 Hi and welcome to the session. I am Priyanka and I will be helping you with exercise 1.2 question number 12 on page 14. The question says, solve the following systems of equation by elimination method that is by equating coefficients of x or of y. So let us start with our solution by rewriting the equations given to us in the question and by taking all the constants from left hand side to right hand side. Let this be the first equation and this be the second equation. Now, since we need to equate the coefficient of x or of y, let us multiply the second equation by 3. On doing it, we have the first equation as it is and now the whole second equation will get multiplied by 3. So by this, we are able to equate the coefficients of y. Now, let this be the third equation and this be the fourth equation. Now, on subtracting 4 from 3, we have 4x minus 3y is equal to 8, 18x minus 3y is equal to 29 and now all the signs will get changed as we are subtracting this equation from this equation and on doing so, we have minus 14x is equal to minus 21. So the value of x is 3 by 2, right? Let us name this as the fifth equation. Now, on substituting the value of x from 5 in any of the first two equations, let us say in 1, we have 4 in place of x. Now, we have 3 divided by 2 minus 3y is equal to 8. That is, 6 minus 3y is equal to 8, minus 2 is equal to 3y or we can write that the value of y is equal to as minus 2 by 3, right? So therefore, the solution of the given equations, the value of 3 by 2 and the value of y as minus 2 by 3, right? So, hope you enjoyed this session. Do take care of all your calculations and bye for now.