# What is the value of k for which the system of equations KX 2y 5 and 3x Y 1 has no solution?

 No worries! We‘ve got your back. Try BYJU‘S free classes today!No worries! We‘ve got your back. Try BYJU‘S free classes today!Right on! Give the BNAT exam to get a 100% scholarship for BYJUS coursesNo worries! We‘ve got your back. Try BYJU‘S free classes today!No worries! We‘ve got your back. Try BYJU‘S free classes today! 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