00:00 Hi everyone, I am Atolagbe Michael, your mathematics teacher.

00:20 The topic today is simultaneous linear equation.

00:26 At the end of today's lesson, you should be able to do the following.

00:30 One, recall and identify simultaneous linear equations.

00:35 Two, solve simultaneous linear equation using substitution method and elimination method.

00:44 simultaneous linear equation. How do you identify a linear equation? Take for instance X plus 4 equals to 7, X squared plus 4 equals to 7.

01:00 I have two equations on the board, X plus 4 equals to 7, X squared plus 4 equals to 7.

01:08 The power of the unknown here, that is X, is 1. Any time you have an equation, which asks a power of 1.

01:15 So, we'll call such equation a linear equation.

01:19 When the power of the unknown is more than 1, then you should know that you're not dealing with a linear equation.

01:25 So, what we're dealing with today is an equation that asks the power of the unknown to be 1.

01:31 There are two methods we'll be looking at, the first method is the substitution method, why the second is elimination.

01:38 Example 1.

01:43 Alright, use the substitution method to solve the following PS of simultaneous equation.

01:51 Example 1.

01:54 Use the substitution method to solve the following PS of simultaneous equations.

02:00 A.

02:01 Q plus 2R equals to 8.

02:04 R minus Q over 2 plus 1 equals to 0.

02:08 I'm to use substitution. Don't forget I told you, we are going to concentrate on two methods.

02:13 Method 1, substitution and method 2, elimination, substitution now, question 1.

02:21 Q plus 2R equals to 8.

02:28 That is equation 1.

02:31 Equation 2R minus Q over 2 plus 1 equals to 0.

02:43 So, we're able to solve this very well. You start from equation 2.

02:49 From equation 2R minus Q over 2 plus 1.

02:58 In equation 2, I have a fraction. The first step is eliminating your fraction.

03:03 I'd rather eliminate my fraction here. What I have is the denominator here is 2.

03:08 It means multiply through by 2 in order to eliminate the fraction.

03:12 So, we are multiplying by 2.

03:16 2R minus 2 times Q over 2 plus 2 times 1 and equals to 0.

03:26 That is 0 times 2. We also give it 2.

03:29 So, what I have is 2R minus 2, I have that to be Q plus 2 equals to 0.

03:42 Then what do I do now? Take the constant to the other side.

03:45 If I take the constant to the other side, I'm going to have 2R minus Q plus 2 equals to 0.

03:55 Now, 2R minus Q. Make you the subject of the equation here.

04:02 Multiply through by 2. Now, we have 2R minus Q plus 2 equals to 0.

04:09 Call this equation 3. Go back to your equation 1.

04:16 In equation 1, I have Q plus 2R equals to 8.

04:21 So, I'm writing that. Don't forget that Q plus 2R.

04:32 Q plus 2R equals to 8 equals to 1.

04:38 From equation 1, make you the subject of the equation.

04:44 Q equals to 8 minus 2R. You can call this 4.

04:52 Now, you have Q equals to 8 minus 2R.

04:55 Substitute equation 4 into what you have in equation 3.

04:58 That is the value of Q. In equation 3, changes to the new one I have here.

05:04 But I will give you what? 2R minus instead of Q. That changes to what?

05:10 Open bracket 8 minus 2R plus 2 equals to 0.

05:17 Now, we have an equation.

05:19 What we are going to have in negative outside of bracket?

05:22 Be careful of negative. Negative times 8 will get you answer.

05:26 The negative times negative of another number. Be careful of that.

05:30 So, we have this to be 2R minus 8 minus minus 2R.

05:36 That changes to what? 2R.

05:39 Then plus 2 equals to what? 0.

05:46 Now, what do I do now?

05:47 Collect my terms. 2R, 2R.

05:50 It constant, it constant.

05:52 Take those constant to the other side. Don't forget your other side.

05:55 Your other side is your R, RHS. That is right-hand side.

06:02 And LHS left-hand side.

06:06 So, I am going to collect light terms now. Light terms are the LHS.

06:09 Light terms are the RHS. Light terms are the LHS.

06:12 I have that to be 2R plus 2R equals to RHS.

06:18 Now, take minus 8 to the other side.

06:21 Minus 8 to this side. That changes to what?

06:24 8. Then, I have constant 2 to the other side as well.

06:28 That changes to what? Minus 2.

