 Okay, cool. Welcome to your session number eight, which is our last session before your exam preparation, before you go write your exams. So you all all know what and which date you need to be here on a Monday based on your timetable, your exam timetable. I think we start with BNU and then the last the other Monday then it will be QMI. Okay, so today's session we're going to be looking at the basic skills on how to solve different types of equations. I'm going to rush through because there are different types of equations and I hope here we do have, I just want to have a feel. Do we have anyone doing QMI? Yes, we do. So wait, let me see. Let me go back there because it's very important to know who is, how many people are only doing BNU, only BNU, nobody. So you both you, some of you are doing BNU and you're also doing QMI. That is good because most of the things we're going to be discussing now are related to QMI mostly than the BNU. The first part of the session will be based on the BNU as well as QMI, but at the later stage we will move into the QMI space when we're solving different types of equations. I'm not even going to ask you if you have any questions or comments for today because I will have to sprint. So by the end of the session today you should learn how to manipulate an equation, how to change a subject of the formula. We've been doing the changing of the subject of the formula almost every week now, but I just want to also introduce that concept again. You should be able to solve linear equations and here I'm talking about one linear equation and you should be able to solve two simultaneous linear equations to solve inequality equation, quadratic equation, I forgot to mention there, quadratic equations and simultaneous inequality equation. As you can see that the list is long and we only have one hour 30 minutes. So first off, when you manipulate equation that's the base of everything you do in meds. In meds we try to manipulate an equation and we know that when manipulating an equation, for example if I have to solve for x in this equation then it means I must make x the subject of the formula or x must be alone on the left hand side or on the right hand side. It must just be on its own. We can apply different methods. What you do on the left you must do on the right. If I subtract on the left I must also subtract on the right. If I multiply on the left I must also multiply on the right. If I divide on the left I must also divide on the right. If I add on the left I must also add on the right or sometimes you don't even have to use whatever you do on the left you you on the right when you are subtracting or when the the value that is ending or subtracting and it needs to move across the equal sign because this is an equality sign if it needs to move to the equal side side the other side of the equal sign then the sign will change if it was minus or it will become plus if it was plus it will become minus if it is multiplying or dividing then it's different if it's multiplying we to get rid of that value we divide if it is dividing to get rid of that value we multiply by that value and you will see when we do some activities now so in order for us to solve this x we need to make sure that x is on its own and we're going to apply the method whatever I do on the left I do on the right so I have seven on this side which does not have an x so it needs to move across the equal side so whatever I do on the left I must do on the right to get rid of seven I must subtract seven on both sides subtract seven on the left subtract seven on the right seven and seven plus seven minus seven will be equals to zero therefore on the left hand side I will be left with four x and on the right hand side 23 minus seven is 16 now four is multiplying with an x and you remember if it's multiplying to get rid of that value we divide so it means we're going to divide by four on the left hand side but we must also divide by four on the right hand side and four and four will cancel out you will be left with x and four goes four times into 16 therefore our x is equals to four and that's how you manipulate the equation and you can validate this because this is an equality side it says whatever is on the left should be the same as whatever I have on the right on the right I have 23 on the left I have four x plus seven so I must make sure that four x plus seven is the same as 23 so and I can take this four you don't have to do this method you can take this x is equals to four substitute it back into the equation of four x plus seven so that we can check if the equation balances out so we go into substitute where we see x which we put four and then it will be four multiplied by four plus seven and that will be equals to 23 therefore it means the equation balances out and that is an extra step that you can use but it's not necessary if you're given this kind of a question in the exam or in your assignment and you need to answer this question but always remember the Bodma's rule as well because we are solving and manipulating equation and using the basic operations as well Bodma says brackets first so we need to remove the brackets so what you do best is to remove this bracket so two multiplied by two y it's four y two multiplied by minus one we distribute two into the bracket which is minus two and we just write the equation as we see it four minus y because I only need to solve the equation and I look at my option it says I must find an answer which says y is equals two so therefore it means I am solving for y so it means everything that it has and y must go to the other side so I'm not going to use whatever I do on the left I do on the right I'm going to move things across the equal side so I'll start with the two to move it over so that will be four y minus three y and I can see that we have a y on this side and a y this side is negative when I move it across the equal side this is my equal side when I move it across because it's minus it's going to be a plus y equals four I have minus two this side when I move it across it will be plus two and I can solve the equation for four y minus three y is y plus y is two y four plus two is six and I'm looking for y not two y so therefore I'm going to divide this side by two and whatever I do on this side I must do on the other side two and two cancels out you left with y and two goes one time into two and it goes three times into six and therefore you have y is equals to three which is option two and that's how you will manipulate equations I'm not unfortunately I can't give you exercises to do because like I said I'm going to sprint so that's how you manipulate equations how do we change the subject of the formula changing the subject of the formula means we need to move one variable to be on its own like we had one variable as such the variable y come on like we had the variable y on its own on the left hand side that's the aim of changing the subject of the formula our formula at this point is v is equals to u plus at and we want to make you the subject of the formula because we are given v we are given a we are given t therefore it means we need to calculate u so we can move u to the left hand side so moving u to the left hand side because u it's my it's adding on it's got a positive sign on the left hand side on the right hand side if I move it I must subtract I must subtract u from both sides so I subtract u from this side from the left hand side I must also do the same on the right hand side and then I get v minus u and then u and u will cancel out on my right hand side and the only answer I have is at and then I'm interested in u not v minus u so I'm going to move v as well to the right hand side so to move v because v it's plus so I must subtract v this side I must also subtract v on the right hand side and the answer will have its minus u is equals to at minus v I'm not looking for minus u I'm looking for positive u so to get rid of a negative a negative u so negative u is the same as negative one times u because with a one we don't write it is the same thing as u divided by one we don't write it we write it as u so the same thing one has no you don't have to show the value of one in some cases for example as well u to the power of one we don't write it as u to the power of one we just write it as u so these are those circumstances that you always need to remember when you are manipulating your equations as well so we're going to multiply by negative one so we multiply or you can say you are dividing by negative one it will still work out the same so multiplying by negative one will negative times negative becomes positive and because we're multiplying everything on the left so you need to make sure that you put it into brackets and not only multiply the first one that you see it also has to multiply the all the variables that you have so we put it into bracket and multiply by negative and therefore we need