 Welcome to your first sessions of your exam preparations sessions. My name is Elizabeth Boye. I will be your facilitator for the next four sessions that we're going to have. If you can hear me, please look out and come back in. It might be that you, it's your connection on your other side, because from my side I can, everything, I think it's fine. Let me know if all of you can hear me, just unmute and let me know. I can hear you. Okay, then that's good. Every week you come in to the session. Please make sure that you complete the register in the chat. I've already placed it in the chat. Please make sure that you complete that. If you have any technical issues, you are unable to connect or you don't know where the recordings are and so on. Please send an email to ctntat at unisa.ac.za and you can copy me at unisa.ac.za. If you have content related questions, let's say you are struggling with your module content, you don't know how to answer certain things, you can just send an email directly to me, but also copy CC in that email ctntat at unisa.ac.za. Otherwise, the recording and the things that we go through, you will find them on the link that they've sent you to join this session, as well as the notes or the exam paper that we will go through, they are also uploaded onto the folder. So if you don't know how to access them, send an email to ctntat. They should be able to tell you where everything is at. Before we start with today's session, are there any questions or comments? Please always remember when you join the session to switch off your video and also mute your microphone. However, during the session, as we go through the activities and I ask you questions, I expect you to unmute and talk to me and then go back and mute yourself. So it's going to be an engaged session. I don't want to be the only person talking. I want to make sure that you all understand the content by you talking to me. It also gives me an idea if you are struggling with some of the things or not. Please feel free to ask any question to stop me as I explain things. If they are not clear, just stop so that we don't have to continue or I don't have to continue and then only ask later. Stop me and ask that question that you have so that we can clarify and then move on. We're going to have four sessions. This is the first of the fourth four sessions that we're going to have now. Excuse me. Excuse me, Mrs. Boy. Oh, you can call me Elizabeth. Elizabeth, I just wanted to know where do I complete, where do I write in the register because I don't know where the check is. I only see it's the video with the camera and then there's more in hand and participants in. So you joined as a guest, right? Yes. You joined as a normal with your UNISA logins. So those who joined as guests, you can see the chat function. You need to log into teams, UNISA teams via your MyLife email so that you are able to see everything that we see and you are able to access everything that we have. So, must I go out again and then in? Because I just, I logged on the link that it was shared via email. That's why I am on this, like, yeah, I don't know. Yeah. Okay, let me just go out and see if I can come back in. Fine. You won't be able to complete the register because you won't, you will not be able to see the check function. Probably by next week, make sure that when you join teams, use your UNISA login details to sign in on to Microsoft so that you are able to see everything that we see. And it also allows you to also communicate with us via the chat because now you can't. The only way you will be able to communicate with us is by unmuting yourself as well. So, okay, which is good for me because then it will force you to talk to me than to write in the chat. But it's bad because UNISA requires the register to know. Yeah, so we require the register to know how many students are participating in this and whether we should cancel them or not. They will look at the register and if only one person have registered, therefore it means there was no one in the room. I mean, when you record this recording. No. Remember, for us to submit to the head office, we require the register to support. Oh, yes. Okay, I understand. So, if there is no register, and there are things that are done, quality assurance things, if there is no register, it means there was no participation, whether there was a video or not. So, there are certain things that are important, which is the register. Okay. But that does not stop us from continuing and giving you a lesson. I would really like to go out and log in with my details because I don't want to do it. No, don't worry for now because then by the time you try to figure out how to solve your problems, then it will be too late. Okay, so where was I? Okay. Yeah, we're going to have four sessions. In all the four sessions, we're going to cover topics, different topics. But I also want to know what are the key topics that you think these are the difficult topics that we need to spend more time with or on? Because when I completed the topic list for the four sessions, I just did it generic. On one, two, I used my common sense actually as well, but also for the past two months, we have had, we have been having sessions week in, week out, every Monday. But there are not a lot of sessions that we cover that covers almost the bulk of the QMI session. So if you haven't been attending those sessions, then you've missed out. And I'm not with this session. I'm not going to be repeating any of those things because we have recorded them and we expected students to come and participate in those so you can go find the recordings. If you don't know where you can also send an email and they will tell you where to find those recordings. So today we only going to look at certain concepts that relate to equations, which will cover linear equation, quadratic equation, simultaneous equation by looking at the past exam papers. So we only have two hours, which already of the two hours we spent 20 minutes just doing the admin work. Okay, so is there anything that you want to highlight or comment on before we start with today's session? Now is that chance to unmute and tell me what is that one topic in the whole QMI subject that you are struggling with. This will also help me to prepare for the next sessions. Hi, Elizabeth. Yes. I feel actually like stupid for, but since my mother first assignment was okay, but I did very bad on the last two assignments. I was struggling a bit with understanding actually everything because I haven't been on the video. I haven't been locked on these Fridays or was it the Saturdays when the doctor was giving lessons and stuff. So I actually have a lot to catch up on and I would like to sit in on every of the four sessions that you are going to do. Alice, what are your comments? Are there first sections that you want us to concentrate on? Yes. Yes, you can talk. Yeah, I have a problem with functions and the linear systems. I'm struggling with both those two chapters. Okay, those are the things that we're going to be doing today. Okay. Hi. I've been struggling with the financial aspect of this. So like the amortization, I don't really understand. Yeah, let's see. Do we have any topic on finance amortization? Yes, which will be the last session that we have so we can concentrate more on that on the last session. So you can't miss that session, right? Which will be on the 22nd of October. Okay, thank you. Hi, Elizabeth. I'm the same, also for the last session, the financial part. So I will be joining them as well. Okay, so those sessions, like, no, the last session we did on basic numeracy skills development or the facilitation that we did, I think two weeks ago, we covered financial meds, but we didn't cover amortization because the Monday that we were supposed to cover amortization, we didn't have enough time. I couldn't share my screen and record the session. So we didn't have that session. So we will cover that during that session on the 22nd in more detail. Okay, so those are the only chapters that you struggle with. So I shouldn't even have included like data handling and differentiation because those are easy, right? You guys, you are able to sort those things out. So where we have placed the past exam papers on the notes section. So you should be able to go through the same exam paper. If you don't have them, then you, you can download them after the session. So we're going to go through one exam paper at a time and go through all the four sections that we need to look at all the, or any question that relates to linear equations. Simultaneous equations, inequalities, and also quadratic equations. So it means we've got a lot to cover. And because you've never attended any of the sessions, so I'm just going to firstly do, every time we do a question and if it's a new question that covers a new topic, I will just give you a brief or summary. Of that content. And then we do some activities and then we'll take it from there. So let me share my screen again. So I have three past exam papers that I'm going to try and find questions based on the topics that we're going to be using. So the one that we will base most of the activities on is the 2016 E1. I haven't