Lesson video

Welcome and well done for making the decision to learn using this video today.

My name is Miss Davis and I'm gonna be helping you as you work your way through this lesson.

There's lots of opportunities for you to really think today and test your understanding.

Make sure you're pausing the video to give yourself that chance to think.

And then I'll help you using any hints and tips as we go through.

If you've got everything you need, then let's get started.

Welcome to this lesson where we're looking at non-solutions to linear simultaneous equations.

By the end of the lesson, you'll be able to identify what happens to the equations when a point other than the intersection is substituted into the equations.

So equations which represent different relationships between the same variables are called simultaneous equations, and an inequality is used to show that one expression may not be equal to another, we're starting to bring our inequality skills in today.

Let's start by investigating points.

To solve two equations simultaneously, we need values of the variables which satisfy both equations.

For example, if I had y equals 2x minus 1 and y equals x plus 2, Sam reckons there's a solution when x equals 3 and y equals 5.

We can check by substituting.

If we look at that first equation, we've got 5 equals 2 lots of 3 minus 1, which is 5 equals 6 minus 1, which is 5 equals 5.

So it does balance that first equation.

Let's look at the second equation, y equals x plus 2, so that would be 5 equals 3 plus 2, which is 5 equals 5.

So it also balances that second equation.

So these values work for both equations, so this is a solution to these equations when solved simultaneously.

Now if we were to plot these on a graph, these would be the x and y values of the intersection point.

There's our graph, we've got the two equations plotted, and then we can see the point of intersection is 3, 5.

So when x is 3 and y is 5.

If two linear graphs have different gradients, then there'll be exactly one intersection point.

No other value of x and y will satisfy both equations.

That value is gonna be unique when they have different gradients.

Sam says, "If this point is unique, a different point like 1, 1 will not lie on either line." Pause the video.

Do you agree? Right, let's have a look at substituting 1, 1 into our equations and see what happens.

You should have 1 equals 2 lots of 1 minus 1, so 1 equals 2 minus 1, 1 equals 1.

So 1, 1 satisfies that first equation.

However, 1, 1 does not satisfy that second equation.

So the point 1, 1 satisfies one equation but not the other, which means it'll be on one line but not the other.

So it's not the point of intersection but it is gonna lie on the line y equals 2x minus 1.

Pause the video.

Can you find a value of x and y which satisfies y equals x plus 2, but not y equals 2x minus 1? Off you go.

You are looking for a coordinate pair which is on the line y equals x plus 2, but is not also on the line y equals 2x minus 1.

So any coordinate on the line y equals x plus 2, apart from 3, 5.

So you could have gone x equals zero, y equals 2, x equals 5, y equals 7 and there's infinitely many others.

What should happen if we substitute the values x equals 6 and y equals 4 into both equations? Pause the video, use the graph to help you answer that question.

Well, x equals 6, y equals 4 is that coordinate there, 6, 4.

It's not on either line, so it should not satisfy either equation.

Let's have a look.

So 4 equals 2 lots of 6 minus 1.

Well that doesn't balance.

4 does not equal 11.

Let's try the other equation.

4 equals 6 plus 2, and that's not true either, so that coordinator is not on either line, so that is not a solution to either equation.

Quick summary then.

If two linear equations with different gradients are plotted, then they will intersect exactly once.

The coordinates of the intersection point will be the only values that satisfy both equations.

A point whose values satisfy one equation but not the other will lie on that line but not the other.

And a point whose value does not satisfy either equation will not lie on either line.

Quick check then.

Which equation is x equals 2 and y equals negative 1 a solution to? What do you think? There's our coordinate 2, negative 1.

And well done if you spotted that it's y equals 3 minus 2x.

It's not on the other line, so it's not a solution to the other equation.

Time for a practise.

I'd like you to use the graph to sort these values of x and y into the correct parts of the Venn diagram.

You can use the letters to represent the different solutions if you wish.

So are they solutions to one equation? Are they solutions to the other equation? Are they solutions to both, or are they solutions to neither? Where's it gonna go in this Venn diagram? When you're happy with your answers, we'll look at the next question together.

Right, this time I'd like you to sort these coordinates into the correct part of the Venn diagram.

I haven't drawn the graph for you, so you might want to think about how you are going to work this one out.