to remove the bracket then it becomes minus at negative times negative is positive then it's positive v or we can rewrite this as u is equals to v minus at so you can still rewrite it look at your options if your options are rewritten you can rewrite them as well there is no harm as long as you use the right signs and you don't drop anything you don't change the value of the equation and that's how you manipulate an equation let's look at an example if p is equals to m squared minus x divided by 2 make x the subject of the formula we can see where x is at is there so in order for us to make x the subject of the formula let me rewrite this equation m squared minus x divided by 2 so I can move minus x divided by 2 to the left hand side and it will be because it's negative it will be positive x over 2 and m and m squared p is positive this side when I move it across it will be minus p now I'm not looking for x over 2 I'm looking for x as a subject of the formula and therefore I can also say I'm going to multiply the side by 2 therefore it means I must multiply or soft on the other side by 2 a bracket means multiplication as well 2 and 2 will cancel out we will be left with x and I can just distribute my 2 into the bracket that will be 2 m squared minus 2 p which one is the correct answer which is number one that's how you will manipulate the equation maybe when we have enough time at the end I do have some extra additional exercises at the end that we can go through and that is how you manipulate you change the subject of the formula any question I'm going to now questions from here before we move on to the next are there any questions is that easy to do clear very clear thank you now we're going to look at how we manipulate linear equations so linear equations are like your straight lines so we're going to look at how we manipulate those equations in terms of linear equation you should be able to know how to draw this the straight line based on the linear equation you should be able to calculate the slope and you should be able to determine the equation of a straight line which is your linear function if given two points to draw a straight line you also just need two points and the point is made up of two coordinates so to draw a straight line you just need two points and a point is made up of coordinate so one point is made up of coordinate x and y and the other point you will need to have the coordinate x and y and that is why we say you need only two points and this one will be x one y one corresponding to one another and x two corresponds to y two when you have the two points you can draw on a Cartesian plane you can put the two points and connect the two points and draw a straight line and this will be your straight line y is equals to ax plus b which will be our straight line and that is all what we're going to be doing in terms of the slope we're just going to calculate the change in that value and the change in this value so we're going to look at the change in the value of y divided by the change in the value of x because this is your x one and x two and this is your y one and y two so we're going to take those to calculate the slope and we can also calculate this equation of a straight line which is y is equals to ax plus b okay so how do we do that if we have two points one x is of one and y of four which is my first point I can relabel this as x one and y one coordinate and I can relabel this as x two and y two points I can then plot them on this Cartesian plane as you can see there my point one where x is one y is four and that is my point where x is four and y is two that will be my point and I join the straight line I drew the straight line on this and this graph passes through some points on the y axis and the x axis based on the slope or based on this graph I should be able to tell what type of a slope this is whether the slope is descending or ascending those are the things that you also need to know especially when you are doing b and u and q mi as well some questions they might ask you those so this graph we say because of the way it looks it states that when the values of x are increasing the values of y are decreasing so this is a descending slope or a negative slope or a downhill slope if you have a slope oh sorry not like that like that it means when the values of x are increasing the values of y are also increasing and this we say it is an ascending slope it is a positive slope and it is also an up what slope and you also get slopes that looks like this which are constant slopes they do not have any influence because the value of x stays constant when the value of sorry the value of y stays constant when the value of x changes or you can have a graph that point like this where it means the value of y changes and the value of x stays constant and that there is no slope the slope there will be equals to zero okay so you need to know all those definitions of the slope as well but we can also calculate the value of a slope because a slope can be calculated using the formula since we know that the slope is the ratio of the change in the values of y to the change in the values of x so it's because we do have two points where x1 and y1 and x2 and y2 we can find this slope by substituting the value of y2 minus y1 divided by x2 minus x1 and that is the change that we are calculating there so let's calculate the slope of this following to calculate the slope of the line or the straight line that passes through these two points point zero and ten and point five and zero we can label both of our points so point one will be x1 y1 zero and ten and point two will be x2 y2 will be five and zero and then we can then substitute into the formula and find the slope calculate and find the slope because y2 is zero y1 is ten so it will be zero minus ten the equation has a negative net right if this value was negative if ten was negative you must put it into bracket and put the minus ten on there so you must also be wary of that when you do some calculations substituting x2 is five x1 is zero so it will be five minus zero zero minus ten is minus ten five minus zero is five and we can still simplify this because it's a fraction we need to simplify to the simplest form five can divide into ten it goes two times therefore the answer will be negative five and therefore the slope for this point is negative two and this is a negative slope therefore it means the slope of this graph if we had to plot this graph it would have been a negative downhill descending slope now when it comes to qmi if we need to find the equation of a straight line using the same point let's assume that we're going to use the same point zero and ten let me write them down zero and ten and five and zero so let's go find the equation of a straight line and the equation of a straight line we can define it as y is equals to bx plus a or y is equals to ax plus b it depends on how you write your equation so depending on your book how they define it so there is always the b will be your slope and your a will be the intercept and some equation it's y is equals to ax plus b where b is your intercept and a is your slope so the value that multiplies next to the x is your slope it is the value that we just calculated so let's calculate this so to determine the equation of this straight line that goes through the point zero and ten we can also define our points because it's very important to go forward in terms of how we're going to plot them into the equation as well so to do that we first calculate the slope oh sorry we first calculate the slope we did calculate what the slope is if i'm using my equation the slope we calculated if it was minus 2b remember that the slope is this one that we calculated here because i'm using the same point i'm not going to come back and calculate it again so the slope is two so i can go back to my equation which is y is equals to bx plus a and substitute my value of my slope which is minus 2x plus a and i can choose any of the two points to substitute into the value of x and the value of y because i do have my x and y when i do that i need to make sure that i don't take the value that is too big like 10 and 5 one of them is big i must stay away from negative numbers i must stay away from fractions but if they are all there on the point you need to just make a decision and choose one what i'm trying to say is when you're given two points choose the points that is the easiest to substitute and calculate don't choose the one that is more complex so i'm going to choose x2 and y2 to substitute into this so it means everywhere where i see x i'm going to put five where i see y i'm going to put zero so my y is zero equals to minus two times my x of five plus a making a the subject of the formula means i must move a to the other side or i can leave it to the right hand side when you're working with equality sign it doesn't really matter whether you leave your