The system of linear equations $5x + ky = 5$, $3x + 3y = 5$ will be consistent if:

$k \neq 5$

The correct answer is Option (4) → $k \neq 5$ ##

We have, $5x + ky - 5 = 0$

and $3x + 3y - 5 = 0$

where $A = \begin{bmatrix} 5 & k \\ 3 & 3 \end{bmatrix}$, $X = \begin{bmatrix} x \\ y \end{bmatrix}$ and $B = \begin{bmatrix} 5 \\ 5 \end{bmatrix}$

Let $AX = B$

Then,

$A = \begin{bmatrix} 5 & k \\ 3 & 3 \end{bmatrix}$

$|A| = 15 - 3k$

The linear equations will form a consistent system if

$|A| \neq 0$

$15 - 3k \neq 0$

$3k \neq 15$

$k \neq 5$