00:00 Hello, my name is Oyase Ose Sunday. I'm taking a maths JSS-3 class, but we have an interest

00:21 in topic for today. The topic says simultaneous linear equations, we'll be looking at the

00:27 graphical method of simultaneous linear equations. Now, the objective is to be able to solve

00:34 simultaneous linear equation using graphical method and understand how the axis works. How

00:41 do you locate your various points on the axis? You have the y axis and the x axis. How do

00:47 you locate your points on the axis? That is what we'll be doing in the class. So, what

00:54 is the system of equation that has two unknown variables? That's xy. I'm really, really

01:01 knowing xy, not because xy is the only variable. It could be p, q, h or any other variable. So,

01:14 it's not only xy, xy. It's okay. We always know it could be. For the purpose of the

01:20 graph, we'll specialize on x and y. Now, there are two linear equations, x, 2x plus y equal

01:27 to 10 and x minus y equal to 5. We'll solve them simultaneously. Move on. Now, there are

01:34 some basic rules for simultaneous equation graphically. Basically, following these

01:40 rules makes it easy for you to be able to plot your graph. And before you can even plot

01:46 the simultaneous equation graph, you'll need to have a graph sheet. Now, in order to solve

01:52 a pair of simultaneous linear equation graphically, number one, you draw the table of values for

01:58 both axis. So, you need to have a specific table of value for both axis, a table of value,

02:04 where you have x and the y axis, table of value. I'm still going to explain when I get to

02:10 that point. You need to identify the graph of the two equations. Identify the graph of

02:16 the two equations. Then, totally, you plot the two straight lines and intersecting each

02:24 other at the common point. When you plot the first line, you plot the second line, the

02:29 point at which they intersect. You bring out the points. Then, the common point intersecting

02:35 the two lines gives the solution of the pair of simultaneous equation. Like I said, when

02:40 you draw the first line, when it intersects with the second line, the point of intersection

02:46 on the y axis and on the x axis gives you your pair of simultaneous equation. Having

02:56 these four rules in mind, we can move straight up into the graphical method. Like I said, you

03:01 work with your graph. And your graph is always on its box. You see the box on the graph, and

03:07 it has two centimeters. So, you just need to work with your graph sheet. You can't be using

03:13 a plane, if you are expected to solve using graphical method. There's a need to work with

03:19 the graph. Now, solve graphically the system of some of those three line equations. x plus

03:26 y equal to 5. x minus y equal to 3. Now, we have x plus y equal to 5. That's for equation

03:41 1. And I have x minus y equal to 3. That's equation 2. We're not going to be solving with

03:51 elimination or substitution method. No. We're going to be using the graphical method. Now,

03:57 in the graphical method, there is a need for you to revive this form of two equations

04:04 in the form of y equal to mx plus b, making the y the subject of the formula. So meaning

04:12 there's a need to be, we write these both equations 1 and equation 2, making the y the

04:18 straight on the formula. So you could get the range of value for x and the range of value

04:24 for y. Check note, some questions will give you the range of value for x. They will say x

04:31 has the range between 1 to 4. And some questions we just want you to use an imaginary. So, whatever

04:40 the question lies, you need to get the range of value for x, the really challenging point

04:46 in simultaneous equation graphically is the range for the value of y. Not that of x because

04:52 x is always kind of straight forward. It's between 1 or minus 2 to the range of the value

04:58 of y. So we have to express the two equations in the form of y. So let's take to the three

05:04 equations. Now, expressing it in the form of y, we have the first one is what? x plus y equal

05:13 to what? 5. Let's take it out, expressing in the form that it should be. So when we take

05:20 it out, we're going to be having y equal to, we have plus x here. It causes the sign of

05:25 equality, it becomes a minus, as minus 5 minus x. Now, if you look at this, the x is expected

05:34 to be in front. According to your expression, the x is expected to be in front, y is constant

05:40 to be at the back. In the way you write it on your graph. So it cannot be in this form. So

05:44 you rearrange. Now, when we rearrange this, bringing the x to the front and the 5 back, we

05:52 are expected to have y equal to minus x plus 5 in this form. Now, this, I have to suck

06:04 it, it's what we're working with. That becomes the center point. Not everything will be working

06:12 around this on our graph. That will now be for equation 1. You will pick the same process

06:19 for equation 2. Now, good and fine, there's a need for us to go ahead with the table. I said,

06:29 in the rules, remember I said you're not able to identify your table. Now, let's just imagine

06:36 we're given the value of x. We're not given the value of x. Let's work with, we're not

06:41 given. First, we're not given the value of x. I want to assume a value to which x can

06:46 be equal. So, there's a need for us to apply the trend error method. The trend error method

06:51 is you find it is all four pair of values that can fit in. Now, when you fit it in into

06:58 this expression, it can finally give you a 1. So, you have 1, 2, 3 and 4. Just a trend error.

