Answer :
2 x + ( k - 1) y = 6
3 x + (2 k + 1) y = 9
Multiply the equation 1 by 3 and multiply equation 2 by 2, and subtract,
(k + 5) y = 0 , so y = 0 or k = -5
Possibility 1 : y = 0 => x = 3 => unique solution
=> k can be any real value
Possibility 2 : k = -5, 2 x -6 y = 6 and 3 x - 9 y = 9
=> x - 3 y = 3 and there is no unique solution. There are infinite solutions.
Hence, the system has unique solution only if y = 0 and x = 3 and it is true for any real value of k.
3 x + (2 k + 1) y = 9
Multiply the equation 1 by 3 and multiply equation 2 by 2, and subtract,
(k + 5) y = 0 , so y = 0 or k = -5
Possibility 1 : y = 0 => x = 3 => unique solution
=> k can be any real value
Possibility 2 : k = -5, 2 x -6 y = 6 and 3 x - 9 y = 9
=> x - 3 y = 3 and there is no unique solution. There are infinite solutions.
Hence, the system has unique solution only if y = 0 and x = 3 and it is true for any real value of k.