3x+2y=5 AND 3x-2y=7.  If I add these two equations together, which unknown will be eliminated?

3x+2y=5 AND 3x-2y=7. If I subtract the two equations, which unknown will be eliminated?

3x-5y=17 AND 2x-5y=10. Subtract the first equation from the second equation.

2x-5y=10 AND 3x-5y=17. Add the two equations together.

3x-5y=17 AND 7x-5y=20. How can I eliminate y from these simultaneous equations?

Eliminating a variable: add or subtract?

﻿Hi, this is Ms Bridgett. And in today's lesson, we're going to be looking at manipulating systems of equations in order to be able to solve them. You're going to need a pen. You're going to need some paper. Take a moment to remove any distractions. Okay, let's go. I'm going to show you two equations. Three X plus two Y is equal to P. And X plus Y is equal to Q. I would like you to use the information in those equations to see if you can fill in the blanks in the boxes below. Once you've done that. See if there's anything else, any see if there's anything else that you can find pause the video and I'll keep going. Okay. Let's have a look at the answers to these. So for the first box four X plus three Y, I can see that four X plus three Y can be created by adding those two equations together. So it's equal to P plus Q, Two X plus Y can be created by looking at the difference between those two equations. So P subtract Q. Six X plus four Y is double the first equation. Therefore it must be equal to two P. And six X plus six Y we can create by multiplying that second equation by six. So we know that's equal to six Q. I would love to know what else you found, but here are some of the things that I found in addition, 18 X plus 12 Y is equal to six P. So, I took that first equation, and then multiplied it by six. And then I also found this one, 12 X plus 12 Y is equal to 12 Q I created that by multiplying this final equation, the second equation by 12 on the screen, you've got a pair of simultaneous equations. Now, so far when we've been solving simultaneous equations, we've been eliminating one of the unknowns by either adding equations together or subtracting them. Now look carefully at this pair of equations, without them, neither of your unknown's going to be eliminated and if we subtract them neither of the unknowns are going to be eliminated. What i'd like you to do is to pause the video and have a think about a strategy you could use to eliminate either the X or the Y. So pause the video and have a think about it. Okay. There are quite a few ways that you could go about this. Now, what we're going to do is we are going to firstly thinking about eliminating one of the variables, and then we're going to eliminate the other one. Now, what I've done to eliminate the Y is I've kept the first equation exactly as it is. So the three X plus two Y is equal to 11. I kept it exactly as it is. But look what I've done to the second equation. So the second equation was X plus Y is equal to three. What I've done is doubled it, I've multiplied the whole thing by two. So it's become two X plus two Y is equal to six. Now the impact that that's had on my system of equations is that if you now look at the coefficients of Y you can see that they're the same. So at this point we can now subtract those equations, and that will eliminate the Y or we get left with an equation, just with one unknown in it. Once we've got an equation with one unknown in it, we know we can go back. We can substitute, we can find the other value. We can check any of the equation so we can manipulate one of the equations to make the variables, to make the coefficient variables the same. How might we manipulate these equations to eliminate the x instead? Pause the video and have a think Okay. Again, there are several different ways to go about this, but this is the way that I've done. So I kept the first equation, exactly the same three X plus two Y is equal to 11. And then the second equation X plus Y is equal to three. I've multiplied it by three. So that turned into three X plus three Y is equal to nine. It's transformed into that. Now at this point, if you look at those equations, you can see that the coefficients of X are now the same. Because the coefficients of X are now the same, when I subtract them, I eliminate those variables. So I've done two Y subtract three Y and 11 subtract the nine. That will give me an equation with just one known in it. I can find the Y. Once I've got the Y, I can substitute it to find the other value. Now, what's really great about doing this is I've got two different methods of doing it. So I can do a triple check to see whether I've got this correct. So I can eliminate the Y first of all and get one set of one set of solutions. I can eliminate the X and get another set of solutions. Should have the same solutions. And then I can put those back into those original equations to triple check it, to check that I have actually got the correct solutions, for the Independent task, I'm giving you two pairs of simultaneous equations. I would like you to do two things with each of these pairs. So for each pair, I'd like to eliminate the X in order to solve it. And then I would like you to eliminate the Y in order to solve it. So you're solving each set of equations twice while you're doing this, keep one eye on yourself as you're doing these and be thinking about which of these, which of the two methods is easiest. And why is it easy? Why is it easiest? You can check your solutions using substitution. Once you've got to the end. Pause the video and have a go. Let's have a look at that first pair of simultaneous equations. The first thing i asked you to do is to try to eliminate the Xs. We've got three X and two X. Now I need the coefficient of X to be the same. So what I'm going to do is I'm going to try and find the lowest common multiple of three and two, and make that the coefficient of the Xs. So let's just call these equations one and two. So it's a little bit easier for us to refer to them. I'm going to take that first equation, three X plus four Y is equal to 45 and I'm going to multiply it by two. Now, what that's going to do is creating a new equation. Six X plus eight Y is equal to 90. I'm just going to call that equation three. And then I'm going to take the second equation. Two X subtract Y is equal to 19. And then I'm going to multiply that one by three to create six X subtract three Y is equal to 57. So I've made the coefficient of the X six and I've chosen six because six is the lowest common multiple of three and two. Let's call that equation four. Now, rather than using my original set of equations. One or two. I can now use these transformed equations three and four. I've lined them up next to one another. Now, if you look at these, you can see that we've now created a scenario where the coefficient of the X is the same, and that means I can subtract these equations and it will eliminate the Xs. So if I subtract them I'd end up... And solve them, I'd end up with X is equal to 11 and Y is equal to three. And you can go back and substitute these and solve them. I haven't written all the workings out here because there isn't huge amounts of space, but obviously you would. Let's look at this a second way. So that's the method using by eliminating the Xs, that type of look at what happens when we eliminate the Ys. So this time I want to make the coefficients of Y the same. Now at the moment I've got plus four Y and I've got a subtract Y. So what I can do is I can leave the first equation completely alone, and I can just tamper with the second one two X subtract Y is equal to 19 if I multiply that by four, that will transform the equation into eight X subtract four Y is equal to 76. Let's just call that equation three. I don't need to do anything to that first equation. I can leave it as is. And I can just put my new equation three next to it. If you look at that, you can see the absolute values of those coefficients of Y is the same. I've got a plus four I've got a subtract four. So at this point I can add the equations and it will eliminate the Ys and I can then go on to solve it. Now, just take a moment to think about that. They both got us to the solution. There are both perfectly good ways of getting to the solution, but I think one of them was slightly nicer than the other one. And I think in this particular case, we had less work to do, to eliminate the Ys. So that would be my preferred method here. I'm going to choose the one where I've got less steps to go to eliminate one of those variables. I'm not going to talk through the whole of the two different methods for solving the second equation, but hopefully you ended up with the solution X is equals to five and Y is equals to one. And you were able to double check that. Now again, I don't know if you agree with me, but in this set of equations, I think it was easier to eliminate the X rather than the Y I think there was less work to do to eliminate the X than it was the Y. So I think if we just doubled that first equation, it would make the absolute values of those coefficients of X the same. And therefore that made it a little bit easier to eliminate than the Y. Okay. For your final test this lesson, I'd like to take a look at these, this pair of simultaneous equations, and I'd like to solve them. Just take a minute to do that. So after one minute, come back and unpause the video. Off you go. Now what happened when you tried to solve the equations? Could you actually solve them? Now i'm sorry to say. I just gave you an un-solvable pair of equations. You couldn't find solution to these. So what I would like to think about now, is what happened when you tried to solve them? Why were they unsolvable? And finally, can you come up with your own system? Your own un-solvable system of simultaneous equations. Pause the video and off you go. Okay, let's have a think about why this particular system of simultaneous equations couldn't be solved. Now it's because it isn't actually a system of simultaneous equations. It's the same equation repeated. I gave you three X plus two Y is equal to 10 as your first equation. Now if I take that and I multiply it by three, I end up with nine X plus six Y is equal to 30. It's exactly the same equation. It's just been transformed. Now we taught quite a few lessons ago about the fact that in order to solve a simultaneous equation with two unknowns in, we need two equations. We need two sets of equations for two unknowns. If we've got two unknowns, one equation, there isn't enough force to solve it. So here I haven't given you that second piece of information to allow you to find out the, find out the unknowns. So creating your own un-solvable system of equations. Here's some that I came up with. So I started off with X plus Y is equal to two. I doubled it. I can multiply it by 10 or I can be very exciting and divide it by two and any pair that I pick after that will be unsolvable. And it will be unsolvable because it's the same equation, just written in a different way. Now, if you are able to, it would be great. If you could take a picture of your unsolvable equations and share them with your teacher, or if you want to ask your parent or carer and they can share them with @OakNational on Twitter. Otherwise thank you so much for your work today. And next lesson, we're going to go on and look at some other ways and some other contexts that might need simultaneous equations.

