Question

Find the value of $k$ for which the system of equations $kx-y=2$ and $6x-2y=3$ has no solution.

Answer 1

Let $kx-y=2$ ....(1)
and $kx-2=y$ .....(2)

Substituting the value of $y$ in (1) from (2), we get
$6x-2(kx-2)=3$
$\Rightarrow 6x-2kx-4=3$
$\Rightarrow x(6-2k)=7$

For no solution, we know
$6-2k=0$
$\Rightarrow 6=2k$
$\Rightarrow k=3$