Question

For what value of k the following system of linear equations has no solution?
$3x+y=1$
$(2k-1)x+(k-1)y=2k+1$

Answer 1

We know that the system of equations
$a_{1}x+b_{1}y=c_1$
$a_{2}x+b_{2}y=c_2$

has no solution, if
$\frac{a_1}{a_2}=\frac{b_1}{b_2}\ne \frac{c_1}{c_2}$

Here, $a_1=3,b_1=1,c_1=1,a_2=2k-1,b_2=k-1,c_2=2k+1$

So, the given system of equations will have no solution, if

$\frac{3}{2k-1}=\frac{1}{k-1}\ne \frac{1}{2k+1}$

$\frac{3}{2k-1}=\frac{1}{k-1}$ and $\frac{1}{k-1}\ne \frac{1}{2k+1}$

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

Clearly, for $k=2$ we have $\frac{1}{k-1}\ne \frac{1}{2k+1}$

Hence, the given system of equations will have no solution, if $k=2$.