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Let $\text{k}$ be a constant. The equations $\text{kx + y} = 3$ and $\text{4x + ky} = 4$ have a unique solution if and only if

1. $|\text{k}| \neq 2$
2. $|\text{k}| = 2$
3. $\text{k} \neq 2$
4. $\text{k} = 2$

Given that, the equations.

• $kx + y = 3 \; \longrightarrow (1)$
• $4x + ky = 4 \; \longrightarrow (2)$

For a system of equations.

• $a_{1}x + b_{1}y + c_{1} = 0$
• $a_{2}x + b_{2}y + c_{2} = 0$

to have unique solution, the condition to be satisfied is

$$\boxed{\frac{a_{1}}{a_{2}} \neq \frac{b_{1}}{b_{2}} \neq \frac{c_{1}}{c_{2}}}$$

We can re-write the equation $(1)$ and $(2)$ as follows:

• $kx + y – 3 = 0$
• $4x + ky – 4 =0$

For unique solution $: \frac{k}{4} \neq \frac{1}{k} \neq \frac{-3}{-4}$

Taking first two terms,

$\frac{k}{4} \neq \frac{1}{\text{k}}$

$\Rightarrow k^{2} \neq 4$

$\Rightarrow k \neq \sqrt{4}$

$\Rightarrow k \neq \pm 2$

$\Rightarrow \boxed{|k| \neq 2}$

Correct Answer$: \text{A}$

$\textsf{PS:}$

$1.$ For a system of equation $a_{1}x + b_{1}y + c_{1} =0 ; \; a_{2}x + b_{2}y + c_{2} = 0$ to have no solution, the condition to be satisfied is,

$$\boxed{\frac{a_{1}}{a_{2}} \neq \frac{b_{1}}{b_{2}} \neq \frac{c_{1}}{c_{2}}}$$

$2.$ For a system of equation $a_{1}x + b_{1}y + c_{1} =0 ; \; a_{2}x + b_{2}y + c_{2} = 0$ to have infinitely many solutions, we must have

$$\boxed{\frac{a_{1}}{a_{2}} = \frac{b_{1}}{b_{2}} = \frac{c_{1}}{c_{2}}}$$

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