subject of the formula on the left or on the right so i can leave a on this side so the answer here will be zero minus two times five is minus 10 plus a and i can move 10 to the other side and therefore it becomes positive because that is zero it means nothing and the answer here will be a is equals to 10 or i can write it as a is equals to 10 i can rewrite it because it's an equality sign it only works when you do in equal when you have an equal sign when we get later on please please please move the value to the left always move it to the left when we do the inequalities and now i have my intercept intercept is the value that passes through the y axis where x is equals to zero the value of y will be the same as the value of a so now i have my a i can then complete my equation of a straight line because that is what they said the only value i need to substitute is b and a y and x these as they are so y is equals to and my b we did calculate it was minus two x and my a plus 10 and that is the equation of a straight line as it is enter right there i'm going to use the same the same question or example to do the same equation of a straight line but from the b and u perspective so qmi you use this formula that's what is in your study guide you use y is equals to a x plus b or y is equals to b x plus a or y is equals to mx plus c that is the formula that you use to calculate or find the equation of a straight line for b and u to find the equation of a straight line which is the same as what we have y is equals to b x plus a we are going to use the same point but your equation of a straight line is defined as such so this will give you the equation of a straight line so what does that mean it means you don't even have to go and calculate the slope whereas with the qmi before you calculate the equation of a straight line you need to first calculate your slope then substitute choose two points and substitute in order to get there in b and u you don't have to do that you use this formula y minus y1 divide by x minus x1 equals to y2 minus y1 divide by x2 minus x1 we know what our points are and we have defined them before right we know that our first point is 0 and 10 and our second point is 5 and 0 where we see x1 y1 we substitute them on to the where we see x2 and y2 we substitute the values on there so let's go on and substitute our y1 is 10 and our x1 is 0 so that is right on the left hand side we substitute like that and i'm not going to talk about this side because this is the slope equation that we calculated previously we know how to substitute that now we need to solve this before we solve this we can solve the right hand side because those are numbers and this side we've got variables again so let's solve the numbers we know that the slope gave us minus two when we solved this previously so it's minus two in order for us to get rid of x minus zero because it's dividing here at the bottom so this is the same as x so we can multiply by x on this side or we can take both of it and multiply but this will still remain zero and then we take minus 10 to the other side let's move it one step back so we know that two times x is minus two x two times zero minus two times zero will be plus zero which means this value yeah will just be minus two plus zero which is the same as minus two x and if we have y y minus 10 is equals to minus two x all what we need to do is move 10 to the other side it's positive and the answer we get will be y is equals to minus two x plus 10 which is the same as what we got previously so you can either use both of them so in the exam you can use both of them to save time you can use this method if you are doing qmi but you need to know the equation this is the equation to find the equation of a straight line are there any questions if there are no questions let's look at this determine the equation of a straight line that passes through these two points and they give us the equation of a straight oh that is our option so we need to determine that I'm going to do it two ways I'm going to start with the qmi so the first step is to find let's assume that in this instance we're going to use y is equals to ax plus b so therefore we need to find a first so a is equals to y2 minus y1 divide by x2 minus x1 so substituting into the formula we need to come back and define this so this is our x1 y1 x2 y2 y2 it's minus six y1 minus minus two because it's minus two and our equation has a minus in between so I must also keep that minus x2 is five minus x1 is three minus six minus minus two so it will be minus six plus two divide by two which will be equals two I can write it just here which will be equals two minus two plus two it's minus four divide by two it's minus two right am I calculating it right I'm calculating it by by heart that is minus two so since we have minus two so we can determine now we can rewrite the equation so I'm going to write it here so our equation is y is equals to ax plus b so we can substitute back this equation there so we have minus two x plus b which is still missing so we need to find b we need to choose any of the two stay away from big numbers so I can see that this one has smaller numbers than this one so I can choose x1 so I'm going to use that one to substitute because this one has big numbers they both have negative numbers right but this one the numbers are smaller than those ones so where we see y we're going to put minus two where we see x we're going to put three plus b minus two minus two times three it's minus six plus b and moving minus six to the other side minus two plus six is equals to b and therefore b is equals to minus b is equals to four because it's minus two plus six so it takes the sign of the bigger number therefore our equation of a straight line is y is equals to our a of minus two x plus four so which is option four so let's do the this is q mi right I need to always remind you that let's do the b and u one so b and u we say y minus y1 divide by x minus x1 is equals to y2 minus y1 divide by x2 minus x1 I'm not going to do this step again because I don't I'm not going to substitute these values again because I know we've calculated it so you will substitute you will substitute these values on your step so I'm not going to do that because we calculated it it's minus two and I can just substitute the values on the left hand side which is minus x1 y1 we're going to use y1 it's minus two divide by x minus minus three because we're using y1 and x1 so we can take this value and multiply it on the other side as well so we'll be left with plus two is equals to minus two times x minus three and we can simplify this whole thing minus two times x is minus two x minus two times minus three it's plus six y plus two y is equals to moving two to the other side we are left with two x plus six minus two because it's positive when it comes over it becomes negative and you left with y is equals to minus two x six minus two it's oh plus four and as you can see this is bnu1501 you can find the same answer doing using two different methods as well and that is linear equations I hope we're still right on track oh yes we are yeah now let's look at how we solve simultaneous equation unless if you have a question because the way I'm flying I'm sprinting I'm like who who's saying bolt you say bolt the way I'm sprinting and if you don't ask and let me explain again you will get lost so moving to the next section going to look at the simultaneous equations so with simultaneous equations it means they give you two systems of linear equations and you can use two methods to do that you can use the method of substitution or the method of elimination you choose the one that you feel comfortable with because different people solve problems differently me I prefer the method of substitution others prefer the method of elimination because it's quick and easy but you need to choose and decide which one works best for you because at the end of the day when you are writing your exam you want to save up on your time as well so how do we solve equation of simultaneous equations so remember now with linear equation oh now I'm drawing different graph so remember with simultaneous equations so if this is our y come on 10 don't disappoint me this is our Cartesian play where we have our x and y and in the middle there is zero so with simultaneous equation there are two two straight lines so let's say one straight line goes like this and the other straight line goes like this with simultaneous equation we want to find this point the point where both of these two equations y is equals to mx plus c and let's let's call this y is equals to mx let's call this minus mx because the slope here is negative so it means the value here in front of m is negative this one the slope is positive and this one our c on this one is positive so that is why it passes through the positive side of y and this one it passes through the let's not use