 Hello and welcome to the session. In this session, we will discuss how to solve a system of linear and quadratic equations graphically and algebraically. Now, consider the following system of equations. Now we have to solve this quadratic and linear system. It means we have to find those values of x and y which satisfy both the equations. In simple words, we have to find the point of intersection of the given equations. Now let us solve the system of equations graphically. Now the first equation is y is equal to x square plus 2x plus 1. Now this is a quadratic equation and it is a parabola and it is of the form y is equal to ax square plus vx plus c. Its x is of symmetry is x is equal to minus v upon 2a. Now here, b that is coefficient of x is 2 and a that is coefficient of x square is 1. So, x is of symmetry is x is equal to minus 2 upon 2 into 1 which implies x is equal to minus 2 upon 2. This implies x is equal to minus 1. Now let us make table of values to draw its curve. It means we will put different values of x in this equation and we will get corresponding values of y. First of all, let us put x is equal to minus 3 in this equation and for x is equal to minus 3 we get y is equal to 4. Then for x is equal to minus 2 we get y is equal to 1. Then for x is equal to minus 1 we get y is equal to 0. Then for x is equal to 0 we get y is equal to 1. Then for x is equal to 1 we get y is equal to 4. Then for x is equal to 2 we get y is equal to 9. Now let us plot these points on the coordinate plane. Now let us plot the first point on the coordinate plane. So we have plotted the point with coordinates minus 3 4 on the coordinate plane. Then let us plot the next point and this is the point with coordinates minus 2 1. Clearly we will plot all the other points on the coordinate plane. So we have plotted all the points on the coordinate plane. Now joining all these points we get graph of the quadratic equation. y is equal to x square plus 2 x plus 1. This is a parabola whose axis of symmetry is the line x is equal to minus 1. Now let us draw the graph of the linear equation. y is equal to x plus 3. For this we will make its table of values. First of all let us put x is equal to minus 3. Now for x is equal to minus 3 we get y is equal to 0. Then for x is equal to minus 2 we get y is equal to 1. Similarly putting all these values of x we are getting these corresponding values of y for this linear equation. Now we will plot all these points on the coordinate plane. First of all let us plot the point minus 3 0 on the coordinate plane. Now this is the point minus 3 0. Similarly we will plot all the points on the coordinate plane. So we have plotted all these points on the coordinate plane. Now joining all these points we get a straight line. And this is the graph of the linear equation y is equal to x plus 3. From the graph we can see that the line intersects the parabola at two points. And these two points are minus 2 1 and 1 4 like this point v a and this point v b. So the solution set of the given system that is this system is given by the points minus 2 1 and 1 4. So we have solved the given system of equations graphically. Now let us solve the system of equations algebraically. For this we make use of substitution method. Now in this method we will follow the following steps to solve the linear and quadratic system algebraically. Now since there are given two equations in terms of y is equal to f of x. So substitute the value of y from one equation into another. Secondly write the new formed equation that is the new formed quadratic equation in standard form and set it equal to 0. Then solve the quadratic equation then next. Substitute in the linear equation to find the corresponding value of y. Then in the next step write the solution as ordered pairs of numbers. Then in the next step check the ordered pairs in each of the original equations. Now we will solve the given system of equations algebraically by using substitution method. Now let this be equation number one and this be equation number two. Now the first step substitute the value of y in equation number one from equation number two. So substitute in the value we have plus three is equal to x square plus two x plus one. Now let us solve this equation for x. Now this implies zero is equal to x square plus two x plus one minus x minus three. Which further implies zero is equal to now combining the life terms we have x square plus two x minus x is plus x plus one minus three is minus two. And we can write it as x square plus x minus two is equal to zero. Now this is a quadratic equation in x. So we will solve it by factorization. Now as written in the middle term this implies x square plus two x minus x minus two is equal to zero. This further implies now from these two terms x is common so it will be x into x plus two the whole. And from these two terms taking minus one common it will be minus one into x plus two the whole is equal to zero. This further implies x plus two the whole into x minus one the whole is equal to zero. Now solving we get two values of x that is x is equal to minus two and one. Now we will find values of y for x is equal to minus two and x is equal to one. So we will put x is equal to minus two in this linear equation. So for x is equal to minus two y is equal to minus two plus three that is equal to one. Now we will put x is equal to one and for x is equal to one we get y is equal to one plus three that is equal to four. So for x is equal to minus two we get y is equal to one and for x is equal to one we get y is equal to four. Now let us write this solution as ordered pairs of numbers. So we get two solutions of the system of equations. First is the ordered pair minus two one and second is the ordered pair one four. Now lastly we have to check that whether these ordered pairs satisfy both the equations or not. Now let us take the first ordered pair that is minus two one. So let us put x is equal to minus two and y is equal to one in the first equation. So we have one is equal to minus two whole square plus two into minus two plus one which implies one is equal to four minus four plus one. And this implies one is equal to one which is true. Now let us put x is equal to minus two and y is equal to one in equation two and we have one is equal to minus two plus three which implies one is equal to one which is also true. So the ordered pair minus two one is the solution of the given system of equations. Now let us take the second ordered pair that is the ordered pair one four. Now let us put x is equal to one and y is equal to four in equation one. And we have four is equal to one square plus two into one plus one which implies four is equal to one plus two plus one that implies four is equal to four which is true. Then let us check it for the second equation. So here we have four is equal to one plus three which implies four is equal to four which is also true. It means the ordered pair one four is also the solution of the given system of equations. Thus the solution of the given quadratic linear system is ordered pair minus two one and the ordered pair one four. Now if we are given a system of equations in which one equation is quadratic in two variables like the equation of a circle. For example x square plus y square is equal to r square and a straight line then also we will follow the same procedure as rapidly and graphically to solve the given system of equations. Now let us discuss times of solutions of quadratic linear system. Now a quadratic linear system can have exactly two solutions exactly one solution or it can have no solution. Now we can know this algebraically from substitution method when we substitute the value of y we get a quadratic equation in x that is we get equation of type a square plus bx plus c is equal to zero. Now here we check value of discriminant that is b square minus four ac. Now if b square minus four ac is greater than zero then the system has exactly two solutions. Graphically line will intercept the curve at two distinct points when b square minus four ac is greater than zero and if b square minus four ac is equal to zero then system has exactly one solution. It means graphically line will be tensions to the curve and corresponding system of equations will have two identical roots and we can say one double root and if b square minus four ac is less than zero then the system has no solution. Graphically it means line never intercepts the curve for example in above question after substituting value of y we have equation x square plus x minus two is equal to zero which is a quadratic equation in x. Now let us find value of discriminant that is b square minus four ac. Now here b that is coefficient of x is one a that is coefficient of x square is again one and c that is the constant term is minus two. So b square minus four ac is equal to one square minus four into one into minus two which is equal to one plus eight that is equal to nine. Therefore b square minus four ac is equal to nine which is greater than zero and we know that if b square minus four ac is greater than zero then system has exactly two solutions. So here the system has exactly two solutions. Now earlier we have drawn the graph of these two equations. Now graphically we can also see that the line intercepts the curve at two distinct points. This means when b square minus four ac is greater than zero then the line will intersect the curve at two distinct points. So in this session we have discussed how to solve quadratic linear system of equations and generally as graphically and this completes our session hope you all have enjoyed the session.