Question

For which value of $k$ will the following pair of linear equations have no solution?
$3x+y=1$ and $(2k-1)x+(k-1)y=2k+1$

Answer 1

We know that if $\frac{a_1}{a_2}=\frac{b_1}{b_2}\ne \frac{c_1}{c_2}$ then the sysytem of linear equations has no solution.

where $a_1=3,b_1=1,a_2=2k-1$ and $b_2=k-1$

$\Rightarrow \frac{3}{2k-1}=\frac{1}{k-1}$

$\Rightarrow \frac{3}{2k-1}-\frac{1}{k-1}=0$

$\Rightarrow \frac{3k-3-2k+1}{(2k-1)(k-1)}=0$ for $k\ne 1,\frac{1}{2}$

$\Rightarrow k-2=0$

$\therefore k=2$