soujanyareddy13
asked
in Quantitative Aptitude
Sep 17, 2021
edited
Mar 23, 2022
by Lakshman Patel RJIT

240 views
1 vote

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

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