 Hello guys, so today we are going to start our another chapter which is again very crucial chapter This is linear equations in two variables So I just waiting for a couple of more minutes to For people to join in and then we'll start. Okay, so let's start. Okay, so we are going to discuss in this Session Linear equations so the agenda is to discuss linear equations and you know linear equations especially in two variables and after that what we'll try to do is we'll also see How to solve linear equations Solving linear equations, then we also talk about consistency consistency of solutions or Right, and then towards the end. We'll also see what problems and for autonomous syllabus Specifics will discuss towards the end Okay, so let us start so basically, you know the definition will quickly run through the definitions and all the basic Concepts and then we'll try to solve as many problems as as possible So a pair of linear equations in two variables x and y can be represented as a brightly as follows a1 x plus b1 y plus c1 equals to 0 and a2 x plus b2 y plus c2 equals to 0 Where a1 a2 b1 b2 and c1 c2 are real numbers such that Yeah You have to you have to see that a1 a1 and e2 both together cannot be 0 so a1 is not equal to 0 so basically do both together cannot be 0 a1 and and b1 Not equal to 0 Simultaneously cannot be possible right so hence all of them together cannot be zero hence this this particular this particular expression has been given Okay, so this is my definition of linear equations now next is graphically or geometrical a pair of linear equation in two variables represent a pair of straight lines Okay, so we know and let me Just draw a graph so hence how does linear equation? Look like in a graph, okay, so let's say this is x this is why Okay, so linear equation will be of this form where this is a1 x plus b1 y plus c1 equals to 0. This is not equal to this is plus here Okay, so a2 x plus b2 y plus c2 equals to 0. So another equation could be something like this so Right, so this this one is let's say this one is a2 x plus b2 y plus c2 equals to 0 Okay, and this point of intersection. This is the point of intersection Okay, this point here is the point of intersection, right? So hence we say that These two lines are intersecting. This is the first condition, you know And there will be questions around that you know this that those two lines are intersecting What is the criteria criteria is a1 upon a2? That means coefficient of x in the first equation divided by coefficient of x in the second equation must not be equal to coefficient of y in the in the in the first equation and Sorry, this is not here. So this is b1 by b2. So this is the condition for a Unique solution or one solution one pair of solution that is pair of solution means what you will get x and y One one set of x and y you will get right if they are parallel The other condition could be if they are parallel. Let's say Let's say this is parallel. Okay, so parallel lines So this is x this is y and again if this is equation number one Let's say this is a1 x plus b2 y plus c sorry b2 b1 y plus c1 equals to 0 and this line is a2 x plus b2 y Parallel by a2 Is equal to b1 by b2 Sorry for parallel line this must not be equal So a1 by a2 is equal to b1 by b2 is not equal to c1 by c2 Okay, so this is for parallel lines and for a coincidence. This is case number one This is case number two and case number three is this coincidence. So coincidence is this that means all the ratio All the ratio must be equal. So for coincidence or overlapping coincident lines Coincident lines or coincidence coincident lines the ratio should be all equal a1 by a2 is equal to b1 by b2 Is equal to c1 by c2 Right, this is our linear equation You know consistency criteria Okay, so Okay, now so a pair of linear equations in two variables can be solved by two methods What are two methods graphical method one is graphical method Another is algebraic method, right and in algebraic method. You also know there are three processes one is called substitution Then there is something called elimination And third is cross multiplication Right, so we'll see all these methods one by one, right? And uh, please be very very careful in linear equations If the question categorically asks you to use one particular method Then do not solve the question using some other method So hence read the question carefully Many times they would mention which particular method they are asking for You can use your other methods to check whether the solution which you have arrived at is correct or not Okay, so to solve a Pair of linear equation in two variables by graphical method We first draw the lines represented by them If the pair of lines intersect at a point then we say that the pair is consistent And the coordinates of the point Is the unique solution is the unique solution Now if the pair of lines are parallel then the pair has no solution and is called inconsistent pair of equation It's called inconsistent. So be Careful about these keywords. They will be used. They will be they will be using so consistent Then inconsistent and then if the pair of coincident lines, then it has infinitely many solutions So there are three things each point on the line being one solution In this case, we say that the pair of linear equations is consistent with infinitely many solutions. So, yes What all one is you have unique pair solution or unique solution This is then there are infinitely many solution Infinitely many and then third is no solution in these cases is called consistent the system is consistent consistent Consistent And in this case it is called inconsistent Okay, so you must remember these because in very last year also in 2018 there has been one question Where they had asked a question on consistency of the two equations So let's start Yeah, so You know, this is the algebraically. Let's see and understand how to solve. Let's say in this case if two equations are given So, let us say a x a 1 x plus b 1 y plus c 1 equals to 0 