For what integral value of k, the system of equations kx – 5y + 6 = 0 and 4(k – 1)y – 12x + 3 = 0 has no solution?

Rewrite second equation: –12x + 4(k–1)y + 3 = 0. For no solution, coefficient ratios must satisfy: k/(–12) = –5/(4(k–1)) ≠ 6/3. From first equality: 4k(k–1) = 60, so k² – k – 15 = 0, giving k = (1 ± √61)/2, which is not an integer. Also check inequality with constants: 6/3 = 2, and k/(–12) must not equal 2, i.e., k ≠ –24. Since no integer k satisfies the consistency condition for no solution, the answer is 'No such value exists'.