For what value of $k$, the system of equations $kx+4y-k+4=0$ and $16x + ky=k$, has an infinite number of solutions?
Answer & explanation
Correct answer: option 3
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