 Welcome friends, so in this session again, we are going to solve a pair of linear equations but this time we are going to use the method of elimination, so as the name suggests elimination means you have to eliminate one of the two variables, one of the two variables and then you have to reduce the equation in with only one variable and then it becomes a linear equation in one variable and then it becomes much easier to solve that equation, so how do we use elimination, now if you see this is equation number one and this one is equation number two, now how do we start you know this process is first of all we have to notice the we have to decide between x and y which we which one of the variables we want to eliminate, so you can pick any one of them, so let us say I want to eliminate x, so to eliminate x I have to notice or I have to take note of the coefficients of x in both the equations, so let us see here the coefficient is 2 and here is the coefficient 3, okay then what do we do is we take this coefficient and multiply this equation the second equation by you know the coefficient of x in the first equation, so in this case it is 2, so hence I will multiply this equation by 2 and then I will multiply the first equation by 3, so if you see I will take 3 here and multiply, so let us say I am multiplying it by 3, okay now rewrite the two equations after multiplication, so if you see the first equation will now become 6x plus 9y and minus 15 equals 0, let it be equation number 3, okay, now multiplying the second equation by 2 you will get 6x again minus 4y and then plus 4 equals 0, let it be equation number 4, now clearly if you see what do we notice the term containing x is 6x here and 6x here, so I have equated the coefficients of x, now what is left is just subtract the two equations, what is meant by subtracting the two equations, so you have to subtract LHS of 1 minus LHS of 2 and then RHS of 1 minus RHS of 2 and then you have to equate them, so 6x minus 6x, now 9y minus minus 4y will become 9y plus 4y and this will become minus, simply reverse the sign and add, so 6x minus 6x is 0, here 9 plus 4 13y and this will be minus 19 and this will be 0, right, so hence what is y, so y will be nothing but 19 upon 13, so clearly we got one of the variables 19 upon 13 that is how y was obtained, now to find out x what you can do, like the method of substitution, you can take any one of these equation and solve for x, let us take the first equation, so it becomes 2x plus 3 into 19 by 13 minus 5 equals 0, is it not, so 2x will be equal to 5 minus 19 theta 57, 57 by 13, so it will be nothing but 65 minus 57 by 13, okay, so x will be nothing but this is and here if you see this is 8 upon 13, so hence x is 8 upon 13 by 2, which is 4 upon 13, right, so this is what do we get as a solution for this pair of equations, so x is 4 upon 13 and y is 19 upon 13, this is one way of doing it, other way of doing it is instead of substituting y in one of the equation you could try and eliminate y, let us see how, so let us rewrite the equations once again, so we had these equations 2x plus 3y minus 5 equals 0 and the other one was 3x minus 2y plus 2 equals 0, now this time we will try and eliminate y, so to eliminate y we will have to equate the coefficients of y, okay, so if you see if I multiply this by 2 which and how do I get this 2, so simply I am using this 2 to multiply the first equation and I will multiply the second equation by 3, why this 3 because 3 is here, okay, simple, so now what will you do, so let us see this is 4x and then this is 6y minus 10 equals 0 and this will be 9x minus 6y plus 6 equals 0, so now you do not need to do anything but simply add LHS to LHS and RHS to RHS, so 4x plus 9x is clearly 13x and then plus 6y minus 6y gets you know 0, so that is what elimination is y gets eliminated, here is minus 4 equals 0, so from here also you get 4 upon 13 which we just verified by substitution as well, correct, now the last thing to do would be for any whenever you solve linear equation in 2 variables there you must check whether you have done it correctly or not, so let us take the second equation in this case, so 3 into x what is the value of x, x is 4 upon 13 isn't it and minus 2 times y what is the value of y, 19 upon 13, if you calculate and then add 2 to it what will you get, you will get 13 as LCM then this is 12 then this is 38 and this is 26, so if you see this comes out to be 0 which is equal to RHS of 2, isn't it, so that means our solution is correct, so x equals to 4 upon 13 and y equals 19 upon 13 is the correct solution to this pair of linear equations into variables, I hope you understood the process of elimination, thank you.