06:31 2R, 2R.

06:33 8 minus 2. What do we have? 6.

06:38 4R and 6.

06:40 Divide 3 by 4. Divide 3 by 4.

06:43 R equals to 4 divided by 4 will give you 1.

06:49 Then, you have your R. 2R is what? 2R is what?

06:53 Now, I have my value of R to be 3 over 2.

06:59 Don't forget we are using some decision method.

07:03 Now, that you have the value of R to be equals to 3.

07:06 Going back to your equation 4. Remember, in the equation 4, what we had there was what?

07:11 Let me take you back to that.

07:15 We have Q equals to 8 minus 2R.

07:20 Now that I have the value of R, I substitute back into this equation.

07:28 Q equals to 8 minus 2R.

07:32 Q is not equals to 8 minus 2 open bracket or multiplied by what is the value of your R now?

07:41 3 over 2.

07:44 This goes. This goes. The Q equals to 8 minus 3.

07:50 Q equals to 5.

07:54 Now that I have the value of Q to be 5, I have been able to apply some decision method in solving for the value of R and the value of Q.

08:04 Q equals to 2.

08:13 Q equals to 2. I have y minus x over 4 equals to 3 and 3y plus x equals to 23.

08:23 That's what I want to solve. I will solve this.

08:32 y minus x over 4 equals to 3. Equation 1.

08:39 I want it to be the second equation.

08:48 The seventh equation now, 3y plus x equals to 23. I have that to be 3y plus x equals to 23.

09:06 Okay.

09:08 Just like the first example, I'm having a fraction here.

09:12 I can solve it with a fraction. The first step is clear your fraction.

09:16 So where do we have the fraction? I have the fraction in example. That is equation 1.

09:21 Then go to equation 1, clear the fraction.

09:24 We have y minus x over 4 equals to 3.

09:29 Multiply through by 4 in order to clear the fraction and that will be what?

09:34 4y minus x equals to 12. Don't ask me. Where is that 4? Don't forget I've multiplied by 4.

09:41 4 divided by 4 will give you what? x. So 4y minus x is equals to 12.

09:46 Call this equation what? Equation 3.

09:50 So you cannot choose between equation 3 or equation 2.

09:57 They want to make the stability of the formula.

09:59 What we're careful? Always go for the one that is linear.

10:03 Or a variable that has no coefficient that is greater than 1.

10:09 I have 3y, the coefficient is 3. The coefficient of x is 1.

10:14 I will prefer 1. That would be easier to solve.

10:18 Go to that. Equation 2 or equation 3?

10:22 Form equation 2 now.

10:29 I have 3y plus x equals to 23.

10:35 Make x the subject of the equation.

10:40 x equals to 23 minus 3y. What have I done yet?

10:45 You take this to the other side.

10:47 And I will become 23 minus 3y.

10:51 Not that I have this. I can substitute now.

10:54 Going back to this equation that I have here. Equation 3.

10:58 I have 4y minus x equals to 12.

11:01 And I have x equals to this. You can level it as well.

11:04 Level it equation 4.

11:06 So that would be very easy for you to substitute.

11:09 So I will stop here.

11:11 Substitute.

11:16 Equation 4 into equation 3.

11:24 Go to equation 3. What do we have in equation 3?

11:27 4y minus, I am not going to write x this time around.

11:31 What am I writing? Open brackets 23 minus 3y.

11:36 And that would be 12.

11:39 4y. Open the brackets minus 23.

11:47 Minus nice minus 3y.

11:51 Then equals to 12.

11:54 That's like I said earlier. Call it the light times now.

11:57 Take one to the RHS, take the order to the LHS.

12:01 LHS, RHS. Left hand side, right hand side.

12:05 4y, 3y. Group them together.

12:11 12 plus 23.

12:19 4y. Higher 4y plus 3y. Good.

12:22 Then equals to 12 plus 23, fine.

12:25 Take it off.

12:27 4y and 3y. We are hiding.

12:30 That gives you what?

12:33 7y. 12 and you are 23.

12:37 That gives us what? 25.

12:39 The battery by 7. The battery by 7.

12:42 YA will give you 5. Now you have the value of y.

12:46 Then we move to the next.

12:53 From here, x is 23 minus 3y.

12:55 So to do the value of y very fast.

13:00 x equals to what?

13:03 23 minus 3y.

13:06 The value of y is 5.

13:08 Put that into the equation.