seen what the questions looks like. So I'm going to take a step at it. So the first question since we're talking about linear equations and equations. I'm going to assume that some of the topics are easy and straightforward. So we're not going to draw too much on those, like for example, how to solve equations. You should be able to know how to solve equations, right? By manipulating the equations. If you don't, because you didn't say you don't know how to do that. So those are the basic things that you should be able to do. So for example, like this question for where they're asking you to solve for x, which you need to know the basics that if it's dividing, you multiply to get rid of something. And if you move things across the equality sign, the sign will change. And if it's multiplying, you need to divide and so on. So those are the things that you need to know about equations. And then what we're going to be looking at is linear equations. Let's see if we can get there. So linear equations. So with linear equations, there are a couple of things that you also need to remember. So with linear equations, you will be given two points and the two points have coordinates. They have the two coordinates, x and y. So because there should be two points in order for you to draw a straight line. So the first coordinate, the first point will have x and y, which will be x1 and y1. They correspond to one another. And the next point will also have x and y, which you will label x2 and y2. And once you have those two points on a Cartesian plane where your x value will go on the bottom and your y value will go horizontal or vertical. I don't know which one is which. So if I have the first point, which is this, it has coordinate x1 and y1. Therefore it means it corresponds to x1 and y1. The next point, if it's there, this will be my x2, which will have a coordinate of y2. And when I have the two points, I can just draw the line that joins the two points. And that is the equation of a straight line. We can find it because this straight line has an equation called y is equals to ax plus b or y is equals to mxc. So since we know that that is the equation of a straight line and that is our straight line and the straight line has two points that correspond. Between those two points, we can calculate what we call a slope. And the slope in this instance is what the value of a represent. And also because the Cartesian plane looks like this, it has the positive and the negative. On the left of the horizontal line, you have the positive and on the right you have, on the left you have the negative, on the right you have the positive the same. At the top you have positive values, at the bottom you have the negative values. And where your straight line passes through the y-axis at the point, we call that an intercept. And an intercept is the same as our b value. It is where x is equals to zero, then that will be the point. It is where it passes through the y-axis. So if they ask you to calculate the slope, which is a, the slope is given by the changes in the values of y, divided by the changes in the values of two, of x, sorry, x2 minus x1. And that's how you will find the slope. Finding the equation of a straight line in b and u, it's easy to find it. In QMI, you need to first calculate the slope, pick any of the two value or the two points, whether you pick the first one or the second one, and substitute them as your x and y into the equation of a straight line because you would have had your a value. You will substitute your x and y based on any of the points that you choose. If you choose point one, you will use half and one over three as your x and y. You will just substitute them into the equation. If you choose point two, you will use your x and y of point two and substitute there and find the value of b. And once you have the value of a and the value of b, then that is your complete equation of a straight line. That's how easy and straightforward like that. So how do we choose which point to use as well? There are some guidance that I can give you. Always stay away from fractions, stay away from big numbers, stay away from negative numbers. But if there is no way that you can stay away from them, you can then use them. For example, this is fractions. It's fraction plus there is a whole number. Then I would prefer to use the one way. It's not a fraction, but I also can say I prefer to use the fractions. Then I can solve everything as fractions. So it's up to you how you choose that. So if the question says the slope of this line y is equals to a x plus b passing through the points. And they give you the two points point one as your x one is a half and y two. Sorry, y one is one over three. That is your first point. And your second point, which has x two and y two coordinate as two and three over four. What will be the slope? So you first needs to always remember them to identify your points by marking them x one y one. So if you would have been part of my sessions previously, you would know that in most of the sessions that I give. We always use what we call Newman's problem problem solving method. What it does is it says in your own understanding, tell me what the question is asking you to do. So in this instance, the question is asking us to find this loop. That is what the question is asking if I understand it in my own words. Number two, it says what are the facts or things that are given to help you answer that question. So we are given two points and I'm already identifying them point one has x one and y one coordinate point two has x two and y two. So I'm able to visualize and identify the facts given in this. Step number three of that it says you need to identify what formula I need to be using in order for me to answer this question. Is there a formula or a table or some sort of a calculations that I need to do or anything that will help me like including the bot mass rule. Can I apply bot mass rule? Can I you need to ask yourself those type of questions. So identifying the formula. I know what formula I need to be using and since I've already highlighted and identified my point, I then can come into my formula and start substituting. After I've substituted the values into the formula, I do my calculation. Right. So let's do that. So our Y two we go back to that because we identified it. It's easy for us to find which value we need to be substituting way. Y two is three over four minus y one. And also remember the equation has a negative side. If the point has a negative, you also need to put the negative value onto the formula. So on this one, our Y one is one over three. One over three divided by X two is two minus X one is one over two. Now we need to, I just want to remove, just want to remove this. So then I'm going to run out of space and it is very difficult to erase a PDF. The same as when I'm working on a PowerPoint slide and just bear with me while I try to clean up the page to clean everything. Just leave it like that. Okay, so now let's solve. So you can solve this as a fraction. You can take your calculator. Those who have a cashier calculator, it's easy to solve your fraction. Those without a cashier calculator, you need to know how to solve fractions manually. If you are subtracting or adding equations, know that with your fractions, when you add or subtract your fractions, you need to find the common denominator. If you multiply, you can simplify your equations. So in this instance, we have a minus, which is a subtraction. So it means we need to find the common denominator. So I'm going to solve the top part first. Finding the common denominator between four and three, the common denominator is 12. Four goes a many times into 12. It goes three times three times three is nine minus three goes a many times into 12. It goes four times four times one is four. And then I go to the bottom. Any value without a fraction, we can determine it as two, as a divide by one. We don't have to write the one, but because I'm solving it as a fraction, then I have to include the one so that then I can find out which what is my common denominator. So my common denominator between the two, two and one is two. So one goes a many times into two. It goes two times two times two is four minus two goes a many times into two. It goes one time. One times one is one. So simplifying the equation. I'm just going to write it on this side from nine minus four is five over 12 divide by three minus. Just give me a second. I need to admit someone. If you join using your my life email, it makes it easy. I don't have to admit anyone in you join automatically. If you join with a guest, it makes it difficult because when I'm doing the explanations and I need to move away and go and help you come in. So please next time join with your my life email. Sign in onto your teams with your my life emails. Okay. So four minus one is three over two. So three over two. And yeah, I've got a division. So with a division. When we simplify fractions, we keep the first we change the sign and we flip the second. So we're going to keep the first fraction, which is five over 12. We're going to change the sign from a division to a multiplication. We're going to flip the