Lesson video

Hello, Mr. Robson here.

Welcome to Maths.

Today, we're solving a quadratic and linear pair of simultaneous equations graphically, and you don't meet many people who don't love graphs.

This should be wonderful.

Our learning outcome is that we'll be able to relate the solutions (0, 1, or 2) of a linear and quadratic pair of simultaneous equations to the graphs of these equations.

Keywords today: linear and quadratic.

The relationship between two variables is linear.

If, when plotted on a pair of axes, a straight line is formed.

Quadratic, a quadratic is an equation, graph, or sequence whereby the highest exponent of the variable is two, the graph of which forms a parabola.

Three parts to our learning today, and we're gonna begin by solving graphically.

It's possible to solve a pair of simultaneous equations where one is quadratic and one is linear using substitution.

If I call 'em equation one and equation two, I can substitute equation one into equation two, and I get that.

By substituting equation one to equation two, we can create a single quadratic, which can be solved to find the x-values.

We can make this quadratic x-squared minus x minus 2 equals 0 by rearranging.

From there, we can solve by factorising, and we'll find that x equals positive 2, and x equals negative 1.

But we're not done there.

We need to find the y-values.

So we substitute 2 back into one of the original equations and find that y equals 3 when x equals 2, and when x equals negative 1, we find that y equals negative 3.

It's possible to solve the same pair of simultaneous equations graphically.

If we're going to solve it graphically, we need to plot the graphs.

We can do that with a table of values.

If I want to populate this table of values for y equals x-squared plus x minus three, I'm gonna start by substituting in the easiest x values: x equals 0, x equals 1, x equals 2, and I find those corresponding y coordinates.

I did the easiest x values first, because when I start to substitute in x equals negative 1, x equals negative 2, and those numbers come out as y values, I know I'm right, because we know a parabola will have a line of symmetry.

So there's no surprise to see symmetry in those y values.

When I'm populating the table for y equals 2x minus 1, and again, start with the easiest x values, x equals 0, x equals 1, x equals 2, because from there, we know a linear graph is gonna have a constant difference, so no surprise.

I see a constant difference in those y values.

From there, I need to plot this, and plot those linear coordinates and draw the straight line through them, and I can plop the coordinates of the parabola and draw a curve through those.

Our solutions are the points of intersections of the two lines.

In this case, the coordinates of the intersections are minus 1, minus 3, and 2, 3.

That's the point x equals 2, y equals 3, and the point x equals negative 1, y equals negative 3.

You'll notice x equals 2, y equals 3, x equals negative 1, y equals negative 3, the same pair of solutions as when we solved algebraically.

Quick check, you've got this.

I'd like you to use these graphs to solve this pair of simultaneous equations.

Pause and choose your solutions.

Welcome back.

Let's see how we did.

I hope you went for B, x equals negative 1, y equals positive 7.

That's the first intersection.

And C, x equals 2, y equals negative 5.

That's the second intersection.

Those are our two solutions for this pair of simultaneous equations.

In the case of A, the coordinate 1 negative 5 is on the parabola, but it's not on the straight line graph, so it only satisfies the quadratic equation.

And D is the opposite way round.

The coordinate 1 negative 1 is on the straight line, but it's not on the parabola, so it only satisfies the linear equation.

The solutions won't always be entered your values.

When solving graphically, sometimes we have to estimate.

In this case, y equals x-squared minus 3x plus 3, and y equals 1/2x plus 1.

The graphs look like so.

You can see that those intersections are not integer value coordinates.

That doesn't mean we can't find a solution or an estimate for the solution.

If we look at the intersections, we can estimate accurately to one decimal place with the scale on these axes.

The first intersection is x equals 0.

7, y equals 1.

4.

Our second intersection is x equals 2.

8 and y equals 2.

4.

Those solutions are accurate to one decimal place.

Quick check, you can do that now.

I'd like to estimate the solutions to this pair of simultaneous equations, giving your answers to one decimal place.

Pause and give me those two solutions now.

Welcome back, let's see how we did.

Hopefully, you identified the first intersection and the second intersection to B, at x equals negative 1.

5, when y equals negative 1.

3 and x equals 1.

2, and y is equal to 0.

7.

We can't read the exact answer from this graph, but we can read these solutions which are accurate to one decimal place.

Practise time now.

Question one, I'd like you to complete the table of values for the graph of y equals x-squared minus x minus 6, and the graph of x minus y equals 3.

Once you've completed the table of values, part B is to plot the graphs of the two lines.

And then part C is to find solutions which satisfy both equations simultaneously.

Pause and give those things a go now.

Question two.

This is the graph of y equals 2x-squared minus 6x plus 1.

Part A is for you to draw the graph of y equals 5 minus 4x.

You could do a table of values if you wish, but at this moment to remind you, y equals 5 minus 4x is a linear graph.

You only need two coordinates to draw that straight line.

