Question

If the system of equations $3x+y=1,(2k-1)x+(k-1)y-(2k+1)$, has no solution, then the value of $k$ is

• A. $2$
• B. $3$
• C. $-2$
• D. $1$

Answer 1

The correct option is A $2$

$3x+y=1$
$(2k-1)x+(k-1)y=(2k+1)$
$\because$ It has no solution
$\therefore \frac{a_1}{a_2}=\frac{b_1}{b_2}\ne \frac{c_1}{c_2}$
$\frac{3}{2k-1}=\frac{1}{k-1}\ne \frac{1}{2k+1}$
$\Rightarrow 3(k-1)=2k-1$
$\Rightarrow 3k-3=2k-1$
$\Rightarrow 3k-2k=3-1$
$\Rightarrow k=2$