second one, which is the bottom one as two over three. And we can then simplify two goes one time into two and it goes six times into 12. One times five is five over six times three. It's 18. And the answer is five over 18, which is option three. And that is the thing that you need to know and be able to find when you enter linear equations. Are there any questions, comments, query? Let's look for more exercise. But before we go there. Is there any question? No questions. Are you all good? Remember, you need to talk to me. Hi, Elizabeth. Yes, I'm all good. Thank you. Yes, thank you. For me to be hyped up about the session is if you guys talk to me and I feed off your energy. If you are all quiet and then two hours, it will only be me talking. And I don't even know if you understand or not. And if you are quiet, it makes things even worse. I just wanted to ask, will the African-speaking, can you speak African-speaking? Yes, okay. Me, I'm one person. I do understand a little bit. But you can explain in English. It's just difficult to me because I am a trans person. Okay. Talking in Afrikaans, I know Melissa is here and Karin and Nolan. Yeah, they can explain if I don't understand. Okay, Nia, do you know how difficult it is for me to live out of the sessions if it's too many English people and I understand the minority of it. And so what you clearly make sense, and I want to know, all the expressions, you know, the slope of the line, why? No, what do we know, the points. It's always going to be like this. Okay. Okay. I will try, if I confuse you even further, then it's my fault because I'm not, Afrikaans is my fifth language. Yes. Yeah, no, it's fine. When you need to calculate the slope, they will always give you the two points. Right? But always, it's not always going to be in a fraction format. So you just need to know the concepts, the basic concepts in terms of how do you solve the fractions? Number one. Number two, you need to know how to substitute by first identifying them. It's very important that you don't confuse your X1 and Y1 and your Y2 and X2. They always go together. And always remember the following, right? They might give it to you in this format, or they might give it to you in a weight format. Let me see if we can get another exercise on this. Where they give you weights, where you need to identify from that weight statement, which one is your X, which one is your Y. But you must always remember that they always go together. Your X and Y will always be together. So I can see that when we get to the inequalities, they give it to you in weight format. Like, for example, this one is in a weight format. Can you see that? But you need to identify from this statement what is your X, what is your Y. I was trying to see if we can find the linear one. So in this question paper, there is none. So let's see on the other one. Also on this one, you see here now. They want you to identify which one of these are the two points from this statement. So it's always not going to be as straightforward as the way I've shown you. But also you just need to know how to calculate certain things, the way they've given it to you. So for example, this is another linear equation question. So it says a shop owner has a special offer on peaches. He sells one packet of peaches, 33 rent, and three packet of peaches for 90 rent. Let X be the number of packets sold and Y be the price. Now, you must also always remember that because they give you that your X is the number, which is the quantity and your Y is the price. So it means a quantity goes with the price, right? So when you go back and read the sentence, he sells one packet, which is the quantity for 33. So already from here, you know that this is your X and this is your Y. You can clearly say, for my first point, because that is the first statement that they gave you, this will be your X1, this will be your Y1, because they go together. X1 and Y1, they go together, they are MH. Three packets for 90 rent. Your three packets is your quantity, which will be your X and 90 will be your Y. So three packets is your X2 because it's the second thing that they mention and the price for that second thing that they mention. So already you have your X1 and Y1, right? So in order for you not to get confused, you can go back and write X1, Y1. Right? And X2, Y2, those are your points. And then you can just substitute the values that we have. One packet goes with 33 rent. And X2, these three packets goes with... I don't like that. Three packets goes with 90 rent. And you can go and find out which one satisfies that, which is number two. And you can see that you can get very confused quickly because of all the other options that they gave you. So you just need to make sure that you understand how to classify or identify your two points. And from there, you just need to know how to calculate the other things like the slope, like the equation of a straight line and so on. Hi, Elizabeth. Yes. Yeah, I actually wanted to mention, I think that's what confuses me sometimes. They give so much information that I start getting confused about what they actually ask. So sometimes, especially during exams, I struggle to answer what they're actually asking in the question for me to do. Yeah. The other thing, like I said, when I mentioned the Newman's Prompt, it's always easier to use that because the first thing that you need to clarify in your mind is what is it that they are trying to ask in this? Right? Because if you read the question, I didn't read the entire question, but I went and I looked at the options and I can see that the options is identifying two points. And I also, in the question, I could have read the full question and in there, I have the key thing that the question is asking. So it says the two points following directly from the information given that will satisfy these relationships. Ah, so therefore, in your own understanding, you need to have already clarified it in your mind that you are looking for two points from the statement update, right? And in your mind already, you should say, what are the next thing? What are the facts given that will help me answer this question? But why don't you go and read again the sentence again and say, OK, I know if we're talking about the points, the point has an X and Y coordinate that go together. Then go and read the question because you've read it the first time. Then you go back and identify the facts given. And the first step will be to read the first line. And from that first line, identify the first key things. Then go to the next one. And then the next one until you have satisfied everything. And that is how you should be able to answer the questions. What I've also picked up and realized previously with some of the students is that the minute they see something, they just want to go in and start calculating. And you are very quick to go onto your calculator and start calculating. Take your time to understand. Take your time to unpack the question so that you understand. If it means writing out, like I've written out what my X1 is, my Y1 is. Do that before you even take your calculator and start calculating. Let your calculator or calculations be the last thing that you do with if you have five minutes. Take the first three minutes to understand the question. The last two minutes to do your calculation because once you have put your building blocks in place, calculations will be easy because you have everything in front of you is just slicking on your calculator and then voila. Okay, so moving on from linear equations, we can go to what we call simultaneous equations. So linear equation is when you are given only one equation. simultaneous equation is when you have two linear equations. Right. So there are a couple of things that you need to know about linear equations. So when you have a linear equations, you can simplify it in multiple ways. The first one is by using a method of substitution. The other one is by using the method of elimination. Find the one that you feel it's comfortable enough for you to use and apply when you answer the question. I always think the method of substitution is the easiest, but some people find it difficult. So I'm going to show you the method of elimination and the method of substitution to answer the same question. Right. So let's first start with the method of substitution. So with method of substitution, you need to go to your equations and you need to identify them and call them equation one and go and identify the next one. Equation two. The reason why you identify them is so that you don't have to use the same equation again and again and again and again. You are able to now clearly know which one you have used, which one you need to use now. Okay. So looking at the two equation, you need, when we use the method of substitution, we want to make sure that we are left with only one variable on one side so that we can take that one variable and substitute it into the other. So for example, we can look at this and say maybe we need to use equation two. So let's use equation two because we just want to make y the subject of the formula. So we're going to use equation two and write it in this format and make y