Lesson video

Hello! I'm Mr. Langton and in this lesson we are going to look at forming and solving linear equations.

Don't panic! Just make sure you've got something to write with, something to write on and ideally, you are in a quite place without any distractions.

Are you ready? Then let's begin.

Okay, we'll start with the Travis activity.

The shapes in the grid represent unknown values.

The numbers show the sum of each row and column.

How many different equations can you form and how can you write your equations in different ways? What I want to do is, you want to pause the video and give you the chance to have a go at this and I want you to write down as many different equations as you can.

So you're going to pause and have a go in 3 2 1 Go So many did you come up with? Let's go through a few together now.

Look at the top row, there are two quadrilaterals and a circle and all together that makes sixteen.

Now writing that as a sentence is quite long so I'm going to try and break it down a little bit and try to answer it algebraically.

So I'm going to write the quadrilateral just as Q.

I've got two of those plus the circle, see? And that makes sixteen.

Underneath that I've got a hexagon and a quadrilateral and another hexagon and that makes twenty six.

Now to simplify that the two hexagons plus the quadrilateral makes twenty six.

At the bottom I've got a pentagon and I've got a quadrilateral and I've got a triangle.

Then combined they make thirteen.

We look way down, look at the columns we know that we've got a quadrilateral and we've got a hexagon and we've got a pentagon and we can add those three together to get twenty eight.

We know that three of the quadrilaterals together make twelve and we know we've got a circle and a hexagon and a triangle.

If we find the sum of those three you get fifteen.

And you could probably round more as well.

You could have for example: You could take the three quadrilaterals and you could write it as Q + Q + Q = 12 So how many of these did you get? Some that I didn't get that you've got.

If we have an equation with just one unknown we can usually solve it to find it's value.

So looking at that top row we've got to quadrilaterals and a circle and that makes sixteen.

There are two unknowns so at the moment we can't solve it.

If we look down, that left column, that first column.

We've got a quadrilateral, we've got a hexagon, we've got a pentagon.

That's three unknowns.

And although it makes twenty eight we don't know what each individual one is.

We look down that middle row, we've got three quadrilaterals that equal twelve.

And we can solve that we can do the inverse and multiply it by three.

Three divided by three.

Then we find Q equals four.

Quadrilaterals are each worth four.

And we can start to use that now to solve different problems. We can go back at the top row: two quadrilaterals plus a circle are worth sixteen.

So eight plus the circle is sixteen.

So the inverse of adding eight is subtracting eight, so one circle must be worth eight.

That doesn't help us with this middle one at the moment the quadrilateral, the hexagon and the pentagon.

But there are other things that we can start to solve now.

If we look at this middle row here, we've got two hexagons and a quadrilateral which equals four and that makes twenty six.

So we start to solve that now, it's going to take two steps this time.

So we're currently two hexagons plus four makes twenty six.

We do the inverse of adding four if we subtract four and move up the two hexagons.

Subtracting four that gives us twenty two.

So two hexagons are worth twenty two, so one hexagon must be worth eleven.

We can now start to look at some of the others.

So going down this first column here we've got the four for the quadrilateral, we've got the eleven for the hexagon and we've got the pentagon.

Four plus eleven is fifteen and fifteen plus the pentagon makes 28.

Now we do the inverse of the fifteen, the inverse of adding fifteen is subtract fifteen.

And we get the pentagon is worth thirteen.

That puts us in a really strange position for finding this triangle in this box here because we've got thirteen plus four, plus something else twenty is back to thirteen.

Let's make a little bit of space here We are going to work this one out.

Now thirteen plus four plus the triangle needs to make thirteen.

Thirteen plus four is seventeen, seventeen plus the triangle is thirteen.

We need the same method.

We need to do the inverse of adding seventeen, we need to subtract seventeen.

So just leave T.

Then if we subtract seventeen there we actually get negative four.

So in this case the triangle is worth negative four.

So what I'd like you to do now is pause the video and have a go at the task.

When you're ready unpause the video and we'll go through some of the answers together.

Good luck! So how did you get on? Should we have a look at the answers? Okay, so one last task.

The diagram shows a perfectly balanced hanging mobile.

The shapes on the mobile weigh 360 grammes altogether.

How much does each shape weigh? What I'll do, I'll give you the chance to have a go so pause it, try it without me.

If you're stuck unpause it, and I'll give you a little bit of a hint and you can have another go.

So pause in: 3 2 1 Go! Okay, so here's a hint, the diagram say's it's perfectly balanced and that means that this half here must weigh the same as this half here.

Does that help? Pause if you want, carry on.

Right, Okay.

And we start to go through it now.

So we've said that this side, this left one side here is evenly balanced and if it weighs three hundred and sixty grammes altogether then this bit combined must weigh one hundred and eighty grammes and this bit combined must weigh one hundred and eighty grammes.

Now, this left hand side is also misplacing it's two pieces: So that means that these three here, these three triangles must weigh the same as the two stars and the.

is it a rhombus or is it a kite? Let's call it a kite.

So that means, our three triangles must weigh nighty and our two stars with a kite also weigh ninety.

So we can start to solve the equations now, we've said that we've got three triangles that weight ninety, and so one triangle is thirty grammes.

So that's that labelling that on the diagram to make it look a little bit easier for us.

Now, that's not really helping us at the moment with the stars and the kite but we can look over at the right hand side now and where this is balanced.

So we've got two circles that are equal to a triangle, a circle and a star.

And because we know that in total it's one hundred and eighty grammes, that means that the two circles must weigh ninety grammes and it means that there three here must weigh ninety grammes.

So we can write down the equation, yeah? We know that two circles weigh ninety grammes, so one circle is forty five grammes.

Let me write on the diagram forty five forty five, forty five.

So now let's look at this right hand side, here we've got the triangle and the circle which together make seventy five grammes.

I'm writing on to that the star and that one with ninety grammes.

So if we subtract seventy five from each side we get that the star weighs fifteen.

Remember to be a different colour here outside, so I think my purple is going to show up.

So the star is fifteen grammes and let's do that overhere as well.

Fifteen grammes for the star, fifteen grammes for the star.

So all that's left now is to find the weight of the kite.

We know that two stars plus the kite weigh ninety grammes and a star is fifteen so two stars are thirty.

So thirty plus the kite is ninety grammes.

So we do the inverse of adding thirty, if we subtract thirty.

Then we are going to check K equals sixteen.

And then so a kite weighs sixty grammes.