mx let's use ax b ax minus b because it passes through a negative y so this side is negative y this side is negative x so those two lines they pass each other at some point and at that point on it is that point that's the purpose of simultaneous equation we want to determine at one point these two equation are passing through each other they are cutting each other through okay so let's get to that first yes you lost my screen yeah okay so I was explaining to myself let's get it back I don't know why it disappeared okay I don't know where I lost you and I think these things happen almost with every presentation okay so what I was explaining in terms of this Cartesian plane we have two lines we have this line which is this line y is equals to minus mx plus c I was explaining that one minus mx plus c because our slope of this one is negative is declining but our c which is our y intercept it is positive so that is our c because the graph passes through the y the y axis at that point and I was also explaining that this line y is equals to ax minus b because the slope is positive therefore it means a will be positive but our y intercept passes through the y axis in the negative side on the negative side of the y axis so that is why my b there is negative right I'm saying when we solve simultaneous linear equation we are trying to find this point where both of these two lines pass each other and we want to find this point which has x and y coordinate that is the point of simultaneous equations so given two x plus five y is equals to 12 and four x minus y is equals to two let's solve this simultaneous equation and it says I must plot the graph I might not plot the graph because I'm in a marathon so you can plot the graph because the graph will be the same way as you plot the graph when you have your linear equations and I didn't show you when you have any linear equation how do you plot your graph but we will get to that it's fine we will need we just need two points so we'll have the c and the point that we are having and then I can just show you how to plot it it's gonna be easy so with method of substitution which is the first one that we're going to use what you need to do is to label your plot so this is graph we're gonna call it equation number one and this is equation number two because we're doing the process of substitution so it means we need to take one equation and substitute it into the other equation looking at this two equation I must take the one which looks simple and equation two looks simple so I'm going to say change equation two into y is equals to ax plus b that's what we know most we know how to do that so let's do that we have four x minus y is equals to two we're going to move four to the other side minus y is equals to I'm gonna it's positive so it moves this side it becomes minus four x and plus two because I'm trying to rewrite it the same way as they have it so I'm going to multiply by a negative number this side therefore it means I must multiply by a negative number that side negative times negative becomes positive and on this side negative times negative is four x negative and positive it's minus two so that is my equation two a I'm gonna call it two a because I've got a two there so now I can say substitute equation two a into since equation two a is the same as equation two I cannot substitute it to itself I'm gonna say substitute it to equation one so it means where I see x where I see y I must put four x minus two because I'm doing the substitution so we have two x plus five y is equals to 12 two x plus five my y is four x minus two is equals to 12 remove the bracket two x plus five times four is 20 five times two it's minus 10 is equals to 12 and I must move 10 to the other side two plus 20 is 22 x is equals to 12 plus 10 which is the same as 22 x and on the right hand side is 22 because 12 plus 10 is 22 now I must get rid of 22 it means I must divide by 22 this side I must divide by 22 that side 22 and 22 cancel out 22 and 22 goes one one time one time so therefore x is equals to one now I have my value of x I can then substitute x is equals to one into I can substitute now into either equation two a or two remember two a and two are one and the same so two a was easy because it's already in the format that I need so I'm gonna substitute it to equation two two a two a not just two a because two a and two are the same thing so we have y is equals to four x minus two therefore y is equals to four times one minus two y is equals to four times one is four minus two which is equals to I'm trying to leave space here for the graph which will be equals to two so therefore my coordinates x and y of one and two let's draw the graph now drawing the graph at least equation number two I know what my y intercept is let's change the pen the color of my pen for these peoples let's use blue so we know that for this equation our y intercept which is our y intercept which I don't know what I can call it but which will be the value that passes through the y exists that is minus two so it means that graph passes through somewhere this side if this is minus one this is minus two it passes through at this point and we also know that the slope is positive so slope is positive so since the slope is positive therefore it means the graph will move like that so but we also know that both these two graphs one and two they pass through x of one so I must one two three so if x of one one two three and y of two so therefore it means they pass through somewhere that point so I have I don't have a ruler so bear with me if my line is crooked it's fine it's fine because everything is not to scale it's just for demonstration peoples so this is equation for x minus y is equals to two so you should be able to identify it if they give you multiple graphs so that will be that equation now let's do equation one sorry change my pen to green I use the green now so we come to this one so as you can see this one the the equation is not in the format that we want so you cannot take this number and say that is the value so we need to also write it as for y is equals to four x a x plus b so this if I rewrite it it will be five y is equals to minus two x plus 12 and I know that this will divide by five and this will divide by five and therefore the site divide by five it will get rid of it so I know that the slope of this is negative so it means it go it's going this way it's going this way somewhere and it passes through what is 12 divided by five it's two 10 two over over five in decimal let's see uh 12 divide by five 12 divide by five it's 2.4 so if this is two so it's somewhere it passes through somewhere there somewhere there and I know that the other line is there anyway because I'm I'm not drawing it to scale so it's fine as long as you understand the concept that I'm trying to give you and this is two x plus five y is equals to 12 and that's how you will identify your graph so there is your graph for this mountainous equations so there are the two of them and we know that both of them they share one point and that point is at that let's go back to my right pin you share that point there okay and that is the mountainous equation so now let's look at if we do method of substitution again I think so with this method of substitution I want to substitute one equation to the other equation but not the way I've done it so I do have these two equations so I must make both of them y is equals to ax plus b change them we already changed them you saw that right I did that remember so I don't have to explain it again oh come on I did equation number one I changed it to y is equals to ax plus b equation two we changed it to this so it's the same thing that I'm trying to get to on the next slide so you change both of them to y is equals to ax plus b so they all need to look like y is equals to ax plus b so therefore it means they both are equals to so they all have y as a common so I can take one equation and substitute it into the other equation as well so I can make those two I can make these two equation equals to one another my substitute team into the second equation I can substitute the y value of the first equation and that's what I'm doing right here so my first equation is the same as my second equation which is equal to one another and then I solve move for x to the other side then move 12 plus 12 divide by five to the other side and you will have minus two over five x minus four x is equals to minus 12 divide by five minus two and solve the equation you will get minus minus 22 because you simplify the fraction right the common factor between four x and five minus two x over it's five and then you say five goes how many times into five it goes one time one times two is minus two uh one goes how many times into five it goes five times five times four is 20 minus 20 minus two is minus 22 and that's how I got minus 22 the same thing here the answer will