the subject of the formula two x minus y is equals two minus three. And to make x the subject of the four or y the subject of the formula, I need to move to x to the other side. So on the left, I will be left with minus y is equals to minus three. And then when I move minus two this, sorry, I move to x to the other side across the equal sign, the sign will change. It's positive right now. The sign becomes negative two x. Otherwise you can do whatever I do on the left. I must do on the right. What I mean is, if I need to get rid of minus, oh, sorry, if I need to get rid of two x, I can subtract two x on the left, but I also need to subtract two x on the right as well, which is the same thing that I have done. So we are looking for y not minus one. So it means we need to get rid of the negative number in front of y to get rid of it because it's multiplying by its minus one actually. So we can divide by negative one or we can multiply by negative because negative times negative gives you positive. So if I multiply by a negative value on the left, I must also multiply everything on the right with a negative value. So I'm going to multiply by negative one. So negative times negative will be positive. Then I will be left with y equals negative three times negative will be positive three negative two times negative two times negative two times negative will be positive two x because I've multiplied by a negative number. Now I have my equation 2A. I'm going to relabel this to 2A. So this is my equation 2A. As you can see that equation 2A is the same as equation 2, but now we only have a variable y. So we say y is equals to or y is the same as three plus two x. So if my y is equals to that then on equation one, I can substitute the value of y by replacing the y value with three plus two x. So that's what I'm going to do. Substitute equation 2A into equation one. So it means I'm going to take my y value and substitute it into equation one. So let's do that. We have three x minus two times and I know that now my y is three plus two x. So instead of writing y, I'm going to write three plus two x. And that is equals to both. I put it in the bracket because y is equals to three plus two x is not and y is multiplying to minus two. So you need to put everything in the bracket when you're substituting it because everything that is inside the bracket is the same as y. So now we can apply the Boltzmann rule. The Boltzmann says brackets first, so it means we must get rid of the bracket. To get rid of the bracket, we're going to keep three x. We say minus two, we distribute it into the bracket. So minus two times three, it's minus six, minus two times positive two, it's plus four x. So that equals four. Now, we need to simplify this equation by making x the subject of the formula so that we can have our x value. So the other thing I forgot to mention, when you solve what you call this equation of systems of equation or two simultaneous equation, you are trying to get the x and y value where they both meet. So if I can draw this again. So a system of linear equation is where you have two lines. So where they both meet there, that's what you do to solve the equation to just identify what is the common x and y value. Sorry, Elizabeth. Yes. The minus two, but isn't it also supposed to be like minus four x if it's multiplied by two x? Oh, sorry, yes, you are right. Yes, so okay, so let's continue. I just forgot to mention that one thing which I just did now and just messed up my view. Okay, so continuing. Let's make x the subject of the formula so we can find the x coordinate. So go find the x coordinate. So our x. Anything that like temps things that looks the same stays together. So three x and four x they look the same. So three x minus four x stays the same. And four and six will move across. It will be plus six and we can solve the equation. Four plus six is 10 and three minus four is minus x, right? Because it's minus one. It will take the sign of the bigger number to get rid of minus. Now we multiply by negative number throughout. This will be x is equals to minus 10. Now the equation it says we need when substituting and solving this substituting the value of x of the second equation into the value of x in the first equation, but I didn't read that. I just want to explain the linear equations in a way in a nutshell. But they give you instruction in terms of answering this question. So once we have our x, we can then go back and substitute this value of x into the second equation. So we're going to say substitute x is equals to minus 10 into the second equation because we use the first equation yet to substitute into we cannot go back and substitute it into that again. So what we do is in the second equation we can either use 2a or 2. So I prefer to use 2a because 2a and 2 are the same thing and already my y will be the same, right? So let's do that. So let's use 2a instead of 2 because then my y is equals to 3 plus 2 times and the value is minus 10. So my y will be equals to 3 minus 20 because 2 times minus 10 is 20 and 3 minus 20 is 17. So my y is equals to 17 and the answer I'm looking for is option 4. So then you can follow what they have asked you to do, which they say when substituting to solve the systems of linear equation, the results of substituting the x value of the second equation into the x value of the first equation will be. So instead of finding y, you need to go find x first, you could have done that first. It will give you your answer. So that is using the method of substitution. The method of elimination, we also do the same because with method of elimination we say we need to make sure that whatever we do, we are left with at least one variable. So if I subtract this from that value, I should be left with only x. And looking at this, I think it might take us forever because looking at these two equations that we have here, it might take us long. So but there is nothing like long. So I'm going to do that. Do you have a rough work? Yes, we do. I'm going to write both of these equations on here. 3x minus 2y equals to 4 and 2x minus 1. 2x minus y equals and minus 3. I'm still going to say this is equation 1 and this is equation 2. And yeah, we're using the method of elimination. So with method of elimination, we're saying if we want to get rid of one of this and be left with one variable, the next thing is to look at what I have. So if I look at this, if I subtract 3 minus 2, I'll get x and 2 minus, I'll still get minus y. So I'll still be left with two variables. So I need to make sure that one of these equations doesn't give me two variables when I leave them. So what I will do is I will multiply the top, the second equation. So multiply equation 2 with 2. I'm going to multiply equation 2 with 2. So let's do that. So I'm going to be left with equation 1 will be 3x minus 2y is equals to 4. And this I'm multiplying it throughout with 2. So 2 times 2x is 4x. 2 times minus 2y is minus 2y. 2 times minus 3 is minus 6. Now I'm going to subtract equation 2 from equation 1. So I've got my equation 1 and my equation 2. I'm going to call it 2a. So let's do the same because it's not the same as the first one. So this now becomes 2a. So I'm going to subtract my equation 2a from equation 1. So what I do is I'm going to add the minus there. So 3 minus 4 is minus x. And minus minus minus 2 is 0. So it will be 0. So I'm not going to write 0 there because that will be minus 2 plus. Because minus so it minus 2 minus minus 2, which is the same as minus 2 plus 2, which is equals to 0. So I don't have to write it there. Equals minus 404 minus minus 6 is equals to 4 plus 6, which is equals to 10. And I can just also let me write it on top here. So that I don't run out of space. So it's minus x is equals to 10. And therefore multiply throughout. I will get x is equals to minus 10. So now I've got x is equals to minus 10. What I need to do right now is to take my x value and substitute it. And I can now with this process, I can take my x and substitute it in any of the equation. Whether it's equation 2 or equation 1, it doesn't matter which equation I choose. I can just use any of them because originally I used both of them. So let's use equation 2. So it means we're going to follow from what they said. Oh, sorry. We're going to substitute the value of x into equation 1. So we can use the same concept that they used there. So let's do that. So we're going to substitute x of minus 10 into equation 1. So where we see x, 3 times minus 10 minus 2y is equals to 4. 