 Good afternoon, everyone. I'll wait for next two minutes so that each one who wants to join the class can join it and then I'll start the class. Okay, so I'll start the class now. So today I'm going to take the topic from mathematics which is pairs of linear equations. The objective of this session would be to take you through different kinds of linear equation and different varieties of solving linear equations. So we would be looking at the type, I mean different types of linear equations, I mean consistent, inconsistent. What is the meaning of consistent linear equation in consistent linear equations? How the solutions of a pair of linear equation changes depending on the value of their coefficients. Secondly, when we are done with these things, what we will do is we will look at the different types of solving the linear equations, which are graphical method, which are substitution method, elimination method, cross multiplication method, everything. And at the end of the session, in the last 15 to 20 minutes, what I will do is I have selected a few questions which are based on application of pairs of linear equation. Primarily, they are word problems related to time, speed and distance. If time permitted, I shall be taking one or two more questions apart from what I have included in this PPT. Now, what I want you guys to do is that this topic is not very difficult but remain patient with me so that I can cover all the different type of questions. If you have any doubt, you can let me know during the presentation of this particular topic and I shall be taking your doubts. So, I am starting with linear equation in two variables. So, first of all, how this is presented. So, any linear equation in two variables is presented as ax plus dy plus c is equal to 0. I have assumed that x and y are two variables here. So, x and y are two linear, two variables here. And what are a and b? a and b are coefficients of x and y respectively. So, number coming in front of this number coming in front of this variable is known as coefficient of that variable. Now, we should know that a and b should be real numbers only. They cannot be equal to 0. Why they cannot be 0? Because if I put 0 here, 0 multiplied by y will make this variable y without any relevance here. So, if you look at the equation ax plus dy plus c is equal to 0. And if you put b is equal to 0, this equation turns to be ax plus c is equal to 0. So, this meaning of two variables does not hold true when one of them becomes 0. I mean one of the coefficients of x and y becomes 0. So, any linear equation can be represented as ax plus dy plus c is equal to 0 where x and y are your two variables. a and b are coefficients of x and y and c is the constant over here. What we should need to understand over here is a should not be equal to 0 and b should not be equal to 0. Otherwise, the second variable will become without any relevance. So, relevance of second variable will not be there in the equation. Now, how do we solve these linear equations? So, solution of linear equation. So, suppose I take any equation ax plus dy plus c is equal to 0. And what I do is I replace x with any number p and y with any number q such that ap plus bq plus c becomes 0. x is equal to p and y is equal to q is known as solution of this equation. Now, what do I mean by a solution of equation? So, if I need to put it in word the value of x and y such that upon putting x and y or replacing x and y with p and q in the linear equation ax plus by plus c is equal to 0, ap plus bq plus c becomes equal to 0. It means that replacement of x and y with corresponding values of x and y in equation ax plus by plus c if the answer given to me is 0, that becomes solution of linear equation. So, suppose I have 2x minus 3y plus 1 and if I put x is equal to 1 and y is equal to 1 here, what happens is 2 into 1 minus 3 into 1 plus 1. So, this comes out to be 2 minus 3 plus 1 which is 3 minus 3 and that is equal to 0. So, what comes out over here is that if x and y both are replaced with 1 respectively, this particular equation 2x plus 3y plus 1 gives me an answer 0. Hence, x, y, 1, 1 is solution of this particular equation. So, what I am trying to mean in this particular scenario is that if x and y are respectively replaced by any 2 values p and q such that ap plus bq plus c gives you an answer 0, p and q becomes solution of this equation. So, it is as simple as this. Now, let me go to the next topic which is simultaneous linear equations into variables. So, what do I mean by simultaneous linear equations into variables? So, here I will have a pair of linear equations and that pair of linear equation would be represented as a1x plus b1y is equal to c1. This is my first linear equation and the second pair would be a2x plus b2y is equal to c2. So, this is called simultaneous linear equations into variables. I have two variables x and y here and just by changing coefficients of x and y two times, first case I am taking a1 here, second case I am taking a2 here, b1, b2 here in first and second case and c1, c2 here respectively in first and second case. I am trying to make two linear equations and together these two linear equations would be called simultaneous linear equations or a pair of linear equations. So, two linear equations into unknowns x and y are said to have set to form a system of simultaneous linear equations if each of them is satisfied by same pair of x and y. So, what I do is it becomes simultaneous equation when you have same value of x and y. So, if I write it in the previous format that a1x plus b1y minus c1 is equal to 0 and a2x plus b2y minus c2 is equal to 0 and I take one value x is equal to p and y is equal to q such that a1p plus b1q minus c1 is also 0 and a2p plus b2q minus c2 is also equal to 0. In that scenario, these two equations becomes simultaneous linear equations. That is how it has to be understood. Now, let me go to another topic. So, what I have selected next is consistent system of linear equations and inconsistent systems of linear equations. What do I mean by consistent systems of linear equations and inconsistent systems of linear equations? So, see what happens when we select, when we have a pair of or simultaneous linear equations. So, what happens there are actually three possibilities that suppose I am taking a1x plus b1y is equal to c1 and a2x plus b2y is equal to c2. What it does is actually that I have three cases possible. Case one is that we will have unique solution. Case two is that we will have infinite solution. Unique solution means one value of x and y, one value of x and y which satisfies both the equations. Now, what comes over here in infinite solution is if both the lines, if a1 and a2 are corresponding, b1 and b2 are corresponding similarly c1 and c2, I will discuss these things in later slides. What happens is that the two lines are coinciding with each other. So, if two lines are drawn like this, this is line l1 and this is suppose like l2, line l2. What will happen is that line l1 and l2 are coinciding with each other. If they are coinciding with each other, then at each point or on infinite points, the infinite values of x and y will be satisfying these two equations. That is why I am saying that we will have infinite solution. And the third case is no solution. So, two linear equations can be or these are linear equations. What happens in a linear equation if you have a little bit of idea about coordinate geometry? Just for the sake of understanding, I will be discussing it. Not for, from the perspective of your exam, what happens is if you understand the equation of line, what happens is equation of line is given by a1 x plus b1 y is plus c1 is equal to 0. So, any linear equation represents in coordinate geometry a line, something like this for an example. And what happens is when two lines are found to be parallel, they do not meet each other at any point or for any common value of x and y, they would not be meeting at each other because there would be no intersection of these two lines. So, what happens when two lines appear to be parallel or when two lines are parallel, no value of x and y is satisfying, no value of x and y satisfies the two equations simultaneously. So, in that particular case, the third condition which is no solution would be observed here. Now, what do I mean by no solution is these two kind of cases. So, consistent means either one unique solution or infinite solution. So, if you read what I have written over here, that system of consistent linear equation. So, when system of two linear equations in two variables have at least one solution. So, I will have consistent system and inconsistent system means we do not have any value, I mean any x and y value satisfying those systems. So, we will have no solution, simple as that. So, I hope you understood consistent and inconsistent systems, consistent linear systems of linear equation has either one unique solution or infinite solution, inconsistent system means you have no solutions. Now, with this, I move to another topic which is solving simultaneous linear equations which tells me that the three possibilities that I was discussing. Now, suppose I have a pair of linear equations a1x plus b1y is equal to c2, c1 and a2x plus b2y is equal to c2. Now, case one is saying that the two lines intersect with each other. So, if I draw x and y axis like this and my two lines are like this. So, you see here the two lines are intersecting at one particular point, suppose this points are alpha and beta. This point the alpha and beta is common to both the lines a1x plus b1y is equal to c1 and a2x plus b2y is equal to c2. Hence, I can write that a1 alpha plus b1 beta is equal to c1 and a2 alpha plus b2 beta is equal to c2 which tells me that alpha and beta is common and only solution to the pair of these equations. So, hence whenever the two lines are intersecting with each other, it will provide me one such point such that it is common to both the equations and hence we will have only one solution. So, I will have only one solution here. Now, what happens is, suppose the two lines are coinciding as I discussed and if two lines are coinciding, so this is my x axis and this is my y axis and the two lines are coinciding like this. So, suppose this is line n1 and this is line l2 and what happens is, on line l1 and l2 there would be infinite x and y points which will be coinciding these two equations. Hence, I will have infinite solutions here and here two parallel lines. So, suppose one line is like this and the second line is like this. So, there would be no x and y which will be common to these two lines hence I will have case of no solution. Now, how do I do these things graphically? So, I have taken a few questions that how to solve these questions graphically. So, my first question which I have taken is 4x plus 2y is equal to 2, 3 and 4x plus 3y is equal to 2. So, how do I solve these questions? So, look at here, to solve this question what I do is first I will take equation 1 and I represent y in terms of x. So, if it has been told to me that x plus 2y is equal to 3, y comes out to be 3 minus x divided by 2. So, what I will do is I will take a few values of x and I will try to find out a few values of y corresponding values of y. So, I will take x in such a manner that I don't get any decimal points here. So, I start with a negative value, I take x is equal to minus 3, when I take x is equal to minus 3 it gives me 3 minus minus 3 by 2 which is 6 by 2. So, y comes out to be 3, I take x is equal to minus 1. So, if I take x is equal to minus 1 this will give me 2 here. If I take x is equal to 1 this will give me 1 here and if I take x is equal to 3 this will give me 0 here. So, what I do is I take x and y axis like this. So, this is my y axis and this is my x axis and what I do is I start plotting my values. So, this is x is equal to 1, 2, 3 and here I take minus 1. I take minus 2, I take minus 3. So, on x is equal to 1 this is 1, 2, 3. So, on x is equal to minus 3 this is 3. On x is equal to minus 1 this is 2. On x is equal to 1 this is 1 and on x is equal to 3 this is 0. if I need these lines these four points can be joined by a line and this becomes my line l1. So suppose I am also writing it as line l1. Now what happens is I do the same thing for the second equation. So the second equation y comes out to be 2-4x divided by 3. Again I will take a few values of x and I try to find out a few corresponding values of y. Now here I have 3 in the denominator. So what I would like to do is that I would like to take a few values of x such that I don't get any decimal value of y. Then there is no problem in getting decimal value of y just for the sake of making things easier and I am trying to take only such values of x which will give me non-decimal values of y. There is absolutely no problem if you are finding any value of y which is in decimal. Just that in the graph paper you have to select such point properly and then you have to draw that line. So look at here this is so what I will put here is I will put x is equal to minus 2. So sorry x is equal to minus 4. So if I put x is equal to minus 4 this becomes minus 4 into minus 4 16 plus 218 divided by 3 I will get y is equal to 6. Now if I put x is equal to so I know that up now the value has to be in multiples of 3. So I will start decreasing it by 3. So this becomes minus 1 so increasing it by 3 sorry. So this becomes minus 4 into minus 1 is plus 4 plus 2 is 6 6 divided by 3 is 2. Now I again increase it by plus 3. So you see here when the denominator was 2 I increased it with a common difference of 2 now I am increasing it with a common difference of 3. So just select one value and look at the denominator keep on increasing it with same frequency you will you will find the correct answer. Now you put 2 here so 4 into 2 is minus 4 into 2 is minus 8 plus 2 is minus 6 and you get minus 6 divided by 3 as minus 2. So if I have these points what I get is at y is equal to minus 4 I get 1 2 3 4 5 6 I get one value here so I get one value here then at y is equal to 2 y is equal to x is equal to minus 1 I get y is equal to 2. So you see here these two points y x is equal to minus 1 and y is equal to 2 here and x is equal to minus 1 and y is equal to 2 here is common to both the equations. So it means that my line is passing something like something like this so here the common point the point of intersection which is x is equal to minus 1 and y is equal to 2 is common to both of these lines and in this case what happens is just by plotting the lines I can see that I can write that the two graphs what I write at solution is the two graphs the two graph lines intersect at minus 1 2 hence they are or this is the solution of the given simultaneous equation. So this is how I write at the last after drawing so what I have to do the process is you take equations one by one you take the you you represent y in terms of x you put few values of x so that you get corresponding values of y you plot the first line then you go to the second equation you do the same thing representing y in terms of x and then what we do is we find out corresponding values of y after plotting and then we plot the second line we just check at what point these two lines are intersecting and then we write this that the two graph lines intersect at this point hence that would be the unique solution of of given pair of equations so that is how we have to do do this question if if somebody is asking you that how to solve these kind of questions you can draw the graphs easily now let me go to another equation I have to show that the system so that the above system of linear equations is inconsistent what do I mean by inconsistent so just give me a moment inconsistent so meaning of inconsistent means the pair of linear equation has no solution I have already explained that the no solution condition will come only when the given pair of lines are parallel to each other so if they are parallel to each other it means that my question becomes now very clear that the two lines which I am plotting now has to be parallel to each other and they cannot be cutting or intersecting each other at any particular point now what I do is I take a few values so you look at here I take first equation and I represent three y is equal to five minus two x and y is equal to five minus two x divided by three now what I do is I take look at here I take x and corresponding values of y now I take x is equal to minus two so if I took take minus two minus two into minus two is four four plus five is nine divided by three is three so at x is equal to minus two I get y is equal to three now I told that look at the denominator increase it by plus three this becomes one so at x is equal to one this will give me one at x is equal to four this will give me five minus eight so this will give me minus one so what I'll do is I'll draw this particular line in the graph paper given so I take x is equal to minus two here and here one two three so this is one point the second point is at x is equal to one so at x is equal to one I get it equal to one and at x is equal to four which is two three four I get this as minus one so this is the line system which has been going like this so this is how my line and one will look like so and one look like looks like this and let me go to the second pair of equations so what I get here is two i is equal to x or y is equal to three plus four x divided by six so now again I do the same thing x and y and I take values of x which can satisfy so see here choosing a value of x here so that y is divisible by six uh prime of a c because if I take one two two it will not give three it will not give so but on x is equal to three what I'll get is I'll get y is equal to half something so so three in plus four into three which is 15 divided by six which is nothing but 2.5 so I'm starting with so what what what I do over here is I start with x is equal to minus three if I put x is equal to minus three this will give me 1.5 if I put x is equal to three this will give me 2.5 and if I put x is equal to nine this will give me 3.5 sorry if I put x is equal to three then I get 2.5 if I put x is equal to nine I get nine into four plus three divided by six so this will give me six and a half and if I put x is equal to minus three so this will give me minus nine by six which is one and a half so what happens over here is I take x is equal to minus three and this is my one and a half somewhere here if I take x is equal to three have I made any this this line graph is this first line graph I'll draw it once again the first line graph is so in the first line graph y is equal to five minus two x divided by three if you put minus two here so at x is equal to minus two so this is actually a negative sign here so these values will change so what happens as x is equal to this this this becomes minus three at x is equal to three one this becomes minus one and x is equal to three this becomes four so lying would be something like this where I put the values that x is equal to minus two this this x is equal to minus through this this is going out to be three minus three at x is equal to one this will be four and x is equal to minus one this will be nearly one now what happens over here is at minus x is equal to minus three I get one and a half here so this will be this point at x is equal to three I get one two three I get two and a half here so what happens is a line parallel to this like this will come so what happens over here is this this this line should actually go go from here this line l1 because at x is equal to one I'm getting a minus one here so this probably gives me a better picture of how the lines has to be drawn so the lines has to be drawn like this so at minus two I'm getting a minus three at one I'm getting a minus one and at four I'm getting a one so line is like this l1 and l2 would be at x is equal to minus three I get one and a half like like this so on y is equal to positive one and a half at x is equal to three it will be two and a half so x is equal to one two three so it will be two and a half somewhere here so you see the line line would something like this so l1 and l2 looks parallel so I write that as l1 and l2 is parallel given simultaneous equation it does not intersect anywhere hence no solution and hence inconsistent so first we will have to write that it has no solution and then we will write that it is inconsistent so this is how it is now let me go to another example and which tells me that I have taken this this particular equation purposefully because I have to find area bounded so what I do again I write y is equal to 11 minus 3x here there is no denominator so I take few values of x and I find corresponding values of y so I take x is equal to minus 2 so if x is equal to minus 2 then y is equal to 11 minus 3 minus 2 is equal to 17 and similarly if x is equal to 0 so y comes out to be 11 so I have taken 0 and 11 and then I have taken x is equal to 2 so then y comes out to be 11 minus 3 into 2 which is 5 so x is equal to 2 y comes out to be 5 now I plot this line like this so what happens is I let me plot this line for you so I will take this as this as minus 2 and I am taking 2, 4, 6, 8, 10, 12, 14, 16 and this will be 17 so somewhere here and then I am taking x is equal to 0 so I am taking then I will have at x is equal to 0 y is equal to 11 so somewhere here and then at x is equal to 2 which is here I will have y is equal to 5 so somewhere here so I am getting one line like this which is my line l1 so I am saying line and l1 is like this now y is equal to 1 minus x in the second case so what happened is x and y so this again I take as x is equal to minus 2 so I get 3 here and I take x is equal to 0 so I get 1 here and I take x is equal to 2 so I get minus 1 here so what happens in this situation is at x is equal to minus 2 x is equal to minus 2 y is equal to 3 so this is this is somewhere like this so x is equal to minus 2 y is somewhere here now at x is equal to 0 y is how much y is equal to 1 so which is which is which is so once again let me check the values once again so if x minus y is equal to 1 so y comes out to be x minus 1 so this is this is at x is equal to minus 2 this becomes minus 1 so this becomes minus 2 at x is equal to minus 2 I'll take minus 2 so this becomes minus 3 this becomes minus 1 and at 2 this becomes 1 so at minus 2 this is this is somewhere here which is which is minus 3 at x is equal to 0 this is minus 1 and at x is equal to 2 this is 1 so the line goes something like this and it intersects here so this particular area I'll graph I'll put a graph like this and what I'll do is I'll shade it like this now I'll I'll try to if if somebody asks asks me that what are these points so I'll put x is equal to 0 in the because at y axis x is equal to 0 so I know this is 11 and at x is equal to 0 here y is equal to minus 1 so this point would be y is equal to minus 1 if somebody asks me that where it intersects x axis so for x axis I'll put y is equal to 0 or the line itself if I draw a line with the help of a scale and all those things so what happens is automatically you get a line which is which is intersecting at such point so you can you can draw that point and you can you can get the coordinates over there only so you just have to shed this part and you'll find the answer now what happens is in this particular case these these questions would be three marker or four marker or five marker so so specifically this question would be something something nearer to four marks question three marks four five marks question so please pay required attention in solving these kind of questions now I've taken substitution method of solving questions and the substitution method the methods are like this I'll explain the methods and then I'll solve this question the method is like this that express take one equation take any one equation and express y in terms of x or vice versa x in terms of y in any any one equation then what I need to do is that I need to substitute this value of x or y substitute x or y in second equation and similarly if if I'm substituting x then find value of y or x by substitution if you get one value the second equation is already available to you from and find x or y by putting value in first step so these are the four steps to solve these questions and what happens is I will solve this question and and and and tell you how this these things happens so I'm taking this equation and what I do over here is y is equal to seven minus two x from first equation I'll put it in the second equation so four x minus three seven minus two x plus one is equal to zero so four x minus 21 plus six x plus one is equal to zero so 10 x minus 20 is equal to zero 10 x comes out to be equal to 20 so x comes out to be equal to two now if x is equal to two I'll put it in this equation so seven minus two into two is equal to seven minus four which is equal to three so y is equal to three so this is how you have to solve solve by substitution method what I did was that I found the value of y here and then I put the value of y in the second equation found the value of x and then by finding the value of x I found the value of y once again so what I need to do in this equation this equation elimination method is I write the steps to solve elimination method so two steps to solve elimination method is first what I need to do is multiply the given equations I need to multiply the given equations and then I'll multiply look at here if I look at the coefficients of x and y none of them are same if they are not same then in that condition what happens is I need to eliminate either x or y the objective is to eliminate one of them to eliminate one of them I need to make the coefficients of either x or y equal if in in that particular scenario what what is the nearest number which is equal to a1 and a2 which is here 10 and 6 so 10 and 6 I take lcm of 10 and 6 that comes out to be 30 it means that if I need to eliminate x I need to make it 30 or if I take 3 and 5 the lcm is 15 so I need to make 3y and 5 by 15 y so so what I do that's what I'm writing here that multiply the given equations by suitable number why I need to do multiply it with suitable number so as to make coefficients of one unknown equal or one variable equal the second one is look at the sign if the signs are same then subtract so you subtract it if they are different then add it so it means that if both are positive or both are negative then you subtract it if they are different then you add it and then what I do is find the value of one unknown which is not eliminated so eliminated one would be gone and by knowing this value put this value in any of the equation to get second unknown so this is how you have to solve so what I'm trying to say you is that I'll I'll I'll take an example and I'll do it so I'm writing here 10x plus 3y is equal to 75 and 6x minus 5y is equal to 11 what I do here here is I'm eliminating y so to eliminate y I just told that 3 and 5 needs to be made same and LCM of 3 and 5 is equal to 15 so if I have to make this 15 I multiply this by 5 and I multiply this by 3 because 5 into 3 is 15 so this becomes 50x plus 15y is equal to 75 into 5 which comes out to be 375 and multiply this with 3 you get 18x minus 15y is equal to 33 now I need to eliminate y because the coefficients of y are same in this case what happens is as the signs are different I just added after addition this will get eliminated so this gives me 68x is equal to this becomes 80408 so how much it comes x comes out to be 408 divided by 68 how much it will be so 50 plus this is 375 so x will come out to be 16 to 6 plus 8 into 6 so this will come out to be 6 now if x is 6 I'll put this value in in one of the equations here so 6 into 6 minus 5y is equal to 11 so 5y is equal to 36 minus 11 which is equal to 25 and y is equal to come 25 by 5 which is 5 so x comes out to be equal to 6 and y comes out to be equal to 5 so that is how you have to solve this equation now let me go to the next topic so the next topic is let me go to the next topic the next topic is I have taken a few few questions over here and I have not given by which method I should solve these questions so the first equation over here is 2ax minus 2by is equal to a plus 4b plus a plus 4b is equal to 0 and the second equation is 2bx plus 2ay plus b minus 4a is equal to 0 so how to solve this question so first I write it in terms of a1x plus b1y is equal to c1 so I write it 2ax minus 2by is equal to minus a minus 4b and I write this 2bx plus 2ay is equal to 4a minus b now if you look at this equation substitution would be a bit difficult if I start substituting so I find that minus 2by is equal to minus a minus 4b divided by 2a so y will come out to be a plus 4b divided by 2ab sorry 4ab now to put this value and do all these things would make the question or the process tedious what I do is you look at here if I multiply this equation with b so this this x would be equal to 2abx and if I multiply this equation with a this x coefficient would be 2ab so it means that I can make the coefficient of x same at both the places which looks like a feasible solution so what I do is that I multiply this equation with a sorry b and this equation with a so what I get over here is 2abx minus 2b square y is equal to minus a b minus 4b square and if I multiply this this I will give me 2abx plus 2a square y is equal to 4a square minus a b now what I do is I will eliminate x so negative negative this becomes negative and this becomes positive so this is gone I get 2a square minus 2b square y and this a b a b gets cancelled out I get 4 here I get minus 4 because both of them are minus so I am taking minus 4a square plus b square now this is also negative here because this this this comes out to be minus 2a square minus 2b square so I will take negative out now this negative negative will get cancelled out so what I get is 2a square plus b square y is equal to 4a square plus b square so a square b square gone y comes out to be 4 divided by 2 which is equal to 2 now if y is equal to 2 what I will do over here is I will put it in this equation so what I get over here is 2ax minus 4b because y is equal to 2 is equal to minus a minus 4b so this minus 4b minus 4b gone x comes out to be minus a divided by 2a which is equal to minus 1 by 2 so x comes out to be minus 1 by 2 so if in this kind of question where you find that you don't have to select the values of x and y sorry you can't substitute the values of x and y so in that particular all substituting the values of x and y is a bit tedious what what we need to do in this particular condition is that we need to use the substitution method so wherever the method is not mentioned please identify that which method is suitable for you and the same method has to be used now let me take this this equation so this equation is I have taken a special case so what does this special case means the special case is 47x plus 31 y is equal to 63 and 31x plus 47 y is equal to 15 what this means is I am interchanging the coefficients of x and y in both the equations so when the coefficient of x and y are getting interchanged in both the equation the first step is add both the equations so first step is add both the equations and by addition what I get 47x plus 31x plus 31 y plus 47 y is equal to 63 plus 15 so what I get is 47 plus 31 is 68x and similarly 31 plus 47 would be 68 y sorry this this is 78x plus 78y is equal to 78 so x plus y is equal to 78 divided by 78 and that is equal to 1 so x plus y is equal to 1 what what will I get here equal to 1 I will get this so and now I will do substitution so y is equal to 1 minus x I put here 47 plus 31 1 minus x is equal to 63 so I get 47 plus 31 minus 31x is equal to 63 so what do I get I get 47 plus 31 is equal to so 47 68 minus 31x am I doing something wrong over here so I have got equation x plus y is equal to 1 so y would be equal to 1 minus x so y I am keeping as 1 minus x so this is 47x that's what I'm saying so 47x minus 31x is is equal to this this this makes 16x plus 31 is equal to 63 16x is equal to 63 minus 31 is equal to 32 and x is equal to 2 now if x is equal to 2 this gives me 1 minus 2 is equal to minus 1 so x comes out to be equal to 2 and y comes out to be equal to minus 1 what we need to take care in this particular case is that whenever the x and y coefficients are getting interchanged we need to add it this is the one thing we need to take care and after that we can do it with either substitution or elimination method now let me go to another type of question and in this type of question what I have selected is that the denominators are in sorry variables are in denominator so if variables are in denominator what will I do is what are the variables here x and y but they are coming as so I can write this as 3 a into 1 by x minus 2b into 1 by y is equal to minus 5 and this I can write as a into 1 by x plus 3b into 1 by y is equal to 2 now 1 by x is equal to u and 1 by y is equal to v so what I do over here is 1 by x I replace 1 by x with equal to u so I get 3 a u minus and 1 by y with v so minus 2 b v is equal to minus 5 and I get here a u plus 3 b v is equal to 2 so now what I do I multiply this with 3 why because I want to cancel out u so multiply this equation with 3 what will you get so I am writing first equation here I get 3 a u minus 2 b v is equal to minus 5 and when you multiply this you get 3 a u plus 9 b v is equal to 6 now you subtract it what you get is this 3 a u 3 a u gets cancelled minus 11 b v is equal to minus 11 so v comes out to be minus 11 divided by minus 11 b so this is gone v comes out to be 1 by b so you look at here I write 1 by y is equal to 1 by b so y comes out to be equal to b now in this equation only in the in the in the in the second equation which was a u plus 3 b v equal to 2 I put v v is equal to 1 by b so a u plus 3 b into 1 by b is equal to 2 so what I do is this b and b gets cancelled out a u plus 3 is equal to 2 a u is equal to 2 minus 3 is equal to minus 1 and u comes out to be minus 1 divided by a so if you I put over here 1 by x is equal to minus 1 by a so x comes out to be equal to minus a so y came came to be equal to b and x is equal to minus a this is the solution of equation over here so I hope you understood this question I've taken one more question like this which is which is a difficult question from this particular topic and this is my question which I have selected so here instead of x and y both the variables are coming in the denominator of both the places so this this this method remains the same so what I assume over here is that 1 by 2 x plus 3 y is equal to you and 1 by 3 x minus 2 y is equal to v and then I'll start replacing it so this gives me 1 by 2 u plus 12 by 7 y sorry 12 by 7 b so this I write as 12 by 7 b and this comes out to be 1 by 2 and this I write as 7 u plus 4 v and that is equal to 2 now even after doing this this equation looks like a bit tedious equation so am I right by by by by by taking this this this equation correctly so what I get over here is so see what I do is I try to take this particular because here denominators are coming and which is 2 7 and 2 I'll try to remove these denominators first so what I do is I'm writing this equation 1 by 2 u plus 12 by 7 b is equal to 1 by 2 and as the lcm is 14 so I'm writing 14 everywhere and this comes out to be 7 u plus 24 v is equal to 7 so this 14 14 gets cancelled out my second equation comes out to be 7 u plus 24 v is equal to 7 so what I get over here is this negative this negative this negative minus 20 v is equal to minus 5 and v comes out to be 1 by 4 so I put it here this is equal to 1 by 4 so 1 by 3 x minus 2 y is equal to 1 by 4 so I write here 3 x minus 2 y is equal to 4 and second equation if v is 1 by 4 now let me put it in this equation so I write here 7 u plus 4 into 1 by 4 is equal to 2 so 7 u plus 1 is equal to 2 so 7 u is equal to 1 u comes out to be 1 by 7 so this is equal to 1 by 7 which gives me 2 x plus 3 y is equal to 7 now I need to solve these two equations to get the value of x and y so what I do is I get let me go to another slide so what I get over here is 3 x minus 2 y is equal to 4 and 2 x plus 3 y is equal to 7 now multiply this equation with 7 and multiply this equation with 4 I get here 21 x minus 14 y is equal to 28 and I get here 8 x plus 12 y is or sorry what what what I can do is that I can rather than doing this I can multiply this equation with 3 and this with 2 so what I get is 9 x minus 6 y is equal to 12 and 4 x plus 6 y is equal to 14 so add it you get 13 x is equal to 26 x comes out to be 2 you put it in this value 3 into 2 minus 2 y is equal to 4 so this comes out to be minus 2 y 4 minus 6 is equal to minus 2 y comes out to be equal to 1 so if I write x is equal to 2 here and y is equal to 1 here this is your answer so I had taken a very difficult question of from from this topic and if what what you need to do in this case is that the only way you can make any mistake in this question is calculation mistake because so what you need to do in this question is write every step very clearly there should not be any correction errors and even if you are doing correction errors make it in such a way so if you are doing any correction errors suppose you have done something down just cut it by one line one slash line and then you write everything below it not start writing there only so that every step that you write over here is visible to your eyes very clearly and then you can match the answers now one particular thing which we which we need to take care in case of linear equations is that whatever value of x and y I am getting should be checked with the answer so what what what I do is I get x is equal to 2 here and y is equal to 1 here let me put it in in the values and try to get the answers so if I put x is equal to 2 and y is equal to 1 so what happens is this is 1 by 2 x plus 3 y plus 12 by 7 3 x minus 2 y is equal to 1 by 2 and I got x is equal to 2 and y is equal to 1 I put the values here 1 by 2 2 into 2 plus 3 into 1 and do it on a rough paper so 12 by 7 3 into 2 minus 2 into 1 so what I get over here is 1 by 2 and this comes out to be 4 plus 3 plus 12 divided by 7 and this 6 minus 2 is 4 so what I get over here is 1 by 14 plus 3 by 7 so this is nothing but 14 1 plus 6 so this gives me 7 by 14 which is equal to 1 by 2 this is the right hand side here it means that whatever values of x and y I've got over here is actually correct so you in this kind of question don't just leave the answer by just getting it on the rough paper you should definitely put the values of x and y as I did and you should find out the values so this is how it is so now condition of solvability I have already discussed condition of solvability with you that when lines are parallel so unique solution you get in when intersection of line is possible infinite solution in case of coinciding and no solution in case of parallel lines but how will you decide that when lines are parallel when lines are coinciding and when lines are intersecting so for that I'm giving you conditions now and what I do over here is I take two equations a1x plus b1y is equal to c1 this is question 1 and a2x plus b2y is equal to c2 this is 2 so what I do over here is look at here so I can I can take corresponding for each for x the ratio of coefficients would be a1 by a2 for y ratio of coefficients is b1 by b2 and for constant coefficient is c1 by c2 now there are three