be 22 over five so it's minus 22 over five to get rid of minus 22 over five we can multiply by the inverse multiply by negative five over 22 this side you do the same multiply by negative five over 22 this side they both will cancel out this will also cancel out you will be left with and negative and negative will be positive and you will be left with x is equals to one and we substitute x into the formula and we can choose any of the two equations because now we made both of them equals to one another we can choose anyone you can choose equation one or equation two doesn't really matter you choose any one of them substitute the value of x of one so that you can find the value of y is equals to two that is another method of substitution that you can use right the last method that you can use is the method of elimination with the method of elimination it works different because we have two equations you also need to label them equation one and equation two because you're going to tell yourself certain things that you need to refer back to you want to make sure that one of these two equation can eliminate another equation and make sure that only on the left hand side you are left with only one variable either you are left with an x or you are left with a y you cannot be having the two equations so you can either divide equation number one into equation number two or you can subtract equation number one from equation number two or vice versa you need to make sure that one of those two equations subtract or divide or multiply or whatever it does make sure that you have one variable left so here we have two x plus five is equals to 12 and for x minus y is equals to two so I've labeled that the next step is to multiply equation one by equation two oh sorry equation one by two so it means I'm going to make this equation double so two multiplied by two x is for x two multiplied by 10 is 10 y 12 multiplied by two sorry two multiplied by five is 10 y and 12 multiplied by two is 24 now this is my equation one a with my equation one a I can eliminate the equation two because at the moment as it was I couldn't do anything because even if I subtract or divide or multiply I will still be left with two a variable x and y so now if you look at equation one a which is now converted I can subtract equation two from it and that's equation two and I'm going to subtract equation two from equation one a so it means I'm just applying a minus there so four minus four x is zero 10 minus y it's plus 11 because that will be minus times minus it's positive so it will be 10 plus y is 11 24 minus two is 22 and I'm left with y 11 y is equals to 22 oh I don't know why it's there 11 y is equals to 22 and I can divide by 11 on the side on the left hand side you divide by 11 on the right hand side and you will be left with 11 and 11 cancels out you are left with y and 22 and 11 11 goes two times into 22 and y is equals to two I don't know why my numbers are far right there should be somewhere here so but yeah the answer there it will be y is equals to two and we can take y and substitute it into any of those three equation it doesn't really matter you can choose equation one or equation two to substitute so we're taking equation one so we substitute the value of y is equals to two into equation one and two x plus five times two is equals to 12 and that gives us two x plus 10 is equals to 12 and moving 10 to the other side we left with two x is equals to 12 minus 10 and the answer will be 12 minus 10 is two and we divide both sides by two and x will be equals to y so whether you use the method of substitution or the method of elimination you should get the same answer you can practice both methods and see which one you feel comfortable with when you go write your exams you don't have to use both to solve you just need to find the one that you feel comfortable with and that is systems of linear equations any questions and I can see that we run we running against time so I have 30 minutes to do two sections three sections let's see if I can I can get that quadratic equations so with quadratic equations because it looks at the relationship between two variables but it takes the ballcage shape which the linear equation was y is equals to ax plus b quadratic equation has a square to it the value of a can never be equals to zero because then when the value of a is zero then it becomes a linear equation so it's not a quadratic equation so a quadratic equation should always have a square value on it okay and the value that we we with quadratic equation okay so with quadratic equation you do not have to calculate the slope and all that but you can calculate certain values and I'm going to tell you just now so with quadratic equation a b and c are your constant values so they will be given to you to say this is your quadratic equation calculate the turning point the x intercepts and so forth and so forth so we do things differently so at this point be a new student those who are doing pure b and u you are excused you can if you want you can leave you can stay for 30 minutes until the end of the session it's up to you quadratic functions you need to know how to answer questions relating to quadratic equations based on the value of a if the value of a is greater than zero we say that value it's at minimum and the graph opens up when the value of a is less than zero we say it is at maximum and the graph opens down so you need to remember this opposite less than means maximum means it doesn't mean smiley it means set less means set maximum less let's let's let's take it this way if the value of a is less than zero therefore it means the temperature is very cold and remember what the when the temperature is cold how does it make you it makes you feel sad right so you just need to remember that if we know that the temperature is at maximum at that point oh when it's very very hot anyway it doesn't really matter whether it's very very cold or very very hot it's at maximum it means it makes you sad so you will always remember that right when the temperature is just fine like it's greater than zero so anyway this time the temperature won't work because here we're saying when it's greater than zero so we need to find another way of oh let's let's say the bank balance now so when your bank balance is more than zero you need to remember oh no it cannot work with the bank balance because when it's greater than zero therefore the bank balance is not at minimum um how how would you remember this i don't know how you will remember this but you need to know that when it's uh greater than zero it is at minimum or let's say you are at minimum of being bankrupt so it it needs to make you happy let's put it that way so when a it's positive it means you are positive your your bank balance is positive it's at minimum of you being broke so you it should make you happy right so you should remember that so less than zero means happy because your bank balance is not low it's not minimum it makes you happy when it is less than zero think of a temperature it is cold it makes you sad so let's remember that okay so you can be asked to identify whether the quadratic equation is at minimum or at maximum remember where you will see at minimum maximum and minimum you will just use the value of a if the value of a it's it's less than zero if it's less than zero it is at maximum right so when it's negative we say it is at maximum because a is less than zero at that point and therefore the graph will also be set so you just need to remember that so if the value of a is negative what does that state it means our a is less than less than zero it means our a is less than zero and that's what I just said there so okay so I've got I'm just gonna bring all of them up because I don't know why I didn't remove the animations so in terms of quadratic equation there are several things that you need to know and how you need to know how to calculate them so the vertex which is the turning point the turning point which is that point there where the graph tends or if it's a set face it will be that turning point will be at that point that is the turning point is your vertex and the formula is xm is equals to b minus b divided by 2a and it also has the y point and the y point it is your original formula which is ax squared plus bx plus c depending on the formula that they give you so let's say for example I'm gonna go back one slight back so if I'm giving this so I'm just gonna use ym and my x I'm gonna change it to m because I would have calculated my vertex xm which is minus b over 2a I would have calculated this I will take this value of my vertex my x coordinate and substitute it back into the formula so that's what you will do with when you calculate the vertex or the turning point and especially if they ask you to find the x and y value of your of your vertex you must remember that your y intercept it is where x is equals to zero same