edited by
562 views

1 Answer

1 votes
1 votes

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 

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

edited by
Answer:

Related questions

3 votes
3 votes
2 answers
1
soujanyareddy13 asked Sep 17, 2021
926 views
Dick is thrice as old as Tom and Harry is twice as old as Dick. If Dick’s age is $1$ year less than the average age of all three, then Harry’s age, in years, is
1 votes
1 votes
1 answer
3
soujanyareddy13 asked Sep 17, 2021
528 views
If $x_{1} = \;– 1$ and $x_{m} = x_{m+1} + (m + 1)$ for every positive integer $m, $ then $x_{100}$ equals $ – 5151 $$ – 5150 $$ – 5051 $$ – 5050 $