is given And then a 2 x plus b 2 y plus c 2 is equal to 0 given excuse me So, let us say let us see through Algebra geojibra Okay, so here is the Two equations. So let's say y equals to 3 x plus 2 here And y is equal to 2 x minus 5 these two equations are given. So what you need to do is You need to First of all plot the graph of first equation then plot the graph of second equation And wherever they are intersecting for example, this is the point here. So if you see This point this point is the point of intersection a minus 7 minus 19 is the solution to this pair of System of equations. So how do you plot? I will just briefly give you a Briefly give you a process. So let's say you had 2 x plus 3 y plus 1 equals 0 and you had 3 x minus 2 y plus 2 equals 0 then you convert or let's say transform them in such a way that it is in y equals to mx plus c form So, how do I write this? I can write this as 3 y equals negative 2 x negative 1 And y hence is equal to minus 2 upon 3 x minus 1 upon 3 correct Then what you need to do is you need to just simply need to you know Put random values of x usually we start from 0 1 2 and all that and some negative 1 and negative 2 and all that And then plot at least 2.2 points are good enough to plot a linear equation. So So let's say if you you know get 2 points 2 points x and y So let's say if you put x equals to 0 here, what will you get? You will get y equals negative 1 by 3 so put that so hence first ordered pair is 0 and minus 1 by 3 Okay Right and simply if you put x equals to 1 then y so when x equals to 1 then y is equal to minus 2 by 3 minus 1 by 3 is minus 1 Okay, so another point is 1 comma minus 1. So what next you need to simply need to plot x and y in a graph paper They will give you a graph paper and then what do you need to do plot 0 comma 1 point minus 1.3 I'm not doing it with scale you guys when you do it you have to do it using scale and graph paper So 0 comma minus 1 by 3 will be this is x equals to 0 and this is let's say minus 1 by 3 Okay, so 0 comma this is x equals to 0 and y minus 1 by 3 and The second point 1 comma minus 1 will be 1 and maybe this is minus 1 Let's say this is 1 and this is minus 1 to this point. So hence you have to draw it freehand And using a scale you should do so you will get this line Similarly in this case you will get You know positive slope value. So hence it is nothing but y equals here I'm writing in one step itself. So this is 3 by 2 x plus 1 Okay, so this line is somewhere like that Okay, so this is so hence You know this line and this was the line initially. So hence, whatever is the This is the point. So hence you have to draw perpendicular from here perpendicular from here. So basically you have to If you see this perpendicular has to be dropped from here. So this is x And this is y Right, so hence, whatever is the x and y coordinate here you'll have to Write that as a solution. So this is a graphical solution. Okay Fair enough Now let's go to the algebraic form. So there are three substitution elimination and cross multiplication method And I think we should be taking examples to Explain these in detail. So just let me take an example so that we can solve One one question so that all the three methods are All the three minutes are covered Okay, so let me take a question 2x minus y minus 3 so the question is So let us go down to our gap. So let me just explain it here Okay, so we are going to Take one question to explain All the three methods. So let's say we have to solve the question is solve Solve this pair what pair 2x minus y 2x minus y minus 3 equals 0 and 4x plus y minus 3 equals 0 So 2x minus y minus 3 equals 2 0 4x plus y minus 3 equals 2 0 and we'll see each method one by one Now clearly this is of the form of a 1x plus b 1 y plus c 1 equals 0 And the second one is a 2x plus b 2 y plus c 2 equals 0 And clearly a 1 is equal to 2 b 1 is equal to negative 1 and c 1 equals negative 3 Then a 2 equals 4 here b 1 and c 2 equals negative 3 again right now Let's us first take a substitution What is substitution? You have to first express one of the variables in the Form of other variables and you can use any one any one of the variables But but usually we try to avoid fractions So hence instead of expressing x in the form of y I would have gone for expressing y in terms of x because that would have eliminated any Appearance of fractions in the equation. So if you see It's always a good practice to keep numbering the equations and then mention the equation in terms of numbers So you can say from 1 from equation 1 you can say y equals Be very very careful with sign guys. So if you're writing y that means y is going to the right hand side here So y is equal to 2x minus 3 It's always a good practice to go a little slow here Uh, because if you miss any sign then the entire set entire equation is Going to be wrong. So hence y equals 2x minus 3 And now this is 3 This is equation number 3. So you'll write Using using 2 and 3 using 2 and 3 Okay using 2 and 3 what can you say you can say put y back into the second equation So 4x plus y now I can substitute y for this thing which we just found out So let us say let us say 2x minus 3 here minus 3 here equals to 0 and then it is 6x Minus 6 equals to 0 which becomes 6x equals to 6. So hence x equals 1 Right x equals to 1. So if x equals to 1 then clearly what is y? y I will use equation 3. So from 3 from 3 y equals 2 times 1 minus 3 Which is nothing but negative 1 Correct, let's check them. Let's check it whether it is correct or not. So equation was 2x minus y minus 3 equals 0 So if you put x equals to 1 You will get 2 here and then y is equal to minus 1. So minus minus 1 is plus 1 So this is 3 minus 3 which is 0 So hence this particular equation and second equation is also satisfied if we check 4 into 1 minus 1 minus 3 equals to 0 So both x equals to x equals to 1 And y is equal to minus 1 is the correct solution So we just check this is called substitution. What did we do? We express one of the variables in terms of the