07:07 The most question would always give you a pair of value to work with for the x. But let's

07:13 do the trend error method. We have between 1, 2, 3, 4. Now, infuse it, when x equal to 1 into

07:23 the equation, the equation is y equal to minus x plus 5. So, when x equal to 1, the y equal

07:31 to 1 minus 1 plus 5. So, minus 1 plus 5, that gives us 4. Why is it 4? Remember I said, negative

07:42 down positive or your final answer is positive. Negative or positive down your final answer

07:50 is negative. Don't forget that. Now, that's we continue x. We have x equal to 2. That's

07:59 why. Let's have y equal to minus 2 plus 5. Minus 2 plus 5, that gives what we have as

08:10 what? 3. Remember, it's positive. It's still on the upside. Now, we continue down the

08:17 system. We have x equal to 3. We have y equal to minus 3 plus 5 equal to the addition

08:29 sign instead of the upside. So, I have it as 2. Now, I continue. I have x equal to 4, x equal

08:38 to 4. And I have y equal to minus 4 plus 5 equal to minus 4 plus 5. That gives me 1. Now,

08:47 with my final answer, you carefully observe that my final answer is stopping at 1. Because

08:51 I put x equal to 5. That's minus 5 plus 5. Now, give us 0. And I'm looking at the point

08:59 where in my plotting of the graph, I will not try to avoid the point of 0. So, I have

09:05 to look at the positive and the negative. Now, I have minus 4 plus 5 equal to 1. So, if

09:13 I collect them together in a table, the look at the table, we have 1, I have 4, 3, 2 and

09:20 1. So, the y. My x is already 1, 2, 3 and 4. I repeat the same process. So, the second equation.

09:30 Now, my second pair of equation, like I said, I will put the same process for my second pair

09:35 of equation. What's my second pair of equation? My second pair of equation is x minus y equal

09:42 to 3. Making the y the subject of the formula, you move the y, you have, yeah, you have what

09:49 you have, you have what? You have minus y equal to what? Minus x plus what? Plus 3. There's

10:01 it. Additional sign in front of the tree. So, yeah, it's moving to become. Now, you develop

10:06 a side by minus, you develop a side by negative. So, when you develop a side by negative or, so

10:12 let's see, let's have variables are by negative. So, what will I would have in at the end of

10:21 the day, we are expected to have y equal to negative negative visor's positive of x. Negative

10:29 negative is positive of x. That's x. The negative positive, remember the four laws. Negative

10:35 positive is negative of 3. Now, this, circle it again. It's my system of equation for 2. My

10:45 system of equation for 2, the system of equation for 1. So, if I'm going to be substituting into

10:51 the value to get my value of x, I will be working with this. And remember, I said you have to have

10:58 two tables, first table for equation 1 and second table for equation 2. So, you could plot

11:04 the two tables on your graph. That will give you the range. Now, back a bit, a little, is if you look

11:11 at the equation, the table for the equation 1, for the equation 1, you will see that I already

11:17 have my y. My y is 4, 3, 2. So, on the graph, this is already a number of this equation, 1, 2, 3, 4.

11:26 Now, my x, this is it. So, substitute it in, I substitute when x equal to 1, when x equal to 1, x

11:36 equal to 1. I will have y equal to minus, y equal to, we have 1 minus 3, 1 minus 3. Remember,

11:51 negative all positive down, negative all positive down and so becomes what negative. That becomes

11:59 minus 2. Now, I have the next one, I have x equal to 3. So, I have y equal to substitute it in,

12:11 x equal to 2. Sorry, x equal to 2. If I would come to 3, x equal to 2. So, I have 2 minus 3.

12:20 Now, give me, negative is still on the high side. That's minus 1. Now, I have the next one, where I

12:28 have x equal to 3, x equal to 3. And I have y equal to 3 minus 3. That's 0. Now, I have 0.

12:43 I have zero point. Now, I need to work with the fourth one. Again, let me have the, since I have

12:51 worked with x equal to 1, 2, 3, and 4 with my equation 1. I need to repeat the same process for my

13:02 equation 2 for x. So, it could be on the same page, when you are plotting your graph. If you don't do

13:08 that, your straight line will not intersect. And if your straight line will not intersect, then you're

13:14 not going to get a pair of equations. Now, of x equal to 4, x equal to 4. We have y equal to

13:22 4 minus 3. Now, look at it very well. You see, the positive is greater than the negative, positive,

13:30 greater than negative. So, as I become positive, I have y. Now, put it on the graph, put it on the table.

13:39 On the table, I have my x already, 1, 2, 3, 4. But the values for y are quite different.

