 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 given are 4x square plus y square is equal to 13 and x plus y is equal to 2. Now let us start with the solution of the given question. We are given the system of equations as 4x square plus y square is equal to 13. Let us mark this equation as equation number 1 and x plus y is equal to 2. Let us mark this equation as equation number 2. Now we will use substitution method to solve these equations. Now equation 2 can also be written as y is equal to 2 minus x. Now we put this value of y in equation 1 and we get 4x square plus y square that is 2 minus x whole square is equal to 13 which implies that 4x square plus now 2 minus x whole square can be written using this formula that is a minus b whole square is equal to a square minus 2ab plus b square. So we write 2 square minus 2 into 2 into x plus x square is equal to 13 which further implies that 4x square plus 2 square that is 4 minus of 2 into 2 into x that is 4x plus x square is equal to 13. It further implies that 4x square plus x square that is 5x square minus of 4x plus 4 minus of 13 is equal to 0 which implies 5x square minus of 4x Now plus 4 minus 13 is minus 9 is equal to 0. Now we get a quadratic equation in x. Now we solve this quadratic equation by factorization method which implies that 5x square now minus of 4x can be written as plus 5x minus 9x minus 9 is equal to 0 which further implies that now taking 5x common from first two terms we get 5x into x plus 1 the whole and taking minus 9 common from last two terms we get minus 9 into x plus 1 the whole is equal to 0 which implies that 5x minus 9 the whole into x plus 1 the whole is equal to 0 which implies that x is equal to minus 1 and 9 upon 5 so we have got two values of x and now we obtain corresponding values of y for this we put the value of x either in equation 1 or equation 2 let us use equation 2 to obtain value of y when x is equal to minus 1 we have minus 1 plus y is equal to 2 which implies that y is equal to 2 plus 1 that is equal to 3 for x is equal to minus 1 y is equal to 3 so we have ordered pair minus 1 3 when x is equal to 9 upon 5 then we have 9 upon 5 plus y is equal to 2 which implies that y is equal to 2 minus 9 upon 5 which further implies that y is equal to now taking the LCM we have 5 in the denominator and in the numerator we have 10 minus 9 which implies that y is equal to 1 upon 5 so for x is equal to 9 upon 5 the value of y is 1 upon 5 so the second ordered pair is 9 upon 5 1 upon 5 so we have two solutions of the given system of equations given by the ordered pairs minus 1 3 and 9 upon 5 1 upon 5 we find that this blue line intersects the curve that is the ellipse at these two points with the coordinates minus 1 3 and 9 by 5 1 by 5 thus we say that the solutions of the given system of equations are given by the ordered pairs minus 1 3 and 9 by 5 1 by 5 this is the required answer this completes our session hope you enjoyed this session