Question

The system kx-y=2 and 6x-2y=3 has a unique solution only when

(a) k=0 (b)$k\ne 0$ (c) k=3 (d) $k\ne 3$

Answer 1

$kx-y=2$
$6x-2y=3$

equation are written as
$kx-y-2=0$....(1)
$6x-2y-3=0$....(2)

Compare the above equations with

$a_{1}x+b_{1}y+c_1=0$ and

$a_{2}x+b_{2}y+c_2=0,$
we get

$a_1=k,b_1=-1,c_1=-2;$

$a_2=6,b_2=-2,c_2=-3;$

Now ,

$\frac{a_1}{a_2}\ne \frac{b_1}{b_2}$

[ Given they have Unique solution ]

$\frac{k}{6}\ne \frac{-1}{-2}$

$\frac{k}{6}\ne \frac{1}{2}$

$k\ne \frac{6}{2}$

$k\ne 3$

(d) $k\ne 3$ is the correct answer