 Hello and welcome to the session. In this session, we will discuss the following question and the question says, solve x upon x minus 3 plus x minus 3 upon x is equal to 5 upon 2. Let's start the solution now. We have given this quadratic equation and we have to solve it. So the given quadratic equation is x upon x minus 3 plus x minus 3 upon x is equal to 5 by 2. Now we will take ncm on the left hand side. So taking ncm on left hand side, we get the numerator becomes x square plus x minus 3 whole square whole divided by the denominator which is x into x minus 3 the whole is equal to 5 by 2. Now we will open the brackets. So this implies numerator at the left hand side becomes x square plus x square plus 9 minus 6x whole divided by x square minus 3x is equal to 5 by 2. Now we will do the cross multiplication. So on cross multiplication, we get 2 into 2x square plus 9 minus 6x the whole is equal to 5 into x square minus 3x the whole. Now we will open the brackets. So this implies 4x square plus 18 minus 12x is equal to 5x square minus 15x. Now we will transpose 4x square 18 and minus 12x to the right hand side. So this implies 5x square minus 15x minus 4x square minus 18 plus 12x is equal to 0. This implies x square minus 3x minus 18 is equal to 0. So we obtain this quadratic equation. Now we will factor the quadratic equation by splitting the middle term. Now the quadratic equation that we have is x square minus 3x minus 18 is equal to 0. The middle term is minus 3x. Now minus 3x can be split into minus 6x plus 3x and we can see that the product of these two terms that is minus 6x into 3x is equal to minus 18x square which is same as the product of first and last term of the quadratic equation. So we can split the middle term in this way. This implies on splitting the middle term we get x square minus 6x minus 3x minus 18 is equal to 0. This implies we take x common between the first and the second term so we get x into x minus 6 the whole. Now we take 3 common between the last two terms so we get plus 3 into x minus 6 the whole is equal to 0. This implies x plus 3 the whole into x minus 6 the whole is equal to 0. So in this way we have factorized the quadratic equation into these two factors. This implies x plus 3 is equal to 0 as x minus 6 is equal to 0. This implies x is equal to minus 3 from the first equation or x is equal to 6 from the second equation. So we obtain these two possible solutions for the quadratic equation. Now we will check if these two values of x satisfy the given quadratic equation. First we will substitute x is equal to 6 in the quadratic equation. So the left hand side of the quadratic equation becomes 6 upon 6 minus 3 plus 6 minus 3 upon 6. This is equal to 6 upon 3 plus 3 upon 6 which is equal to 2 plus 1 upon 2 which is equal to 5 upon 2 which is the right hand side of the quadratic equation. So x is equal to 6 satisfies the quadratic equation. Now we will check that x is equal to minus 3 so we will substitute x is equal to minus 3 in the quadratic equation. So the left hand side becomes minus 3 upon minus 3 minus 3 plus minus 3 minus 3 upon minus 3. This is equal to 3 upon 6 plus 6 upon 3 which is equal to 1 upon 2 plus 2 which is equal to 5 upon 2 and this is the right hand side of the quadratic equation. So x is equal to minus 3 also satisfies the quadratic equation. So we can say that the final solution of the quadratic equation is x is equal to minus 3 and x is equal to 6. With this we end our session. Hope you enjoyed the session.