way where x is equals to zero therefore it means your c value this is your y intercept your x intercept you calculate using these two formula it's got the minus and the plus sometimes it's written like this you need to be careful the minus and the plus the minus it's for the minus side the plus is for the positive side but sometimes it can be both on the same minus side it can be like this what am I drawing now it can be like this they are both on the negative side but you just need to know that one is the lower side and the other one is the upper side lower side upper side lower side upper side don't swap them because you will get some of the questions that has both of the values swapped around you just need to always make sure that you always say minus and plus regardless of whether you draw it on that side or this side minus and plus the minus first plus the second before you calculate your x intercept you need to evaluate your discriminant which is this value you need to calculate the value underneath the square root first before you can calculate whether you have two intercepts or one intercept the reason for that is if your discriminant is greater than zero then you can calculate both of those two x x intercepts if your discriminant is equals to zero therefore it means if this is equal to zero you can see that this one exists and this formula is the same as your vertex and if you have calculated the vertex therefore it means it is the same as the vertex if it is less than if it's less than zero if you calculate this value and you find that it is negative when it's negative then there are no intercepts so it means this graph never touches so let's let's start with this first one so the first one it says if it is if it is greater than zero the vertex is greater than zero therefore you will have two x intercepts so you will have two x intercepts on all of these graphs that I have in front of you there are two x intercepts so they and they right if it is equals to zero when it's equals to zero therefore it means your you have only one turning point so it means your graph touches your y your x axis at one point even if it comes from the bottom it will go only at one point regardless number three which is that one when it is negative therefore it means your graph never touches the x axis anywhere so your graph will look like that or it can look like that because it doesn't touch the y axis anywhere or it can look like this I don't know how to draw I felt you're drawing at school so so that will be your graph will look like that so let's look at an example a pool is treated with a chemical and this treatment follows a quadratic function and they gave us the quadratic function I always forget to fix my slide this is a minus the minus is at the top so I can rewrite this equation into the normally quadratic equation because we know that the quadratic equation is y is equals to a x squared plus b x plus c because I can multiply remove the bracket multiply 30 times t 30 times t is 30 t squared 32 times minus 10 is 300 t squared plus 700 now because if I need to solve this I need to find what is my value of a what is my value of b and what is my value of c you need to define that so a is 30 in this instance and b you take the sign as well which is minus 300 don't leave the side and c is 750 in honor for me to calculate the vertex which was because I see that my a is taking which is greater than zero so therefore it means my amount of algae is at minimum remember when it's greater than zero it's at minimum um so I can calculate my vertex and I know the formula for the vertex is here we're using t not x so if I was using x this will be x so t is equals to minus minus b divided by 2 a because the formula has a negative and my b has a negative so it's minus minus 300 divided by 2 times a is 30 substitute minus times minus is positive 300 2 times 30 is 60 300 divided by 60 is 5 so the amount of algae will be at minimum after five days I need to calculate my y vertex my y vertex because yeah I've got my x vertex I need to calculate my coordinate so I've calculated my x m now I need to calculate my y m take the same formula and substitute where I see t because now my t is five days where I see t I'm going to put five days so my a t which is five is equals to 30 t squared which is five squared minus 300 t which is five 300 times five plus 700 and you solve the equation you will find that it's equals to zero if I need to calculate my x intercept I need to first evaluate the discriminant in order for me to first find out whether I need to calculate my x intercept so calculating my x intercept oh my discriminant b is minus 300 squared minus 4 times a is the tc times is 750 and if you solve this minus 300 squared is 90 000 4 times the t times 750 I think probably it will give you 90 000 and the answer is zero and what do we know if the discriminant is equals to zero then therefore it means I've already calculated my turning point it's five and you can draw the graph if you want to draw the graph by using your turning point and your y intercept so if I have to draw the graph my y intercept is 750 so let's assume that there is 750 of my y intercept and my turning point it is five and zero so where y is zero five and x is five so let's assume that five is somewhere here because of my graph you can assume that this is five and I can draw the graph like this okay maybe probably my five is here and there is your graph to scale okay so that is that is quadratic equation so we won't find time to do the others let's assume that they gave us this question in the exam so consider the quadratic function below choose the correct graph so here we need to identify what is the value of a a will tell us whether the graph opens up or opens down so the value of a here it is less than zero because it's negative so therefore it is at maximum and when it's at maximum it makes me sad it makes me sad so my graph should look like that so process of elimination this will not be the graph and this will not be the graph because already I've looked at the value of a the next one is to look at the value of my y intercept the value of my y intercept will tell me whether the graph passes through the y-axis on the positive or the negative so I can try quick quick it passes through at y-axis and positive three so that passes at positive three and positive three that does not solve our problem we need to solve our problem by looking at the telling point so let's look at the telling point what is the value of a this the value of b what is the value of c a is minus one b is minus two c is three x is equals two x is equals two minus b divide by two a minus minus two divide by two times a of minus one therefore I have two divide by minus two which is minus one that does not help me because there is my oh it does help me actually x should be in the negative side x on this side it's on the positive side so therefore through my processor of elimination this also is not correct so my answer is option two if they both were on the negative side like for example this one is on the negative side your x is on the negative side then you're going to calculate your y m because this is x m and you're just going to substitute the value of negative one into minus negative one squared minus two times negative one plus three which will be negative one plus two plus three which is minus two plus one is one plus three is four so we know that y will be positive and this one the y is in the negative and so forth and that's how you will do your process of elimination when you're looking at the graph can I have your permission to continue until eight o'clock you tell me before we do that let's do this do I have your permission you need to tell me now because me I can go and sleep right now do I have your permission please make sure that you complete the register as well let's see let me copy the register nobody wants to say anything so I can stop right here right there I can are you guys alright if we go on until eight o'clock it's only additional 30 minutes no okay okay I'm okay thank you Courtney for saving the rest of the group okay let's continue so that is quadratic equation and I'm gonna stop right here with quadratic equation because I think I've covered almost most of the things now we can move on to systems of linear inequalities with systems of linear inequalities we don't have equal sign so it means one side is not equal to the other it might be bigger than or equal or it might be less than or equal so it's inequalities so we're going to first look at when we're solving equation with only one variable so it means when we are given x or we are given y on his own and we need to solve this equation of linear inequalities when solving the equations you must remember if you multiply or divide by a negative number the sign will change so if you had minus 2y it's less than 4 if you have it like this and you're