possibilities if a1 by a2 is not equal to b1 by b2 then you you will get unique solution and why because if these are my two parallel lines and they are intersecting at each other it means that their slopes are not equal to each other if their slopes are not equal to each other and what are the what is the slope so in this line the in in first condition if you know the coordinate geometry the slope would be minus a1 by b1 and here the slope would be and this will not be equal to slope here minus a2 by b2 negative negative a1 by a2 should not be equal to b1 by b2 this is what the logic is so if two lines are intersecting each other the logic is that they should not be parallel to each other they should not be coinciding with each other in that particular case if the line is two lines are not coinciding or parallel to each other then what happens is that their slopes are not equal so it means that slopes not equal so a1 by a2 not equal to b1 by b2 in in in exactly opposite case when slopes are equal there are two possibilities so if slope a1 by a2 is equal to b1 by b2 it means that either the lines are coinciding like this l1 and l2 or the lines are parallel to each other so when the lines are parallel to each coinciding it means that c1 by c2 everything has to be corresponding so the second case is a1 by a2 equal to b1 by b2 equal to c1 by c2 which will give you infinite solution and the third case is a1 by a2 is equal to b1 by b2 and not equal to c1 by c2 what do what do i mean by c1 and c2 are y intercepts so if y intercepts are not equal it and slopes are equal it means that two lines are parallel to each other and if they are parallel to each other you will not get any solution so no solution here so these are the three conditions just write it down before i start solving the question on this this concept so i've taken a question on this concept so let me go to that topic so look at here what i need to do is that find the value of k for which system of equations 3x plus y is equal to 1 and 2k minus 1x plus k minus 1y is equal to 2k plus 1 so what do i need here i i need no solution here so in this case which which particular case a1 by a2 should not should be equal to b1 by b2 that should not be equal to c1 by c2 so i'll apply that in this question so a1 is equal to 3 a2 is equal to 2k minus 1 b1 is equal to 1 b2 is equal to k minus 1 c1 is equal to 1 c2 is equal to 2k plus 1 now a1 by a2 is equal to b1 by b2 that should not be equal to c1 by c2 if you don't write this you don't get the answer you don't get the marks so you need to first write this so what happens over here is let me write it replace the values 3 by 2k minus 1 is equal to 1 by k minus 1 and that is not equal to 1 divided by 2k plus 1 so if these two are equal let me equate it so let me cross multiply it 3 into k minus 1 1 into 2k minus 1 so this becomes 3k minus 3 is equal to 2k minus 1 so this gives me this goes here 3k minus 2k is equal to 3 minus 1 so k comes out to be equal to 3 minus 1 2 so what is this ratio then i need to find out this ratio so at k is equal to 2 this ratio is equal to 3 divided by 2 into 2 minus 1 and this is nothing but 1 divided by 2 into 2 minus 1 2 minus 1 so this is 3 4 minus 1 is 3 and that is equal to 1 by 1 so this is also equal to 1 by 1 and and what is this if i put 2 here so this is 2 into 2 plus 1 so which is 1 by 5 so now i can say that 1 by 1 which is 1 is not equal to 1 by 5 hence k is equal to 2 value i can accept it so this is how you have to solve the question now let me take another question of of this type i have not included it here but i would like to take that question here for better understanding of this concept so this is a cvse 2003 question which i am going to write now and in this question is x x plus k plus 1 y is equal to 5 and k plus 1 x plus 9 y is equal to 8k minus 1 and what do i want here i want infinite solutions so i write here for infinite solution a1 by a2 should be equal to b1 by b2 and that should be equal to c1 by c2 so a1 by b a2 is 1 by k plus 1 and a1 b1 by b2 is k plus 1 divided by 9 and c1 by c2 plus 5 divided by 8k minus 1 now i have three cases if i assume to be p this this complete thing this to be q and this to be r so or let me write it capital r now i have completely three cases so i will make p is equal to q so this is my first case so p is equal to q this gets multiplied here so k plus 1 square comes out to be 9 so k plus 1 comes out to be plus minus 3 so if i take plus 3 k comes out to be 2 if i take minus 3 k comes out to be minus 4 then i keep p is equal to r so in that case i get 8k minus 1 is equal to 5 k plus 5 so i get 3 k is equal to 6 and k is equal to 2 comes out now if i keep q is equal to r in that particular case i get this k plus 1 multiplied by 8 k minus 1 is equal to 9 into 5 45 so i get 8 k square plus 8 k minus k minus 1 is equal to 45 so what do i get here if you solve this equation this is 8 k square plus 7 k minus 46 if you solve this by minus b plus minus b square minus 4 ac divided by 2a you will find that it will come out to be k is equal to 2 and minus 23 by 8 now what i will choose only that value of k which is common to all three conditions so k is equal to 2 is common to all three conditions so as i write here as k is equal to 3 is common to all three cases hence k is equal to 2 is the answer so this is how we have to take questions and we have to solve in this particular chapter now let me move to another topic which is cross multiplication method now what is cross multiplication method i will just explain to you so first of all cross multiplication method a 1 x plus b 1 y is equal to c 1 and a 2 x plus b 2 y is equal to or or or write it plus c 1 is equal to 0 and plus c 2 is equal to 0 and i say that when a 1 by a 2 not equal to b 1 by b 2 what do i mean by this that i have unique solution so in this particular case x is given by b 1 c 2 minus b 2 c 1 divided by a 1 b 2 minus a 2 b 1 if you will solve this equation by elimination or substitution we will find out this value of x and y will come out to be c 1 a 2 minus c 2 a 1 divided by a 1 b 2 minus a 2 b 1 now you need to understand that c 1 and c 2 over here has been taken before this equal to sign if if somebody gives you equation like this a 1 x plus b 2 y is equal to c 1 and a 2 x plus b 2 y is equal to c 2 when i take this c 1 and c 2 on the other side to find out this formula this comes out to be a 1 x plus b 1 y is minus c 1 is equal to 0 and a 2 x plus b 2 y minus c 2 is equal to 0 so wherever c 1 and c 2 is coming over here that would be replaced by a negative sign over here so this c 1 will become negative this c 2 will become negative here and here c 1 will become negative and c 2 will become negative when when the c 1 and c 2 has been taken on the other side of the equal to sign it if it had been taken on this side of the equal to sign then this particular formula has to be taken as i have written on the white board but if this c 1 and c 2 is on the other side of the equal to sign then it has to be interchanged with a minus c 1 and minus c 2 in each case so for this equation complete thing this a 1 x plus b 1 y what i do is a 1 x plus b 1 y the better way plus c 1 is equal to 0 and a 2 x plus b 2 y plus c 2 is equal to 0 it can be written as x divided by b 1 c 2 minus c 2 b 1 so if x is there i am taking coefficient of y and constant if y is there i am taking so c here if y is there so i am starting with first equation so see here if x is there so i have taken b 1 and then i have multiplied here and then i have taken b 2 c 1 if y is there so i am taking first equation constant so c 1 and then i go to here c 1 a 2 minus c 2 a 1 and then i write 1 divided by a 1 b 2 minus a 2 b 1 so this is how you can write it or you can write like this so you can write x y and 1 and you can write this as b 1 b 2 c 1 c 2 a 1 a 2 b 1 b 2 this is how you have to remember so for x you multiply this