 Hello and welcome to the session. In this session, we will discuss y, the x coordinate of the points where the graphs of the equations y is equal to f of x and y is equal to g of x intersect, are the solutions of the equation f of x is equal to g of x. And we will also discuss how to find solution of system of equations approximately using graphs. Let us consider the following system of equations y is equal to f of x is equal to x square plus 2 and y is equal to g of x which is equal to x plus 2. Now, let us see graph of these two equations. Here we see that the equation y is equal to x square plus 2 represents an upward parabola with vertical shift of two units. So, this represents the equation y is equal to x square plus 2. Now, the function g of x that is equal to x plus 2 represents a straight line and here this blue straight line is graph of the equation y is equal to x plus 2. And from the graph we see that the two curves intersect at two points that is point a which coordinates 0, 2 and point b which coordinates 1, 3. And these points are the solution set of the given system of equations. It means at these two points the values of both the functions are same. Thus we say that at point a which coordinates 0, 2 f of x is equal to g of x. Also at point b which coordinates 1, 3 f of x is equal to g of x. That is when the value of x is 0 then the value of y is equal to 2 for both the functions similarly when the value of x is equal to 1 then the value of y is 3 for both the functions. Now let us put values of f of x and g of x in equation f of x is equal to g of x. We get f of x that is x square plus 2 is equal to g of x that is x plus 2 which implies that x square plus 2 minus of x minus of 2 is equal to 0 which further implies that x square minus of x is equal to 0 that is after simplification we get this equation as x square minus x is equal to 0. Now we will see that x is equal to 0 and x is equal to 1 will satisfy this equation. Now when we put x is equal to 0 in this equation we get 0 square minus 0 is equal to 0 which implies that 0 minus 0 is equal to 0. That is 0 is equal to 0 which is true also when we put the value of x as 1 in this equation we get 1 square minus 1 is equal to 0 which implies that 1 minus 1 is equal to 0 that is 0 is equal to 0 which is true so we say that both x is equal to 0 and x is equal to 1 are solutions of equation f of x is equal to g of x. Now let us discuss how to find the solutions of the system of equations approximately. Now consider the following graph of the system of equations y is equal to 2x and y is equal to square root of 1 minus x square. That is this blue straight line represents the graph of the equation y is equal to 2x and this red coloured semicircle represents the graph of the equation y is equal to square root of 1 minus x square. From graph we can see that the two graphs intersect at one point if we carefully see this point of intersection we have the value of x approximately as 0.4 the value of y as 0.9 approximately thus the approximate solution of the system of equations is given by the coordinates 0.4 0.9. Now let us see its actual solution by substitution method. Now here we are given two equations y is equal to 2x and y is equal to square root of 1 minus x square. Now we put y is equal to 2x in the equation y is equal to square root of 1 minus x square and we get 2x is equal to square root of 1 minus x square. Now squaring both sides we get 2x whole square that is 4x square is equal to 1 minus x square. Which implies that 4x square plus x square is equal to 1 which further implies that 5x square is equal to 1 which implies that x square is equal to 1 upon 5 which further implies that x is equal to plus minus of square root of 1 upon 5 or we can write it as plus minus of 1 upon square root of 5. So we get x is equal to 1 upon square root of 5 and minus 1 upon square root of 5. Now we put these values in the equation y is equal to 2x that is first we put x is equal to 1 upon square root of 5 and we get the value of y as 2 into x that is 2 into 1 upon square root of 5 which is equal to 2 upon square root of 5 also when x is equal to minus 1 upon square root of 5 this implies that y is equal to 2 into x that is 2 into minus 1 upon square root of 5 which is equal to minus 2 upon square root of 5. So we have 2 points with the coordinates 1 upon square root of 5, 2 upon square root of 5 and minus 1 upon square root of 5 minus 2 upon square root of 5. Let us check these 2 points in the 2 equations. Now for point with the coordinates minus 1 upon square root of 5 minus 2 upon square root of 5. Now we put this point in the equation y is equal to square root of 1 minus x square and we get y that is minus 2 upon square root of 5 is equal to square root of 1 minus x square that is minus 1 upon square root of 5 whole square which implies that minus 2 upon square root of 5 is equal to square root of 1 minus minus 1 upon square root of 5 whole square is equal to 1 upon 5 which further implies that minus 2 upon square root of 5 is equal to square root of now taking LCM we get 5 in the denominator and in the numerator we have 5 minus 1 that is 4. So we have minus 2 upon square root of 5 is equal to square root of 4 upon 5 which implies that minus 2 upon square root of 5 is equal to 2 upon square root of 5 as square root of 4 is equal to 2. So here we have got minus 2 upon square root of 5 is equal to 2 upon square root of 5 which is not true. So this point is not the solution of the given system that is the point with coordinates minus 1 upon square root of 5 minus 2 upon square root of 5 is not the solution of the given system of equations. Now we shall take the same for the second point with the coordinates 1 upon square root of 5 2 upon square root of 5. We put this point in the equation y is equal to square root of 1 minus x square that is we put x is equal to 1 upon square root of 5 y is equal to 2 upon square root of 5 in this equation. And we get y that is 2 upon square root of 5 is equal to square root of 1 minus x square that is 1 upon square root of 5 whole square which implies that 2 upon square root of 5 is equal to square root of 1 minus 1 upon square root of 5 whole square is equal to 1 upon 5. 5. This further implies that 2 upon square root of 5 is equal to square root of, now taking LCM we get 5 in the denominator and in the numerator we have 5 minus 1. So we have 2 upon square root of 5 is equal to square root of 5 minus 1 whole upon 5 which implies that 2 upon square root of 5 is equal to square root of 4 upon 5 which further implies that 2 upon square root of 5 is equal to 2 upon square root of 5 as square root of 4 is equal to 2. So we have got 2 upon square root of 5 is equal to 2 upon square root of 5 which is true. Now again we put this point with the coordinates 1 upon square root of 5, 2 upon square root of 5 in the second equation that is y is equal to 2x and we have y that is 2 upon square root of 5 is equal to 2x that is 2 into 1 upon square root of 5 which implies that 2 upon square root of 5 is equal to 2 upon square root of 5 which is true. So exact solution is given by the ordered pair 1 upon square root of 5, 2 upon square root of 5. Thus we see that the approximate solution of the system of equations is given by the ordered pair 0.4, 0.9 which is obtained graphically and exact solution is given by the ordered pair 1 upon square root of 5, 2 upon square root of 5 obtained using substitution method. We should note that sometimes we have such non-linear system of equations which cannot be solved algebraically so we can only find their approximate solutions by graphing. It is always better to draw the graph of non-linear function to know the number of solutions of the given system of equations. Thus in this session we have learnt why the x coordinate of the points where the graphs of the equations y is equal to f of x and y is equal to g of x intersect at the solutions of the equation f of x is equal to g of x. We have also discussed how to find the solutions of the system of equations approximately using graphs. This completes our session. Hope you enjoyed this session.