 Okay friends, so today we will be first describing one of the methods of solving a pair of linear equations So we are going to deal with substitution method, so we are going to talk about substitution substitution method to solve a pair of near equations Linear equations and that doing two variables Okay, so we will first take an example and then understand how a system of linear equations Is solved now, let us say we have a pair of linear equations. Let's say 2x plus 3y Minus 5 equals 0, okay, and another one maybe let's say 3x minus 2y Plus 2 equals 0 So this is equation number 1 And this is equation number 2 So what we are going to do is we are going to use substitution method as the name suggests Substitution method to solve these equations how to do that So first step in that is always try to express one of the variables in terms of the other What does it mean? Let me show you so let us take I am picking up this particular This particular equation now. This is how do I express this? I am trying to express y in terms of x so can I not write this as 3y equals Minus 2x plus 5y because I took this 2x onto the right-hand side and This 5 minus 5 also to the right-hand side. So the signs will change, okay Now so hence similarly I can write y equals minus 2 upon 3x plus 5 upon 3 and why is that I divided this equation by 3 divide by 3 Okay, so that I get this expression this Equation sorry now What is to be done next now you use equation number 1 to arrive at this right? So now what you do is as the name suggests sub substitute y But not in the first equation, but into the second equation That will reduce the second equation in terms of only one variable. What do I mean? I'll just see so let us pick up the second equation. So this is 3x Minus 2 now there is y so instead of that y I will use the y which I just Figured out from here and hence I can write that as minus 2 upon 3x plus 5 by 3 This is what? The y was which was derived from the first equation. So and that much that is equal to sorry There is another minus 2 here if you see, I'm sorry. This plus 2 is there. So hence I have to write Plus 2 and this equals 0. So what did I basically do basically? I? Substituted this y with this value Okay, now you'll ask why don't you just put it in the first equation itself Now if you can try it out if you do it you will get 1 equals to 1 or 0 equals to 0 something like that, right? So it will not serve any purpose So I substituted the y which I obtained in equation number 3 Into equation number 2. Okay. Now, let's say if you would have started with equation number 2 then whatever Expression you would have got for why you'd have substituted that in 1. Okay, so now Solving it you'll get 3x minus and then minus into minus minus 2 into minus 2 is plus 4 upon 3x and Then this minus 2 into 5 by 3 will give me What would it give me minus 10 upon 3 plus 2 equals? 0 so if you see now this equation is having only one variable and that is x Right, and you know how to solve a linear equation in one variable. So now let us simplify further So if you see x I can take common from the first two terms. It becomes 3 plus 4 by 3 and The constant term I can transfer onto the right-hand side. It becomes 10 by 3 and then minus 2 Isn't it now if you see Let's simplify further. So it is x x and then if you see this is 9 plus 4 upon 3 and This is equal to 10 minus 6 upon 3 taking the LCM. Now if you simplify further, this is nothing but 13 x upon 3 Equals 4 by 3. So this means x equals 4 upon 13 Okay, so this is what I want to do calculate now you can use this value of x and use equation number 3 using Using equation number 3 using equation number 3 This three equation number 3 What do we get? So what is this? This is equation number 3. So let me solve this here Okay, so now what do I get using equation number 3? I can put this value of x there So y is equal to minus 2 upon 3 into what see x was 4 upon 13. So let's put 4 upon 13 over there plus 5 by 3 Correct, this will give me the value of y. So hence y is nothing but let's simplify So it's you know if you take minus 8 by 39 Plus 5 upon 3 which can be further written as this is 39 as LCM and this is minus 8 and this is 13 into 5 that is 65 which will give you 57 upon 39 which can be further reduced to so there is a 3 as 3 is a factor for both. So this is 19 upon 13 19 upon 13 so we get x equals 4 upon 13 and y equals 19 upon 13 let us say whether we have got it correct or not So let us deploy in any of these equations and see whether the solution is correct or not So let us take the first equation 2x plus 3y minus 5 equals to 0 so 2x Plus 3y minus 5 equals to 0. So what are we doing now? We are checking We are checking whether we have got the correct solution. So let us put put the values of x and y so x was 4 by 30 and y was 19 by 13 and This is minus 5. So what is the value of the LHS? So we'll start with LHS if you see this is 13 and Then this is 8 and this is 19 times 3 is 57 and minus 5 so If you see this is 65 by 13 minus 5 which is 5 minus 5 which is equal to 0. So hence which is equal to LHS that means Whatever solution we obtain is correct So it is it is a nice idea always to check whether the solution which you have obtained is correct or not So what did we learn? We learned that in the method of substitution We have to express first take one of the equations and express One of the variables in terms of others here. I have expressed y in terms of x But it's not necessary you can also express x in terms of y and then substitute accordingly into the second equation And make the second equation Equation in one variable solve for that variable. You will get the first value Then you put this value in any of the equations obtained above to get the other value So you can find out You can put in this value in any of the equations above and you can find out the other value, right? And then it's a better idea always to check the answer what you have obtained to see whether you have done it correctly or I hope you understood the process