 Hi friends, welcome to the session I am Alka. Today we are going to solve the sum on the basis of a pair of linear equation in two variables. Our question is the sum of the digits of a two digit number is 9. Also 9 times this number is twice the number obtained by reversing the order of the digits. We have to find the number. Now let us begin with the solution. Let us first assume the tens digit number and the unit digit number. So, let the tens digit of the number be x and let the unit digit number, unit digit of the number be y. Now this becomes, therefore our number becomes, therefore number is equal to 10x plus y and we are given that sum of the digits of a two digit number is 9. This means that sum of the digits, sum of the digits x and y, x plus y is equal to 9. This is our first equation. Now we are given that 9 times this number is twice the number obtained by reversing the order of the digits. 9 times of the number is 9 times, 9 times, 9 times the number is equal to 9 multiplied by 10x plus y which is the number give us 90x plus 9y. Now according to our question when the digits are reversed because we have to, we have to reverse the order of the digits. When the digits are reversed that is when the unit digit come on the tens place and the tens digit come on the unit place, so it becomes when, when the digits are reversed the number is equal to 10y plus x. Earlier the number was 10x plus y. Now since the digits are reversed the number is 10y plus x. Now according to question we see 9 times the number which is 90x plus 9y, 90x plus 9y becomes twice of the number obtained by reversing the digits is twice of 10y plus x. This implies that 90x plus 9y is equal to 20y plus 2x. This gives us 90x minus 2x plus 90y minus 20y is equal to 0. This implies 90 minus 2x is equal to 88x and plus 9y minus 20y is equal to minus 11y is equal to 0. On taking 11 as common we get 8x minus y is equal to 0 and this implies that 8x minus y is equal to 0. When we divide both the sides LHS, when we divide both sides LHS and RHS by 11 we get 8x minus y is equal to 0. This is our third equation which we get after dividing LHS and RHS by 11. So our equations are our first equation is x plus y is equal to 9 and second equation is 8x minus y is equal to 0. Hence the equations are x plus y is equal to 9 and 8x minus y is equal to 0. Further we see that if we add equation number 1 and 2 we get on adding these we get 9x. These will get cancelled out is equal to 9. This implies 9x is equal to 9. This implies x is equal to 9 upon 9 therefore x is equal to 1. Now on substituting the value of x is equal to 1 in equation number 1st we get our first equation is x plus y is equal to 9. Now we will keep x here as 1 which gives us 1 plus y is equal to 9. This implies y is equal to 9 minus 1 which gives y is equal to 8. Now therefore the required number is on the 10th place it was x which is 1 and on the unit's place it was y which is 8. So the required number is 18. Hence we see the equations are x plus y is equal to 9 and 8x minus y is equal to 0 where x and y are 10s and units rigid and the required number is 18. Hope you got the problem and understood the solution. See you and goodbye.