(3)

Dependent Consistent

(4)

$3x+y=1$  ...(i)

and $(2k-1)x+(k-1)y=2k+1$ ...(ii)

Comparing eq. (i) with $a_{1}x+b_{1}y+c_1=0$ and eq. (ii)

with $a_{2}x+b_{2}y+c_2=0,$ we get

$a_1=3, a_2=2k-1, b_1=1, b_2=k-1, c_1=-1$ and $c_2=-(2k+1)$

Since, system is inconsistent, then $\frac{a_1}{a_2}=\frac{b_1}{b_2}\ne \frac{c_1}{c_2}$

$\Rightarrow \frac{3}{2k-1}=\frac{1}{k-1}\ne \frac{-1}{-(2k+1)}\Rightarrow \frac{3}{2k-1}=\frac{1}{k-1} \text{or} \frac{1}{k-1}\ne \frac{1}{2k+1}$

$\Rightarrow 3k-3=2k-1 \text{or} 2k+1\ne k-1\Rightarrow k=2 \text{or} k\ne -2$

Hence, the value of $k$ is 2.

(4)    inconsistent but can be made consistent by extending these lines.

(3)     intersecting at (2a, 3b)

(2)

Given equation are $3x-y+8=0$ and $6x-ky+16=0$

Here, $a_1=3, b_1=-1, c_1=8$

$a_2=6, b_2=-k, c_2=16$

For Infinite many solution, $\frac{a_1}{a_2}=\frac{b_1}{b_2}=\frac{c_1}{c_2}\Rightarrow \frac{3}{6}=\frac{-1}{-k}=\frac{8}{16}\Rightarrow \frac{1}{k}=\frac{1}{2}\Rightarrow k=2$