Question

The value of k for which the system of equations $kx - y = 2$, $6x - 2y = 3$ has a unique solution is ,
A) 3
B) $ \ne 3$
c) $ \ne 0$
D) $0$

Answer 1

Hint:We have given two linear equations & have a unique solution.Linear equations have unique solution if $\dfrac{{{a_1}}}{{{a_2}}} \ne \dfrac{{{b_1}}}{{{b_2}}}$ with the help of this we can find the value of k.

Complete step-by-step answer:
We have a system of equations.
$kx - y = 2$
$ \Rightarrow kx - y - 2 = 0……(1)$
Second equation is,
$6x - 2y = 3$
$ \Rightarrow 6x - 2y - 3 = 0 ……(2)$
We have to compare these two equations with the standard form of linear equations.
Compare the first equation.
${a_1}x + {b_1}y + {c_1} = 0$
$kx - y - 2 = 0$
Here we get,
${a_1} = k,{b_1} = - 1,{c_1} = - 2$
Compare the second equation.
${a_2}x + {b_2}y + {c_2} = 0$
$6x - 2y - 3 = 0$
Here we get, ${a_2} = 6,{b_2} = - 2,{c_2} = - 3$
Now, to find k we have to put all these values in$\dfrac{{{a_1}}}{{{a_2}}} \ne \dfrac{{{b_1}}}{{{b_2}}}$
$\dfrac{{{a_1}}}{{{a_2}}} \ne \dfrac{{{b_1}}}{{{b_2}}}$
$ \Rightarrow \dfrac{k}{6} \ne \dfrac{{ - 1}}{{ - 2}} = \dfrac{1}{2}$
To find k, multiply both sides by 6.
Then we will get,
$ \Rightarrow \dfrac{{6k}}{6} \ne \dfrac{{ - 1}}{{ - 2}} = \dfrac{1}{2} \times 6$
$ \Rightarrow k \ne 3$

So, the correct answer is “Option B”.

Note:We conclude that, for all real values of k, except $k \ne 3$, equations have a unique solution. Students should remember the condition for linear equations to have a unique solution.