had to fill in the missing term as 2a plus 10.

Well done if you've got these right.

It's now time for the second part of our lesson, and that's on simplifying expressions.

Find an expression for the perimeter of this rectangle.

So what I'd like you to do is pause the video, remind yourself of what we mean by perimeter and then try writing an expression for it.

Pause and do this now.

Welcome back.

Let's see what you put.

Well, I started by writing down the length of each side.

So I have 2a, plus 5x, plus 6a, subtract 2x, plus 2a, plus 5x, plus 6a, subtract 2x.

Now, of course, there's nothing wrong with that, but it's going to be easier if we can write this in a simpler way.

So what I've done is I've grouped all the terms that are like, and this means I get the result of 16a plus 6x.

Well, that's a lot easier to write and gonna be a lot easier to use.

Now, of course, I could treat equivalent expressions as like terms, or what do I mean by that? I mean this.

I could say I have two lots of 2a plus 5x, and I could say I have two lots of 6a subtract 2x.

And by adding those two expressions together, that gives me an expression for the perimeter.

How could we write an expression for the perimeter of this regular hexagon? What I'd like you to do is pause the video and have a go at this.

See if there's more than one way you can do this.

Pause and do this now.

Welcome back.

Let's see what you wrote.

Now, a hexagon has six sides and because it's regular, we know that all of the sides have the same length.

So I could write out 3y plus 6x, six times, remembering to add all the expressions together.

Or I could say it's six lots of 3y plus six.

And this is one way to write the perimeter.

Did you go for one of these? Maybe you expanded those brackets and said it was 18y plus 36.

Absolutely fine if you did, it's still equivalent.

Quick check now, which of these could be an expression for the perimeter of this kite? Remember there may be more than one correct answer.

Pause the video and work this out now.

Welcome back.

Which ones did you go for? Well, B is definitely correct.

I know that in this kite, there are two sides which have the same length.

That's one of the properties of a kite.

So I could say that I have two lots of the length, 5x subtracted 2y, and I have two lots of the length, 2x plus seven.

I could of course also group the like terms. So by collecting like terms, I reached the expression 14x subtract 4y, add 14.

Treating an expression as a term can be useful for simplifying far more complex expressions.

For example, eight lots of x minus two, add two lots of x minus two.

I can see here that the expression in the brackets for both of these terms is the same.

And because they're identical expressions, I can treat them as like terms. So I can say what I actually have here is 10 lots of x minus two.

Now how could you write each expression as a multiple of one bracket? So here we've got some more examples you can practise with.

Pause the video and have a go at writing these as a multiple of just one bracket.

Welcome back.

Let's see how you got on.

In the top one, you should have said that we have 11 lots of 2x minus five.

For the second one, you'll see that we have three lots of five plus 3y.

Now the next one down, did you spot that you could put brackets around the x plus y? That will give you eight lots of x plus y in total.

And then for the last one, well, seven add four is 11, takeaway two gives us nine lots of a plus b.

How can we write this expression as a multiple of just one bracket? Ah, so it's different now.

I don't have a numerical coefficient in front of both brackets.

Well, Jun says, "I can see the brackets are equivalent," but he is not sure he can add five and x.

What do you think to this statement? Well, Jun can add x and five, but as they're not like terms, they can't be written as one single term.

So what he could do is this.

He could say I have x lots of x plus eight and I have five lots of x plus eight.

So altogether I have x plus five, lots of x plus eight.

So Jun's now looking at this expression, it's a little bit longer than the last one.

And he's wondering can he write this as a multiple of the expression 2x plus y? What do you think? Well done if you said he can.

This expression is the sum of the multiples of 2x plus y, and we can add the coefficients to write this as an expression multiplied by 2x plus y.

So in fact what we'd have here is x take away five, add two, lots of two x plus y, and in fact we can simplify that first bracket so it becomes x takeaway three, lots of two x plus y.

Time for a quick check now.

The expression five lots of 2x plus one, subtract x lots of 2x plus one is equivalent to which of the following expressions? Pause and make your choice now.

Welcome back.

Did you pick A? Well done if you did.

You've got the coefficients of five and negative x in front of our brackets here.

So in other words, they're going to be the ones that are combined.

Now, x lots of x minus y plus y lots of x minus y is equivalent to which of the following expressions? Pause the video, make your choice now.

Welcome back.

You should have picked C.

I'm just adding the coefficients together remember.

So that's x plus y in the first bracket.

And second bracket, remember, is the part that was the like part of the two expressions.

So that was the x minus y in brackets.

Four lots of seven minus x, subtract x lots of seven minus x plus two lots of seven minus x is equivalent to which expression? Pause and choose now.

Welcome back.

Did you get C? Well done if you did.

Remember, you've got four add two, make six, and you're still taking away x.

So six takeaway x, lots of seven minus x.

It's now time for our final task.

For question one, I'd like you to match each shape to an expression for its perimeter.

Now there may be multiple expressions for each shape, so do take your time and be careful.

Pause and do this now.

Welcome back.

Let's look at question two.

For question two, you need to write each of these as an expression multiplied by x plus one.

Pause and do this now.

Welcome back.

For question three, fill in the gaps in these expressions so they are equivalent.

Pause and do this now.

Welcome back.

Let's go through our answers.

For question one, you can see here I've grouped the expressions for perimeter underneath the respective shape.

So for the square, you should have the two expressions, four lots of 2x minus three and 8x subtract 12.

For the rectangle, we should have two lots of 2x, subtract one, add two lots of 4x minus two, and also 12x subtract six.

And then for the kite on the end, we should have two lots of 5x minus six, plus two lots of x plus two and also 12x subtract eight.

Question two, you had to write each of these as an expression multiplied by x plus one.

So for 2A, we have eight lots of x plus one, for 2B, three lots of x plus one, 2C, negative three lots of x plus one, for D, x plus two lots of x plus one, for 2E, three takeaway x, lots of x plus one, for F, 5x plus two, lots of x plus one.

And then for G, y plus 2x subtract five lots of x plus one.

Question three, you had to fill in the gaps in these expressions.

As you can see, some of the gaps were quite long and there's quite a bit to fill in.

What I suggest you do is pause the video now so you can compare your working to what you can see on the screen.

Welcome back.

You should have now checked all your work and made sure that what you've got has been marked correctly.

Well done if you've got these right.

We're just gonna sum up what we've done today in our lesson.

Like terms have the same variables and corresponding exponent.

Like terms, as we saw, can be collected to simplify expressions.

Identical expressions can be unitized and multiples of the same expression can be treated as like terms. And that's what we're doing for question two in task B today.

Well done.

You've done a great job today, and I look forward to seeing you for more lessons on algebraic manipulation.