other and then Okay, so you use one in terms of the other and then Then substitute in the second equation and then Find the other variable. Okay Second is elimination How do you use elimination same set of equations? I'll use the question is 2x minus y minus 3 equals 0 And second one is 4x Plus y minus 3 equals to 0 This is equation number 1. This is equation number 2. So what do we do? We need to Eliminate so elimination as the name suggests you'll have to eliminate one of the variables Either you can eliminate x or you can eliminate y and how do you eliminate? We manipulate the coefficients of x and y so that the coefficient of x or y becomes same or with opposite sign Right and then either depending upon the sign of the Proficient will add or subtract. So I'll do both case in this case. So let us say I want to eliminate y so if you see if you simply add the two equation Adding two equation means add lhs to lhs add rhs to rhs. So if you simply add all of them So 2x plus 4x will give c 6x Minus y plus y will give you 0 And minus 3 minus 3 is minus 6 equals 0. So Very simply x comes out to be my 6 upon 6 which is 1 Okay Now what you can do is you can deploy x back into the any of any of these equations and find the value of 1 So for example, if I put x putting x equals to 1 Putting x equals to 1 in let's say 1 So it will become 2 into 1 minus y minus 3 equals 0 So ends y is clearly 1 see if you see this is matching with our previous solution Whatever we found out here if you see x equals to 1 and yeah So this is the solution we obtained in the previous case through substitution The same thing we are getting through elimination as well Right another way of let's say in this case we use substitution also But we can use direct elimination here also how so let us say I have to eliminate x So I have to eliminate x. So what I'll do is I'll multiply the first equation So let me write it once again. So 2x minus y minus 3 equals 0 And 4x plus y minus 3 equals 0 now I'm saying I have to eliminate x So how do you eliminate x first try to you know match the coefficient? Okay, match the coefficient. So if you see coefficient of x here is 2 and here is 4 So if I multiply this equation by 2 they will match Okay, so after so 2 into 1 will give me what? 4x minus 2 y minus 6 equals 0 Let us say this is equation number 3 And you write second equation once again 4x plus y minus 3 equals 0. So this is equation number 2 Now what you do you subtract the second equation from the first one. So hence this will become Negative this will become negative this will become positive and hence if you add now you will get 0 minus 3 y Is minus 3 equals 0. So from here y equals to negative 1 Yeah And again you put back y equals to minus 1 in any of the equations. Let's say putting y equals to putting putting y equals negative 1 In let's say equation number In equation number 1 What will you get 2x minus minus 1 minus 3 equals 0 So which is nothing but 2x minus 2 equals 0. So x equals 1. So this is the solution. So we saw both the Both the ways right. So what all did we see so far? We saw first is substitution. You know what to do Again, I'm telling you please be very careful if they have mentioned any particular method do not use any other method second is Elimination these two we learned and the third one which we are going to learn is the cross multiplication Yeah, so the third one is Third one is cross multiplication Now in cross multiplication, there are few things to be taken care of Yeah, in which form you are writing the equation is also important because If you miss the you know one particular form has one particular format of form as in, you know The pattern in which you have to remember the formula Okay So I prefer that Formulate where all the All the terms are on one side Okay, what do I mean? I mean, let us take the same two equations. So the two equations are again if you see 2x minus y minus 3 equals 2 0. So this is one is One is 2 2x minus y minus 3 equals 0 and the other one was 4x plus y minus 3 equals 2 0 now here if you see There is nothing on the right hand side only 0 right another way of writing this could be 2x minus y equals 3 and 4x plus y equals 3 so make sure That you follow one because if you have this particular format ready Then you have another Yeah, just a minute guys There's some issue yes, so Yeah, so what I was saying is you have to have one particular format Yeah, so let's say if you are if you're using this format this left hand side Okay, then you have to have one one type of formula if you are having this particular type Format then another type of formula. So hence I generally use this one And hence in this case what will be the formula to be used so I write it like this So make sure that all of them are In the same side and only zero is there on the Okay, so hence the formula goes like this x upon You have to write b1 b2 and then you have to write c1 c2 And then this equals to y upon c1 c2 and then a1 a2 and this is equal to 1 upon a1 a2 and then b1 b2 And then you have to multiply top left to bottom right first top left bottom right always Top left bottom right always first Okay, and then you can write this as x upon Top left bottom right right so it will go in this direction top left to bottom right you have to multiply So you will get b1 c2 minus b2 c1 And this is equal to y again same thing top left bottom right so c1 a2 minus c2 a1 And this is equal to 1 upon a1 b2 minus a2 b1 Okay, so hence here what will happen this implies x will be equal to simply If you equate this part with this part All three are equal so I can say x equals to b1 c2 minus b2 c1 divided by a1 b2 minus a2 b1 And similarly y is equal to y is equal to you have to now equate these two this this to this Right, so hence you will get y is equal to c1 a2 minus c2 a1 upon a1 b2 minus a2 b1 Right, so let us deploy the formula and see whether we are getting the correct answer or not So let us say what was p1 and all that so let us Let us first find out a1 b1 and all that