 Hello and welcome to the session. In this session we will discuss how to solve linear equations with more than one unknown for the specified variable. First of all let us discuss what is the system of linear equations. Now a system of linear equations is two or more linear equations that are being solved simultaneously. In this session we will be looking at the systems that have only two linear equations and two unknowns. Now two linear equations in two variables or two unknowns x and y are set to form a system of simultaneous linear equations if each of these equations is satisfied by the same pair of values of x and y. Now the general form of system of simultaneous linear equations in two variables x and y is given as a x plus b y plus c is equal to 0 p x plus q y plus r is equal to 0 where a, b, c, p, q and r are constants. Now let us discuss what is the solution of system of linear equations. Now a solution of a system in two variables is an order pair that makes both equations true. You can say that a pair of values of x and y that satisfies each of the equations in a given system of two linear equations in x and y is called solution of the system. Now let us discuss methods of solving simultaneous linear equations. First is substitution method, second is elimination method and third is bluffing method. First of all let us discuss substitution method. Suppose we have two linear equations in x and y then in the first step write y in terms of x from one of the equations then in the next step substitute this value of y in the second or we can say in the other equation and from here we obtain linear equation in x then in solving this linear equation we get the value of x and in the next step we substitute the value of x in the equation that we have obtained in step one and from this we get the value of y. Now you can interchange the role of x and y in the above method. Now let us discuss an example. Here we have to solve the system of linear equations by using substitution method. Now let this be equation number one and this be equation number two. For the first step we will write y in terms of x by using any of the two given equations. So here let us express y in terms of x by using equation number one. So here we can write y is equal to 1 minus x and let this be equation number three. Now we will substitute this value of y in the other equation that is equation number two and we get x minus y that is x minus of 1 minus x the whole is equal to 3. Now this implies x minus 1 plus x is equal to 3. Now combining the like terms we have x plus x that is 2x. Now adding one on both sides we get 2x is equal to 3 plus 1 which implies 2x is equal to 4. Now dividing both sides by 2 this implies x is equal to 2. Now we will substitute this value of x in the equation which we have obtained in step one that is equation number three and we have y is equal to 1 minus x that is y is equal to 1 minus 2 which implies y is equal to minus 1. Hence x is equal to 2 and y is equal to minus 1 is the solution of the given system of linear equations. Now let us discuss elimination method. Now in this method we eliminate one of the variables. In the first step we make the coefficients of one of the variables equal by multiplying the given equations by a suitable number. In the next step we either add the equations as a threat one equation from the other whichever is appropriate to form a new equation that contains one variable only. Then in the next step solve this new equation to obtain the value of one of the variables and in the next step substitute the value found for the variable in one of the given equations and solve it for the other variable. Now let us discuss an example and here we will solve the system of linear equations by elimination method that is by using elimination method then we will solve the same system of linear equations which we have taken in the previous example. Now let this be equation number one and this be equation number two. First of all let us see coefficients of x in both the equations. Now in first equation coefficient of x is one and second equation coefficient of x is again one. Similarly coefficient of y in first equation is one and coefficient of y in second equation is minus one. Now here you can see coefficients of x are of same sign and coefficients of y are of opposite signs. So in this method either we can add the two equations or subtract. So here let us subtract equation number two from equation number one. Now here subtracting the left hand sides we have x plus y by whole minus of x minus y by whole is equal to one minus three that is subtracting the right hand sides we have one minus three. This implies x plus y minus x and minus of minus y is plus y is equal to one minus three that is minus two and this implies now here x cancels with minus x and y plus y is two y and this is equal to minus two. Now dividing both sides by two we get y is equal to minus one. Now we will substitute this value of y in one of the given equations. So we put y is equal to minus one in equation number one and we get x plus of minus one is equal to one which implies x minus one is equal to one and this implies x is equal to one plus one which implies x is equal to two. So x is equal to two and y is equal to minus one is the solution of the given system of linear equations. Now let us discuss graphical method. In the first step we draw the graph for both the given equations. Since these are linear equations so we obtain two straight lines then we find the coordinates of the point of intersection of these two lines and these coordinates give the common solution of the given equations. Now if the two lines intersect at a point then this point is the unique solution of the given equations and we say that the given equations are consistent and independent. If the two lines coincide when we have many solutions and we say that the given equations are consistent and dependent. If the two lines are parallel then the given equations have no solution and they are called inconsistent. Now let us solve the same system of equations graphically. Now for drawing the equations graphically we will make the table of values for both the equations. Now equation one can be written as y is equal to one minus x. Now here we have made a table of values and here we will put different values of x in this equation and we get the corresponding values of y. First of all we will put x is equal to zero in this equation and for x is equal to zero we get y is equal to one minus zero that is equal to one so for x is equal to zero y is equal to one. Similarly for x is equal to one y is equal to one minus one that is zero and for x is equal to two y is equal to one minus two that is minus one so we have obtained this table of values for equation number one. Now let us plot these points on the graph. Now for x is equal to zero y is equal to one so the first other pair that we will plot is zero one. Now for x is equal to zero y is equal to one so this is the point that coordinates zero one. Similarly we will plot the other two points also. So we have plotted all the three points on the graph. Now we will join these three points and joining these three points we get a straight line which represents the equation x plus y is equal to one. Similarly we will draw the graph of the second equation also. Now the second equation can be written as y is equal to x minus three. So for this equation also we will make the table of values and here by putting x is equal to zero one and minus one in this equation we get y is equal to minus three minus two and minus four respectively. So let us plot all these points on the graph. So we have plotted all the three points on the graph. Now we will join these three points. Now joining these three points again we get a straight line which represents the equation x minus y is equal to three. Now from the graph we see that the two lines intersect at this point that is the point which coordinates two minus one. This point of intersection is the solution of the given equations that is x is equal to two and y is equal to minus one is the solution of the given equations. So in this session we have discussed how to solve linear equations with more than one unknown for the specified variable and this concludes our session. Hope you all have enjoyed the session.