x+y=6. Multiply this equation by 3.

x+y=3 AND 2x-y=5 This pair of equations has been transformed.  Which of the following are equivalent to the original pair?

x+y=3 AND 2x-y=5 This pair of equations has been transformed. Which of the following are equivalent to the original pair?

2x+5y=9 AND 5x+4y=14. Solve the simultaneous equations.

10x+5y=5 AND 3x-5y=34. Solve the simultaneous equations.

Manipulating systems of equations

In this lesson, we will manipulate systems of equations in order to eliminate an unknown. We will look at cases where one equation can be altered to enable elimination by addition or subtraction.

Maths

Y9 Maths

Key Stage 3

Year 9

Solving linear simultaneous equations algebraically

Algebra

THEME

UNIT

Reviewing solving equations

Equivalent equations

Linear relationships

Combining algebraic relationships

Exploring expressions with two variables

Exploring systems of equations

Solving simultaneous equations

Writing simultaneous equations algebraically

Checking the solution of simultaneous equations

Solving a simultaneous equation algebraically

Eliminating a variable

Solving word problems

Using substitution to solve simultaneous equations

Substitution or Elimination

Simultaneous equations by trial and error

Let p be the number of protractors bought. If each protractor costs £2, which of the following represents the total amount spent on protractors (in pounds)?

If x=2 and y=3, what is the value of 2x+3y?

Use substitution to decide which of the following is the solution to x+y=3 and 5y+3x=15.

What number am I thinking of, if half of the sum of my number and 4 is the same as 3 lots of my number subtracted by 2?

Fill in the gaps: The "n" in the nth term represents the ____.

Which sequence is formed as a result of adding "4, 9, 14, 19, ..." and "12, 8, 4, 0, ..." together?

I want to buy 3 rulers and 3 protractors. If they cost 40 pence each, how much will I spend?

If p+1=6 and p+q+1=7, what is the value of q?

4x +8y=116 AND 4x +4y=68. Solve

Is there a solution with x value between 0 and 1?

Look at the picture below and then fill in the gap: If two equal expressions are both __________ by the same quantity, then the two resulting expressions will be equal.

Lesson overview:Manipulating systems of equations

Core Content

Solving linear simultaneous equations algebraically: