 Hello and welcome to the session. In this session we discussed the following question which says, find the value or values of k for which the pair of linear equations kx plus 3y equal to k minus 2 and 12x plus ky equal to k has no solution. Consider the linear equations a1x plus b1y plus c1 equal to 0 and a2x plus b2y plus c2 equal to 0 Then this pair of linear equations has no solution if we have a1 upon a2 is equal to b1 upon b2 is not equal to c1 upon c2. This is the key idea that we use for this question. Let's proceed with the solution now. Consider the given pair of linear equations which is kx plus 3y equal to k minus 2. Let this be equation 1 and 12x plus ky is equal to k. Let this be equation 2. Let's move the constant terms from the right hand side to the left hand side. So we get on rearranging the equation 1 as kx plus 3y minus k minus 2 is equal to 0. This is equation 3. Then on rearranging the equation 2 we have 12x plus ky minus k is equal to 0. Let this be equation 4. Now the pair of linear equations 3 and 4 have no solution. Therefore using the condition stated in key idea we have a1 upon a2 that is k upon 12 is equal to b1 upon b2 that is 3 upon k is not equal to c1 upon c2 which is minus of k minus 2 upon minus of k. Or you can say we have k upon 12 is equal to 3 upon k is not equal to k minus 2 upon k. Now considering this we have k upon 12 is equal to 3 upon k which means we have k square is equal to 36. This gives us k equal to plus minus square root of 36 or you can say that k is equal to plus minus 6 so that we have got the values of k as plus minus 6. This is a final answer. This completes the session. Hope you have understood the solution of this question.