The system of linear equations $kx+5y=5, 2x + 3y = 5$ will be consistent if

$k ≠\frac{10}{3}$

The correct answer is Option (3) → $k ≠\frac{10}{3}$

Given system of equations:

$k x + 5 y = 5 \quad \text{(1)}$

$2 x + 3 y = 5 \quad \text{(2)}$

For the system to be consistent (i.e., to have at least one solution), the two lines must intersect (either at one point or be coincident).

The condition for consistency is that the ratios of coefficients of $x$, $y$, and constants should not all be equal:

If $\frac{a_1}{a_2} \ne \frac{b_1}{b_2}$, the system is consistent and independent (one solution).

If $\frac{a_1}{a_2} = \frac{b_1}{b_2} = \frac{c_1}{c_2}$, the system is consistent and dependent (infinitely many solutions).

If $\frac{a_1}{a_2} = \frac{b_1}{b_2} \ne \frac{c_1}{c_2}$, the system is inconsistent (no solution).

Here:

$\frac{a_1}{a_2} = \frac{k}{2}$, $\frac{b_1}{b_2} = \frac{5}{3}$, $\frac{c_1}{c_2} = \frac{5}{5} = 1$

To be consistent:

$\frac{k}{2} \ne \frac{5}{3}$

$\Rightarrow k \ne \frac{10}{3}$

Hence, the system is consistent for all values of $k$ except $k = \frac{10}{3}$.