Lesson video

Hello, Mr. Robson here.

Welcome to maths.

Today we're problem solving with graphical representations.

And what is maths for if it's not for problem solving? Let's get stuck in.

Our learning outcome is that we'll be able to use our knowledge of graphical representations to solve problems. Many key words we're gonna see in today's lesson.

Linear.

The relationship between two variables is linear if when plotted on a pair of axis, a straight line is formed.

Gradient.

The gradient is a measure of how steep a line is.

Intercept.

An intercept is the coordinate where a line or curve meets a given axis.

We'll also see the keyword parabola.

Parabola is a beautiful word and parabolas are beautiful things.

A parabola is a curve where any point on the curve is an equal distance from a fixed point called the focus and a fixed straight line called the directrix.

The line of symmetry goes through the focus, at right angles to the directrix.

Two parts of today's lesson today.

We'll begin by using sketches to locate intersections.

Andeep and Jun are solving equations.

They solve 68 - 3x = 2x + 18.

Andeep says, "I've got the solution x = 10." Jun says, "But I've got the solution x = -50." Our teacher has told us that one of us is right but we're not sure who.

How would you help them? Pause, make a suggestion to the person next to you or perhaps say it aloud to me on the screen.

Welcome back.

Substitution, did you suggest that? If so, well done.

That's one way to help identify who is right.

When we substitute, we find that only Andeep's solution makes the equation balance.

When we substitute in the x value of 10, we get 68 - 30 on the left-hand side, 20 + 18 on the right-hand side, they're both equal to 38.

We've got balance in our equation.

X = 10 must be the solution.

You would not have found that same balance if you'd substituted in Jun's solution, x = -50.

But there is another way, and we mathematicians like multiple methods for solving the same problem.

A simple sketch of these two equations will help us to approximately locate the solution.

If I try to sketch the line y = 2x + 18, there's things I know about that line.

It'll have a y-intercept of 0,18, I'll approximate that to be there, and a gradient of 2.

I'll approximate a gradient of 2 to look like that.

For the line y = 68 - 3x, we know it's got a y intercept of 0,68.

I'll approximately place that there.

And a gradient of -3.

That's a negative gradient, slightly steeper than the gradient I drew earlier.

Our sketch shows us that the lines intersect in the first quadrant, therefore, our solution must be positive.

The x-coordinate of that intersection is positive.

Andeep's delighted.

Andeep says, "We can see that the solution has to be positive.

It must be mine, x = 10." Remember, Jun's had a negative solution.

It can't be that one.

Quick check if you've got that now.

This sketch tells us what about the solution to this equation? 8x + 128 = 39 - 4x.

Does it tell us A, the solution is positive, B, the solution is negative, or C, there is no solution.

Pause this video, have a think about this problem.

Welcome back.

I hope you said it's option B, the solution is negative.

X values in the second quadrant are negative, and we can see the intersection is in that second quadrant.

A sketch can also help us to approximate the location of the intersection when solving pairs of simultaneous equations, such as 10x + 6y = 30 and 2x + 5y = -10.

If I wanna sketch one of these equations, I need to find some coordinates.

For the first equation, when x = 0, 6y = 30, y must equal 5.

I'll plot the coordinate 0,5 there.

When y = 0, 10x must equal 30, x must be 3.

I'll plot the coordinate 3,0.

Note how my coordinate 3,0 is closer to the origin than my coordinate 0,5.

Once I've got those points sketched, I can draw that line.

For the second equation, when x = 0, y = -2.

I'll plot that there.

Notice closer again to the origin.

And when y = 0, x = -5.

And you'll notice I've sketched that at the exact same distance from the origin as my coordinate 0,5.

I can join those two with a line.

And there we have an intersection of the two linear equations.

Our sketch locates the intersection in the fourth quadrant.

That tells us the solution to this pair of simultaneous equations must be a positive x value and a negative y value.

You can also read from here that that x value must be greater than 3 and the y value must be less than -2.

That's a useful piece of approximation.

Let's check if you've got that.