multiplied by divided by this one a 1 b 2 for y what we need to do is that we need to multiply this divided by this so denominator is always right hand side sorry and numerator for x is just below it numerator of of y is just below it and we have to write b 1 b 2 c 1 c 2 and a 1 a 2 and again we have to write b 1 b 2 that's how we have to remember and this is for which equation for equation where this c 1 and c 2 is on the left hand side of zero if it is on right hand side of zero c 1 and c 2 has to be replaced here with a negative sign as simple as that so you need to check this very clearly when you are solving solving equations so what i'll do is now i'll take an example and the example in front of me is 2x plus 3 y is equal to 17 and 3x minus 2 y is equal to 6 so what you what we see here is in in the equation given over here x y z which i had written in front of you and i had written b 1 b 2 c 1 c 2 a 1 a 2 and b 1 b 2 the c 1 c 2 over here was positive in that case because c 1 c 2 was taken here here it would be negative so what i write over here is i will replace x y z so b 1 b 2 is 3 minus 2 c 1 c 2 is minus 17 minus 6 why i'm writing this minus because this is on the other side of the equal to sign and this is 2 3 and again this is 3 minus 2 now x is equal to this multiplied by 3 into minus 6 18 minus multiplication of this so minus 34 divided by you multiply this minus 4 minus 9 so this comes out to be minus 52 divided by minus 13 so x comes out to be 4 now y would be this so minus 17 minus 3 is minus 51 and minus minus is plus 8 divided by this remains minus 13 so this is minus what happens over here is this is minus minus plus 8 so minus 51 and this is minus 12 plus 12 so this comes out to be minus 39 divided by minus 3 so this comes out to be 4 so this is nothing minus 6 into minus 2 is plus 12 so let me write it as plus 12 and this comes out to be this comes out to be plus 12 so this is minus 39 divided by minus 3 is 13 now let me go to another topic which is word problem so this is i have taken two questions in the word problem what i do is i take the first question and then i'll go to the second question and the word problem is like this so the first question which i have taken is from time speed and distance and this tells me that that two places a and b are 160 kilometer apart on highway one car starts from a and another car so i'll make this these are the two places a and b and distance between them is 160 kilometer now car c1 starts from here and car c2 starts from b so if they travel in same direction so it means that what i need to do is suppose the speed of car 1 is x and the speed of car 2 is y so it travels for how many hours 8 hours so it travels for look at here 8 hours so distance traveled would be how much distance traveled would be by car a would be 8x so distance i'm writing the formula here distance is equal to speed into time and distance traveled by car b would be suppose the speed is y so 8 y now what happens is if if if they are going in the same direction if a is moving towards y and they are meeting here so distance traveled by b is 8 y and distance traveled by x is 160 plus 8 y so this 8x is equal to 160 plus distance traveled by b which is 8 y so 8x minus 8 y comes out to be 160 or x minus y comes out to be 20 this is my first equation now they say that when they travel in opposite direction so what happens is in this case 8x and plus 8 y distance traveled by a which is this is 8x and distance traveled by b which is this 8 y and total distance is 160 so 8x plus 8 y is 160 so x plus y no this is this will not be 8x this will be 2x basically because the time taken is 2 hours only so i'm i'm just removing this so 2x plus 2y so x plus y comes out to be equal to 80 so i write here x plus y is equal to 80 so 2x is equal to 100 x is equal to 50 and y gets how much so y would be here 80 minus 50 which is equal to 30 so that is how you have to solve these questions now let me move to the second question which is this question so this question say that a boat goes 16 16 kilometer upstream and 24 kilometer downstream in six hours also it covers 12 kilometers upstream and 36 kilometers down downstream in same time find the speed of the boat and in still water and that of stream so try to understand if speed of the boat vb in stand still water is suppose x and speed of the stream which is i say speed of the river is y so downstream means down the stream so if you are coming down speed would be added up so v downstream speed is equal to x plus y which is speed speed of boat in stand still water plus speed of river and upstream means going against the stream this is with the stream with the stream means speed of the stream gets added up going upstream means going against the stream so what happens speed of the stream does not get added up so it gets subtracted you are going against the stream so stream will oppose the navigation of the boat so hence x minus y vbs minus r please remember these things very properly now what happens is 16 kilometer upstream and 24 kilometer downstream in six hours so time taken here is t1 and here it is t2 so total time t is equal to t1 plus t2 and time is how much distance divided by stream so speed so 16 kilometer upstream so i'm writing d1 divided by upstream speed and this is d2 downstream speed and that comes out to be 16 so i'm writing here 16 divided by x minus y plus 24 divided by x plus y is equal to how many hours six hours so this will this will not be 16 this will be six hours similarly 12 kilometers upstream so 12 divided by x minus y plus 36 divided by x plus y is equal to same time so same six hours so what happens is now i'll assume i'm taking here 1 by x minus y is u and 1 by x plus y is equal to v so what i get over here is 16 u plus 24 v is equal to 6 and what i get over here is 12 u plus 36 v is equal to 6 now this can be cut down by 2 so because 2 is common so this gives me 8 u plus 12 v is equal to 3 and this gives me 6 u plus 18 v is equal to 3 now this can be cancelled out easily so i multiply this equation with 3 and this with 2 if i get multiply this equation with 3 so this gives me let me make some space for you this this this gives me multiply this with 3 and this with 2 so this gives me 24 24 u plus 36 v is equal to 9 and and this gives me 12 u plus 36 v is equal to 6 now this will give me how much 12 u is equal to 3 u is equal to 1 by 4 now if u is equal to 1 by 4 what will be value of v so 24 into 1 by 4 plus 36 v is equal to 9 so 36 v is equal to this this comes out to be 6 so 9 minus 6 is equal to 3 v comes out to be 3 by 36 which is equal to 1 by 12 so it means that u was 1 by x minus y so x minus y is equal to 4 and x plus y is equal to 12 from here x will come out to be come out to 8 and y will come out to be one second so x 2 x is equal to 16 so x is equal to 8 and y would be equal to 4 so x is it means that x i have assumed to be v v s so this is 8 kilometer per hour and this is 4 kilometer per hour so that is how you have to solve these kind of questions with this question i am done with two topics i am i made a few calculation mistakes while doing this questions so the basic problem in this topic is you will never find these questions around because there is not much concept involved the only way i am again repeating this which i repeated nearly five six slides back is that the only way you can get these questions wrong is once you have made calculation mistakes so anyway in maths question you will have few calculations errors here and there which you will identify once you identify it don't do like this that you are writing 2x plus 3y is equal to 5 and you find that this 5 is 7 so you do like this and then you write like 7 don't do this if you identify this that 2x plus 3y is equal to 5 and this is incorrect you just cut it like this right here 2x plus 3y is equal to 7 so this will tell me that specifically that this is not