For what value of $k$, the system of equations $kx+4y-k+4=0$ and $16x + ky=k$, has an infinite number of solutions?

8

The correct answer is Option (3) → 8

For infinitely many solutions, the two equations must represent the same line, i.e.

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

Given equations

1. $kx + 4y - k + 4 = 0$

$\Rightarrow kx + 4y = k – 4$

2. $16x + ky = k$

Compare coefficients

$\frac{k}{16} = \frac{4}{k} = \frac{k-4}{k}$

From $\frac{k}{16} = \frac{4}{k}$​

$k^2 = 64 \Rightarrow k = 8 \text{ or } -8$

Check with constants ratio

$\frac{k}{16} = \frac{k-4}{k}$​

For $k = 8$:

$\frac{8}{16} = \frac{4}{8} = \frac{1}{2}$​

Satisfied

For $k = -8$:

$\frac{-8}{16} \neq \frac{-12}{-8}$​

Not satisfied

Correct answer: 8