going to divide here by negative 2 you don't do that because your sign needs to change so you say so that you don't get confused you're going to say negative 2y divide by negative 2 and then you change the sign and you say 4 that is why you cannot leave anything with a variable on your right hand side it's very important that you move everything with a variable to your left hand side so always move your variables to your left hand side so when you divide by a negative number the sign change oh when you have negative y is equals to 4 because it's multiplying and you want to divide you want to multiply by a negative you're going to say oh am i putting equal not equal let's say it's greater than you're going to say negative y negative y but then you're going to change the sign because the sign was greater than because you are multiplying so this side you were dividing the side we multiply we are multiplying because we multiply my negative number the sign also changes and then it becomes y is less than negative 4 always remember that that is very important right let's assume that we need to solve the following equation also remember that whatever you do on the left you must do on the left whatever you do on the right must do on the right or otherwise when you move things across the equal inequality side the sign will change as well what i'm not going to cover as well is wet problems of inequalities probably when we do the exam preparation i will bring some of those questions to the front especially the ones with wet problems when it comes to inequalities okay so if we need to solve for a then we need to move everything with an a to the left hand side the left hand side is very important so we move 4a to the left hand side whatever i do on the left i must do on the right i must subtract 4a subtract 4a and i'll be left with minus 6a on the left and 4 on the right because 4a and 4a will cancel out or they will be equals to zero which leaves me 4a i'm going to divide by negative six because i'm looking for a i'm solving for a not negative six so because i am i'm going to divide by a negative number my sign will change it is greater than it becomes less than divide by negative six divide by negative six on the right hand side as well six and six will cancel negative and negative is positive two goes into four two times and it goes into six three times and the answer will be negative two over three so a will be negative two over three as the answer and that's how you will solve systems of linear equalities now with linear equalities as well you can be asked to use the number line to answer some of the questions what do i mean which one of the following graphs represent the solution to linear inequalities now you need to remember the following when your sign i'm going to start with the sign when the sign is greater than or greater than or equal right when the sign is greater than or greater than or equal therefore the arrow will point or the arrow will point to the right when it's greater than or greater than or equal when it is less than or less than and equal the arrow will point to the left because it points to the less than so it says the values less than the values that are below right that is the arrow then we have a dot when the sign is greater than or it's less than when it's greater than or the sign is less than and it does not have the equal sign to it the sign the dot will be open because it says it does not include that point when it is greater than or equal and less than or equal then then the cycle will be solid always remember that right so which one of the following graph represent this linear inequality the first thing you need to do is solve this equation so let's first solve the equation we have two x we have two x plus nine it's greater than three x plus 17 and we move three x to the other side two x minus three x it's greater than 17 minus nine and we solve this two x minus three x it's minus x greater than 17 minus nine if my mind was still fresh eight o'clock in the morning i will know what 17 minus nine is is eight i need to remove the negative so it means i can multiply or divide by a negative number so when i multiply by a negative number my sign change from greater than to a less than and therefore x is equals to minus eight what they are not equal x is less than minus eight what that means it says the value of x is any value that is less than negative eight it does not include negative eight how do we represent that value now we first look at the sign what is the sign the sign it's less than so therefore it means the cycle needs to be open closed closed closed it is not the answer the answer is option number two process of elimination let's assume now that this one was also open let's assume that this is also open then it means the two of them are still confined we know that we eliminated number one and number three we are assuming that this one also is open right so when we assume that it means there is additional thing that we need to know what do we know the arrow the arrow where must it point because it says less than the arrow needs to point to the left so the arrow needs to go there because it is less than so this would also be eliminated as such and the only answer here will be option number two that's how you will use the process of elimination when answering questions relating to systems of linear equalities there are several exercises I've uploaded already the notes for today so you can go through the additional exercises that I have after you've went through the video and you can practice with those ones most of the questions come from the past exam papers I did not recreate anything I don't come up with new questions just use your material so you can use them to practice that is system of linear equality when we have only one variable do you have any question before I move because the same concept that we just learned right now we're going to apply it to the next section are there any question is there still something that we don't understand okay so it means you understand everything I just explained remember the arrow remember the dots but we're going to use different things but the same concept we're still going to take it into consideration with systems of simultaneous linear inequalities it means we are now given two equations with two variables which complicates our life but it's easy to do the same way as we've learned with simultaneous equations for linear equalities we're looking for x and y but with inequalities we're not only looking at the point x and y we're not only looking at which point both of these three equations pass through we are also looking at the area that they both share the values of their areas they both share so how do we do that to solve the equations of linear inequalities especially when you have two variables the easiest way to do that is to use a graph it's not like finding this simultaneous equation and doing the method of elimination and the method of substitution no we use the graph now when we use the graph it's the same thing we still need to we still aim to find the x and y coordinates which is the one point that both of these two equations share or satisfy but also we also need the area that satisfies both of these two systems now how do we do that and this is the process that you need to remember always you're going to use it for the during your process of elimination the first step you do is make sure that the lines or the equations you change the sign if it was less than or equal or greater than or equal it doesn't really matter you change it to an equal sign so that you can treat it as a system of linear equation because when it is a system of linear equations it makes it easy for you to find your y intersect which will tell you where the graph passes through the x axis so we just for that purpose for us to find where the graph passes through the y axis we just need to change the sign to an equal sign step number two we need to also to differentiate based on the original sign because remember in step number one we say change the sign we change the sign for that purpose only to make to find our y intercept so we change the sign to find the y intercept and then once we have found the y intercept then we go back to our original function which has the side of less than or equal or greater than or equal or you can replace it back now once you have changed the sign you can then now determine whether the line that you will draw that will pass through the y intercept is it going to be a dotted line or a solid line and that is based on the equality sign if it's greater than or equal or less than or equal the line will be a solid line so it means it will be hard it will be a solid line if it is less than or greater than the line will be a dotted line which we call it a reference line so you just need to remember that right easy right last thing we also need to know is which area