okay. Let us read it again So the two equations were the two equations were this So clearly our a1 is equal to two And b1 equals to minus one and c1 equals to negative three a2 similarly equals to four b2 is equal to one and c2 equals to negative three Right, so let us deploy these values here x x equals to now v1 c2 so negative one into c2 Which is negative three then minus b2 which is one into c1 which is negative three divided by a1 b2 so two into one so two into one Minus a2 b1 so a2 is four into b1 is negative one So hence if you see this is nothing but three plus three divided by two minus Two plus four which is nothing but one And in this case c1 c1 c1 a2 so c1 is negative three and a2 was four so this into four minus c2 is negative three into a1 which is two divided by same thing two into one minus four into negative one And if you see this comes out to be minus 12 Plus six divided by six which is nothing but negative one So if you see in all the three cases we got the right answer Is it x equals to one and y equals to negative one in all the three cases what case is in substitution in elimination and third is cross multiplication Okay, so now we will come back to our This thing solution of the other solutions, sorry the other important topics So again, we are coming back to the consistent consistency part of it So if you see there are two equations a1 x and Plus b1 y plus c1 equals to zero And a2 x plus b2 y plus c2 equals to zero our standard form And then we say that if a1 by a2 is not equal to b1 by b2 Then the pair of linear equation is consistent. Please remember Inequality of the first ratio first two ratio is consistency. That means that you will get one pair of solution here Here in in the second case a pair of linear equation is inconsistent when a1 by a2 is equal to b1 by b2 But it is not equal to c1 by c2. Okay next and the third is a1 by a2 a1 by a2 is equal to b1 by b2 is equal to c1 by c2 in this case These are these lines are parallel lines. So you can remember like that So Chris cross that is intersecting line means not equal to write and then parallel line is the third the second and the third ratios should not be equal and coincident line are When all the ratios are equal Okay, so we'll be seeing this these applications a little later. Yeah Now here is the question now. Now. Let us start solving questions So the question is represent the following pair of equations graphically and write the coordinates of points Where the lines intersect y axis? Okay, so we will use the software here Usually you need you know, you know how to do it if you have a graph paper So let me just take this graph paper. Okay So let us say this is like a graph paper field now. Let us first write down the two to x plus 3 y is equal to 6 and 2 x minus 3 y is equal to 12 x plus 3 y is equal to 6. So hence I have x plus and the other one is 2 x minus 3 y equals to 12 So this is 2 x These are the two equations I have to solve graphically. So if you see there is a kind of graph paper already here So we'll use that and that's how you have to do there Also, I'm not using a scale because I'm using a tab. So let us say, uh, first of all We draw a graph paper or let's say x and y Right, so my drawing will be a little rough. So this is x This is y. Okay, and this is negative x You don't do full negative x and negative y as much as The question demands now. What is it? So x plus 3 y is equal to 6 So x plus 3 y is equal to 6. This is the first right, so hence Let us first Let us first plot the first equation. So hence if you see the first equation is x is equal to 6 minus 3 y This is one. So let us plot or let us find out the values of x and y 6 minus 3 y right now. So hence let's say let us say this is x and this is y. So when y equals to 0 You should take one of them as 0 then it becomes easier to find out when y equals to 0 x equals to 6 And when x equals to 0 You will see if you put x equals to 0 you will get y is equal to 2 Right, so when y is equal to 0 x equals to 6 And when x equals to 0 y is equal to 6 by 3 that is 2. So now you can easily plot this So first point is 6 comma 0. So 1 2 3 4 5 6. This is 6 Comma 0. So let me use this color. So this is my first point 6 comma 0 you write it 6 comma 0 and then 0 comma 2 will be somewhere here 0 comma Okay, then what you need to do Right, so hence what you need to do is you need to just free with a scale We join these So this is the first equation now second equation Second equation is 2 x minus 3 y equals to 12. So yes, again you can write From here I am writing now here. So it can be written as x equals 12 plus 3 y 12 plus 3 y by 2 12 plus 3 y by 2 So which is nothing but 6 plus y by 2 No, 6 plus sorry 6 plus 3 y by 2. So hence again you can plot this we'll say x and y So when y is 0 When y is 0 x is again 6 And y is 0 x is again 6 And when x is 0. So y will be nothing but negative 12 by 3 So here you can you can say when when x equals to 0 When x equals to 0 y will be negative 4 Let us plot it. So 6 comma 0 is the same point and 0 comma minus 4 So this is 0 comma 1 1 2 3 4 so this is And hence the line is something like that So clearly the point of intersection is 6 comma 0 because 6 comma 0 satisfies 4 Let's check whether it is actually true. So if you put 6 comma 0 here It is satisfying the first one and if you put 6 comma 0 2 into Yeah, second or it is satisfying. So then 6 comma 0 is Okay, so this is how graphically you have to solve this Now now comes this kind of how we are solving Linear equations now if you see this is not a linear equation here This is not a linear equation not a linear equation Why because the power of x is negative 1 and power of y is also negative 1 both of them But it can easily be transformed into linear equation how by Assuming or substitution let us say we will say let us say u is equal to 1 upon sorry Yeah, 1 upon x and v is equal to 1 upon y Now the moment you say that the equations are transformed as 2 u plus 2 by 3 v equals 1 by 6 So simply multiply the entire equation by 