3 times minus 10 is minus 30 minus 2y is equals to 4. Like times 30 and 2y are not like times. So we're going to be left with 2y on this side is equals to 4. Moving 30 to the other side, it becomes plus 30 and minus 2y is equals to 4 plus 30 is 34. To get rid of minus 2, we're going to divide this side by minus 2, divide that side by minus 2. And minus 2 and minus 2 cancels out. You are left with y. And 34 divide by minus 2. It's equals to minus 17. Did we get minus 17 previously? Oh, yeah. Why did I have positive 17? It should be negative. It should be negative 17. It should be negative 17. So there was an issue with that answer then. It should be negative 17. But that is how you will answer the question. Are there any questions? Elizabeth. Yes. Hi. So will the answer to this question then be none of the above because there isn't a right answer for it. Then it's not actually y equals 17. Yes. You will choose the none of the above because even if I look at the options, let's see the options. Even if I rewrite the options in this manner, let's see in state of where is the answer that we have. In state of going down this road, let's see. If I divide everywhere by minus 2, because minus 2 will be able to divide into all of them, right? Yeah. So if I divide everywhere by minus 2, just to see if we can get one of the options that are at the top there. Let's see if that is the case. Minus 2 goes how many times into 30. So 2 goes 15 times into 30, right? And minus 2 goes how many times into minus 2? So it will be plus y. It will be plus y. And minus 2 goes minus 2 into that. So if I look at the options there, there is none of them that looks like that. Even if I use positive 2, if we divide by positive 2, that will be minus 15 minus y and 2, which also is still not one of the options there. So none of the above would be the correct one to choose. But I guess they made a mistake there with the positive. They would have used the negative. Okay, so but that is how you will find the simultaneous equations. Let's see if we can find one question on simultaneous equations. None. Only this one. And the next one, simultaneous equations. None, none, none, none. We'll get to this one shortly. Okay, so those are simultaneous equations. I don't see any question. This is linear equation. And this is solving equations. Okay, so there are no more simultaneous equations, but that gives you an idea in terms of simultaneous equations, right? So now let's move into inequalities. So let's see, are we still on track? We left with, okay, so inequalities. So with inequalities, then you will also need to be able to know when they give you equations and ask you to solve inequalities, or they give you a paragraph and they ask you to find the linear inequality in that. You need to be able to do the same. So for example, I'm just going to look at more examples instead of just going in. So this one, you need to find, solve for at least one variable, right? Sometimes you will be given two variables where you need to find the linear inequality by reading the question and answering what linear equality looks like. Okay, so that is that. And then let's look at another option or another way of finding linear inequalities. I've seen one which I also want to highlight with. So this is another one where you are given three equations. Now you can see that here they've got three equations. So therefore it means from this paragraph you need to identify which equation deals with this four. So actually there are four equations, one, two, three, four, four equations. You need to be able to answer that and you need to be able to answer the questions that looks like this. And there was one inequality equation that I saw. Let's see if we go up. This is quadratic. We'll get to quadratic just now. Okay, so this is one inequality equation. We can always look at that as well. The last one, let's look at this exam paper. There is the, yeah, here is another linear inequality. So with this one, you need to be able to know after solving for X how to represent the inequality on the number line. So I'll use this one as an example instead of using this straightforward one because then you should be able to do the same thing. We can always do that. So now the key thing that you need to think about when we talk about one variable inequality is the following. There are two key things. You need to know that if the sign is greater than or it's less than or equal. And the dot that you will be using will be a solid dot. And we're going to get to the simultaneous inequalities as well in terms of the graph visualization. They also apply the same concept. If it's less than or greater than, your dot will be an open dot. So if I look at this, if I apply a process of elimination, already I should be eliminating all of these values that are on here because they are all open. So it's a mistake that they made on this question paper, but we can forgive them for that. You need to be able to identify the greater than and the less than by. There should be a difference between that because a greater than an equal means from that number onwards and a less than in that number means from that from not including this number. So an open dot means not including an close dot will mean including this number. So you cannot see that they all start at five right. So it cannot be that it says excluding this number. So but that is some of the sometimes previous years because you can see that this is 27 previous years. Your lecture has made a lot of mistakes as well, but don't worry about that. Probably the exam that you're going to write doesn't have this kind of errors. Okay, so that is one thing. The other thing is in terms of the sign where the arrow needs to go. So a greater than means the values are bigger. A less than means the values are smaller. The same way this will mean the values are smaller greater than the values are bigger. So it means in terms of the arrow, you should be able to know where the arrow needs to go. Now the other thing it says when you solve linear inequalities systems of linear inequalities as and when you multiply or divide by a negative number. The sign changes. So if you are dividing or multiplying by a negative number, if the sign was greater than or equals to the sign will change to a. Oh, sorry, if it was a less than or equals to the sign changes to a greater than or equals to if it was a less than it will change to a greater than if it was a greater than it will change to a less than. That's what the rule says when solving systems of linear inequalities. Okay, so now. We need to solve this so that we can identify which formula or which number line represent this. So we have 6x plus 45 is greater than or equals to 12x plus 15 right. Like types moving across things so we will have 6x and when we move 12x to the other side it becomes negative 12x. It's greater than or equals to 15 moving 45 to the other side it becomes minus 45. What is 6 minus 12? It takes the sign of a bigger number that will be minus 6x. It's greater than or equal 15 minus 45. It will be equals to. Right. Now, if I'm going to divide, so it will be minus 6x divided by minus 6 I'm dividing by a negative number, then my sign will have to change to a less than. And do the same thing to divide by minus 6. I think that's minus 30 isn't minus 30. Minus 30 takes the sign of a bigger and bigger number. Yes, minus 30. Thank you for being wide awake. It's been a long morning. So minus 6 and minus 6 cancels out you are left with x of less than or equals to minus 30 divided by minus 6 is equals to positive 5 right. Is it option 1, 2, 3, and 4. Look at the arrow. Look at the sign. So the sign says it's less than. So where should your arrow points? Your arrow should point to the left because it says x is any value which is less than 5, which is a lower value. x is any value which is greater or less than or equals to 5. So that will be. And there is some sort of a mistake on number 1 and number 2. They look exactly the same, right? And number 3 and number 4. Hence I'm saying somewhere somehow there is an error with this in the exam. I don't know. But I don't think that you will have this many errors like where you have to always choose none of the above. There should be an answer. None of the above is a rare situation right here. So number 1 and number 2 are actually almost exactly the same. The other times because this is something that is what you call this now. It's scanned copy. There might be some differences in terms of the line when one of them is a solid line. For example, if they didn't use, let me just correct this as well. If they didn't use a solid block like that, sometimes they use what we call a solid line. Let's change it this way because I'm going to discuss that right now. So it's either the line can look like this with an open open, right? Oh, they can say this will stay as it is. But when it's a less than, when it's a less than, then the line will be, it can still be open. But the line will be dotted, dotted like that. And this will be dotted. And for a solid, it can be open, but solid and open, but solid. So there is a difference in terms of how they can give it to you. But normally with one variable, it's always useful because it like, yeah, probably this is the dotted line and this is the solid line. But I can't say for sure because in this view, everything looks exactly the same. But yeah, one of them should be. So the graph should be actually a closed. I'm going to use this one. It should be a closed dot and an arrow going there. Or it should be an open with a solid line like that. And because I can't see the difference between the two, then I don't know how to choose two on that one. So it's just an error that they have there. So let's look at another example. So this is for you to calculate. Let's see if you are able to calculate this. So it says minus two is less than two minus six X