(b) k \(\ne\) – 4
In the given system of equations,
a1 = 2, b1 = 3, c1 = –5
a2 = k, b2 = –6, c2 = – 8
For a unique solution.
\(\frac{a_1}{a_2}\)\(\ne\)\(\frac{b_1}{b_2}\) \(\Rightarrow\) \(\frac{2}{k}\) \(\ne\) \(\frac{3}{-6}\) \(\Rightarrow\) 3k \(\ne\) -12
\(\Rightarrow\) k\(\ne\) -4