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Hello, I'm Mr. Langton.

Today, we're going to be forming and solving linear equations.

Before we start, you're going to need something to write with and something to write on.

Try and find yourself a quiet space without any distractions, and when you're ready, we can start.

Okay, we'll start off with our try this activity.

Zaki has made a quadrilateral using some sticks of three different lengths.

Write an expression for the perimeter of the quadrilateral.

And if the perimeter of the shape is 60 centimetres, how could you find the value of a? What I'd like you to do is pause the video and have a go, and when you're done, we'll unpause it, and we'll talk through it together.

So I'd like you to pause the video now in three, two, one.

So how did you get on? Let's go through it together now.

I'm going to start off by labelling the sides.

We know that the blue side is eight centimetres long, so I can label that a.

Now, the length of the green stick, we're told is six bigger than the blue one.

So it's a plus six.

So let's label that there, a plus six, and over here, a plus six.

Finally, the longer line, which we're told that length is a, this length here is six, and we've got the nine at the end, a plus six plus nine, which is the same as a plus 15.

So what we need to do now, we need to write that expression, don't we? For the perimeter.

So we need to add all those sides together.

So we've got the a plus 15 for the pink one.

We've got a plus six for one at the green ones.

We've got another a plus six for another green one, and we've got the blue one, which is a.

And if we collect together all the like terms, we've got four a's, and we've got 15 plus six plus six is 27, four a plus 27.

So an expression for the perimeter is four a plus 27.

Did you get that? The second part of the question says that if the perimeter is 60, how could you find the value of a? So what we're going to do is we're going to start that out, set it out as the equation.

We know that the perimeter can be written as four a plus 27, and if we're told that the perimeter is 60, we've now got an equation that we can solve to find the value of a.

We'll have a go at that on the next slide.

So we came up with the equation four a plus 27 equals 60.

Now I'm going to attempt to solve that in two different ways.

The first one is using a bar model.

You can see I've represented the four a's as four different green blocks.

And I've also got 27 added onto the end, and that's equivalent in total to 60.

I can write it algebraically as well, 60 equals four a plus 27.

Now, if I'm trying to solve this, I need to think about what these things have got in common and what we can do.

So if we've got four a plus 27 makes 60, if we subtract this 27, if we take that away, then we've got the four a's left over and they're no longer equal to 60, because we've taken 27 off.

So those four a's are equivalent to 33.

And we could do that algebraically.

We don't need the diagram.

We can do the same thing.

We can subtract 27 from each side.

And if we do that, we've got 33 equals four a.

So our next step now, if we know that four a's have got the value of 33, so these four a's here have got a value of 33, we can divide by four to get the value of one a.

And there are quite a few ways we can write that.

We can write it as 33 over four.

Nothing wrong with that.

That's correct.

We can look at it and say, well, how many times does four go into 33? Eight times with one left over.

And because we're dividing it by four, that one left over is 1/4.

We could write it as a decimal.

And if you had a calculator, you could type it in, 33 divided by four to get 8.

25.

And any one of those three answers is absolutely spot on.

And it's the same over here.

We've just done the same thing, haven't we? We've divided it by four to get 33 over four is equivalent to a.

Okay.

What I'd like you to do now is have a go at these questions.

You can pause the video and there's a worksheet that you can access.

You can print it if you like.

Have a go and see how many you can do.

If you're struggling, unpause the video and I'll go through a couple with you.

So you can pause the video on this next slide when it comes up now.

If you want some help, just leave it running a little bit longer and I'll go through a couple with you.

Would you like a hint? I'll tell you what, what I'm going to do is I'm going to go with this second one here, so I'll solve that with a bar mode, and then I'm just going to draw a bar model for this one, and then you have a go at solving it yourself.

So on the left-hand side, we said that 10 plus two a equals 13.

So if we subtract 10, let's write that down, 10 plus two a equals 13.

If I subtract this 10, take that off there, then I've just got two a left over, and that must have a value of three.

If it's 13 all the way along and I take 10 off, then this bit here must be worth three.

So if altogether it's worth three, if two of them are worth three, I need to divide by two.

So you have one a, and that's going to give me three over two, or 1.

5.

Does that help? I hope so.

Now let's just draw a bar model for this one.

Six c plus 3 1/2.

So I'm going to start off by drawing six c's.

One, two, three, four, five, six.

And that bit there is six c's, and then I add on 3 1/2.

And altogether, that's equivalent to 27.

5.

Okay, have a go now and see if you can solve that.

And these are the answers.

How did you get on? Here's the final activity.

Use the shapes below to work out the lengths of the sticks x and y.

Pause the video and have a go, and when you're done, we'll unpause it and we'll work through it together.

So you can pause this in three, two, one.

Hi, guys, I'm Ms. Jones, and I'm going to be going over the solution to this final let's explore task.

Okay, so we're going to start by filling in the information we already know and annotating our diagrams here, okay? So we know the value of each of these rods, so let's fill this in.

I know that that blue rod would be equal to x plus five plus four, which is x plus nine.

This rod is x plus five.

This one is x.

This one is x plus five, okay? And we can then write the perimeter as an equation.

We know that the perimeter is equal to 43, and it's also equal to all of these sides added together.

So we can write four x plus 19 is equal to 43.

And I got that by grouping my terms. So I've got four x's and if I add up digits here, nine, five, and five, I would get 43.

Okay, let's look at our second shape again.

I'm going to fill in what I know.

This time, the green rod we've got is y.

Then we've got y plus three, y, and then x plus nine, which I got from over here, and x, okay? And we could write the perimeter here is two x plus three y plus 12 is equal to 36.

Now, in order to solve x and y, I'm going to need to solve the linear equations.

Now, this equation over here has two unknowns at the moment.

So I can't solve this just yet, but I can solve this equation.

So we've got four x plus 19.

If I take away 19 from both sides, I know that four x would equal 24, which makes x equal to six.

Aha.

Now I know that x is equal to six, I can add that information into my equation over here to make this an equation with one unknown.

So instead of having two x, I can put 12.

So I've got 12 and then three y plus 12.

So altogether, I've got three y plus 24.

Now, if I take 24 from both sides, I'm left with three y is equal to 12 centimetres, which means y has to be equal to four centimetres.

Okay, I'll pass you back over to your teacher to finish the lesson.

If you'd like to share your work with us, please ask your parent or carer to share it on Twitter, tagging @OakNational and #LearnwithOak.