over eight. Now, when you work with linear inequalities, always remember to have your variable on the left hand side. So your X should always be on the left hand side before you do anything else. So this is dividing. You first need to multiply by eight on both sides. So you're going to multiply by eight this side and multiply by eight that side to get rid of this eight. You will have two minus six because eight and eight will cancel and six and eight times minus two will be minus 16. So this is six X. Take six X to the other side. It becomes six X is less than or equals to take minus seven minus 16 to the other side. It becomes positive 16. Six X is less than or equals to two plus 16. It's 18 divided by six divided by six. We divided by positive number. So the sign does nothing. It stays as is. So six and six cancels out. You are left with X is less than or equals to 18 divided by six. How many number of times six goes into 18? It goes three times, right? If I still know math and the answer would be option number four. And that is how you will solve one variable with the systems of linear inequalities. So now in terms of systems of linear inequalities, you can also have what we call simultaneous linear inequalities. And with simultaneous linear inequality, we usually use or they usually use a graph to show that. But before I get to that one, let's look at if they give you a question like this and asks you to state which one is the correct one. Now, when they give you a sentence like this always, they will tell you what is X and what is Y, right? So the boy has 440 to spend and then amusement pack, which is the total, right? It will cost him 26 rent and then entrance to the pack, 23 rent per ride and 40 rent to play some games. So there are three things on there. It will cost him 26 rent to get into the pack, 23 rent per ride and 40 rent to play some games. In that sentence, you can see that some of the things have split things like the game and the right. It's pay. So we can always use some X and Y on that. And then your entrance is a once-off thing that you only pay once. So that is another thing to consider. Which inequality will represent the number of rides and games that they can play and ride? Now, at the moment, you need to also take into consideration that they would have used from the 420, they would have used their 26 rent. And the balance thereof will be shared across the games and the rides. And that is what you need to be solving. So the initial step that you need to do is say they will have 420 if they pay. So 420 minus 26, that will be the amount that is left to be shared across the two. So 420 minus 26, which is 394. So that 394 needs to be shared between the rides and the game. So we know that the ride is R and the game is G. Now, the other thing that you also need to take into consideration is that they cannot spend more than 394. So what does that mean? It means it can only be less than or equal 394. And we know that it will be split between rides and games. But we want to know how many rides and how many games. So we can go and substitute. So what is the price for the ride? It's 23. What is the price for the games? It's 40. And Bob is your uncle. Happiness, are we good? And the answer would be option number one. And that is how you will answer the system of linear inequality. Let's look at another. Yes, if you look at the number, it's greater. Oh, it's less than. Oh, yes, it's three. Oh, yeah, you're right. Thank you. Like I said, it's been a long day. Yes, it's option three. I didn't look at the sign. And that's what happens actually in the exam. Most of the time we are exhausted by the time we get to question number 18. So it means before you write the exam, especially meds, you need to have a good nice. Good, well, rest a night sleep so that you are fresh and not tired because slight mistake. You get everything wrong. Okay, so let's look at another. Inquality question. Oh, I said, I said we will use the other one, but we can use this one is also the same. So on this one. You are given two equations, right? A track rental company has two types of tracks. Type A is 20 meter square meter cubed of the refrigerated space. 40 meter cubed of non-refrigerated space type B has 30 meters cubed of refrigerated space and 30 meter cubed of non-refrigerated space. A food plant must transport at least, there is the other thing, at least 900 meter cube of refrigerated food and at most 10,000 meter cubed of non-refrigerated food. Let X be a number of track A and Y be the number of track B to be used for transportation of food. So now they told us what our A and B are. So A is X and B is Y. So we need to be able to say what is our track O? What is our track X? And what is our track, sorry, our track A and what is our track B? And we need to say how much in total will each one of them can transport? And in between we need the sign, right? This is the template that we can use. So if I'm going to do track A first, right? Because track A is that and track B is that. So both of the two tracks, they've got two spaces, right? The first space that we're going to do is the refrigerated. And then the next one we're going to do is the non-refrigerated, right? So we're going to have two equations, one for the refrigerator and one for the non-refrigerated. What we know is for the, I'm going to exclude the other questions. I'm going to start from here, only the bottom because this is easy to do. A food plan transport at least 90. So what does at least mean in terms of a sign? If I break this down like this. In terms of the sign, at least means greater than or equal. That is bigger than. At most means less than. So if at most the non-refrigerated will be, at most will be the less than. You need to go and Google what in a mathematical sign, what is at most, what is at least? Because you need to know those things before you use them in the equation as well. So this are the at least and at most. I can also include the values because at most 900. At least 900 and at most 1000. So it means they can deliver more than 900 meters squared of food into that transport. And they can also deliver more than, or they cannot deliver more than 1000, 1000 meters squared of food into that. Because the fridge has that capacity up to that. So now we have covered that. Now let's cover the X and the Y. Remember X relates to track A and Y, everything relating to track B. So now because we know that in track A they've given us the space for the refrigerated and the space for the non-refrigerated. So we know for the refrigerator it's 20 and for the non-refrigerated it's 40. You do it that way. So I can put type B which is Y refrigerated, it can take 30 and non-refrigerated it takes 20. Now let's rewrite this in an equation. What we do with an equation, you just put back your X and Y onto that, X and Y. And this will be the plus, plus, plus in between. So let's write the first one. This will be 20 X plus 30 Y. It's greater than or equals to 900. And the second one, 40 X plus 20 Y. It's less than or equals to 1000. Which option? One, two, three and four. This time I'm not going to put myself in pressure. I'm going to ask you to answer that. It's number four. Yes, it's option four. It will be option four. And that is systems of linear equalities. But also before we go into quadratic equation, I see that we're running out of time. Let's look at examples of systems of linear inequality when you are given graphs. I hope we do have. This one is also not clear. Let's see if we can find one that is clear. So you should be able to answer this the same way as I've done it. So you will be given a lot of information. You just need to make sure that you create four graphs. I just want to see if I can find the one with clear. Okay, so none. None. Let's see on this one. Just want to see if there is a clear linear systems of linear inequality. Nope, nope, nope, nope, nope, nope, nope, nope, nope. There are all. Okay, so we don't have a choice, but to use that one that we have, which is not that very clear. But anyway, it doesn't stop us. So when it comes to systems of linear inequality, there are three things that you need to, or four. Four or three or something like that. So in a way, when you have a equation like this where it represents systems of linear inequality based on the shaded area that you see there. It means there is some common area. So we've got one, two, three, four, four equations. You can see that we have four equations right here, right? And this is not what I planned out to do, but in terms of systems of linear inequality. To solve this system of linear inequality, we want to find out which of the four areas they all have in common. They share the same area. So if a sign is greater than or equal, then the line in terms of the line, or let's do it this way. Let's do it this way. I'm going to have the line at the top and we're going to have the area at the top. And then here we're going to have the sign. So this is the matrix that we're going to use to identify things. So the sign, if it's greater than or greater than or equal, less than, less than or equal. So we have four signs that we can use, right? Because it's linear inequality. In terms of a