6 simplify it you will get 12 u plus 4 v is equal to 1 so this can be equation number 1 Secondly in the second equation you will get 3 u plus 2 v equals 0 3 u plus 2 v equals 0 and then this is equation number 2 Now what can you do you can eliminate now let us multiply this equation by 4 So you can 12 u plus 8 v is equal to 0. This is equation number 3 Now what do you do is 1 or Yeah, so you can do 1 minus 3 If you do that You will be eliminated and 4 v minus 8 v is negative 4 v Is equal to 1 minus 0 which is 1 so v is equal to negative 1 upon 4 Okay, and then Another if you put v equal to this thing will get v equal to minus 1 by 4 if you put you will get u But the other way could be you multiply the Um the second one The second one the second equation equation number 2. So I'm multiplying the equation number 2 By 2 so hence you will get 2 into 3 u Plus 2 v equals 0 And this will get you'll get you you will get 6 u plus 4 v equals 0. This is equation number 4 Right now what you can do is 2 1 minus 4 So this will give you the view minus 6 u is 6 u And 4 v and 4 v gets eliminated and then you will get 1 minus 0 as 1. So u is equal to 1 upon 6 Right. So when v is equal to 1 minus 1 by 4 you can write. What was v? v was 1 upon y So y is equal to 1 upon v, which is nothing but negative 4 You can say x equals to 6 You were one upon x so x equals to 6. Let's check whether this is correct or not So first one equation number 1 would be nothing but 2 upon 6 on the LHS Plus 2 by 3 into negative 4 Which is equal to 2 upon 6 and then this will be Uh Yeah, this is not this is somewhere. We made a mistake v equals to y equals to negative 4 2 upon x So we did some mistake somewhere. Maybe hence it is not coming 2 upon x So 2 upon 6 is 1 by 3 and then So it is correct. Correct. No worries. So 2 upon 6 is this is 2 upon 6 and this is nothing but negative 2 upon 12 So which is nothing but nothing but 4 minus 2 by 12. Am I right? 4 minus 2 Ah, so this will be 2 by 12 2 by 12 So I'm doing once again. Sorry for this confusion. So it is nothing but 4 minus 2 by 12 which is 2 by 12 Which is equal to 1 by 6. So which is correct Okay, so the moment you get x So this is this is correct. You can check the second equation also. So I am writing here second equation is 3 upon 6 Minus 2 upon 4 which is nothing but 0 points Solution is correct. So y is equal to minus 4 is correct and x equals to 6 is correct Okay, so this was the first question Let's go to by the way. This was a board paper question Again similar looking question. So here only you need to substitute one y is already formed if you see Here 4 x has to be changed. So then you can assume u to be equal to 1 upon x and then rewrite the equation So it will be 4 u plus 5 y equals 7 And 3 u plus 4 y equals 5 And then again do the repeat of the process you already know You multiply now. How do you so I will go for elimination again substitution would be the last thing I would do So what I'll do is I'll multiply 1 by multiply 1 this entire 1 by 3 and the and the Second equation number 2 by 4 y then my coefficient of u will be same So let us see. So this is 12 u Plus 15 y is it it 15 y equals 21 So multiply by 3 12 y 12 u plus 15 y plus 21 and then second is 12 u plus 16 y 12 u Plus 16 y is equal to 20. So this is equation number 3 And equation number 4 So 3 minus 4 3 minus 4 So what will happen 12 u 12 u? gets cancelled 15 y minus 16 y is negative y And this is equal to 21 minus 20 is 1. So y equals negative 1 Okay, so the moment you put y equals to 1 in any of these So let's say 4 u plus 5 into negative 1 is equal to 7 That means 4 u is equal to 7 minus minus 5 that is 12. So u is equal to 3 That means x equals 1 upon 3 Let's check Why because x equals to 1 upon u? Let's check equation number 1. We'll check 1 is 4 upon x. So 4 upon 1 by 3 Plus 5 y so 5 into negative 1 is equal to 12 minus 5 which is equal to 7 so satisfies an equation number 2 is 3 upon x That is 1 upon 3 plus 4 into negative 5 which is equal to 9 And i'm sorry y is minus 1. So this is minus 1 So hence it is 9 minus 4 which is 5. So this is also true Both are correct. So hence our solution was correct Okay next Okay, so Again the question happens is the moment you see Only variables. There's no numeral here. Then many people actually start panicking. You should not do that Yeah, so if you check there is no numeral here and many people are a worse of these kinds of questions But do not afraid. You know not panic not Do not You know get panicked with such with such questions And you know already how to solve but be very very careful while we are you are handling variables and don't mess up with the signs So let us see how do we do it and we know elimination is one of the uh, you know favorite mechanisms So we'll be using that method only So the equation number one. This is equation number one and equation number two. It always helps if you want you can Uh, simplify this. So hence, let's say P upon ax plus a upon by equals a square plus b square So let us multiply the entire equation by ab. Why am I doing this? Because if you do that then a and b gets eliminated In in both the terms in the first two terms, right? So you multiply ab Alternatively, you can take lcm of the first two terms denominator and then simplify and go ahead like that But here is how we will be doing it in this way So multiplying the entire equation by ab you will get b square x because a will get cancelled and it is a square y and this is equal to a q b plus a b q Okay, so hence, uh, you get this b square x plus a square y Is equal to a cube Okay, so so a q Uh, so b square x plus a square y equals a cube b Plus a b q and the second equation is x plus y is equal to ab So what I can do is I'll adopt elimination. So hence to equate the Proficient of x I will multiply with b square both the entire equation I will get b square x plus b square y is equal to a b q This is my equation number three and this is my equation number four So what I'm going to do is I will do three Minus four y because