To be specific here is the system:
$$x-2y=0 \tag{1}$$ $$x-2(k+2)y=0 \tag{2}$$ $$x-(k+3)y=-k \tag{3}$$
I have already solved it for equations $(1)$ and $(2)$... what should I do with the 3rd equation?
Just to make sure everything goes well here is my method:
$D=-2(k+2)$ and $D_x=D_y=0$
If $k=-2$ then $D=0$ so there are indefinite solutions. If $k\not=-2$ then $D\not=0$ so the solution is $(0,0)$
On
Having the unknowns $x$ and $y$ one can write this system of linear equations as $$ A u = b \iff \\ \left[ \begin{array}{rr|r} 1 & -2 & 0 \\ 1 & -2(k+2) & 0 \\ 1 & -(k+3) & -k \end{array} \right] \to \left[ \begin{array}{rr|r} 1 & -2 & 0 \\ 0 & -2k-2 & 0 \\ 0 & -k-1 & -k \end{array} \right] \to \left[ \begin{array}{rr|r} 1 & -2 & 0 \\ 0 & k+1 & 0 \\ 0 & k+1 & k \end{array} \right] \to \left[ \begin{array}{rr|r} 1 & -2 & 0 \\ 0 & k+1 & 0 \\ 0 & 0 & k \end{array} \right] \to \left[ \begin{array}{rr|r} 1 & 0 & 0 \\ 0 & 1 & 0 \\ 0 & 0 & k \end{array} \right] $$ So after some transformations we notice that only $x = y = 0$ for $k=0$ is possible as solution.
From first equation $x=2y$. Substitute it into 2 and 3: $$ 2y-2(k+2)y=0,\\ 2y-y(k+3)=-k, $$ or $$ y(k+1)=0,\\ y(k+1) = -k $$ If $k\ne 0$, there are no solutions. If $k=0$ then $y=0$ and $x=0$.