For what value of $K$, the system of equations $3x - ky - 3 = 0$ and $2x-3y - 4 = 0$ has no solution?

9/2

The correct answer is Option (4) → 9/2

For a system of linear equations to have no solution, the lines must be parallel and distinct, i.e.:

$\frac{a_1}{a_2} = \frac{b_1}{b_2} \ne \frac{c_1}{c_2}$

Given equations:

1. $3x - ky - 3 = 0$
2. $2x - 3y - 4 = 0$

Compare coefficients:

$\frac{3}{2} = \frac{-k}{-3}$

$\frac{3}{2} = \frac{k}{3} \Rightarrow k = \frac{9}{2}$