Lesson video

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

My name is Ms. Davies 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 are 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 on Checking and Securing Understanding of Simultaneous Equations.

We're gonna look at solving simultaneous equations graphically.

So by the end of this lesson, you'll be able to recognise that the point of intersection of two linear graphs satisfies both relationships and hence represents the solution to both those equations.

If you need a reminder of what simultaneous equations are or what I mean by a linear graph, pause the video and have a read through that now.

So gonna start then by solving simultaneous equations graphically.

Alex is looking for solutions to this equation.

He says, "I think x equals three and y equals 10." How could we check his answer? Pause the video and have a think.

Right, well, we can substitute the values and see if the equation balances.

So we do that now.

We have 10 equals two lots of three plus four, so that will be 10 equals six plus four which is 10 equals 10, so that does balance.

Laura says, "But also x equals one and y equals six would work," let's check that.

So two lots of one would be two plus four would be six, so that would also balance.

When an equation has two variables, so you see here we've got x and y, there are infinitely many solutions.

Lots of values of x and y will make that balance, infinite amount.

If we draw the graph of y equals 2x plus four, we'll be able to see more of the solutions.

So there's the graph.

It's a linear equation, so it is a linear graph.

We've got a straight line.

It has gradient two and y-intercept four.

The coordinates of any point on this line will be a solution to that equation.

So there's 3, 10 which was Alex's solution.

That's 1, 6 which is Laura's solution and we've got some others as well, and that includes non integer coordinates.

There's negative 1.

25 and 1.

5 as well.

Right, so when we have two equations with the same two variables, we can sometimes solve them simultaneously.

So there wasn't just one solution when we had one equation.

There was infinitely many solutions.

But when we have two equations, sometimes there is one solution that satisfies both.

So what we're looking for is values of x and y which balanced both equations.

Pause the video, can you find any solutions which work for both equations? Give it a go.

If you found them, come back.

If you don't think you found them, don't worry.

We'll look at how we can do it together.

Right, so you might have found that x equals two and y equals eight satisfies both equations.

Now these can be really hard to find by just trying values.

I tried to pick a fairly nice one.

If the answers were negative or non-integers, it might have been really, really tough to find them just by picking values and doing a bit of trial and improvement.

Instead we can solve them algebraically or graphically.

You might remember solving these algebraically before.

Today, our main focus is how we can solve these on a graph.

So that was our graph of y equals 2x plus four.

If we also draw the graph of y equals 3x plus two, the point of intersection because you see there's one coordinate where both lines intersect, that shows us a value of x and a value of y which will satisfy both equations.

And that's x equals two, y equals eight and that was the answer that we got on the previous slide.

So this is a solution to this pair of simultaneous equations.

It satisfies both equations at the same time.

It is that point of intersection.

Right, Laura wants to solve these simultaneously.

She says "The solution is x equals one." Pause the video, do you agree? You might have said that the solution is when x equals one, but she should also include the y-value as well.

So we look at the y-coordinate, y is four.

So that point of intersection is 1, 4, so x equals one, y equals four is the solution to both equations simultaneously.

Right, quick check, which solution satisfies this pair of simultaneous equations? Off you go.

We can see from the graph the point of intersection is 0, 2, so it's x equals zero, y equals two.

Here's another one for you to have a go at.

Right, make sure you've got your negatives and your positives correct this time, so we're looking at negative 2.

5, positive 4.

5.

So the solution is x equals negative 2.

5, y equals 4.

5.

So efficiently drawing linear graphs can help to find solutions to pairs of linear simultaneous equations.

So we're gonna do a quick recap on how to draw linear graphs.

So I want to find a solution to these equations simultaneously, so one way would be to use a table of values to draw the graph.

So if I want to draw y equals four minus x, I've picked some x coordinates, negative 1, 0, 1, 2, 3, and then I can substitute into the equation, so y equals four minus x.

I always start with zero and the positive values and then go back and fill in the negative values.

So four minus zero is four, four minus one, and so on.

It's a linear graph, so I should have a linear relationship between my values.

Let's plot those and I get a straight line.

Right, if I want to do a y equals 1/3x.

Now technically we only need two coordinates, although it's always a good idea to have a third coordinate in case you've made a mistake and then you can check if they line up on a straight line.

The coordinates I'm gonna pick here because I'm trying to third the x-value, it'd be really helpful if there were multiples of three.

Then I get nice integer answers.

So I've picked negative 3, 0, 3, and 6.

Zero divided by three is zero.

Then I've got 1, 2, and negative 1 and again, if I plot those and it makes a straight line, I can be confident that I've got my values correct.

The solution to these equations simultaneously then is x equals three and y equals one.

Right, have a think.

What could we do to solve this pair of simultaneous equations graphically? Can you spot any problems we're gonna need to overcome? What would you do? Alex says, "We can make y the subject and use a table of values." What Alex has noticed is at the moment we haven't got them in the form y equals mx plus c which makes them harder to plot.

So if we rearrange the equations to make y the subject, then we can do what we did previously and use a table of values.

So for the first one, if I subtract 3x from both sides, I get y equals five minus 3x.

Our table of values then will be nice and easy to produce.

I just need to do five subtract three lots of the x-coordinate.

For the second one, if I divide all those terms by three, I get x minus y equals three.

Then if I add y to both sides, I get x equals three add y, and then I can subtract three from both sides.

So y equals x subtract three.

A little bit more rearranging to do that time, but this is a really easy equation to draw now.

All I need to do is subtract three from all the x-coordinates.

So there's my table of values for that one, and now I can draw those.

The solutions then is gonna be x equals two, y equals negative one.

We can actually see those from the table.

We didn't need to draw the graphs, but drawing the graphs can be really helpful.

Now a table of values is not always the most efficient method, so let's look at another one.

So I've got these two simultaneous equations.

I don't really want to rearrange them to make y the subject, then fill in a table, then plot them if I can help it, so we can do what Laura suggests.

We can just plot the x and y-intercepts.

So if you look at this equation, if we want to find the y-intercept, that's where it crosses the y-axis, that's when x is zero.

So if I just put x equals zero into my equation, that means I've got 2y plus four lots of zero equals 12, so 2y is 12 and y is six.

So when x equals zero, y is six, so I can plot that coordinate.

Now I can find the x-intercept 'cause that's when y is zero.

So when y is zero, I've got 4x equals 12 and x is three.

Now the only thing with this method is we haven't got a third coordinate to check, so you need to make sure that you're being nice and accurate.

Let's do the same with the other one.

When x is zero, y is gonna be one and when y is zero, x is gonna be negative two, and then we can plot that graph as well.

We can see now that our solution is x equals 2, y equals two.

Quick check to see if you're able to find the y-intercept and the x-intercept.

I've given you a start point.

Can you finish those? Off you go.

Let's have a look.

So the y-intercept is when x is zero, so that means 3y equals 18, so y is six.

Now I've asked for the y-intercept so I should really write the coordinate.

The coordinate is 0, 6.

For the x-intercept, well, that's y is zero, so that means 2x would be 18 and x would be nine.

Again, let's write that as a coordinate, 9, 0.

Right, I'd like you to try and make y the subject of this equation.

What could we do? Off you go.

So if we start by subtracting 2x and then divide in by three, so you get six minus 2/3x.

I wonder if you were drawing this graph, whether you'd use the x and the y-intercept or whether you'd use the form y equal mx plus c and use the gradient and the y-intercept.

Have a little think about which one you would do.

Right, time for a practise.

I have drawn the graphs for you.

All you need to do is find the solutions to these simultaneous equations.

Remember that when there's two variables, you should have a solution for x and for y, off you go.

Good start, this time you need to draw the graphs.

I've given you some table of values in case you want to use those to plot your graphs and don't forget that I want the solution to this pair of equations simultaneously.

Give that one a go.

Right, same idea, you can use whatever method you want to draw these graphs and then find the solution.

And finally, I've drawn one of the lines for you.

I'd like you to draw the graph of 3y plus 6x equals 24 and then again find the solution to those equations simultaneously.

When you're ready for the answers, come back and we'll look at it together.

Well done, so we're looking for the points of intersections.

Just make sure you've looked at the correct lines.

So you should have x equals three, y equals six, for b, x equals one, y equals zero, and for c, x equals negative one, y equals two.

Now go check your table of values and your lines for that question, and your solution should be x equals one, y equals one.

Again, pause the video and check your lines for these.

