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Hello, I'm Mr. Coward, and today's lesson is on solving simultaneous equations by trial and error.

For today's lesson, you'll need a pen and paper, also you'll need to try on a graph, and a calculator.

Please take a moment now to clear away any distractions, including turning off any notifications.

And if you can, try and find a quiet space to work where you won't be disturbed.

OK, when you're ready, let's begin.

OK, so time for the trials test.

Now, what I want you to do is I want you to substitute each of these six values of x into this equation, and into this equation to find the value of y, and then you need to decide if one is bigger or two is bigger.

What do you notice? And when you do this, you'll notice something, and it'll be something that you want to try.

So, pause the video and have a go, pause in three, two, one.

OK, so here are my answers.

I worked out, for x is zero, I started to put my answers in a table, by the way.

I got the two was bigger.

For x equal to three, two is bigger, For x equal to five, two is still bigger, For x equal to seven, one is bigger, one is bigger, one is bigger.

So what happens here? Two seems to be bigger, but then it changes.

Now, what this change tells us is that this change actually tells us that this one and this one must be equal at some point in-between because if you imagine one is always bigger, OK, but then the other one somehow gets bigger, those two values of y, at some point, must cross, at some point, between five and seven, must be equal.

So what value to you want to try? How could we limit and narrow down that area in which we know that they are equal? Well, my natural instinct is to try halfway between them both, and is either it's less than that or it's more than that.

So let's try the value when x is equal to six.

So we have two times six plus 3y equals 24.

So, we have 12 plus 3y is 24, take 12, five by three, and we get y equals four.

OK, what about this one? What is the value of y when x equals six? Five times six, 30 plus 3y.

So, what is y going to be? It's the same.

So what happens? We found our meeting point.

We found the point where they both have the same value.

Everything, since that was six there, up to this point where x is equal to six, this one is bigger, up to that point of six.

So 5.

9, 5.

87, 5.

936, 5.

9982, that one is bigger.

As soon as it gets past six, so 6.

000001, this one is bigger.

So six is our point where they're both equal, and they kind of cross, our solutions cross, and when it starts getting bigger.

And you can check, you can solve this pair of simultaneous equations algebraically, and you will get the answer, is six.

So, this is kind of a trial-and-error approach.

We are trying different numbers, and we are narrowing down the space in which our solution could be.

So this is kind of solving simultaneous equations by trial and error.

So, we're going to use the table of values to find which two integers x lies in-between.

We going to write our answer as an inequality.

So, we'll start off with zero, zero is a nice easy one.

So we got 4x plus y equals 12.

Our x is equal to zero, so we just have y equals 12, very nice.

OK, when x is one, we have four plus y equals to 12, subtract four from both sides, you get y is eight.

OK, when x is equal to two, we have four times two, eight plus y is equal to 12, y is equal to four.

Hmm, do you notice anything here? When x is three, we four times three, 12 plus y equals 12, that means y is equal to zero, and finally, four times four, 16 plus y equals 12, take 16 from both sides, and we get y is negative four.

What is going on there? Well, each time, we're going down by four because if we were to rearrange this line, we'd see that this is decreasing each time, a sequence decreasing by four.

So if we had y on its own, y as the subject, we'd have 12 minus 4x, so it's going down by four every time.

OK, what about this one? This one's a little bit trickier.

So, we have 1x equals zero, we have two times zero plus 3y, so we have zero plus 3y equals 23.

5.

So we get y equals, and I'm going to do that on my calculator because I do not know what 23.

5 divided by three is, 7.

83 recurring, and just the three is recurring, OK? Now, let's do it again, let's do it when x is equal to one, so we have two plus 3y equals 23.

5.

Now, it's actually, it's getting kind of annoying to do this.

So, I'm going to show you a little shortcut.

All right, so this is our equation.

What I'm going to do is I'm going to make y the subject, so we're going to take 2x from both sides.

Got this, then divide both sides by three, so I get this.

And I think, now, this is easiest to substitute the values in on our calculator, and I think this is easier to work with, so we don't need to solve it each time.

