Video: Solving further linear equations

﻿Hello, I'm Mr. Langton. And today we're going to be solving some more linear equations, what lines to do get a pen and paper or something to write on, find yourself a quiet place where you can work without being disturbed. And then when you're ready, let's move on. Good luck. So we'll start with the try this activity, these shapes and then after the same length of sticks, use this information to write an equation. So what I'll ask you to do is pause the video, have a goal set when you done and we'll go through it together. So I have to go now pause you three, two, one. How did you get Tom? Let's have a look at some ideas together. Now the triangle even have made three blue sticks, whereas the hexagon is members six blue sticks. So the perimeter of the triangle must be 1/2 of that, of the hexagon. So if I take the perimeter of the triangle and if I double it, then it will now equal the perimeter of the hexagon. So that one equation I could get. The flip side is saying, if I take the perimeter of the hexagon and 1/2x, that will be equal to the perimeter of the triangle. That's a different again. There's a third one we can get. where if we actually just look at how long one individual length is. So in the case of the hexagon all six lengths, 3x plus 12. So if I divide that by six I'll get one length with the triangle, all three of them, 24 take away 3x. So if I divide that by three, I'll get one of them. So one of the blue sticks from the hexagon is equal to one the blue sticks from the triangle. Now this is a little bit of a play with that a little bit more when simplify that. So, 3x plus 12, all over six. I can break that down into 3x over 6 plus 12 over six. And we said for the size of the triangle, we've got 24 over three subtract 3x over three. So let's break each of those down individually. So 3x divided by six or three over six cancels to 1/2. So it's actually 1/2 of x or x over two, 12 over six is two ,24 over three is eight and 3x over three, cancel down to just x. So that is another equation that we could get. I want you to bear that in mind a little bit, because we're going to be using that on the next slide. So now, we're going to have a go at solving these equations. We'll start off with buttons on the left hand side. Probably a trickier one, because it's got fraction inside. So we'll start off. I would like to write to out myself first eight takeaway x equals x over two plus two. Now I'm not really going to have to draw a bar model for this one, so I'm going to rely on my balancing mathematics. We'll start off. I'm going to subtract two from each side. That's going to leave me with six takeaway x on that side x over two on that side, 1/2x. Now looking at that 1/2 of x, the inverse of divided by two, with 1/2 X, with multiplied by two, I'm going to multiply each side by two. It's going to give me 12, click into 2x is equal to x. And not keep things balanced, I'm going to add 2x to each side, that takes me to 12x is equal to 3x. So you can now see, if I divide both sides by three, x is equal to four. Now, hopefully we'll final translate equation is that we should get the same answer. to do this one find it difficult. First, I always like to write out the question first. It helps me think about it just that little bit more. Even do this in exams, even when I've got the question right in front of me, my first step is to write down it just helps me. So let's expand these brackets, multiply the brackets, two lots of 24, 48, and two lots of negative 3x, negative 6x, so you get 3x, and that's equal to 3x plus 12. So next step. There's lots of steps I could take. I'm going to add 6x to each side gets rid of them. Subtraction there Hopefully it make it a bit easier for me. So in that case, all that will left on one side is 48. If I add to 6x, I'm going to get 9x plus 12. And we're now looking at something a lot more familiar. We can do something with that. I could draw a bar model if I wanted to. I don't really fancy drawing nine boxes, but I could, I can see now, I can subtract 12 from each side. That going to leave you with 36 is equal to 9x. So if I divide by nine, I'm going to be very, very happy. Because once again, I found that the value of x is four. So it's two different ways I can solve the same problem. It's time for your shotgun now. Eight questions each try to pause the video access to worksheet and how it go. If you're struggling, if I play a little bit longer and I'll talk you through a couple of the problems and just start you off on a couple of them. Let's just give you a couple of hands. Let's have a look at question six, instead of stuff in a fraction, as part of the problem, the fraction is not to the whole left hand side is a fraction on the whole, the right hand side is a fraction. So what I would do first, and I think I would multiply both sides by six, because that's going to get rid of that divide by six. That is going to leave me with x minus two on left hand side and six, lots of x takeaway five all over three right inside. And if you want an extra step, something a bit more, obviously what happens when you multiply both sides by three after that? What else could we, haven't gone up. Let's give you a hint on this little trick that we could do for question two. We could expand the brackets and we could try it that way. But what if our first step was to divide both sides by two. In this case, all that you've got left on the left hand side is x plus seven and if we divide by two, instead of four lots of x plus two, we've now only got two lots of x equal to two, I'm make it easier. And let me give you one more hint, 'cause I'm feeling kind of an over it. Over here question eight, my first step will be to multiply both sides by three because of the whole fraction is a whole on the side being divided being divided by three. we're on the left hand side, only the x that's being divided by two. And if you do multiply by three, don't forget, you're going to have three lots over two plus one. Don't get most by both terms by three. So I'll leave that as you can have a go when you're done pause it and we'll get the rest of the answers. Here are the final answers, I hope you did well. So we'll finish off with the, the explore activity. Use the card below to form different equations in the number frame. Solve them and see, how many different values of x you can form? What's the greatest value or the least value of x you can form? Pause the video and have a go and when you're ready, we'll go through it together, supporting it three, two, one. Let's just do a couple of examples. So you might, for example, say that 3x takeaway 2x is equal to three and two. And if we solve that, going to get x on the left and five on the right. So that's one idea. One more idea now, let's start What about 2x take away two equals 3x plus three. So if we're going to do this one, let's add two to each side. That's going to give me 2x is equal to 3x plus five. You start to get a little bit messy. Three is on the same side now, if do that, I'm going to have to subtract 3x from each side. And I don't like the look of that. 'Cause I'm now going to end up with negative x equals five, and now I'm not finished solving it. Then I've got negative x equals five. I actually had the value of positive x, two ways that we'd go around doing that. We could multiply both sides by negative one, which would give us x equals negative five. We could add x to each side. I mean see if it's zero equals five plus x and then subtract five from each side to get an x from the right and negative five on the left. Either way, we've got this value of negative five. Now, I will tell you, is that the greatest value that you can make it's five, the least value you can make is negative five. There are two of the values that you can make, depending on how you arranged your values. You could have got x as a fifth, or you could have get x is negative. Finally, if you'd like to, please ask your parents or carer to share your work on Twitter tagging @OakNational and #LearnwithOak. See you later.

Worksheet: Solving further linear equations

Exit Quiz: Solving further linear equations

Completed: Solving further linear equations

In this lesson we will look at more difficult linear equations and solve them using algebraic methods

Maths

Solving further linear equations

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