Your solutions should be x equals negative one, y equals three, for b, x equals negative four, y equals two, and for c, x equals 4, y equals one.

And finally, I think the most efficient way to draw this would be to use the x and the y-intercepts.

So when x is zero, y is eight.

When y is zero, x is four, and that allows me to draw in my graph.

Other methods are absolutely fine.

My solution then is when x equals three, y equals two.

Well done, we're now gonna have a look at the number of solutions.

So Alex is trying to solve these equations simultaneously.

He says any linear equation in x and y can be plotted on a graph, so any pair of equations can be solved simultaneously.

He decides to solve them algebraically.

So because they're both equal to y, they must both be equal to each other.

So 3x plus four must equal 3x minus two.

He's added two to both sides and then he subtracted 3x from both sides.

But then he gets this, six equals zero.

That doesn't make sense.

What has gone wrong? Pause the video, can you work out why this hasn't worked? Well, I don't know if you noticed that there will be no solution for x and y which works for both equations and that's why there was a problem when he tried to solve it algebraically.

We can't have a value that when you multiply it by three and add four, it's the same as when you multiply it by three and subtract two, just doesn't exist.

We can see this better if we plot them on the graph, so let's see what these equations look like when plotted.

So you've got 3x plus four, that's the graph that crosses the y-axis at four, and we've got 3x minus two, that's the graph that crosses the y-axis at negative two.

Now they both have a gradient of three and we can see that from the equation.

Because they're in the form y equals mx plus c, we know that the coefficient of x is the gradient.

So they both have a gradient of three which means they're parallel.

We can see that on the graph, so they're not going to intersect.

So when two lines have the same gradient, but different y-intercepts, their equations will have no solutions simultaneously 'cause they're gonna be parallel and not have that point of intersection.

Now if the equations are not in the form y equals mx plus c, then rearranging them can make it easier to see the gradients in the y-intercepts.

Pause the video and make a prediction.

Is there a solution to this pair of simultaneous equations do you think? And then we'll check it together.

Okay, let's have a look.

We rearrange them, so I've subtracted 4x from both sides of the first one, then divided by two.

I get y equals three minus 2x.

For the other one, if I add 2x to both sides, I get y equals 2x plus five.

Now they're now in the 4y equals mx plus c.

Their gradients are different.

The first one has a gradient of negative two and the second one has a gradient of two.

Because they have different gradients, they will not be parallel.

So they will intersect and there will be a solution.

Let's have a look on the graphs.

So there's the graphs and you can see that there is a solution.

It's when x is negative 0.

5 and y is four.

Quick check, the equations y equals 2x plus four and y equals 3x plus four will have a solution when solved simultaneously.

Is that true or false? And can you think of a sentence to justify your answer? Off you go.

Well done if you said true.

They have different gradients, so will not be parallel.

Therefore they will intersect and there will be a solution.

How about this one? The equations 2y equals x plus five and y minus x equals two will have a solution when solved simultaneously.

Is that true or false? And again, what is your justification for your answer? You need to do a little bit more work for this one, but it was true.

If we rearrange the first one, we get y equals 0.

5x plus 2.

5 or y equals 1/2x plus five over two.

The second one, if you just add x to both sides, you get y equals x plus two.

So we can see that they have different gradients and they're not gonna be parallel, therefore they will intersect and there will be a solution when solved simultaneously.

Laura is looking at another pair of simultaneous equations.

She says there's a solution to both when x equals one and y equals 12.

We can check this by substituting, so let's do it together.

So if we have y is 12, we've got 12 equals two lots of one plus 10 which is 12 equals two plus 10 which is 12, so that balances.

Let's try the other one.

So two lots of 12 minus four lots of one equals 20.

That's 24 minus four is 20 and that is true, 20 is 20, so the solution satisfies both equations.

Alex says, "But x equals two and y equals 14 would also work for both.

I thought two lines only intersect once." If you want to pause the video and check that Alex is right, do that now.

Right, he's correct, there's two different solutions here that both satisfy both equations.

If we rearrange the second equation, we'll be able to see why we found two solutions that satisfy both.

So that second equation, 2y equals 4x plus 20, y equals 2x plus 10.

Right, can you see what's happened now? The equations will have the same gradient and the same y-intercept when plotted.