applicable to my question don't do like this these things doesn't look good on the maths paper just use this one line to strike down your question properly and once you write like this you have a better idea of what the question is and how what you have to do so try to avoid your calculation mistakes in this question in this topic this is true for all topics in maths but this topic is not at all difficult on conceptual basis you can only make mistake when you have i mean when you do calculation errors so please avoid calculation errors in this case now what happens is i'm taking a last question in this topic and this question is like this so i'm writing here the question is like this 90 percent and 97 percent pure acid solution are mixed to obtain 21 liter of 95 percent pure acid solution find the quantity of each type of acid to be mixed to form the mixture now it's like this so let the given solutions be labeled as p and q a and b anything you can assume and i'm saying that solution can be so what happens is x and y together the total volume has been given as 21 liter so suppose i take x liter of a and y liter of b so first equation over here see you should in this why this question doesn't look simple by reading it is that you have been given only one data 90 percent 97 percent you obtained 21 liter of 95 percent whatever is the concentration of acid if two different liquids have been mixed together to obtain 21 liter of final solution it tells me that addition of volume one and volume two is equal to 21 liter so that is why i'm taking volume and one over here as x and volume two over here as y and x plus y is equal to 21 liters now second equation would be about concentration of liquid so 90 in x 90 percent is acid so 90 into x by 100 and plus 97 percent in y so 97 into y by 100 and this gives me how much 95 percent of 21 so this is 95 by 100 into 21 so what i obtain over here is that this is 9 by 9 x by 10 plus 97 by 100 is equal to if i this becomes how much this becomes 19 this becomes 20 19 into 21 is 9 3 99 divided by 20 and i have x plus y is equal to 21 so i have two equation one equation here and second equation here now i need to solve it so how do i solve it generally what i do is i have already done a question like this that 9x so what i do is in this kind of question i try to eliminate the denominator the the first thing that i do is to eliminate the denominator and how do i eliminate the denominator over here is that 100 10 10 20 the lcm is 100 so this is 90 x plus 97 y is equal to 399 into 5 so this is nothing but 400 into 51995 divided by 100 so this 100 and 100 1 i get x plus y is equal to 21 and 90 x plus 97 y is equal to 195 so multiply this with 90 this gives you 90 x plus 90 y and 90 into 21 is 1890 so what do i get here is 181890 so i get 7 y is equal to 105 so y comes out to be 15 if y is 15 x comes out to be 6 so this is how you need to find the values so i hope you understood everything which i have tried to cover today in today's class i have tried to cover different kind of equations consistent inconsistent equations so consistent equation was where i had at least one solution of x and y so there were two cases possible that either i'll have one solution or i'll have infinite solution in inconsistent this is for consistent in for inconsistent no solution i have already explained that one solution means intersection of two lines infinite solution means coinciding of two lines and no solution means two lines parallel now i told you about graphical elimination substitution and for the for for this case one solution case only i told you about cross multiplication methods of solving the equations then what i did was i took cases for solvability or cases of solvability so in that case i had three cases a1 by a2 not equal to b1 by b2 for unique solution a1 by a2 equal to b1 by b2 equal to c1 by c2 for infinite solution and a1 by a2 equal to b1 by b2 not equal to c1 by c2 for no solution so these three cases i have taken for you in graphical method what i have suggested is one thing which i again suggest is that once you find out point of intersection you just try to put these values in the equation given and check whether that that intersection point which you are writing in the graph paper is correct or not one mistake that people make is once they are solving it with graphical method they don't write the answer in their copy j thus leave it in the graph paper only don't do so if graph papers are provided to you and you are solving it with graphical method it is mandatory for you to write the same thing in in the in the in the copy that this is suppose this is question number 14 so in your answer sheet you write question number 14 all all the xn corresponding x and y value you draw in your in in your notebook in under the heading question number 14 and then you write that graph has been drawn in the graph paper and then you see that as per the graph intersection point is this so hence x and solution of this this pair of equation would be x and y is equal to this and mention that as as per the graph also once you find this x and y value put it in the equation and find out that whether you are finding out right value of x and y or not because most of the time what happens is people find out x and y value but their graphs are not hundred percent accurate so they tend to get wrong answers so it's better that graph may be here and there a little bit one percent mistake here and there but once you solve the equation and find out the values that gives you a surety that what point to mark over here so in in in the rough sheet you should already solve it and find that what corresponding x and y value I get you mark the same point and then write that as per the graph this is your x and y value is so in in in graphical method just be a little bit cautious while writing the answer don't leave things only in the graph paper do write it in the answer answer books also in case of if it has been mentioned that you have to solve any pair of linear equation with elimination or substitution method do apply the same method don't apply your own method over there that I don't know elimination or elimination method is a bit tedious over here or a bit long over there so instead of elimination you are you are doing it with substitution method don't do that mistake if even if your answer is right you will not get the marks you will you will you will simply get zero marks over there and and in the third case cross multiplication do try to memorize what I have already given your cross multiplication a1 plus v1 a1x plus v1y plus c1 is equal to zero and a2x plus b2y plus c2 is equal to zero in this case you should write xyz and you start writing from here b1 b2 c1 c2 a1 a2 and b1 b2 once this c1 c2 goes here replace this c1 c2 by minus c1 c2 so if minus if I write that 2x plus 3y minus 3 and 5x minus 2y is equal to minus 1 here as it is coming after the equal to sign this c1 becomes minus minus 3 so this becomes plus 3 so which is actually the case that this is 2x plus 3y plus 3 is equal to zero and when it is on the left hand side of this equal to sign this c1 c2 is positive so have a clear cut identification of what c1 c2 values you are putting in the equation here because if this is not correct even though the answer I mean most of the people who are writing the equation would get this question right but just don't make mistake here time and again I am focusing on these values that you have to put c1 and c2 values correctly just look at which side of the equal to sign it is it and depending on that you put the c1 and c2 value accordingly so this is it for today's class tomorrow again for the crash course tomorrow you will have a chemistry crash course so I hope you understood everything out here and thank you so much for joining the session so thanks from Centrum Academy once again and wish you all the very best thank you