When you're happy with your answers, come back and we'll look at the next bit.

Well done.

This time I've given you a set of axes, so you may wish to draw the graphs to help you.

So I'd like to know the solution to this pair of simultaneous equations.

So drawing the graphs of y equals 3x minus 4 and y equals 2x minus 1 may help.

Then you've got three other questions to answer.

When you're happy with those, we'll look at the last one together.

And finally, you've got the equations 2x plus y equals 5 and y equals a half x minus 5.

So again, starting by plotting those on the axes may help you answer those questions.

Then I want a point which is on both lines, and then answer B, C, and D.

When you're happy with those, we'll look at the answers.

Pause the video and check you've got your correct solutions in the correct places.

Just draw attention to F: x equals 2 and y equals 6.

That's the intersection of both lines, so that is gonna be a solution to both equations.

When you've checked those, we'll look at the next one together.

For two, the easiest way to do this would be to substitute the values into the equations and see which ones they satisfy.

If you had graphing software handy or you had a set of axes to use, drawing those would definitely have helped.

Again, pause the video and check that you've got the right ones in the right places.

F: 1,5 is a coordinate on both lines because x equals 1 and y equals 5 satisfies both equations.

When you're happy with those, we'll look at the next bit.

Lovely.

So those are our two graphs.

y equals 3x minus 4, has a gradient of 3 and a y intercept of negative 4.

y equals 2x minus 1 has a gradient of 2 and a y intercept of negative 1.

The solution to the pair of simultaneous equations is the intersection point, which is x equals 3 and y equals 5.

They should be written as solutions, not as a coordinate.

For B, the solution for the equation y equals 3x minus 4.

x is one, y is minus 1.

Remember, there's loads of solutions to that equation, that's just one of them.

x equals 1, Yeah, equals negative 1.

For C, for the other graph y equals 2x minus 1, when x equals 1 and y equals 1.

And finally, when x equals 4, y equals 7.

I think the trickiest bit with that was making sure you get the right line.

It might have helped if you'd labelled your line so you remembered which one was which equation.

And question four is a little bit trickier to draw this time.

y equals a half, x minus 5 is not too bad.

It's got a y intercept of negative 5 and a gradient of a half.

To draw the other one you can rearrange.

You have y equals negative 2x plus 5.

That's a gradient of negative 2 and a y intercept of 5.

A point which is on both lines then would be 4, negative 3.

The coordinate on the line 2x plus y equals 5 could be 2, 1 For C you've got 1,3, and for D, you are looking at the other line now when x equals 6, y equals negative 2.

Fantastic.

We're now gonna have a look at regions of a graph.

We've seen how solutions to an equation will lie on the graph of the equation when plotted.

For example, if we've got y equals x, here are some possible solutions.

We know that every coordinator on that line will be a solution to that equation.

How do we know that x equals negative 2, y equals 2 is not a solution to that equation? What do you think? Right, but if we find negative 2, 2 on our graph, it is not on the line, so it's not a solution.

Now this point, the y value is greater than the x value.

2 is greater than negative 2, isn't it? So y is no longer equal to x.

We can say that y is greater than x.

So for that coordinate we can say that y is greater than x.

We can write that using inequality notation.

So this point has coordinates where y is greater than x, and I've read that from left to right using that inequality symbol greater than.

The point negative 2, 2 satisfies the inequality y is greater than x.

I'd like you to write down the coordinate of another point where y is greater than x.

Off you go.

You had loads of options.

If you picked an integer coordinate on the axis, then it could be any of those, or you could have used any not on the grid.

So 25, 27 would satisfy that because y would be greater than x.

Or remember there's infinitely many non integer coordinates as well in that part of the graph.

You can see that I've looked at a particular section of the graph, that's gonna be important in a moment.

I've been looking at part of that graph which is above the line y equals x.

We call this the region where y is greater than x.

So I still have the line y equals x drawn.

Here we have another point which does not satisfy the equation.

This time the value of y is less than the value of x.

What inequality can we write for this point? What do you think? So y is not equal to x, it's not on the line.

This time, y is less than x.

We can say this point satisfies the inequality y is less than x.

And you can use that inequality less than sign as well.

Right, this time I've drawn the line with equation y equals 3x minus 2.