The solution to this pair of simultaneous equations is in which quadrant? Is it the second quadrant, the third quadrant, or the fourth quadrant? Pause, tell the person next to you or say your answer to me aloud at the screen.

Welcome back.

I hope you said it's option B.

The solution lies in the third quadrant.

What does that then tell us? If the solution is in the third quadrant, what does that tell us about our solution? Is it A, the x value is negative while the y value is positive? Is it option B, the x value is negative, the y value is negative? Or is it option C, the x value is positive and the y value is negative? Pause and take your pick.

Welcome back.

I hope you said it's option B, the x value is negative and the y value is negative.

X and y values are both negative in the third quadrant.

It's not just sketches of linear equations that solve problems for us.

Laura sketches this parabola to model her throw of a cricket ball.

A good sketch always contains important values.

Laura knows when she's got the ball in hand, it's at 1.

5 metres above the ground, so she marks that coordinate.

She also records the highest point that the ball reaches.

When it's 16 metres away from her horizontally, it's eight metres high.

She also records that it's landed 36 metres away from her horizontally, so she can mark that coordinate.

We can draw a line of h equals six metres, that's height equals six metres to ascertain how many times the ball was six metres above the ground.

H = 6, we can approximate to be there.

And the two intersections between the parabola and that horizontal line tell us there were two moments when the ball was six metres above the ground.

Quick check if you've got that.

What does this line h = 10 tell us? Pause, have a conversation with a person next to you or a good thing to yourself.

What does that line tell us? Welcome back.

I hope you said no intersections between the parabola and that horizontal line tells us that the ball never reached a point where it was 10 metres above the ground.

Practise time now.

For question 1, I'd like you to sketch two lines for each equation to decide if the solution is positive or negative.

Don't forget to label key features like the y-intercept when you are sketching lines.

Pause and try this problem now.

Question 2, what equations does this sketch help us to solve and what can we see about the solution? I'd like you to write at least two sentences to answer that.

Pause and try that now.

Question 3, place and draw appropriate lines on this graph and identify intersections to answer these questions about Laura's improving cricket throw.

Question A, how many times was the ball at a height of five metres? Question B, how many times was the ball at a height of one metre? And for question C, how many times was the ball at a height of 10 metres? Pause and try this now.

Feedback time now.

Question 1, I asked you to sketch two lines for each equation to decide if the solution is positive or negative.

For part A, the equation 15 - 9x = 6x + 31.

6x + 31, that's a y-intercept of 0,31 and a gradient of 6.

For 15 - 9x, that's a y-intercept of 0,15 and a gradient of -9.

Once we sketch those two lines, we can see the intersection is in the second quadrant, so the solution for x must be negative.

For part B, the equation 9 - 15x = 6x - 31.

For 6x - 31, we've got a y-intercept of 0,-31 and a gradient of 6.

For 9 - 15x, a y-concept of 0,9 and a gradient of -15.

This tells us that the intersection is in the fourth quadrant, so the solution for x must be positive.

For question 2, I asked you what equations does this sketch help us to solve and what can we see about the solution? You might have written, "this sketch shows us that the intersection of the pair of simultaneous equations 5y - 3x = 15 and 2x + 3y = -6 lies in the second quadrant.

For the solution, the x value must be negative and the y value positive." If you wrote something like that, well done.

That's lovely maths.

For question 3, I ask you to draw appropriate lines in the graph, identify intersections, and answer these questions about Laura's improving cricket throw.

Question A, how many times does the ball at a height of five metres? We need to draw the line height = 5.

From there, we can see there's two intersections.

Therefore, twice.

For B, how many times is the ball at height of one metre? There's my approximation for the line h = 1.

There's one intersection.

Therefore, is at that height once.

For part C, how many times is the ball at a height of 10 metres? There's my line h = 10, and we should see one intersection.

Therefore, it was at that height once.

Onto the second half of lesson now where we're going to be matching sketches to real-life models.

Sometimes we need to use our understanding of mathematics to be able to match models to real-life scenarios.

Here's Sofia, and she says, "These models are from a race we do in PE.

