Question

Find the value of $k$ for which the system of equation $3x+y=1,\left(2k-1\right)x+\left(k-1\right)y=2k+1$ has no solution.

Answer 1

We have,

$3x+y=1$

$(2k-1)x+(k-1)y=2k+1$

For resolution

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

Therefore,

$3(k-1)=2k-1$

$3k-3=2k-1$

$k=2$

$\frac{3}{3}=\frac{1}{1}\ne \frac{1}{5}$

$k=2$

Hence, this is the answer.