The given system of equation may be written as kx + 2y - 5 = 0 3x + y - 1 = 0 It is of the form `a_1x + b_1y + c_1 = 0` `a_2x + b_2y + c_2 = 0` Where `a_1 = k, b_1 = 2, c_1 = -5` And `a_2 = 3, b_2 = 1, c_2 = -1` Where, `a_1 = k, b_1 = 2, c_1 = -5` And `a_2 = 3, b_2 = 1, c_2 = -1` 1) The given system will have a unique solution, if `a_1/a_2 != b_1/b_2` `=> k/3 != 2/1` `=> k != 6` So, the given system of equations will have a unique solution, if `k != 6` 2) The given system will have no solution, if `a_1/a_2 - b_1/b_2 != c_1/c_2` we have `b_1/b_2 = 2/1 and c_1/c_2 = (-5)/(-1) = 5/1` Clearly `b_1/b_2 != c_1/c_2` So, the given system of equations will have no solution, if `a_1/a_2 = b_1/b_2` `=> k/3 = 2/1` => k = 6 Hence, the given system of equations will have no solution if k = 6 |