x will get eliminated simply so b square x minus b square x is zero And hence it will give you a square y minus b square y on the lhs And on the right hand side a b q and a b q gets eliminated you will get a q b okay So if you see what is it this what what is it now? So y is equal to y times a square minus b square is equal to a q b So y equals a q b by a q square minus b square right now to get To get y x do not substitute it rather again element you use elimination. So how will you eliminate? So now you multiply um The second equation by by a square right so you will do a square times x plus y equals a b So hence you will get a square x plus a square y is equal to a q b This is equation number five Okay, then how will you eliminate this? You will use three minus five So do three Minus five. So what will you get clearly a square y a square y is getting cancelled So hence you will get b square minus a square x Is equal to an a q b a q will go. So it is a b q So hence x equals nothing but a b q upon b square minus Okay So these are the two solution but other two solutions one is Y is equal to this so If you have time left towards the end of the paper Let's say once you are done with the paper your your time is left Then usually it is a good practice to put a star mark so that when you come back you please deploy these back into equations And c and especially if you see it is already satisfying the second equation. So most tend to give you the solution is correct The solution is Correct. So hence, but yes, nevertheless if you get time again check your calculations Next Second is now we see our question is on the consistency part, right? So determine the value of k these type of questions are there so that following linear equations have no solution So no solution means a one by a two is equal to b one by b two Should not be equal to c one by c two This is the criteria for no solution now So if you see you're the first What is a one here? Let us jot down A one is 3 k plus one right a two is equal to k square plus one right and b one Is equal to three b two is equal to k minus two c one is equal to negative two and c two is equal to negative five these are the Values just check once yes looks like it's correct Very good. So hence, what do we know? We know a one by a two so three k plus one by k square plus one a square plus one Is equal to three k plus uh b one by b two that means three upon k minus two And this must not be equal to minus two by negative five Right, right. So let us solve this part first So if you see cross multiply you'll get three k plus one into k minus two Is equal to three times k square plus one Which is equal to three k square minus six k plus k minus two Is equal to three k square plus three Okay, so three k square three k square gets cancelled So if you see minus five k Is equal to if you take it if you take a minus five k here and then minus two goes on to the Right hand side You will get Five right so k is equal to negative one Okay, k is equal to negative one is the solution Right, so if you put k equals to one negative one then anyways Yeah, so if you put k equals to one negative one then the system doesn't have any solution Okay next Now these are the word problems which are there So the first word problem is the sum of a two digit number and the numbers obtained number obtained by reversing the order Of its digit is 165 if the digit differs by three find the number. So let us say you will say what let us say the digits are x and y Okay Right, so the digits differ by three. So let us say Mod of x minus y will be equal to three Why am I using mod of x minus five because I don't know whether x is more or y is more Okay, so mod of x minus so hence you please keep in your mind that either x minus y is equal to three Or y minus x is equal to three both can be possible And now let us say if my if my number was x y x y then in in actuality the number is nothing but x y is just the position It's not a multiplication x y so the the number will be nothing but 10 x plus y and then they are saying Reverse this so the reverse number will be yx which will be nothing but 10 y plus x So if you add that both of them that is 11 x plus 11 y And the sum is they are saying the sum of two digit number and the number obtained by reversing the order is 165 So you will get x plus y is equal to 15 Correct x plus y is equal to 15 now x plus y is 15 and let us say x minus y was three So hence add both of them together you will get two x equals to 80 So x becomes six Correct, so hence the numbers are 96 and 36 if you add both of them So 96 and sorry not 36 69 96 and 69 so 96 plus 69 165 is the sum this is how you will solve this equation Okay Right, so what is the learning? Please remember number any two digit number is can be expressed as 10 x plus 10 x plus y right Next So a two digit number is such that the product of its digit is 20. So let us again start with x and y so let's say x and y are the digits So the product of its digit is 20. So x y is 20 If nine is added to the number The number would be nothing but what is the number 10 x plus y is the number Right if nine is added to it. So nine is added to this number The digits interchange their places that is this becomes 10 y plus x correct Why because if x y was the number not the product the position y is x y. So hence it is 10 x plus y And if it is y x then it is nothing but 10 y plus x That's what they're saying. So number plus nine The number the digits interchange their places. So 10 y plus x. So hence, what will you get? You will get Um nine x minus nine y plus nine is equal to zero Or y minus x equals to one if you see y minus x y minus x equals to one right this is one And the other equation is x y is equal to 20. So hence if you see this is our question which involves Not only lean soluble linear equation, but also quadratic equation. How let us see So hence I can say y is equal to x plus one from here And then use equation number one and equation number two To solve this so hence x