For part B, you're gonna use your graphs to find solutions to this pair of simultaneous equations.

Pause and do that now.

Question three.

I'd like to use these graphs to estimate the solutions to this pair of simultaneous equations to one decimal place.

Pause and write down those two solutions now.

Feedback time now.

Completing the table of values, we should have got those values for the line y equals x-squared minus x minus 6, and for x minus y equals 3.

Those y values, plotting those on the graph would look like so.

Then your solutions, which satisfy both equations simultaneously will be the intersection when x equals negative 1, y equals negative 4, and the second intersection when x equals 3, y equals 0.

For question two, we have the graph of y equals 2x-squared minus 6x plus 1, and we're asked to draw the graph.

Y equals 5 minus 4x.

You only need two coordinates.

So we could find the coordinate when x equals 0.

y equals 5, when x equals 1, y equals 1.

You might have found another coordinate when x equals 2, y equals negative 3, or you might have found when x equals negative 1, y equals positive 9.

But those coordinates would've been plotted there and given you that straight line.

Once you've got that straight line, you can see the two intersections.

We'll use those intersections to find the solutions to this pair of simultaneous equations.

Our first solution is x equals negative 1, y equals positive 9, and x equals positive 2, y equals negative 3.

Those are the only pairs of values which satisfy both equations simultaneously.

For question three, we can't read an exact answer from this graph, but we can use the graph to find good estimates to one decimal place.

We can see the two intersections, and reading from those intersections, we'll get an estimated solution, x equals 0.

5, y equals 1.

3, and x equals 3.

3, y equals negative 0.

9.

Onto the second part of our learning for today, the number of solutions.

Laura and Lucas are discussing these graphs, the parabola, y equals x-squared plus x plus 2, and the linear graph of y equals 3x plus 2.

We can see them on the graph.

We can see the two intersections and the two solutions.

Laura says, "Because of the shapes of quadratic and linear graphs, there will always be two intersections, and therefore, two solutions." Lucas says, "I disagree.

I think there might be special cases." What do you think? Is Laura right? When we see a quadratic and linear pair of simultaneous equations, will we always get two solutions? Or is Lucas onto something? Pause, have a conversation with the person next to you, or a good think to yourself.

Welcome back.

I wonder what ideas you came up with.

Lucas has an idea.

If we draw the line of y equals 3x plus 1 instead of y equals 3x plus 2, you'll see what I mean.

We can draw y equals 3x plus 1, table of values, coordinates, and the straight line.

Laura has noticed something.

There's only one intersection.

The line of y equals 3x plus 1, only such as a quadratic ones, that is one intersection, and therefore, just one solution for this pair of simultaneous equations.

Well done, Lucas.

Well done, Laura.

One solution at x equals 1, y equals 4.

Laura is onto something else now.

"I've realise something.

Let's plot y equals 3x on the same graph." Can you see what Laura has now realised? Pause and have a think.

What's gonna happen when we plot y equals 3x? Welcome back.

Again, I wonder what you came up with.

You're on the same wavelength as Laura and Lucas here.

When we plot y equals 3x, Lucas realises, "That's right, Laura.

It's also possible that the lines do not intersect.

There is no solution that satisfies both of these equations simultaneously." When one equation is linear and the other is quadratic, it is possible for a pair of simultaneous equations to have two solutions, in the case of y equals x-squared plus x plus 2, and y equals 3x plus 2.

One solution in the case of y equals x-squared plus x plus 2 and y equals 3x plus 1, or as we saw, no solutions whether line does not intersect the parabola at all.

That was the case for y equals x-squared plus x plus 2 and y equals 3x, so sometimes, we see two solutions, sometimes, we see one, and sometimes, we have no solution at all.

Quick check, you've got that now.

True or false, when solving simultaneous equations, if one is linear and one is quadratic, there will always be two solutions.

Is that true or is it false? Once you've decided, can use one of those two statements to justify your answer? Pause and make your choices now.

Welcome back, let's see how we did.

I hope you said false.

I hope you justified that with it is possible that the straight line will touch the parabola once, giving us one solution, or not at all, meaning there are no solutions.

Well done.

Practise time now.

Y equals 3x-squared minus 4x plus 4 is drawn for you.

In each case, draw the linear graph and declare how many solutions for each pair of simultaneous equations.

I don't necessarily want those solutions.

I just want to know how many solutions there are in each case.

Pause and do this now.

Question one part C and part D, very similar to parts A and B, y equals 3x-squared minus 4x plus 4 is drawn for you.

I just want to know when we plot 4x plus y equals 8, how many solutions are there? And when you plot y equals 8, x minus 8, how many solutions are there? Pause.

Tell me the number of solutions now.

Feedback time.

Question one, part A, plotting x plus 4y equals 4.

We can do that with just two coordinates, when x equals 0, y equals 1.

When y equals 0, x equals 4, there is the line x plus 4, y equals 4, no intersection.

