 Hi and welcome to the session. I am Priyanka and I'll be helping you with exercise 1.3 question number 2 on page 20. The question says, solve the following systems of equation by using cross multiplication method. The equation given to us is 3x minus 5y is equal to 20 and 7x plus 2y is equal to 17. But before proceeding on with the solution, we should be well versed with the method we are going to use. The method's name is cross multiplication method. So in this method the solution has a unique solution that is it has exactly one solution. So the pattern of the solution under this method is written as c1. The knowledge of this pattern is the key idea we are going to use to solve this systems of equation given to us. Now let us rewrite the equations given to us by shifting all terms to LHS and introducing 0 in RHS. So now 20 and 17 were positive in RHS so now they are negative in LHS. Here b1, b2 are the coefficients of y that is b1 will be minus 5 and b2 will be a positive 2. c1, c2 are the constants which are given to us in the equation that is minus 20 and minus 17 and a1, a2 are the coefficients of x that is 3 and 7. Now on substituting the values of b1, b2, c1, c2 and a1, a2 in the above given pattern we have x minus 5 to minus 20 minus 17, 3, 7 and then again minus 5 and 2 right. Now to obtain the solution write x, y and 1 separately with an equality sign. Now the first two columns that is minus 5, 2, minus 20 and minus 17 will be written under x. The second and the third column that is minus 20, minus 17, 3 and 7 will be written under y and 3, 7, minus 5 and 2 will be written under 1. Now mark cross arrows pointing downwards from top to bottom and from bottom to top like this in the denominator of x, y and 1. Now to obtain the denominators of x, y and 1 multiply the numbers with downward arrow that is the answer of minus 5 into minus 17 will give us 85 and subtract the product of the upward arrows that is 2 into minus 20 that will give us minus 40. Similarly under y also minus 20 into 7 will give us minus 140 and subtract the product of minus 17 into 3 that is minus 51 then again 1 divided by 3 multiplied by 2 will give us 6 minus the product of 7 and minus 5 that is minus 35. Now to obtain the value of x, y will equate the first expression that is x divided by 125 with the third expression that is 1 divided by 41 and similarly the second equation that is y divided by minus 89 will be equated to the third expression again that is 1 divided by 41. Now to find out the value of x take 125 from the denominator of LHS to the numerator of RHS and similarly minus 89 will be taken from the denominator of LHS to the numerator of RHS. So on doing this we get x is equal to 125 by 41 and y is equal to minus 89 divided by 41. Now as we can see that they are already in the lowest form so therefore x is equal to 125 by 41 and y is equal to minus 89 by 41 is the solution of the given equations. Hope you enjoyed the session and bye for now.