 Hello and welcome to the session. In this session, we are going to discuss equations involving functions. Now, first, there may be two cases in which case one is if the numerical coefficients of variables are fractions and case two, the variables in the denominator. Now, let us discuss in case one if the numerical coefficients of the variables are fractions. For solving these type of questions, step one is to remove the fractions by multiplying both sides of equations by the corresponding ncm of the numbers in the denominator. And in the second step, solve the resulting simultaneous linear equations elimination substitution method. Now, let us discuss an example for this type. In this solved by 3 plus y by 2 is equal to 6 minus 3y is equal to 12. We will remove the fractions by multiplying both sides of the equation by the ncm of the numbers 3 and 2 is 6. Now, multiplying equation number one by 6 that is the ncm of 3 and 2 is 6 into x by 3 plus 6 into y by 2 is equal to 6 into 6. Which implies 2 is 6, so it will be 2x plus, so it will be 3y is equal to 36. Now, let us name it as equation number three. Number three minus 3y is equal to 12 and 2x plus 3y is equal to, now we will solve elimination method. Now for this, we will add these two equations and on adding these two terms will be cancelled with each other and on adding these is equal to 36 plus 12 which is 48 by 4 which is equal to, now we have 4 into 12 is 48, so x is equal to, now this is the equation number three. Now putting in equation number three we get plus 3y is equal to 36 which implies 24 plus 3y is equal to 36. Now this implies 3y is equal to 36 minus 24 which is equal to 12 which implies y is equal to 12 by 3. Now 3 into 4 is 12, so y is equal to 4. Therefore, the solution is y is equal to 4 if the variables are in the denominator. Now let us discuss case two. With the help of an example, we have to solve each by y is equal to 2 by y. Now let us name this equation as 1 and this equation as e that the variables are in the denominators. So, our first step is let 1 by x is equal to u. Now equation number one 8v is equal to 2 plus 12v is equation number three and this as equation number. We can solve these equations by using the elimination method to multiply equation number three by two and equation number four by estimating the coefficients of u equal 14v plus 16v is equal to 4 and 14v is equal to 14 is equation number five and this as equation number six. Now subtracting xv minus 84v the whole is equal to 4 minus 140. This implies 14v plus 16v minus 14v minus 84v is equal to minus 136. Now here these terms are cancelled with each other and this implies 84v is equal to 8v is equal to minus 136 and further this implies v is equal to x by minus 68 we cancel with minus is equal to 2. Equation number three now putting equation number two in equation number three we get plus 8 into 2 is equal to 2 which implies 7v plus 16 is equal to 2 which further gives 7v is equal to which is equal to minus 14 is 14 by 7 which is equal to minus 2. Now we have taken 1 by x is equal to u and 1 by y is equal to v. Now u to u so this implies 1 by x is equal to now putting the value of u here it will be minus 2 which further gives on plus multiplying x is equal to minus 1 by 2 is equal to v is equal to this implies 1 by y is equal to 2 which further gives on plus multiplying y is equal to 1 by 2 and this minus 1 by 2 and y is equal to 1 by 2 all to solve equations involving fractions and this completes the session hope you all have enjoyed the session.