 Hello and welcome to the session. Let us understand the following question today. Is the following pair of linear equations has unique solutions, low solutions or infinitely many solutions? In case there is a unique solution, find it by using cross multiplication method. We have equations as x-3y-3 is equal to 0 and 3x-9y-2 is equal to 0. Before starting with the solution, let us understand what is the unique solution, low solution and infinitely many solutions. Now, if the lines are represented by the equations a1x plus b1y plus c1 is equal to 0 and a2x plus b2y plus c2 is equal to 0, then if a1 by a2 is not equal to b1 by b2, it implies it is a unique solution. And if a1 by a2 is equal to b1 by b2 is equal to c1 by c2, it implies infinitely many solutions. If a1 by a2 is equal to b1 by b2 is not equal to c1 by c2, it implies no solution. Now, this is the key idea to our question. Now, let us write the solution. The given equations are x-3y-3 is equal to 0, that is equation number 1 and 3x-9y-2 is equal to 0, that is equation number 2. Now, comparing the equations with a1x plus b1y plus c1 is equal to 0 and a2x plus b2y plus c2 is equal to 0. So, we have a1 is equal to 1, a2 is equal to 3, b1 is equal to minus 3, b2 is equal to minus 9, c1 is equal to minus 3, c2 is equal to minus 2. Now, a1 by a2 is equal to 1 by 3, b1 by b2 is equal to minus 3 by minus 9, which is equal to 1 by 3 and c1 by c2 is equal to minus 3 by minus 2, which is equal to 3 by 2. Now, we can see from this that a1 by a2 is equal to b1 by b2 is not equal to c1 by c2, 1 by 3 is equal to 1 by 3, which is not equal to 3 by 2. So, by key idea, we have, by key idea, we see that if a1 by a2 is equal to b1 by b2 is not equal to c1 by c2, it implies no solution. So, hence, there is no solution. Therefore, the answer is no solution. I hope you understood the question. Bye and have a nice day.