 Hello and welcome to the session. In this session we are going to discuss the following question and the question says that, find the solution of the given system of equations and the equations are y is equal to x square and y is equal to modulus of x. Let us start with the solution of this question. We are given the system of equations that is y is equal to x square. Let us name this equation as equation number 1 and y is equal to modulus of x. Let us mark this equation as equation number 2. Now we shall use substitution method to find its solution. We know that the solution of equations y is equal to f of x and y is equal to g of x is given by the equation f of x is equal to g of x. So here first equation is y is equal to f of x and second equation is y is equal to g of x. So we have x square is equal to modulus of x and we know that modulus of x is equal to plus minus of x. So this implies that x square is equal to plus minus of x. From here we have two set of equations that is x square is equal to x and x square is equal to minus of x. So first we solve the equation x square is equal to x which implies that x square minus x is equal to 0. Now taking x common from both these terms we get x into x minus 1 the whole is equal to 0 which implies that the value of x is either 0 or 1. Now we shall solve the other equation given by x square is equal to minus of x. Here we have x square is equal to minus of x which implies that x square plus x is equal to 0 which further implies that x into x plus 1 the whole is equal to 0 that is the value of x is either 0 or minus 1. So we get the values for x as 0, 1 and minus 1. Now for these values of x we shall find the corresponding values for y. For this we use equation 1 that is y is equal to x square when the value of x is 0 then the value of y is equal to 0 square which is equal to 0. Similarly when x is equal to 1 then the value of y is equal to 1 square that is 1 also. When x is equal to minus 1 then the value of y is equal to minus 1 whole square that is equal to 1. So the three points are 0, 0, 1, 1 and minus 1, 1. Now let us check these points on both the equations that is on both equations 1 and 2. Now for point with the coordinates 0, 0. Now we put this point in the first equation that is y is equal to x square and we have 0 is equal to 0 square which is equal to 0 which implies that 0 is equal to 0 which is true. Now we put this point in the equation y is equal to modulus of x that is 0 is equal to modulus of 0 that is 0. So we have 0 is equal to 0 which is true. So we say that the point with coordinates 0, 0 is the solution. Now we shall check for the point with coordinates 1, 1. We put this point in the first equation given by y is equal to x square and we get 1 is equal to 1 square which implies that 1 is equal to 1 which is true. Now again we put this point in the second equation that is y is equal to modulus of x and we get 1 is equal to modulus of 1 which further implies that 1 is equal to 1 which is true. So this point with the coordinates 1, 1 is the solution. Lastly we shall check for the point with the coordinates minus 1, 1. Now we put this point in the first equation given by y is equal to x square and we get 1 is equal to minus 1 whole square which implies that 1 is equal to 1 which is true. Now we put this point in the second equation that is y is equal to modulus of x and we get 1 is equal to modulus of minus 1 which further implies that 1 is equal to 1 as modulus of minus 1 is also 1. So this is also true. So we say that this point with the coordinates minus 1, 1 is also the solution of the given system. Thus solution set is the set containing ordered pairs 0, 0, 1, 1 minus 1, 1. Here we should note that we have the following graph of the two equations that is the pink graph shows the graph of the equation given by y is equal to x square and this blue graph shows the graph of the equation y is equal to modulus of x. Here we see that these two graph intersect at three points given by point A with the coordinates minus 1, 1, point B with the coordinates 0, 0 and point C with the coordinates 1, 1. This completes our session. Hope you enjoyed this session.