 Hello and welcome to the session. In this session, we will learn about simultaneous linear equations. Simultaneous linear equations is a collection of linear equations having the same set of variables and equations as there are variables. Equations at least one pair of numbers will satisfy each equation. Now let us see an example and in this x plus 3y is equal to 6 plus y is equal to 7 is a system where equations is in both of these equations. So, linear equations as there are variables. So here as variables are two, so the equations are also equations of numbers simultaneously. Let us name it as 1 and this as 2 is equal to 1. Then number 1, the left hand side of this equation will be equal to x plus which is equal to on putting the values 3 plus 3 into 1 which is equal to 3 plus 3 that is equal to 6 and the arches of this equation is equal to is equal to right hand side both are equal to 6 for the equation number 1. We will take for the equation number 2 the left hand side is equal to now putting these values here this will be 2 into 3 plus 1 which is equal to 6 plus 1 which is equal to 7 and also the purpose equation is equal to 7. Therefore left hand side is equal to right hand side for equation number 2 also. Hence put the 3 and y is equal to 1 is called the solution for both the equations. Now let us discuss the algebraic methods of solving simultaneous linear equations. Now there are two algebraic methods. First is addition or elimination method and second is substitution method. Let us learn addition or elimination method. Now in this the type 1 is when two equations. Now let us see an example for this. In this we will say y is equal to 4 y is equal to 5. Now let us name it as 1 and this as 2. Now in both the equation the coefficient of x is 1. So in this case we will add or subtract the two equations simultaneously so that one of the variables can be eliminated. Now here from 1 minus y minus x plus y the whole is equal to 4 minus 5. x minus y minus x minus y is equal to minus 1 which further implies. Now these terms are cancelled with each other so it is minus 2 y is equal to minus 1 which further gives y is equal to 1 by 2. Now putting y is equal to 1 by 2 equation number 1 is 1 by 2 is equal to 4 which further implies 2 plus 1 by 2 which implies equal to on taking LCM it will be 8 plus 1 by 2 which is equal to 9 by 2. Therefore the solution is y is equal to 1 by 2. Now let us discuss the type 2 when one of the variables in the two equations make the coefficients of the variables two equations equal my dotted line of the given equation. So for this in this we have to solve 2 x plus 5 y is equal to 20 and 5 x minus y is equal to 17. Now these are the equations of the type 2 so let us name it as 1 and this as since one of the variables equal that is x equal you have to multiply these two equations by a suitable numbers the coefficients of y equal by multiplying these equations by the suitable numbers. Now here we will make the coefficients of y equal this we will multiply equation number 2 is equal to 23 the both sides of this equation with 5 we get 25 minus 5 y is equal to 85. Now let us name it as equation number 3 and this as equation number of y in both of these equations. Now to solve this two equations that is equation number 3 and 4. Now adding plus 5 y plus 25 x minus 5 y is equal to 23 plus 85 which is 108. This one implies these are the 27 x which is equal to 100 h this further implies x is equal to y 27. Now into 9 is 27 and 3 into 30 so this is equal to now putting into 4 plus 5 y is equal to 23 which implies 8 plus 5 y is equal to 23. This further means 5 y is equal to 23 minus 8 which is 15. Further implies y is equal to 15 by 5 which is y is equal to 3. Now this question first of all we have equated to coefficients of y by multiplying the given equations with the suitable numbers. Then after that we have added these resulting equations and eliminated y and then at the last we are getting the values of x and y. Now let us discuss the substitution method. Now in this method one equation is solved for one of the variables in terms of the other variable and then this value is substituted for the variable in the second equation. Now let us take an example for this and in this solve is equal to 23 plus y is equal to 19. Now let us name this as equation number 1 and this as equation number 2. Now in this method first of all we will solve equation number 1 for one of the variables in terms of the other variable. So from equation number 1 this 2 y is equal to 23 which further implies 23 minus 2 y and it gives x is equal to 23 minus 2 y whole upon this value of x in the second equation. So substituting 23 minus 2 y whole upon 5 in equation number into 23 minus 2 y plus y is equal to 19. Further implies 19 to minus 8 y whole upon y is equal to 19. This further gives on taking the same here it will be 92 minus 8 y plus 5 y is equal to 19. On cross multiplying further this implies 92 minus 3 y is equal to 19 into 5 which is 95 which implies minus 3 y is equal to 95 minus 92 which further gives minus 3 y is equal to 95 minus 3 which is equal to minus 1. Now let us name this as equation number 3. Now putting y is equal to minus 1 in 3 we get x is equal to 23 minus 2 into minus 1 whole upon 223 and here it will be plus 2 which implies x is equal to 25 by 5 which is equal to the solution is and y is equal to. So this was the substitution method for solving the simultaneous linear equation. You have learnt about simultaneous linear equations methods of solving simultaneous linear equations. This is the session where everyone have enjoyed the session.