You may disagree, I mean, you may prefer to stick to it like this, and that is fine, they'll both get the same answer, it's just personal preference.

So, this one would be 23.

5 minus two times one, so 23.

5 minus two times one, which is just two, so 21.

5, 21.

5 divided by three, which gives me 7.

16 recurring.

OK, this one, well, that's 23.

5 minus two times two, four, so 23.

5 minus four divided by three, which gives me 6.

5.

23.

5 minus two times three, six, divided by three, that gives me 5.

83 recurring.

And then, finally, 23.

5 minus two times four, so minus eight, divided by three, 5.

16 recurring.

So, what does this tell us? Well, here, this one is bigger, here, this one is bigger, here, this one is bigger.

So this tells us that these two equations must cross between one and two.

So the solution to our equation is x is greater than one but less than two, and we know that x is not equal to one because those two are equal, and we know that x is not equal to two because those two aren't equal, but x is somewhere in-between one and two.

OK, just see how they're kind of getting closer as well, so they're getting closer, and then they cross and then we start to get further away.

OK, so what I want you to do is I want you to have a go at the independent task.

So, there are two questions that I want you to do.

So, pause the video to complete your task and resume once you're finished.

OK, welcome back, here are my answers.

So we have x is between three and four, and if you solved it algebraically, you'd get 3.

5 and 11.

Here, x is between negative two and negative one, and if you solved it algebraically, you'd get this and this.

And here, on the final one, well, x is between 1.

7 and 1.

75, so that means that x to one decimal place will round to 1.

7, and some computers do algorithms like this to work out, and you could keep going, so you could do it, too.

You see, you could get in-between, say, for instance, get in-between, I don't know, if you found it was in-between 1.

72 and 1.

73, you could keep going and get more and more accurate with each iteration.

OK, so now, it's time for the Explore task.

So, what I want you to do is I want you to substitute into both equations two and seven and recall what the value of y is.

Do the same for this one, do the same for this one.

Now, you need to decide which lines and which equations meet between two and seven, which ones meet before two, and which ones meet after seven.

Can you describe what happens to the values of y for each inequality? So, can you describe what is going on with y here if x is between these two? Can you describe what is going on with y when x is less than two or bigger than seven? So, pause the video to have a go and resume once you've finished.

OK, welcome back.

Now, here are my values of y, and here are my values, which ones satisfy which inequality.

So just give you more to digest this, and then I'll talk through it.

So why is this one this one? Well, can you see how that one's bigger, sorry, that one's smaller, and then that one's smaller, so, it swaps which one is bigger.

So that must mean it crosses between two and seven.

Or is this one, this one doesn't cross between two and seven.

This one's bigger, and then this one's bigger, so they don't cross.

So we know that this cannot have a solution between two and seven.

This one, they don't cross, that one's bigger, and that one's bigger, so this one can't have a solution between two and seven.

But let's have a look at these numbers.

OK, what's the difference between them two? The difference between them two is 9.

5.

What is the difference between them two? 35.

So they're getting further apart.

The values of y are getting further apart.

So you can kind of think of this as like an inequality sign, this is where they meet, and they're getting further apart.

So because they were closer when x equals two, then x equals seven, they must have met before x equals two.

This one, what's the difference there? Well, it's 5.

83, and this one is two.

So if you imagine this, when x is two, they are 5.

83 apart, and when x is seven, they are just two apart, so they're getting closer, so they must meet after seven.

So because of that, if the highest value of y flips between two and seven, that means it's got a solution there, and if x is less than two, the y values are closer at two than at seven, so they're closer than they are there.

Or here, they're closer if it's bigger than seven, they're closer at seven than they are at two.

Now, this links perfectly to solving simultaneous equations graphically, which you'll go there very soon.

This is just kind of to get you thinking about getting further or getting closer and crossing and what that happens to the y value.

So, hopefully, you found it interesting, I certainly did.

Thank you very much for all your hard work.

If you'd like to, please ask your parent or carer to show your work to Twitter tagging @OakNational and #LearnwithOak.

Thank you very much.