13:48 Values for y here is minus 2, minus 1, 0, and 1, minus 2, minus 1, 0, and 1. So that means on my

14:00 y axis, I am going to have this now. You pick up your graph, you pick up your graph, and you say,

14:09 oh, if this is my graph, my graph is expected to have 2 axes in this pattern. This axis is called my y

14:21 axes, where y forged. This axis is called my x axis, where x forged. This is my positive axis,

14:34 positive of x. This is my negative of x. This up is my positive of y. Down minutes is my negative

14:47 of y. So, it means, if I am going to be doing my plotting, my x is 1, 2, 3, 4. So, what will I just

14:58 simply do on the graph sheet? I will simply pick, I have 2 centimeters. The graph is already given

15:05 in 2 centimeters, 2 centimeters, 2 centimeters, 2 centimeters. So, it's equal. So, I will just simply

15:13 have 1, 2, 3, 4. Simply have 1, 2, 3, 4, 1, 2, 3, 4. This is 2 centimeters. But on my y axis,

15:33 I will just repeat the same process here. So, I will have it. Although my x does not have negative,

15:40 but the graph just starts to be complete. So, now I have, on my axis, I have 1, 2, 3, 4 for the y,

15:48 for the x. I will pick just the same process here. I will have minus 1, minus 2, minus 3,

15:54 I have minus 3. You can stop at minus 3 or minus 4. Anyway, you have it. Then for the y, look at

16:02 the range of y. I have the range of y. I have minus 2 and minus 1, I have zero. This is the point of

16:09 origin. It's called point zero. This point zero is applicable to both y-axis and x-axis. Good.

16:16 But the point below is negative. So, what will I do? Just simply say, the boxes are already in

16:21 2 centimeters. So, I will just say, the first one is minus 1. The first one will give me minus 1.

16:31 Minus 1. The second one will give me minus 2. Third one will give me minus 3.

16:36 Now, putting it minus 3, then the range of the positive value of y. Check the second graph.

16:44 The positive value of y, the highest value for y is 4. So, it means that your value range for y

16:53 has to be 4 or x is 4. It can't be less than that because when you plot, it shows. So, I have

17:00 positive value. I have 1, 2, 3 and 4. So, I can stop at 4 and I can pick it at 5. Now, with this,

17:14 you now plot. And what will you simply do? On the x-axis, you start with the first graph.

17:21 The first graph, that's where you start with. On x-axis, 1, trace to 4. When x is 1, y is 4. So,

17:28 what you just need to do is, you put it dotted, you connect. Just connect. I will show you my

17:35 own complete work. This is the rough sketch. You connect. You connect to star. When x is 2,

17:43 when x is 2, what's y? x is 2, y is 3. You just trace off, you connect. When x is 3, y is 2,

17:55 when x is 3, you can y is 2, x is 3, positive. You trace off, you connect. Now, once you connect,

18:04 when x is 4, x is 4, y is 1, x is 4, y is 1, you trace off, you trace off, you connect. Now,

18:14 you connect. You could say. So, invariably, you would do the top straight line because my is not

18:21 straight. I'm not expecting the straight line. It will form a straight line. This is for the first

18:27 equation. What's the first equation? The first equation is y equal to minus x plus 5.

18:35 Then, you pick up the graph for the second equation. The second equation, you start. When x is 1,

18:43 when x is 1, y is minus 2. x is 1. You can see, y minus 2 is down. So, you trace

18:50 pop, pop, pop, pop, pop. You start. When x is 2, y is minus 1. x is 2, y minus 1,

19:01 you trace off, connect them. When x is 3, y is 0. Look at the point 0. So, you just trace the top

19:09 here and start it. Now, when x is 4, when x is 4, y is 1. So, you just trace 4 to 1.

19:20 Yeah. Oh. So, you could see. Draw to connect. So, when you connect, you have this.

19:26 Want me to show you my graph? Someone said, show me your graph. This is my graph.

19:33 Now, now, if your net graph, you could see vividly that they connect at this point. At this point,

19:41 well, before you put that at the point, you say your y. Your yf for the second variable is what?

19:47 y equal to x minus 3. This, for the first, you can see intersect. At what point did they intersect?

19:55 They intersect here. So, trace it to the y. I have at the point of intersection, I have y equal to

20:04 1 and x equal to 4. Now, looking at this 2, looking at this 2, these are the point of intersection.

20:17 To take it, look at my point of intersection now. Take it up, then my point of intersection is y

20:26 equal to 1 and x equal to 4. Now, my graph, this is my graph. This is the way your graph should look

20:36 like, connecting each. You could see from the graph, the 2 straight line intersect at what? 4 and 1.

20:44 So, therefore, x equal to 4 y equal to 1 is the solution for the given system of what?

20:50 Simultaneous what? Equation. With these 2. Good. And with this, I wanted to pick up the pen and do

21:01 the evaluation. We said, solve graphically the system of simultaneous equation, linear equation.

21:07 I have 3 x minus y equal to 2. 9 x minus 3 y equal to 6. Simultaneous equation,

21:17 solve it with the graphical method. I will come to the end of the class. My name is Oyaset with

21:26 Sondi. Thank you.