times now y is x plus one And this is equal to 20 so it is x square Plus x minus 20 equals zero. So if you see this is we have got our quadratic equation So how to solve this split the middle term x square? It will be clearly five x minus four x minus 20 equals zero So hence if you see it is x times x plus five Minus four times x plus five equals zero So hence it becomes x plus five times x minus four equals zero and hence either x equals to minus five or x equals four But my dear friends x equals to minus five is not a feasible solution Not a feasible solution Or not a possible solution why because digits cannot be negative So what is the correct solution x equals to four? So if x equals to four, then you know y equals to five So let us check So x equals to four and hence y is equal to five because Four into five is 20 so hence the numbers are 45 and So hence if you add nine into it you'll get 54 the numbers the digits reverse you order Okay, so hence our solution is correct Okay, so numerator denominator type the sum of the numerator and the denominator of a fraction is 12 So let us say the let the fraction be So this is very typical Typical problem typical linear equation problem of cvc board So let the fraction be Let the fraction be x upon y Correct, and they're saying the sum of the numerator and denominator of a fraction is 12. That means x plus y equals 12. This is equation number one If the denominator is increased by three Denominator is increased by three. That means y becomes y plus three The fraction becomes one by two. So hence x upon y plus three equals One upon two so you have to find out x and y So let's cross multiply you'll get two x equals y plus three Right and hence you simplify it you will get two x minus y is equal to three. This is equation number two This is equation number two then what remains to do is add one and two One plus two. Why am I adding because if I can see there is a plus one as a coefficient of y negative one as a coefficient of y in the second equation Adding both you'll get three x And this is equal to 15. So x whatever five so x is Five when x is five Then you know y is seven Right. So if you see Five by seven plus three is actually one by two five by ten So hence our solution is correct Okay Great next Okay, now distance speed time type Abdul traveled 300 kilometers by train And 200 kilometers by taxi It took him five hours 30 minutes But if he travels 260 kilometer by train and 240 kilometer by taxi he takes six minutes longer Find the speed of the train and that of the taxi. So whatever is required you take assume that to be the variable So let's say speed of the train speed of the train be you Okay, and speed of speed of Taxi be V Okay speed of the train be you speed of the taxi be v now Abdul traveled 300 kilometers by train and 200 kilometers. So time taken right So time taken In case a so what is case a? Time taken is nothing, but you can write the formula first that is nothing but distance by If the velocity is constant, you can write distance by speed This is time taken. So hence let's first time taken first Let's take the first case right and then please also mention This is kilometers per hour right so that there is no confusion in units We look v kilometers per hour now time taken is distance upon speed So 300 kilometers by train that means time taken will be 300 upon u And then 200 by kilometer by taxi. So 200 upon v isn't it It took him five hours and 30 minutes that means it is 5.5 hours Right, but he travels 260 kilometers by train So 260 by u plus 240 by v Okay This will they are saying six minutes longer. So whatever the initial this thing was 5.5 but six minute is in minutes. So we'll have to convert it into hour, which is nothing but six upon 60 hours Please be very careful. So many people will just add six here and they will not be able to solve it. So be very very careful Now so how do we solve this so six by 60 is one upon 10 so point one This is nothing, but if you see this is nothing but 5.6 isn't it 5.6 now let us solve the um So hence if you see let us assume one by u is equal to x And one by v is equal to y. So the equations Which are left is nothing but 300 x plus 200 y equals 5.5 And 260 x plus 240 y is equal to 5.6 Okay, so now What you can do is you can simplify and then use substitution also So if you subtract these two equations together You will get very simple Relations to subtract these two so you'll get 40 x minus 40 y is equal to negative 0.1 Okay, so if you see this is nothing but x or you can you can write y minus x is equal to 0.1 by 40 I reverse the order of x and y So that I can eliminate negative sign from the right hand side, right? So hence you will get y minus x is equal to 1 by 400 Okay, now what you can do is y is equal to x plus Right, so I have to just now solve this I'm short of space here. So I'll go and Now solve that in the note. So just let me just rewrite the equation 300 by u 200 by v Right, so the equations are what are the equations? So the equations are 300 upon u plus 200 upon v is equal to 5.5 And then we had 260 upon u plus 240 upon v Is equal to 5.6 Okay, and then we transform these equations to 300 u plus 200 v sorry 300 x That's 200 y equals 5.5 and the second equation was 260 x plus 240 y equals to 5.6 and then Let's say this is one and this is two So do two minus one and you will get 40 y minus 40 x is equal to is equal to Is equal to 0.1 and hence I can write y minus x is equal to 0.1 by 40 Which is nothing but 1 upon 400 Right y minus x. So y y can be written as y can be written as x plus 1 by 400 Okay, and then now you can use the first equation. So using the first equation. What can I say? 300 times x plus 200 times x plus 1 by 400 Is equal to 5.5 Right substitution. I'm using so 300 x plus 200 x Plus 200 by 400 Which is half is equal to 5.5. So hence I will get 500 x Is equal to 5.5 Minus 0.5 because 200 by 400 is 0.5. So which is equal to 5 So hence x is equal to 5 upon 500 Okay, so x is 5 upon 500. So which is nothing but 1 upon 