Lesson video

Hi everyone, it's Ms. Jones here.

And today's lesson is all about comparing algebraic and graphical methods.

So putting those two different methods that you've learned to solve simultaneous equations together and comparing them, contrasting them, deciding which one is better for which occasion, or whether they both work just as well.

But before we can start, make sure you have a pen and some paper, you have a nice quiet space to work, if possible, and you remove any distractions.

Pause the video now so that you can make sure you're ready before we begin.

Okay, let's make a start.

So the first thing I would like you to do is solve the following simultaneous equations using an algebraic method.

So this is just a recap of what you have done before with solving simultaneous equations using an algebraic method.

So you could use substitution, or you could use elimination.

It's about going back and thinking about what you would like to use, what's best in this situation, and applying it as a recap.

So pause the video here to do that.

So either method that you used, you should have got x = 4 and y = -1.

So if you decided to do it using substitution, you could have rearranged the first equation to get y = x - 5, and substitute the y into the second equation where this becomes x - 5.

And then expand the brackets, solve the equation by putting the variables and the constants on one side, and eventually you get x = 4, which you can then substitute into one of your original equations to find that y = -1.

Or you could have done it through elimination.

So trying to make one of the variables have the same coefficient.

So we could have changed y - x = -5, multiplied that by 2 to get 2y - 2x = -10, and then add it to our second equation to eliminate our 2x, and then ended up with 5y = -5, y = -1.

And substitute the y = -1 back into one of the original equations to find that x = 4.

But either way you get these solutions.

So really, really well done.

If you managed to recall your prior knowledge and get those solutions there.

So we can solve simultaneous equations using algebraic or graphical methods.

So we've got the same two equations that we had before and we've plotted them onto these axes.

And you can see, in fact, you can tell me and show me where it is that our solution is.

Hopefully you've pointed at the intersection.

Now we can see that yes, x = 4 and y equals -1.

So whichever method that we use, we end up with the same solution.

So there's not necessarily a right or a wrong method.

It's sometimes about preference and sometimes about what the question is originally.

So you might think that to solve these two equations simultaneously, it's going to be quite long to draw out the cross and then find the intersection you prefer to do algebraically.

That's absolutely fine, but you still need to know how to do it graphically.

Because for example, it might be a question where you're already given this graph and you'll be wasting valuable time by going back and solving this algebraically, where you can just literally see the intersection.

It will take you a few seconds to solve that equation and find the values of x and y.

So it is really important that you can do both methods and that you understand when is best to use either one.

So hopefully you can now very easily do both of those methods, having done both of them throughout the course of the last few lessons using algebraic or graphical methods.

So what I would like you to do now is pause the video to complete your independent task, which I will show you on the next slide.

Yeah.

So for the first question you've been asked to solve the following simultaneous equations, both algebraically and graphically.

So with the first one, should be fairly simple to plot.

They're already in the right format in terms of having y = something.

So they're nice and easy to plot and you should have something that looks a bit like this.

And you find your intersection and that will be where your solution is.

Or you could have solved it by substitution or elimination.

But either way, you should be getting these as your solutions.

For the first one, x = 1 and y = 3.

For the second one, x = 2 and y = 8.

And for the third one, x = 2/3 and y = 7/3.

Now this third one is quite interesting.

And you may have noticed that when trying to solve this graphically, using a graph, it was actually very tricky because it wasn't a whole number.

So you kind of had to estimate almost.

And you wouldn't have got the exact answer unless you did a very zoomed in graph, and you used fractions, for example, to do that, which would be unusual.

But if you did that then that's great.

But actually you are going to almost certainly get a more accurate answer to problems that have a fraction or a decimal answer when you solve these algebraically.

So there is a perfect instance of where solving algebraically is better than solving graphically.

But we've seen that both of them are important.

We can check our answers for the first part by resubstituting those numbers back in.

So for example, for this one, if we substitute the x value back in, we'd have 4 x 1, which is 4 - 1, which is 3.

So it works for this one.

And if we substituted it in here, we'd have -1 and 4 which is 3.

So it works for this one as well.

So that's how we can check our answers.

Question four says, why do -4x + 5 and y + 4x = 12 have no solution when solved simultaneously? Hopefully you've remembered that it's because they both have the gradient of -4, so they are parallel.

Because if you were rearrange this one into your y = mx + c, the format of a straight line, you can check that the gradient is in fact -4.

And if you were to plot both of those graphs, you would see that the gradient is -4 and they're parallel.

And so it will never meet and will not have a solution.

Really well done if you got all of those answers.

That's absolutely brilliant and a great recap.

Good job.

What could the coefficients of x and y be in each equation to make the point of intersection ? So we want both of these lines to have an x value of 3 and a y value of -2.

There are loads and loads of solutions to this, so it might be that you want to start in lots of different ways.

And I will be starting off for those of you that need a little bit of extra support.

Because this is actually a very tricky question.

You might want to do some work on graphs, and try and work that way, and use that to, to help you.

Or you might want to work algebraically; it's up to you.

So pause the video if you just want to have an experiment and have a go, which I encourage the lot of you to do.

But if you want that little bit of extra support, you can see what I would do to work this one out.

So we have unknowns here in these gaps.

We don't know what they are.

They could all be the same number.

They could all be different numbers.

But I'm going to call them a, b, c, and d.

Now actually, I know what my y and my x values need to be.

My x value needs to be 3, so I have 3a.

My y value needs to be -2.

So I will have add -2b, or subtract 2b = 15.

I will then have 3c - 2d = 3.

And having these equations makes it a little bit more logical and a more strategic way of finding what those a, b, c, and d values could be.

But you can literally substitute any value into these letters that works.

So any lots of a, lots, any lots of 3.

Sorry, subtract any lots of 2 = 15.

So you could pick any numbers that work with that.

The same with any lots of 3, subtract any lots of 2 = 3.

Pause the video now to have a go at doing that.

So as I've said, you could have got any solution.

There's absolutely loads of answers to this one.

You could have got anything at all.

But as an example, we could choose a = 3, so you'd have 9 here.

What would we need to subtract from 9 to get 15? We'd have to subtract -6.

So we could put b = -3.

So we would be subtracting -6 to get 15.

So you could have had a coefficient of 9, and -6 here.

As an example for the second one, we could have used, let's say, c = 4.

And we get 12 here.

And in order to get to 3, we need to subtract 9.

So I need two lots of 4.

5.

Cause remember it could be a fraction or a decimal as well.

It could be negatives, could be anything.

So those are just some examples but there were absolutely loads.

Really well done if you managed to find some of those.

Great job.

And that brings us to the end of today's lesson.

Please make sure that you complete the quiz to check your understanding.

But really, really well done today.

You did absolutely brilliantly.