For which value(s) of \(k\) will the pair of equations \(kx + 3y = k - 3\) and \(12x + ky = k\) have no solution?
\(k = -6\).
\(kx + 3y - (k - 3) = 0\)
\(12x + ky - k = 0\)
For no solution: \(\dfrac{a_1}{a_2} = \dfrac{b_1}{b_2} \ne \dfrac{c_1}{c_2}\).
First two ratios equal when
\(\dfrac{k}{12} = \dfrac{3}{k}\Rightarrow k^2 = 36\Rightarrow k = \pm 6\).
Check constants ratio \(\dfrac{c_1}{c_2} = \dfrac{-(k-3)}{-k} = \dfrac{k-3}{k}\).
For \(k = 6\): all three ratios are \(\dfrac{1}{2}\) ⇒ coincident, not “no solution”.
For \(k = -6\): the first two ratios are \(-\dfrac{1}{2}\) but \(\dfrac{c_1}{c_2} = \dfrac{3}{2}\).
Hence unequal ⇒ parallel distinct lines ⇒ no solution.