 Hello and welcome to the session. In this session, we will discuss the following question and the question says, So, 2x plus 6 upon 2x plus 4 is equal to 6x minus 13 upon 2x minus 6. Let's start the solution now. We have given the quadratic equation 2x plus 6 upon 2x plus 4 is equal to 6x minus 13 upon 4x minus 6. We will now do cross multiplication. So, on cross multiplication, we get x plus 6 the whole into 4x minus 6 the whole is equal to 2x plus 4 the whole into 6x minus 13 the whole. Now, we will open the brackets. So, this implies left hand side becomes 8x square minus 12x plus 24x minus 36 is equal to the right hand side becomes 12x square plus 24x minus 28x minus 56. This implies plus 12x minus 36 is equal to 12x square minus 4x minus 56. We will now transpose all the terms on the left hand side to the right hand side. So, this implies 12x square minus 8x square minus 4x minus 12x minus 56 plus 36 is equal to 0. This implies 4x square minus 16x minus 20 is equal to 0. Now, we will divide both the sides by 4. So, dividing both sides by 4, we get x square minus 10x minus 5 is equal to 0. Now, we will factorize this quadratic equation by splitting the middle term. The quadratic equation that we have is x square minus 4x minus 5 is equal to 0. And the middle term is minus 4x. We will split minus 4x into two terms in such a way that the product of the two terms is equal to the product of the first and the last term of the quadratic equation. Now, minus 4x can be written as minus 5x plus x. Also, the product of these two terms that is minus 5x into x is equal to minus 5x square which is equal to x square which is the first term of the quadratic equation into minus 5 which is the last term of the quadratic equation. So, the product of these two middle terms is equal to the product of first and last term of the quadratic equation. So, minus 4x can be split in this way. This implies the given equation of splitting the middle term becomes x square minus 5x plus x minus 5 is equal to 0. This implies taking x common between the first and the second term we get x into x minus 5 the whole taking one common between the last two terms we have 1 into x minus 5 the whole is equal to 0. This implies this one the whole into x minus 5 the whole is equal to 0. So, we have factorized the quadratic equation into two factors. This implies x plus 1 is equal to 0 or x minus 5 is equal to 0. This implies from the first equation we get x is equal to minus 1 or from the second equation we get x is equal to 5. So, these are the two possible solutions of the quadratic equation. Now we will check for these two solutions. First we will check if x is equal to minus 1 satisfies the quadratic equation. So, substituting x is equal to minus 1 in the quadratic equation we have the left hand side becomes the numerator is 2 into minus 1 plus x divided by the denominator which is 2 into minus 1 plus 4. This is equal to minus 2 plus 6 upon minus 2 plus 4 which is equal to 4 of them 2 which is equal to 2. Now the right hand side becomes 6 into minus 1 minus 14 divided by 4 into minus 1 minus 6. This is equal to minus 6 minus 14 upon minus 4 minus 6. This is equal to minus 20 upon minus 10 which is equal to 2. So, we have the left hand side is equal to right hand side. Therefore, x is equal to minus 1 satisfies the quadratic equation. We will now substitute x is equal to 5 in the quadratic equation. So, the left hand side becomes 2 into 5 plus 6 upon 2 into 5 plus 4. This is equal to 10 plus 6 upon 10 plus 4 which is equal to 16 by 14. This is equal to 8 by 7. Now the right hand side becomes 6 into 5 minus 14 upon 4 into 5 minus 6. This is equal to 13 minus 14 upon 20 minus 6. This is equal to 16 upon 14 which is 8 by 7. So, the left hand side is equal to the right hand side. Therefore, x is equal to 5 also satisfies the quadratic equation. So, the final solution of the quadratic equation is x is equal to minus 1 and x is equal to 5. With this the end our session. Hope you enjoyed the session.