All right, on this section here we're going to be talking about simultaneous linear equations right there. And we're going to show you how to add or subtract these equations. method of substitution for one unknown and you'll see what we mean as we go through this. So we're going to have their their determinants for two equations and determinants for three and if you don't really understand what I mean by that, let when we get into it, we will so with nothing else said let's get on to adding or subtracting equations. Okay, simultaneous linear equations Okay, so let's let's look at the the bullets here. And linear equation has the unknown First power without any power exponents, and what we're meaning is if you look at the power, right here, we're, we're to the first power.

Okay? So you won't see anything like x squared or y squared, y cubed and so forth. All right. Let's look at the second bullet here. simultaneous linear equations form a group with the same unknown. Yeah.

Okay, in this example, I've got two simultaneous linear equations, and I have the same unknowns on both of them. So we form a group and the last one here, bullet number three, the number of equations must equal the number of unknown so I've got two unknowns, x and y. And I have two equations. Alright, okay. Okay, so now as I said previously with the simultaneous equations, they can either be added or subtracted. So on this on this example when I add them, and if you look at the reason is because I've got a plus y and a minus y.

So if I add them, the y term cancels out. So what am I left with? Well, I'm left, I'm left with two x equals 10. What do I want to do? I want to find the x to solve for x. So I just what do I do I say two x over two.

And what I do on one side of the equation I got to do to another so I got to divide over here. So x equals five. So now I will plug back in, and I'm going to solve for y Because I know what x is. So x plus y equals eight, right? We know that x is five, so five plus Y equals eight. I solve for y, y equals eight minus five, because well again, what I do on this side, I have to do on this side.

So y equals three. And just for the heck of it, we're going to use the, we're going to do the same thing and x minus y, so we've got five minus y equals two, five minus two. Again, I subtracted two over here and I get a subtracted two over there. So five minus two minus y equals zero. And if I do the math, I get y equals three here. So they checked, so when I plug in the variable on both both equations in this example five, are they checked, so we're good.

In this recording, we're going to subtract Ah, the same equation on the previous slide we, we added now we're going to subtract so you can see that right here, and when I subtract minus x from x, I get a zero. And then when I subtract a minus minus y, that y right here, we can think of as positive, so I get a plus two y and that equals six. And so all I need to do is solve for y and I've got a two y so I divide over here. It divides out and then I'm ended up with a To over six, so y equals three. And just to check, ah, we say y equals three, so we plug it in here. So if I plug three and one three for y right here, I get x plus three equals eight.

I get x alone on the left, so I subtract the three right there from each side, and I'm left with x equals eight minus three, and x equals five. Same thing down here, x minus y equals two x minus three equals two. x equals two plus three x equals five. All right, so it checks. So anyways, that said, take a look at that and we're off to the next slide. All right, on this slide here.

We've got something that's that's got something in the colon. efficient right here. I mean, in the in the previous two examples, it was a very simple equation. I think it was x plus y equal five and x minus y equal three or something i. So, it was a, it was simple, it was easy. I just did that to introduce you to simultaneous linear equations.

But what happens if I've got something that I show you here, all right, three i plus i squared equals 14, two plus three equals seven. And what we want to do is we want to eliminate one of the variables. And from the previous examples, we can either subtract or add the equation to get rid of one of the variables. So I'm looking at this here. And I've got i squared three i squared. So why don't I multiply this guy here by three.

All right. And when I do that, look at what happens. Three times three is nine. Three times I two becomes three I and three times 14 becomes 42. Remember what I do on one side of the equation I have to do to the other. So if I multiply this side here by three, then I've got to do the same thing over here.

And we get 42. Okay, so let's stop there. And let me clean this slide. Often we'll go on. So again, I multiply this equation by three and I get this here. Alright, so now what I decide to do, as you as we stated we can either multiply or subtract the equations.

If I subtract this, then I'm going to get rid of this variable. So I've got nine I one minus two i plus three minus a plus 342 minus seven. So nine minus two, I'm left with seven I one, plus three i squared minus three i squared. I'm left with a zero, and 42 minus seven equals 35. So seven out E quills 35 from previous lessons, we need to isolate eyes. So I need to divide by seven, which I show you there.

And I what I do to one side of the equation I do to the other. So I gotta divide 35 by seven, so we know that i equals five. All right. So now if I want to solve for the other variable, I just substitute, I know what I want is, so I substitute I one. So I've got three times five plus i two equals 14. Three times five is 15 plus i two equals 14.

I need to subtract a 15 from each side to get I two alone. Right? So I two equals A minus one. And that's my answer. Alright, so take a look at that. replay this if you need to go it over again.

And I'm going on to the next slide. All right, I want you to stop and do these problems, I give you four of them. And again, we can do it using the addition or subtraction method. If you're looking at the this one here, and this one here, this one kind of falls apart, right? If we use what the addition method on that one, this one here we could use, well, let's see on that one there. We're going to have to multiply one of these equations to get rid of one of the coefficients.

All right, and let's take a peek here at this one. Well, we can use the addition method and then use the addition method here. And then you're going to have to use, you're going to have to use the multiply, you're going to have to multiply this one here. To get rid of actually the best. The best one to get rid of is the why. So I give you some hints.

All right. So right now I'm going to clear the slide you're going to do them, the answers will be on the next slide. Okay, here we are. And here are the answers. I mean, I could call them out and say them, but there they are answers. Alright, so if you had a hard time doing it, or finding these, go through it again, and do it But right now I'm going to show you a website where you can plug these variables in and get the answer.

So let's stop it here. And I'm going to switch switch screens. Alright, here is the website right here. Let me let me draw that. Okay, here's the website right here. And I'm going to show it to you in a minute.

Really neat. And we're going out, let me just swap over to it. I've already got it up here. Here it is. Let me let me just position this. So we can use it.

And I'm going to leave it there even though I'm not showing you the whole screen. But this is the URL right there. There's the URL plus I put it right Right here on the slide. So, I mean, there's different. There's different determinants and linear equation rules. We only scratched the surface here.

This is an advanced topic. But on this example we want to use, gosh, join elimination. And if you look alright, we want to set the matrix right there. So basically, what we want to do is we've got two by three. So let's see what happens. All right.

So I have my X, my X, my Y, my Y, down, x here, x here, two y, two y, and then of course, the answers, four and two. So all I do is put the coffee Send. So I go to two to minus two to put four here, and 12 there, and then I hit where the solve button. And so we should get four. And now that's the answer. What did I What did I give you is the answer four and two.

So let's see if we get that and look at that four into. All right. Now I wouldn't rely on this if you if you need to use it, but do some by hand. So if it doesn't come out, right, you get a feeling for it. Let's, let's just do let's do another one. And let's just do let's do this one.

All right, you can, I mean, you can have, you can have fun with this. You can go to the website, plug some answers in and so forth. Okay, so let's do this 1333 minus 324. Six, it solved. And we got five and three, five and three. So we're good.

So, there you go. Take a look at this. We're going to stop right here on this lecture. And we've got, we've got one coming up on the next section. Alright, see you over there. Now.