13:12 Then that will give us 23 minus 15.

13:17 23 minus 15 will give you what?

13:19 It. So with this, you have been able to substitute using

13:24 substitution method, the value of x and the value of y.

13:27 Now we move away from substitution method to elimination method.

13:31 I told you we have to algebraic methods in solving simultaneous linear equation.

13:35 So the decision and elimination.

13:38 I have two examples here. But I will be solving one.

13:42 Then I will leave you with the other one.

13:44 It is an elimination method.

13:46 I will solve A because of the fraction.

13:49 I will solve A. Now I will leave you with D.

13:56 Don't forget this elimination.

14:02 To eliminate A. Now P plus R equals 1 over 4.

14:11 1.

14:13 5P plus 2R equals to 2.

14:20 Equation 1. I have a fraction here.

14:24 There are two ways to this.

14:27 I will solve and I want you to follow what you have on this slide as well.

14:31 Like the other time, clear your fraction or come to example equation 2.

14:36 Reverse through by 2 in order to have a fraction.

14:39 Like the fraction you have in one.

14:41 Then see how you can go about it using the operations of mathematics.

14:44 Let's start with the number one.

14:46 I will be very fast with this.

14:48 1.

14:52 Eliminate the fraction or clear the fraction.

14:59 How do you eliminate the fraction?

15:01 Multiply by what? 4.

15:03 Now we have 4P plus 4R.

15:07 There are one of our 4 times 4 gives you 1.

15:10 And down here you have 5P called this equation 3.

15:13 Because you have any new equation different from the one you have.

15:16 5P plus 2R equals to 2.

15:20 This T equals to 2.

15:23 I want to eliminate.

15:25 It is either you eliminate P or you eliminate R.

15:28 But you won't be able to eliminate until the coefficients are the same.

15:33 I have the question of 4A to 4A 5A.

15:36 Then you choose the one you like.

15:38 I prefer R. I want to eliminate R.

15:40 The question of R at the second denominator.

15:45 The question of R at the second part is 2.

15:49 Multiply this side by 2.

15:51 The question of R at the upper part is 4.

15:54 Multiply this side by 4.

15:56 With that, you will be able to balance the coefficients.

15:59 Here we have 4P plus 8R equals to 2.

16:05 Then it will give us 20P plus 8R equals to 8.

16:11 Now I want to eliminate this.

16:14 I have a addition here.

16:15 Maybe you have been told that when the signs are the same, you stop right.

16:18 When the signs are the same, it is different.

16:20 It is good.

16:21 But just note this.

16:22 8 and 8.

16:23 In order for 8 and 8 to go, I just need to minus.

16:25 I just need to minus.

16:26 Introduce a minus and 8.

16:28 For minus 20P, that gives you what?

16:32 Minus 16P.

16:34 Then equals to 2 and 8.

16:36 2 minus 8.

16:38 That gives you what?

16:39 Minus 6.

16:40 As Elia said, two methods to be applied.

16:43 Elimination method and substitution.

16:46 We have done substitution now.

16:48 Elimination.

16:49 I have two examples here.

16:50 I will be solving one.

16:52 Then I will leave you with the other one.

16:54 4R equals 1 over 4.

16:56 5P plus 2R equals to 2.

17:02 Elimination.

17:09 I have P plus R equals 1 over 4.

17:14 Equation 1.

17:16 5P plus 2R equals to 2.

17:20 Equation 2.

17:23 Go into equation 2 because I have fraction in equation 1.

17:29 I want to make equation 2 to be in order of equation 1.

17:32 Then how do I do that?

17:34 I have 2A.

17:35 I can divide by 2 in order for this 2R to be R as the R in equation 1.

17:42 So divide 2 by 2.

17:44 Equation 2 now.

17:47 Divide by 2.

17:49 If I divide by 2, I'm going to have 5 over 2P plus 2R divided by 2.

17:56 That gives us R.

17:59 Then 2 divided by 2, I have 1 here.

18:01 Code is equation 3.

18:07 Now I can combine 1 and 3 together now.

18:11 1P plus R equals 1 over 4.

18:16 3, 5 over 2P plus R equals 1.

18:25 C2 and 3.

18:29 The essence of elimination is for you to eliminate either side.

18:34 That is, you are either eliminating P or you are eliminating R.

18:38 Now I have R to be balanced.

18:41 How do I know they are balanced?