they both share based on the lines it will tell you whether the shading should be above the line or below the line so here is the scenario you will have x and y if it is a positive slope that is the other thing why we change it back to the y to also identify whether we have a positive or a negative slope as well if it is a positive slope this will be above this is above and this is below if it is a negative this is above and this is below this is above and this is below now what do I mean by above and below above and below when I mean above it means when the sign is less than or greater or less than or equal it will be below so this will be below less than or it can be less less than or equal or less than or equal if it's above it will be greater than or equal greater than or equal so it means we're going to shade this side or shade this side and we can see that both of these two lines share that area and that's how we will define our graph those are the three things that you need to remember right you got it when it's less than or equal or greater than or equal is a solid line when it is less than or equal or less than it is below I wrote this thing wrong this should be above sorry my bad that should be above above so let's look at an example of how we do this how we identify this so I have two graphs x minus y oh minus x plus y minus one is less than or equals to zero and two x plus y minus four is less than or equals to zero now this does not help me much because it is not even in the format y is equal to mx plus c or something like that so that's where step number one comes in so we're going to do step number one for one so I'll do step number one change this sorry this equation to an equal sign so we say minus x plus y plus y minus one is equals to zero and I move everything except y to the other side so this will be y is equals to x minus one so this is step number one so that tells me my y intercept y intercept is minus one I'm happy with that my slope is positive because it's positive one so it means my graph should actually actually go like this it must light like that so step number one done step number two it says I must go back remember go back to my equation and look at the sign what does the sign say the sign says it is less than so my sign is less than or equal and what does that mean that means my line is solid in terms of the line step number three what does that mean in terms of the area it is less than or equal so therefore it means the value should be below because it's less than so I'm done with number one let's go to number two number two we do the same step one we change the equal sign or the equality sign to equal so it's two x plus y minus four it's less than not less than we change it to equal sign so y is equals to moving minus two x plus four so I know what my y intercept is my y intercept is positive four so it means it passes through my y exit four but my slope is negative so it means my graph should look like this step number two the sign is a less than so it means a dotted line and my sign is a less than so therefore it means they are the points are below so let's draw that magic I drew it and there is our graph I drew our graph so because you are not going to be asked to draw a graph they will give you the graph so you can just use your graphs to determine whatever the values you are looking for or to respond to that so if we were going to solve this as a simultaneous equation we would find that the point they both meet at it's at that point they already defined that so now in the exam you're going to choose they will give you four plots you just need to make sure that you understand how to identify each one of them so let's see if we can identify our graph so the first one our y intercept is minus one minus one is it minus one or positive one I wrote it wrong it's positive one because it's minus there should not be negative one it should be positive one so because minus it's y minus one so when you move it this side it should be positive so it's one so our graph number one should pass through our y intercept at one so let's see there is our come on look with me there is our one what else do we know we know that the graph should be a positive remember how we identify the graph so this is our graph we are able to identify that part is the line solid yes you can see that the line is solid it's not dotted at the points below so the points are those lines they are below the line remember when I did the demonstration below means below the line then we have identified our first equation let's go to our second equation our y intercept is positive four so let's see if it passes through four what do we also know the slope we said it should look like like this and it does surely looks like that and it should be a dotted line and for sure it's a dotted line and the point should be below the line and remember this is above so below the line will be those values the lines below now where are the two areas that both of these two graphs share this is the area where they both share so not there not there so let's look at the example that you will see in the exam this is the example now you need to look at your three graphs and identify which one is which so we use the process of elimination now you remember that I gave you the three steps you don't have to do them in order you can start from the bottom and go up so I like to stop to start with the dotted line or solid line let's start with that one let's define each one of them looking at the sign is this a dotted line or a solid line it will be a solid line so this one is solid line number two this is a dotted line or a solid line it's a solid line because it's got an equal sign to it this one does not have an equal sign so therefore it will be a dotted line so number one done number two I can define whether the values will be below or above I don't have to look at the graph as yet I'm just collating the information I need to enable me to look at the graph and do the process of elimination right so we go number one it says greater than so therefore it means they should be above number two they should be above number three it says less than so they should be below so step number two and three done let's go to step number one step number one says I must change all this graph and make them y is equals to axb so let's do that let's change this first one so we're going to start with the first one which is two y plus five x is equals to ten and we say here what do we say here two y is equals to minus five x plus ten so therefore y is equals to minus five over two x plus ten divided by two it's five already calculated that because it's I'm dividing by five everywhere or why did I divide by two the you guys you must say wake up so divide no we divide by two yes it's right we divide by two so my intercept or I can come back here my y intercept is five so it means that graph should pass through the y-axis anywhere on the y-axis it must pass through five and my slope let's write the slope here the slope is negative so it means my graph should look like this should go like that let's do number two number two says three y minus two x is equals to six three y is equals to two x plus six and divide by three y is equals to two over three x plus six divide by three is two I have my y intercept of positive two so it means it must pass through the y-axis at two and my slope is positive so it means this graph needs to look like this right the last one which is the easy one it says y is equals to four which makes life easy it says my y intercepts my y intercept is four my y intercept is four what is my slope I don't have a slope so therefore it means the line will just pass through the y-axis at the constant because I don't have x-axis it does it will not touch the y the x-axis anywhere so it will just be a line there so it will just be flat like this because this is just the y values if it was x is equals to four then it will be like this this is for y the y it will just be horizontal like that not right okay so now let's answer the question using the process of elimination but the lucky enough is that all these graphs are labeled one two three one two three so but we can start here we're looking for graph number one it needs to pass through the y-axis at five there there's one that passes at five and the slope needs to be like that and it needs to be a solid line and the values need to be above above is here so we can eliminate that graph already using only number one number one we already eliminated our graph let's look at number two number two the slope should pass through the solid line it looks like a dotted line but I'm going to take it as a solid line you can see that it's a solid the graph should look like this yes and the value should be above above is here so we eliminated that one so still on number one we haven't even looked at number two number three and number four on all the x