Therefore, there are no solutions for that pair of simultaneous equations.

For part B, 4x plus y equals 4.

We can find two coordinates to plot that line, when x equals 0, y equals 4.

When y equals 0, x equals 1.

Plot the line and we find one intersection, one solution.

For part C, 4x plus y equals 8.

Some coordinates, we could plot 0, 8 when x equals 0, y equals 8, and when y equals 0, x equals 2.

When we plot that line, we'll find two solutions.

You can see one intersection on our graph, and you can see there's going to be another intersection.

There's two solutions to that pair of simultaneous equations.

For part D, y equals 8x minus 8.

We can plot the coordinates 1, 0 and 2, 8, giving us that line where there is one intersection and one solution.

Onto the third and final part of our learning for today, the roots of the combined quadratic.

That sounds exciting.

Let's take a closer look.

We have many options when solving a pair of simultaneous equations.

For example, x-squared plus 2y equals 4, and 2x plus y equals 2.

We could graph them and see the two intersections at x equals 0, y equals 2, and x equals 4, y equals negative 6.

We could have solved algebraically by substitution if I label them equation one and equation two.

Rearrange equation two to make y the subject, substitute it into equation one, and we get x-squared plus 2 lots of bracket, 2 minus 2x equals 4.

Multiply out the bracket, simplify, we get x-squared minus 4x equals 0.

We call that a combined quadratic.

If we've gone to solve the same pair of simultaneous equations by elimination, again, we label them equation one and equation two.

We can double equation two, because we want a common coefficient.

And then when we do equation one, subtract two lots of equation two, we're left with x-squared minus 4x equals 0.

Oh, look, we got the exact same combined quadratic.

From there, we could factorise.

And we'll find the solution, x equals 0.

When x equals 0, substitute that back in to find the corresponding y value of 2, and x equals 4, which allow a corresponding y value of negative 6.

The same solutions as our graphical solution.

The same pair of simultaneous equations, that's the graph you saw a moment ago with the two intersections, the two solutions.

The combined quadratic that we found when solving algebraically by elimination or substitution, x-squared minus 4x.

X-squared minus 4x would make that parabola.

If I put all those things on the same graph at the same time, what do you notice? Maybe pause.

Have a look.

What's going on here? I wonder what ideas you came up with.

When the graph for the combined quadratic is also drawn, we can see the solutions to our pair of simultaneous equations at those intersections and we can see what we call the roots of the combined quadratic there.

i.

e.

x-squared minus 4x when z equal 0 at those points.

Have you noticed? There in line, a combined quadratic has its roots where the quadratic and linear equations intersect.

That's when x equals 4, and when x equals 0.

By roots, if you haven't heard that word before, I mean, the roots are where the curve intersects the x-axis.

Note that the roots of a combined quadratic only gives us x-values.

Once we have those, we need to substitute them back in to find the corresponding y-values in order to solve our pair of simultaneous equations.

Quick check, you've got that now.

I'd like you to finish this sentence.

When solving simultaneous equations where one is linear and one is quadratic, the combined quadratic, how are you gonna finish that sentence? Will it be with option A, option B, or option C? Pause and make your choice now.

Welcome back.

I hope you said it's options C.

The combined quadratic has its roots where the quadratic and linear equation intersect.

Practise time now.

I'd like you to fill in the blanks.

There's one, two, three blanks in that lovely piece of mathematics there.

You're gonna need to use that graph to support your thinking on this one.

Go ahead, pause, and fill in those blanks.

For question two, I'd like you to estimate the roots to one decimal place for the combined quadratic for this pair of equations.

Pause and make that estimation now.

Feedback time now.

For question one, I asked you to fill in the blanks.

The first part was all about the intersection of the straight line and the parabola x-squared plus 2y equals 13.

You can see intersections at x equals negative 3, y equals 2, and at x equals 1, y equals 6, these two intersections.

The second part, the combined quadratic, x-squared plus 2x minus 3 equals 0 has roots at x equals negative 3 and x equals 1.

The combined quadratic has roots where the quadratic and linear equations intersect when x equals negative 3 and when x equals 1.

You might wanna pause and make your graph look just like mine and copy down that sentence.

For question two, I ask you to estimate the roots for the combined quadratic for this pair of equations.

The combined quadratic has its roots where the quadratic and linear equations intersect, in this case, at those intersections when x equals negative 1.

8 to one decimal place and when x equals 1.

6 to one decimal place.

Sadly, that's the end of the lesson for today.

What have we learned? The number of solutions of a linear and quadratic pair of simultaneous equations could be 0, 1, or 2.

This is determined by whether their graphs have 0, 1, or 2 intersections.

When solving a linear and quadratic pair of simultaneous equations, the combined quadratic will have roots, whether linear and quadratic graphs intersect.

Wonderful stuff.

I hope you enjoyed it as much as I did, and I look forward to seeing you again soon for more mathematics.

Goodbye for now.