# Lesson video

In progress...

Hello everyone, it's Mr. Millar here.

Welcome to the fifth lesson on inequalities, and in this lesson, we're going to be forming and solving inequalities.

So first of all, I hope that you're doing well.

Let's have a look at the "Try this" task to start.

So, here I've got a rectangle, and I'm told that the height of the rectangle is two centimetres longer than the width.

I'm also told that the perimeter is less than 40 centimetres.

So, what could the dimensions be? Okay, so what I want you to do is I want you to try different values for the height and the width, and see what you find.

So, for example, if you think that the width is six centimetres, then the height is going to be two centimetres longer, so it's going to be eight centimetres.

And then, you can work out the perimeter, which is the distance around the whole shape.

So, I can fill in the lengths of the other sides, and if I add all of those up, I get 28 centimetres, so I could have these dimensions.

What I want you to try is other values, and see what you find.

Which values work? Can you find a value that doesn't work as well? Pause the video now to have a go at this task.

Okay great, so hopefully you had a go at this, and you found some values that worked and some values that didn't.

For a value that didn't work, you could have tried, for example, nine centimetres as the width, and 11 centimetres as the height.

And if you added up all the lengths, you would have found that they added up to 40 centimetres, but you were told that the perimeter had to be less than 40 centimetres, so these dimensions would not work.

But, anything at which was, had a width of less than nine centimetres, would work.

So, this obviously does relate to inequalities, and we're going to find on the Connect slide why this relates to inequalities.

Okay, so here's the Connect slide, and I can see, again, I've got the same problem with a rectangle.

But this time, I'm going to use some algebra to see if I can work this one out.

So, what I'm going to do is, I'm going to say that the width of this rectangle is equal to x.

So, I am going to put an x up here, and an x down here.

And I can say that I can let the width equal x.

Now, if the width is x centimetres, and the height is two centimetres longer than the width, can you think of an expression for the height of the rectangle in terms of x? Well, if you're thinking x plus two, then really well done.

That is an expression for the height of the rectangle.

Now, what about the perimeter of this rectangle? How do I work this out? Can you think of an expression for the perimeter in terms of x? Well, what I can do is, I can add these all up.

So, x plus x plus two, plus x plus two plus x.

That is an expression for the width, for the perimeter, sorry, of the whole rectangle.

And now, if I add all of these up, if I collect my like terms, well I've got four x's here, so I've got four x plus two plus two, which is four.

That is an expression for the perimeter of the rectangle.

Now, I'm told that the perimeter is less than 40 centimetres.

So, how could I turn this in-expression into an inequality? Well, because the perimeter is less than 40 centimetres, I can say that four x plus four is less than 40.

And now, I've got an inequality, and we learned last lesson that we can solve this inequality in the same way that we solved an equation.

So, do you remember how to do this? Okay, first step: subtract four from both sides.

So, I have four x is less than 36.

And now, I'm going to divide by four, so I have x is less than nine, which is actually the same thing that I just found out.

So, I found out that as long as x is less than nine, I can have a perimeter which is less than 40 centimetres.

So, what I've done here is, I have set up an inequality, and then I have gone ahead and solved it.

And, in this lesson, we're going to have a look at a few more examples of doing this.

So, when you're ready, I'd recommend writing down this example into your notes, but when you're ready, move on to the Independent task.

Okay, so here is the Independent task.

Two examples for you to have a think about here, and you'll notice that I've got two shapes again, but this time, I am thinking about area.

And, you'll know that area is, of course, length times by width, if we're thinking about a rectangle or a square.

So, have a go at these two tasks.

You may want to try out different values of x, and see what works.

Or, you may want to even try to set up and solve an inequality, which is what we should be aiming for.

So, pause the video now to have a go at these two problems. Okay great, so the first one.

I'm told that the length is x centimetres longer than the height.

So, the height is four centimetres, so the length is going to be four plus x.

It's x centimetres longer.

The area is greater than 40, so I can do length times by height, so four times by four plus x, which is greater than 40.

And then, I can solve this inequality.

I could multiply out brackets here, or I could just divide by four straight away.

So, divide by four on both sides, and I have four plus x is greater than 10.

And then, I have x is greater than six.

So, when x is greater than six, then I have an area greater than 40 centimetres.

For the square, I know that the height and the length are the same, so the length is also going to be x.

And, the area of the square is less than 64.

Now, you should be able to work out quite easily that x has to be less than eight.

But, I could also say that x squared is less than 64.

And therefore, square rooting, x is less than eight.

Now, you do have to be careful when you are square rooting with inequalities, because when you square root, you have to think about the negative values here.

But, and of course, I could have negative values.

So, for example, x could be negative five, because negative five times by negative five does give me 25, which is less than 64.

But, in this example, I don't really need to worry about negative values, because I'm thinking about a shape, and I can't have a negative length on a shape.

So anyway, here are my two answers to these inequalities.

And I worked these out by setting up and solving an inequality to work these out.

So, I'd recommend that you get these examples down, if you haven't already.

And then, we're going to move on to the final slide, the Explore task.

Okay wonderful, so here is the final slide of today, the Explore task.

And here, Antoni is thinking about a rectangle, and you've got three different clues.

So, there's a number of ways you could approach this problem.

The first, of course, is trying out different dimensions and seeing what works.

But, what we've been looking at this lesson is setting up and solving an inequality.

So, you might want to start by calling one of the lengths x, and another one something in terms of x, and then seeing if you can set up and solve some inequalities.

So, pause the video and see how far you get with this one.

Okay great, so let's go through this, and if you needed a little bit of a help, here it is.

And, you could start off by using the middle clue, which says that the longest side is three centimetres longer than the shortest.

So, if we call the shorter side x, the longer side is going to be x plus three.

And therefore, we can set up and solve an expression for the perimeter, because we're told that it's more than 26.

So, if I collect up my like terms here, I will get four x plus six greater than 26.

Then I solve this, so I subtract six from both sides.

Four x is greater than 20, and then divide by four.

So, I have x is bigger than five.

So, I know from looking at the first clue that the value of x must be greater than five.

But, I also have to think about this final clue about the area, to see if there are any more things I need to think about here.

So, the area is less than 70 centimetres, so how do I figure out the area? Well, the area is going to be x times by x plus three.

And, I know that this is less than 70.

And actually, you don't know how to solves this particular inequality, because if I multiply out brackets, it's going to give an x square term, which you won't know how to solve.

But actually, if you just look at this inequality, you can actually kind of figure out what's going on here.

Because if you imagine that x was equal to seven, you would have seven times by seven plus three, which is 10, which is 70.

But, you want what's on the left-hand side to be less than 70.

So therefore, can you think what x must be less than? Well, if you're thinking x must be less than seven, then, of course, you're right.

Really, really well done, because if x was seven, you would have 70, so x must be less than seven, so you have less than 70.

So, that is kind of it, because we have found what x must be.

On the one hand, x must be greater than five.

We found that out when we looked at the perimeter.

But on the other hand, x must be less than seven.

So, x has got to be in between five and seven.

So, it could be six, it could be six and-a-half, it could be five and-a-half, anything like that.

So, that is it for this lesson, and I hope you've enjoyed setting up and solving inequalities in the context of shapes.

In the next lesson, we're going to do something similar in terms of setting up and solving inequalities, but this time in the context of word problems. Thanks very much for watching, and I will see you next time.

Have a good one, bye-bye.