greater than, the line will be dotted. If it's greater than or equal, the line will be solid. Less than, the line will be dotted. Greater than, the line will be solid. Greater than or equal, the line will be solid. Right. That's one. So if you come here onto these graphs that you see the lines, one, two, three, four. Looking at the lines, you can see that number one, the line is solid. I'm going to put this S for solid. Number two, it's this line, the thick line that goes down. It's also solid. So therefore it means this line is either greater than or equal or less than or equal. But we will come there now. Number three and number four, they are dotted. So therefore, they are dotted. Already, I'm able to see and identify those lines. Now, the next one. Just give me a second. I need to call out my daughter. Sianna. Sorry. Apologies for that. Where were we? Okay. So we know that line three and line four are dotted. Now, how do we also identify certain things? So in terms of, I hope you are able to hear me. Okay. I'm not muted anymore. Okay. In terms of the sign, you also need to identify which areas above the line or below the line. So when it is greater than, therefore it means the shaded area will be above. What do we mean by above? If I have a line like this, this is above. This will be below. If I have a line that looks like this, this is above. This is below. If I have a line that looks like this, this is above. This is below. So you just need to be able to identify which ones are above, which ones are below. Right? So the greater than or equals to, it means the line will be above. The less than below. Below. Because it means the values are below the line. The values are above the line. Just like that. So it means if I look at this because the shaded area looks like this, therefore it means line number one should be actually it is above the line. Right? You can see. I said if it looks like this, this is above. This is above and this is below because the line looks like this and the shaded area is above the line. So this is greater than or equal because it is shade. It is a solid line and it is above. So it will be a greater than or equal. Right? Then line number two should be it says it's just a solid line on X. So therefore this line also the area is below the line because oh, I didn't have an example like that. This is above. This is below above below. So this is below because it's smaller. This is bigger. So this line would be and it is solid. Right? We are able to see that clearly that it's solid and it is below. So therefore it means line number two will be less than or equal. Line number three, it looks like this one. It is above and below. So because the shaded area is below the line and it is dotted, therefore it is less than. This looks like that. We said this is above. Right? This is above because the values on top are above the values at the bottom will be minus and that it's below. So this is dotted and the values is above the line or the shaded area is above. So therefore it will be greater than. So now we've identified line one. I just need to write that line one will have greater than. Line two will be less than or equal. Line three will be less than and line four will be greater than. Now with line, line three, not line three, line two. You see line two, it just says X is equals to. So line two, actually line two here will just be X is less than. What is that value? Four. That is what that will be. That is this line line two. It's easy to identify that because it just passes through the X axis and line three. It also just passes through the Y axis. So line three will be Y is less than it passes through at four. So I can only use those two actually to guide me because the others it's going to ask of you to go find what your Y intercept is and X intercept. What do I mean by Y intercept? If I divide this by three and divide this by three, my Y intercept for line one will be equals to three. It goes how many times into 15? It goes five times. So my Y intercept for this line will be five. So if I go there, the Y intercept is five. It doesn't really matter whether I know what the equation looks like or not. What matters for number one and number four would be, or number one and number four, yes. For number one and number four would be for me to identify from that the sign. Okay, so based on that, let's go and look at the two or this equation. So we're going to apply the method of elimination, right? So let's first use the two, line number two and line number three. Is it three or four? Three and two. Line number three and line number two. So we know that line number two should just say X is less than or equals to four. This says greater than. This says less than, so this might be the right one, the right line, right? And if I come here, this also looks like it can be the right one. And this one also looks like it can be the right one. Then I go to the next one, which is line number three. It says it should say Y is less than four. This one says it's greater than or equals to. We know that it's not correct. And this one also says it's greater than or equals to. We know that that is not correct. This one says it's less than or equals to. That means it's incorrect. This one says it is less than. Already I've eliminated three of the graphs. So you can see clearly that this is the correct one. So if I look at option one, the sign is greater than option two. The sign is greater than. So option four will be the correct one. And that is how you will answer questions relating to systems of linear inequality. You go into look at the sign, whether it's dotted line, greater than. Above or below to identify. Sometimes they will give you graphs and say identify from these equations. Which one is the correct one? You just need to know how to do that. Right. So this can be your guide. To say how do you identify the common area of the graphs or the points that you are looking at. Or the lines that you are looking at. And we have 10 minutes. I hope you are able to stay longer than the 10 minutes. So that we can do the last part, which is the quadratic equation. And then we would have sealed it off nicely. So let's see if I can squeeze it in in 10 minutes. Okay, so quadratic equations. Let's see. So this is a quadratic equation. So with a quadratic equations, there are four things you need to be aware of. Or is it five? It doesn't matter. For a quadratic equation, you should be able to identify your constants A, B, C. Those are very important. And the sign in front of them are very important. You should be able to identify them. Because not similar to your linear equation where they give you two points. With quadratic equation, you will be given an equation and you need to find certain things. Like for example, you need to be able to identify your values of A, B and C. You also need to be able to identify whether your A value is greater than zero. Or it's less than zero. When it's greater than zero, it means, is it at minimum or at maximum? Whether it's at, when it's less than, is it at minimum or maximum? You need to be able to identify those. The other thing, you need to be able to calculate your 10 points, which is your XM, which is calculated by means of minus B over 2A. That is the formula you use. Hence, you need to be able to identify A, B and C so that you can substitute into the formula and calculate. Remember, the formula has a negative. If your B has negative, you need to include the negative number as well. Once you have your 10-point, the 10-point has also the Y 10-point, which is your equation. To calculate the Y 10-point, because this is the 10-point, right? When I talk about the 10-point, I'm talking about where the graph curves through, which is that part. At that point, you have your, at that point, you have two coordinates. You have your X coordinate and your Y coordinate. Your Y coordinate, you're going to find it by taking your X coordinate of the 10-point and substituting it back into the equation. And that will give you your Y coordinate. Those are the two points for the 10-point. Now, to find out if your value has the X intercept. Before the X intercept, you need to be calculating your Y intercept. And your Y intercept, which is where it passes through, it is where X is equals to zero. So it means on your equation to find your Y intercept C, you need to substitute zero onto the X value and calculate. So your Y intercept will be the same, which is the same as your C. C value, which is where X is equals to zero. If your X is zero, your Y intercept will be C. So in this instance, your Y intercept is that value. So that will be your C value. The next one is to find the X intercept. To find the X intercept, you need to first apply what we call the discriminant. And what is the discriminant? Because your Y, sorry, because your X intercept, you can find it by using the equation minus B plus or minus the discriminant, which is the square root of the discriminant, which is B squared minus 4AC divided by 2A. So in order for you to find the X intercept, which means if, whether, before you can go and find the X intercept, you need to first evaluate your discriminant. Now, if your discriminant value is equals to