There's y equals 2x plus 10 and there's 2y minus 4x equals 20.

They are the same line.

They're gonna lie on top of each other.

That means any values of x and y which satisfy one will also satisfy the other.

So there are infinitely many solutions, just like that first one we looked at in the beginning of the lesson where we just had one line, we said there was infinitely many solutions.

If two lines lie over each other, then they're gonna have infinitely many solutions simultaneously.

Quick check, I'd like you to select the pair of simultaneous equations which will have infinite solutions, off you go.

If we rearrange the second one, we get y equals 3x plus three and now we can see that it's b and c.

They have the same gradient and the same y-intercept, so they'll have infinitely many solutions simultaneously.

Let's have a go at this one.

Select the pair of simultaneous equations that have infinite solutions, off you go.

Well then if you started by rearranging these, so the top one should be y equal 1/2x plus four, and the bottom one, y equals 1/2x minus two.

They've all got the same gradient.

The ones that are gonna have infinitely many solutions are a and b.

c is gonna be parallel to those, but have a different y-intercept which means there'll be no solutions to a and c and no solutions to b and c.

If we look at it on the graph, that's 2y minus x equals eight and y equal 1/2x plus four, and you can see that they lie on top of each other.

If I plot the other one, we can see that it's parallel with a different y-intercept so we'll have no solutions when solved with the other two.

Time to put all that into practise.

I'd like you to draw the graph of both equations and state any solutions when solved simultaneously.

If you don't think there are any solutions, then you can write "No solutions." For question two, for each pair of simultaneous equations, how many solutions are there? I want you to write a sentence to explain your answer.

For some of them, you might need to rearrange them first to see what's happening, off you go.

Question three, Laura's trying to solve this pair of simultaneous equations.

She's drawn them on a graph and says there are no solutions to these simultaneous equations.

Can you explain why she is incorrect? And then show by substitution that there is a solution when x equals 48 and y equals 39.

Give that one a go.

Then come back for the answers.

Well done, let's have a look at those answers.

So you might wanna pause the video and check your graphs.

For the first one, there's a solution when x equals zero and y equals one.

For b, there's a solution when x equals one and y equals zero and for c, we've got parallel lines.

There are no solutions.

For question two, for a, there will be one solution because the gradients are different.

So they will intersect at one point and that will be the solution to both equations simultaneously.

For b, if you rearrange that second one, you get y equals 2x plus one.

There'll be infinite solutions.

They've got the same gradient and the same y-intercept, they'll lie on top of each other.

That means there's infinite solutions.

For c, you need to do a bit more rearranging this time.

You might have spotted that both rearrange to y equals 2/5 of x plus 3/5.

That means they're gonna have infinite solutions again.

They've got the same gradient and the same y-intercept.

And finally, the first one rearranges to y equals 2.

5 subtract 3x and then y equals five subtract 3x for the other one.

Now there's no solutions to these.

They've got the same gradient, but they've got different y-intercepts.

So they're gonna be parallel lines with a gradient of negative three, but they're not going to intercept.

They've got different y-intercepts.

And finally, Laura is incorrect 'cause the lines are not parallel.

They may look to be parallel if you looked at the graphs very quickly.

But if you look closely and actually draw on your gradient triangles, you'll see that they're not parallel.

An easier way to do that will be to look at the equations.

One of them has a gradient of 3/4 and the other has a gradient of 5.

6, so they're not going to be parallel.

There will be a point of intersection.

It's just not visible on this section of the graph.

The point of intersection is 48, 39 and we're gonna show that now.

So if you start by substituting it into the equation at y equals 3/4x plus three.

We've got 39 equals 3/4 of 48 plus three, so 39 is 36 plus three and that balances, 39 is 39.

For the other equation, 39 equals 5/6 of 48 subtract one.

So 39 equals 40 subtract one, so 39 is 39.

Well done if you practise using your fraction skills to do your fractions of an amount.

Fantastic, we've had a really good review today of all our graphing skills, how to draw graphs, how to look for that point of intersection, and then what that means in terms of the solution to a simultaneous equation.

If you want to read through all the things we've looked at today, feel free to pause the video now.

But you're now in a fantastic place to carry on with your work on inequalities.

This is gonna be really useful when you are graphing inequalities as well.

I really hope you decide to join us for another lesson.