One model is from today, the other from a few years ago." Can you tell which is which? What's your mathematical intuition telling you? Hm, interesting.

I hope you said, "The line with the steeper gradient means a faster speed, therefore, we can assume that one is from today and that one is from a few years ago." Alex also has two graphs to represent his time today and it's time from a few years ago.

At first glance, it looks like there's no difference and Alex has made no improvement, but good sketches in maths include some important values.

When we add these to Alex's graph, we can tell which is which.

I label those distance axes with 50 and 100 metres, but look at how I've labelled the time axes.

This information now shows us that the graph on the left is Alex's new improved race.

Quick check if you've got that.

These models show the time taken for the first 50 metres of a race.

Which one shows the fastest time? Is it line A, line B, or line C? Pause, tell a person next to you or tell me aloud at the screen.

Welcome back.

I hope you said option A.

Why? Because the steepest gradient reflects the fastest speed.

In real-life distance time models, we'll see curved lines.

A straight line represents a constant speed.

We won't always travel at a constant speed, so we'll often see curved lines.

This graph shows a pupil running a 200 metre race on sports day.

What is happening here at that moment in the graph and how do you know? Pause this video, have a conversation with a person next to you or really good think to yourself.

Welcome back.

I hope you said something along the lines of, "The gradient is getting less steep, which tells us that the pupil is slowing down." You can actually see their race go through several identifiable periods.

Right at the beginning, the line is getting steeper, we're accelerating, at its steepest, we're moving very fast, then we're slowing until we're running slowly.

Let's check if you've got that.

A firework being let off is being modelled by the distance time graph.

What is happening to it? Is it A, it has a constant speed, B, it is accelerating, getting faster, or C, it is decelerating, getting slower? Pause, have a think about this one.

Welcome back.

I hope you said it's option B, it is accelerating, getting faster.

How do we know? A constantly increasing gradient on a distance time graph reflects acceleration.

Practise time now.

Question 1, I'd like you to use these blank axes to draw three race events.

Race A is a short distance, fast race run at a constant speed.

Then you'll draw a separate line for B.

That'll be a middle distance, medium pace run at a constant speed.

And then another line for C, a long distance slow race run at constant speed.

A point to note, you need not label these axes.

However, any reader looking at your graph should be able to tell your three races apart.

Pause and try that now.

Question 2, this graph shows two animals running.

One is a human, one is a cheetah.

Identify which line represents which and explain how you know.

I'd like you to point to more than one key moment in the graph to support your explanation.

Pause and try that now.

Feedback time now.

For question 1, I ask you to use these axes to draw three race events.

I'm gonna start by drawing race C.

Why? Because that is the longest time and the longest distance I can draw on this graph.

That one will be my slow speed.

I can then mark on B.

It's a middle distance about half the distance of my race C and a medium speed.

It's faster than race C because it's got a steeper gradient.

I can then mark on my race A.

That's the shortest distance, the steepest gradient.

i.

e.

, the fastest speed.

I hope that your lines look something like mine.

For question 2, I told you this is the graph of two animals running, one a cheetah, one a human, and I ask you to identify which line represents which, and most importantly, explain how you know.

I hope you identified that line to be our cheetah and that line to be our human.

How do we know? You might have pointed to this moment and written, "Faster acceleration, faster top speed," for the cheetah.

The gradient on the cheetah's line gets way steeper than anything we see on the human line.

You might have pointed out this key moment, that is full stop for the cheetah after exhausting their explosive energy.

You might also have pointed out this moment, the fact that the human is a slower animal but able to maintain a steady, constant speed for long periods.

You should have noted that the human did not stop.

This intersection is reassurance that you can in fact outrun a cheetah, provided you last this long.

Sadly, that's the end of the lesson now.

What have we learned? We've learned our knowledge of graphical representations can help us to solve a variety of problems from identifying the locations of intersections on graphs and solving equations to real-life problems involving working with gradients on time distance graphs.

I hope you've enjoyed it as much as I have, and I should look forward to seeing you again soon for more maths.

Good bye for now.