Why is the point 1, 4 not on the line? Right, because the y coordinate is not equal to three lots of the x coordinate minus 2, it doesn't satisfy that equation.

In fact, the y coordinate is greater than the value of 3x minus 2, when x is 1.

We put x is 1 into that expression, 3x minus 2, we get 1.

So 4 is greater than 1, so the y coordinate is greater than the value of 3x minus 2.

This point satisfies the inequality y is greater than 3x minus 2.

Where else will I find coordinates which satisfy the inequality y is greater than 3x minus 2? What do you think? Right, it's gonna be everywhere in that region this time.

Of course there's gonna be infinitely many others with non integer and integer coordinates that's not on our grid, but we're looking at that section of the graph.

Let's look at a point to see what happened.

If we substitute that into the expression 3x minus 2, so 3 lots of negative one minus 2, well, that's negative 5.

So our y coordinate is greater than the value of 3x minus 2 because 1, which is our y coordinate, is greater than negative 5.

It's important to make sure you're getting that the right way round.

Right, can you write down the coordinate of a point which satisfies the inequality y is less than 3x minus 2? Where are you looking on that graph? Right, we're looking in this section.

So you could have named any of these points.

2, 0, 3, 0, 4, 1.

There's loads there.

Of course you could have named any points not visible on the grid where the y value is less than the value of 3x minus 2, or you could have chosen any point with non integer coordinates in this region.

And there's infinitely many to choose from.

The quickest way to think about this is we know the line is y equals 3x minus 2.

So if you pick a value of x on the line and we want where y is less than 3x minus 2, you can just pick a y coordinate below that.

So you are looking below that line for the corresponding x value.

Jun says, "I think the values x equals 1 and y equals 6 satisfy this inequality." How could we check Jun's idea? Right, there's two ways to do this.

You can substitute or you can draw the graph.

Let's start by substituting.

So we put x equals 1 into the expression 2x plus 1, that gives us 3.

And our y coordinate is greater than the value of 2x plus 1.

So that does satisfy that inequality, y is greater than 2x plus 1 when x is 1.

Let's draw the graph as well.

So there's the graph of 2x plus 1.

So any point on that line will have coordinate y equals 2x plus 1.

We want y to be greater than 2x plus 1 and there's the coordinate 1, 6.

You can see that you need to go up from the graph at the point 1, 3 to get to our point 1, 6.

We can make more statements when we graph two equations.

So we've got two equations.

What do we know about this point? Well let's look at this one together.

This point satisfies the equation y equals 2x plus 1, because it's on the line y equals 2x plus 1.

But it also satisfies the inequality y is greater than 3 minus x.

We can check by substituting.

So we do 3 minus x, so 3 minus 2 is 1, and our y coordinate is 5.

So 5 is greater than 1.

y is greater than the value of 3 minus x.

Where can we find integer coordinates which satisfy y is less than 3 minus x and y is less than 2x plus 1? Pause the video.

Where are we gonna be looking? If we start by looking at 3 minus x, that's the line y equals 3 minus x, so we want y to be less than 3 minus x.

We're looking at y coordinates underneath that line for the corresponding x values.

The same for y equals 2x plus 1.

If I want y to be less than 2x plus 1, I'm looking for y values less than the corresponding y value for those x values.

If it needs to satisfy both, then it needs to be in that bottom region there.

Let's check one.

There's the coordinate 2, 0.

3 subtract 2 is 1.

So the y coordinate is less than the value of 3 subtract 2, 'cause our y coordinate is zero and that's less than 1.

Let's try the other equation.

Two lots of 2 plus 1 is 5.

Our y coordinate was zero and that's less than 5.

So we are looking in the correct part of the graph.

Your go.

Which point satisfies both y is greater than 2x plus 1, and y equals 3 minus x.

You've got three to choose from.

Well done if you picked A.

We can see it's on the line y equals 3 minus x so it satisfies that equation.

But also y has to be greater than 2x plus 1, so it needs to be above the line for that value of x.

Here's another one.

Which point satisfies both y is less than 2x plus 1, but also y is greater than 3 minus x? You've got four choices.

Well done if you said C.

You can see that y is less than 2x plus 1 because it's underneath that graph of y equals 2x plus 1, those y values are smaller than the y values on the line.