zero, if this value is equals to zero, therefore your X intercept is the same as your tending point. You don't have to calculate it because if this part of the equation is zero, you can see that it's minus B divided by 2A, which is the same as what you found there, right? It's the same thing. If you evaluate your discriminant and you find that it is greater than or equals to... Oh, sorry, it is greater than zero. It cannot be greater than or equal. So it has to be greater than zero. So if it's greater than zero, therefore it means you will have your X is equals to minus B minus the square root of the discriminant. I'm just going to write it as that, divided by 2A. Or X is equals to minus B plus the square root of the discriminant because you have already calculated the discriminant, divided by 2A, which means there will be two intercepts. There will be this side and that side. As you can see from here, it means the value of your discriminant as you evaluate this equation, it will be positive, and when it's positive, then it means you have two discriminants. What happens when your discriminant B squared minus 4AC is less than zero? It means your graph never touches anywhere on the X axis, so there is no X intercept. If your discriminant is greater than zero, then your graph won't look like this. It probably would be like that, or it will be like that. It will open up there, but never touches anywhere on the X axis, or it will open down on the Y value, but below the X axis. And that scenario will be for where your discriminant is greater than zero. Okay, so based on all these things that I've just said, because it's so much to say in 10 minutes, let's see if we can answer some of the questions looking at quadratic equations. Which one of the following is true? The value of Y at the turning point is negative. The value of X at the vertex is positive. The value of A is negative. The discriminant is negative. Okay, so I can do with process of elimination because I know that those two points exist. Number two is incorrect because the discriminant is not negative. It's positive because I've got two X intercepts. Number three, I cannot answer it as yet because this would mean that the value of A is less than zero. So how do we say the value of A for a quadratic equation? Wait, I must just refresh my mind as well. It has been long. What is A less than zero? Just give me a second. I don't want to tell you lies. I want to tell you the right thing. quadratic equations. It should be something that I always remember. And that is something that you always have to remember as well. Because you only have time to Google something or Google answer your way out. There we go. So the value of A can never be zero because if your value of A is zero, then it becomes a linear equation. But we need to know what will be the value of A. If the value of A is positive, your graph will open up. If your value of A is negative, your graph will open down. So this value of A, the value of A will be negative. So that is what I have here. That's the correct one. And when it's negative, it opens down. So obviously I've already established the correct answer because of that. And if I look at the value of Y, they say at the turning point, the turning point and the vertex are one and the same thing. Don't get confused. Turning point, vertex, they mean one and the same thing. So the value of Y at the vertex is negative. Let's see. It is positive. Y is positive and X is negative. So yeah, it says Y is positive is negative. So that is not correct. And X is positive, that is not correct. So the only value that is correct is option three. But that is one of the other way of answering questions when it comes to quadratic equations. Just hang on a little bit. Don't go anywhere. I know that it's two o'clock so that we can have more other questions to look at. So the next question on quadratic equation, you can see that it asks you to find the discriminant. All they're asking you is to calculate what the value of this discriminant is. So the first important thing to do is what is your A, what is your B, and what is your C? A is for C or B, it's minus 15. Always take the sign that is included on those constants. And C is 10. So B is minus 15 squared minus 4 times A is 4 times C is 10. 15 squared is 225 minus 4 times 4 times 10. And that is 100 and C is 160. And 225 minus 160 is equals to 65. And the answer will be that one. Are we good? Are we happy? Let's see if I can find one more question. So this is the other question. So question 13, 14 probably. Question 15 and 16 relates to the same information. Sorry, I'm going to make it smaller. I hope you are able to see it. So it says question 15 and 16 are based on the graphs below. So it says from question 15, I don't even have to look at the data, the options. Determine the vertex of graph A and graph B. So it means we must go and find what is that value and what is that value there? Which is very difficult to find if you don't have an equation, right? Because they didn't give us the equation unless the previous one gave an equation. No, they didn't. So it's going to be very difficult to find that. So it means we're going to rely on the options to guide us. But other than that, we can always assume things by just thinking that that is the same as that. So because that's what the vertex is. So let's assume that based on the graphs that I just did now, or the points that I just highlighted, we are able to see that. So in the middle here is three, right? So we can use number one or A to find the exact value. Because with number B, we need to find the missing other value. So this should be three in between and here. Either one because it's two, four, six, so it's zero, one. So it's some number. So there should be some value there, which I don't know whether is it one or that. But all the values are positive. So we know that at this point, if I use this, the vertex of this will be our X value is three and our Y value is 16. So let's do a process of elimination A and I've already found my A vertex. Then therefore it means this one is three and one because I can just assume that that one is one because I already made an assumption that that one is three. So our X is three and Y is one, but was able to use that as a reference. So let's go to 16. 16 says we need to find the X intercept. So X intercepts are these values here. So also from here I can see that this is seven and this I can assume that is minus one, minus one and seven. So it's positive and minus one. Those are the things that I can assume with this. On this one, there is no X intercept. Can you see that this graph doesn't touch any way the X exists anyway. So therefore it means there are no X intercepts. So I've already know based on B, I can go and find out which one. So the other thing that you also need to pay in mind when you look at the X intercept, the X intercept correspond with the Y intercept. But always most of the time your Y value at this point is zero. So this has the point minus two and zero and this has the point seven and zero. That's how you can do the point of your X intercept. So let's see. And then this is no. Oh, let's see. They use none. So we can use none as well. So the only option that is correct because on this one it says B has the X intercept as that. So that is not correct. And this is not correct. And this is not correct because the graph never touches the X axis anywhere. And the only option here is four. And you need to go and practice because all this that I've just shown you today is just touching the surface. The more you practice, the more you find more questions relating to this type of equations, the more you will understand it better. Are there any questions before we close off today and see you tomorrow or next week, Saturday, same time, same place. Here is another equation, but I'm not going to go into detail on this one. So you can take a picture of it and try and answer it. So in this instance, where they give you a quadratic equation in a mumbo jumbo, which is not in a format Y is equals to AX squared plus BX plus C. It means you need to solve this equation and make it the replacement like that so that it becomes your quadratic equation. And then after that, then you can identify what is A, B, and C because you would have solved that equation. And once you have solved that equation, this is one question. The following question asks you a turning point if this is the equation that they give you, which actually you also need to always remember to write it as AX squared plus BX plus C. Identify what is A, what is B, what is C and find the turning point. Remember at the turning point, there is XM and YM. How you find XM, you will use minus B over 2A. How you find YM, you will use, I'm going to use the same equation that they have here, right? I'm going to use the same. To find YM, you will use 2XM squared because that will be the answer that you got there plus 3XM minus 4. And that once you solve that, it will give you the answer. And that's it for today. I will see you next week. Are there any questions, comments, query? Before we switch off. Remember to please complete the register. Questions, comments, nothing. Okay. Thank you.