18:43 Question of R is 1, question of R at the second path also is 1.

18:48 Now I can eliminate RR.

18:51 You may be having to.

18:53 Now when you have the science to be equal, you subtract.

18:57 But that doesn't go at the time.

18:59 Just have to know each.

19:00 What I do is, if I do R plus R, that will give 2R.

19:04 That is, I have not eliminated R.

19:06 So the only way R is, introduce a negative sign.

19:09 If I introduce a negative sign, it means,

19:12 I am subtracting P minus 5 over 2P and R minus R.

19:20 Or you will subtract equation 2 from 3.

19:23 Now I will subtract 2 from 3.

19:26 Or 2 from 3 or 3 from 2.

19:29 In our way, then you are OK.

19:34 But now I am going to subtract equation 2 from 3.

19:38 In equation 3, I have 5 over 2P minus P.

19:47 Plus R minus R.

19:51 And that will be 1 minus 1 over 4.

19:56 5 over 2.

19:58 The SEM is 2.

20:00 2 in 2 will give you 1.

20:02 1 times 5 will give you 5P minus P.

20:06 P is sent in as P over 1.

20:08 1 in 2 is 2.

20:10 2 times P will give you 2P.

20:12 Then plus R minus R, that goes to 0.

20:16 And at this side, the SEM of this is 4.

20:20 1, as you all know, can become 1 over 1.

20:23 1 in 4, you have 4.

20:26 4 in 4, 1, 1 times 1, 1.

20:29 5P minus 2P.

20:32 You have 3P over 2, then equals 4 minus 1 will give you 3 over 4.

20:42 Now, you can cross multiply.

20:47 This goes to this side and this comes to this side.

20:50 4 times 3P will give you.

20:53 And 2 times 3 will give you.

20:56 2 over 2.

21:03 P is not equal to Cz is 1, Cz is 2.

21:06 We have 1 over 2.

21:08 Where is we have the value of P?

21:12 Now, I want to solve for the other part.

21:16 I want to solve for R.

21:19 But don't forget I'm eliminating.

21:22 You have pick equation 1, equation 2, or equation 3.

21:26 Even if you have equation 4, you can pick equation 4.

21:29 Substitute the value of P that you have gotten into any of these equations, then you are good to do.

21:35 So I'm substituting into equation 2.

21:39 In our equation 2, we have 1 over 2 plus R equals 1 over 4.

21:47 What is the value of P?

21:51 So let's go back and check that very way.

21:55 P plus R equals 1 over 4, equation 2.

22:08 P plus R equals 1 over 4.

22:12 That is equation 2.

22:15 I know the value of P, what is the value of P?

22:21 1 over 2, substitute.

22:27 What is R?

22:28 R runabi 1 over 4 minus 1 over 2.

22:33 Likewise, the S m of 4 and 2 is what?

22:37 4.

22:38 44, 1, 1 times 1, 24, 2, 2 times 1, 2.

22:47 1 minus 2 is what?

22:49 Minus 1, they have minus 1 over 4.

22:52 What is?

22:53 I have the value of R.

22:56 The value of R being minus 1 over 4.

23:00 I will leave you with a bit to play with.

23:03 Just go through it.

23:05 I have 3 P minus 2 R equals to 21.

23:08 And 4 P plus 5 R equals to 5.

23:11 If you solve that correctly, you should be able to get your R to be minus 3.

23:16 And have your P to be 5.

23:18 And likewise, you can check to know if you are right or not.

23:24 Evaluation.

23:27 Using some decision method, solve the following PS of simultaneous equations.

23:32 4x minus y equals to 4.

23:35 It x minus y equals to 7 over 2.

23:38 x minus y over 2 equals to minus 1.

23:42 2x minus y equals to 8.

23:45 6x minus 2y equals to 0.

23:48 y minus 4x minus 1 equals to 0.

23:52 Equation 2.

23:54 Use elimination method to solve the following PS of simultaneous equations.

23:59 3x plus 2y equals to 2.

24:02 2x plus 3y equals to minus 7.

24:05 x over 2 plus y over 2 equals to 7 over 2.

24:09 x plus x over 4 plus y over 4 equals to 15 over 4.

24:14 And the last question.

24:16 3y minus 2x equals to 21.

24:19 4y plus 5x equals to 5.

24:25 Thank you for watching.

24:27 I hope to see you again next time.