 Hello and welcome to the session. Let us discuss the following question today. For which value of k will the following pair of linear equations have no solution? We have 3x plus y is equal to 1 and 2k minus 1x plus k minus 1y is equal to 2k plus 1. Now let us write the solution. Given equations are 3x plus y is equal to 1 and 2k minus 1x plus k minus 1y is equal to 2k plus 1. Comparing the equations with a1x plus b1y plus c1 is equal to 0 and a2x plus b2y plus c2 is equal to 0. We get a1 is equal to 3, a2 is equal to 2k minus 1, b1 is equal to 1, b2 is equal to k minus 1, c1 is equal to 1 and c2 is equal to 2k plus 1. Now we get a1 by a2 is equal to 3 by 2k minus 1, b1 by b2 is equal to 1 by k minus 1, c1 by c2 is equal to 1 by 2k plus 1. Now the given system of equations has no solution if a1 by a2 is equal to b1 by b2 is not equal to c1 by c2. Therefore 3 by 2k minus 1 is equal to 1 by k minus 1 is not equal to 1 by 2k plus 1. Therefore, 3 by 2 k minus 1 is equal to 1 by k minus 1 and 1 by k minus 1 is not equal to 1 by 2 k plus 1, which implies on cross multiplication we get 3 multiplied by k minus 1 is equal to 2 k minus 1, which implies 3 k minus 3 is equal to 2 k minus 1, which implies 3 k minus 2 k is equal to minus 1 plus 3, which implies k is equal to plus 2. Now, here for k is equal to 2 we have 1 by k minus 1 is not equal to 1 by 2 k plus 1. Hence, the given system of equation will have no solution if k is equal to 2. So, the required value of k is 2. I hope you understood the question by and have a nice day.