But it's above the line y equals 3 minus x, those y values are greater than the y value on the line for that value of x.

How about this one? Which statements are true for the point shown? What do you think? Well done if you said that y is greater than 3 minus x and y is greater than 2x plus 1.

Time for a practise.

I'd like you to write either the equals sign, the less than sign, or the greater than sign in the gaps to make these statements true for each point.

So you'll see there's a coordinate marked A on the grid.

I would like you to fill in the correct symbol in those equations to make it true for A, then do the same for B, C, D, E and F.

Then you'll also notice there's a point G.

Can you write your own statements for point G? Loads to do there.

Come back when you're ready for the next bit.

Well done.

For each pair of statements, which of the labelled points satisfies both statements? You've got some equations and some inequalities.

So for A, which of those points satisfies y equals 5 minus a half x, and y is less than 5 6x minus 1.

You've got the graphs drawn for you and you are picking from that selection of coordinates.

When you're happy with your answers, we'll look at the next bit.

Question 3: I'd like you to say whether each statement is true or false.

I've drawn the graphs of y equals 4x plus 3 and y equals negative x minus 2 to help you.

You might wanna plot some of the coordinates, see if they're in the regions described.

All you need to do then is write whether they're true or false.

Off you go.

And finally, Sofia says, "x equals negative one, y equals 3 satisfies the inequalities y is greater than 2x plus 1 and y is less than x plus 3." I'd like you to use whatever method you like to show that Sofia is incorrect.

Then can you mark a point on the graph which would satisfy Sofia's inequalities? Definitely drawing some lines is gonna help you there.

Give this one a go.

Well done.

Let's look at each point in turn.

So for A, y is greater than 4x plus 7, and y is greater than x plus 4.

For B: y equals 4x plus 7, and y is greater than x plus 4.

For C: so y is less than 4x plus 7, and y equals x plus 4.

For D: y is less than 4x plus 7, and y is less than x plus 4.

For E: y is less than 4x plus 7, and y is less than x plus 4.

And for F: y equals 4x plus 7, and y is less than x plus 4.

You're writing your own statements for point G, so you should have y is greater than 4x plus 7, and y is less than x plus 4.

Just check that you are using those inequality symbols correctly.

Okay, for A then, you are looking at point F.

For B, you're looking at point B.

For C, you're looking at point D, and for D, you're looking at point I.

You may wish to pause the video and just have a look at those and compare them to your answers.

For question three.

So let's plot these as we go.

x equals 1, y equals 3 is there, so it's not a solution to the equation y equals 4x plus 3.

x equals 2, y equals 4 is there.

So it does satisfy the inequality y is less than 4x plus 3.

For C, x equals negative 3, y equals zero is there.

And again, that does satisfy this inequality, which is y is less than negative x minus 2.

For D: x equals negative 2.

2, y equals negative 3 is there.

And that's false.

It does not satisfy the inequality y is less than 4x plus 3.

If we look at x equals 4, y equals negative 5.

5, that is there and that is also false.

y is not less than negative x minus 2.

We can see that y is greater than negative x minus 2.

Only a little bit greater than, but it is still greater than.

And finally, you had choices for how you wanted to show that Sofia was incorrect.

We could substitute into the expression, so there's 2x plus 1, so 2 lots of negative 1 plus 1 is negative 1.

And that bit did work.

The y coordinate, 3 is greater than the value of 2x plus 1.

3 is greater than negative 1.

Here's where the problem is.

Negative 1 plus 3 is 2, and that y coordinate is not less than the value of x plus 3.

3 is not less than 2, and we want to satisfy the inequality y is less than x plus 3.

You might have done this by drawing the graphs as well.

That's absolutely fine.

So there's the graphs of y equals x plus 3 and y equals 2x plus 1.

And her coordinate negative 1, 3, we can see is greater than both lines, and she wanted it to be less than x plus 3.

For B, you can have any of those coordinates.

Of course you can have any with non integer coordinates in that region as well.

Fantastic.

We've really tried to explore that idea of inequalities now, and what the idea of being greater than or less than on a graph looks like, and that'll definitely help you with your inequalities skills moving forwards.

If you want to pause the video and just check those things that we've looked at today, feel free to do that now.

Thank you for all your hard work and I really hope you choose to learn with us again.