100 So what is then what was u? So u is 1 upon x Is equal to 100 kilometers per hour 100 kilometers per hour. So now when x is this much So what is y? y is equal to 1 by 100 plus 1 by 400 Because y is x plus 1 plus 400. So hence you will get 4 plus 1 by 400 Isn't it? So which is nothing but 5 by 400 which is nothing but 1 upon 80 So what is v? v is 1 upon y which is equal to 80 kilometers per hour So these are the two Speeds one is 100 kilometers per hour. Another is So you can check if it is correct. So hence 300 first equation will check 300 by 100 is 3 and 200 by So you'll check in this equation 300 by 100 is 3 and 200 by 80 is 2.5. So 3 plus 2.5 is 5.5 Looks like correct Okay So This is the This is again a bold question Next is Find the value of a for which the pair of linear equation this and this has infinitely many solutions So you know this condition again for infinitely many solutions and a1 by a2 is equal to b1 by b2 Is equal to c1 by c2 always remember for infinitely all equal Infinitely solutions all equal all ratios must be equal now What is a1 by a2? So first of all write the equations very very carefully So you can write 2x plus 3 y and hence it is equal to 7 So it is better idea to write all of them together All right, and the second one is 4x plus ay 4x plus ay Minus 14 equals to 0 So that means a1 by a2 2 by 4 must be equal to 3 by a must be equal to minus 7 by minus 14 Right, that means 1 by 2 must be equal to 3 by a And hence a is equal to 6 Right, so this is not Sorry So actually if you see this particular question was asked in cvse 2018 so this is cvse 2018 problem last year this question was asked in the board exam Okay Next yeah, so this is the question the question is the tense digit of a number is twice its units digit And the number obtained by interchanging the digits is 36 less than the original number find the original number for this is also cvse 2018 same basic to the one equation Okay, so the tense digit of a number is twice its units. This is so let's let the digital or you should write let the tense digit D x and let the units digit dy The number obtained by interchanging the digits is 36 less than the original number. So original number is original number Is nothing but what is original number? 10 x plus y and new number after Interchanging the digits Interchanging the digits will be 10 y plus x Now, what are they saying that the the number obtained by interchanging 36 less so the new number is 36 less that means The original number was 10 x plus y and if you subtract 36 from it you will get 10 y plus x Okay, and another information which was given in the very first line was the tense digit of a number is twice its units digit So hence, I know x is equal to y Isn't it? Tense digit of a number is twice its units digits. So x equals to 2 y and now simplifying this equation you will get What will you get 9 x? minus 9 y is equal to 36 And now put x equals to 2 y so you'll get 9 into 2 y minus 9 y Is equal to 36 So 18 y is equal to 36 So y is equal to 2 And x hence equals to 4 right, so Am I right? 42 and 24 is not No, something I'm sorry 4 this will be 4 this is 9 x minus 9 y is 36 so x is 2 y so 18 y oh, this is 9 y and sorry So hence this is 4 and hence it is 8 Right, so the number is 84 because If you reverse the number it will get 48 and the difference is 36 Yes So hence what is the original number original number is so don't write these things in your paper because this is just to For me to check so you can check it in your rough rough area so x is or original number Number is equal to 18 4 Okay So these are this is also cbc 2018 so you now know the you know the questions of linear equation will not be of That you know the questions will not be such that you will not be knowing it The problem general problem in linear equations You need to take care is But what should you take care of one is? Linear equation you must be very very careful in terms of calculation mistakes Most of you would be losing marks because of this Right Second this thing is re-check The solution Please re-check the solution Okay, so you can buy by deployment by deployment of the actual value Please check it. Okay third is remember the three criterion remember criteria For consistency Consistency means whether the solution is there or not remember the three can three cases So one a1 by a2 is not equal to b1 by v2 This is for one solution Right, then a1 by a2 is equal to b1 by v2 is not equal to Is not equal to is not equal to c1 by c2 This is parallel line or inconsistent and third is a1 by a2 is equal to b1 by v2 c1 by c2 infinitely many solution many This is also called no solution So, please remember These three do not get confused and do not use wrong condition then solve the question by the desired method or the or the method Mentioned These are the common errors Do not just jump on solving this equation Just after looking at the question Okay, so you please read it read the question carefully Okay, and then very importantly spare some time to go through the solution and check whether the obtained solution satisfy the given set of Okay, so I think this is good enough for you for you know for the entire summary of the chapter So I think you'll be able to solve it and the questions will be within reach Do not The only thing would be that you will be making mistakes while calculation and so do not do silly errors Please be mindful of this and I think you should be able to solve linear equations are very um, you know very low hanging fruits in your exam paper So guys if you have any other problem, you can post it here. Otherwise, I think you are good to go and So we will meeting again with the another set of Let's say another topic Okay If you have any other question, you can please post it here. Otherwise, you can always reach out to me separately So if you don't have any other queries, then we'll call it a day and then we will take up New topic tomorrow Anyone is having any problem? So thanks a lot guys. See you then tomorrow. Thank you