Answer
under $s\ne0, and\,s\ne1$ condition;
$ x_{1}=\frac{-7s}{3s(s-1)}$
$ x_{2}=\frac{4s+3}{6s(s-1)}$
Work Step by Step
Viewing equation in matrix form $A\textbf{x}=\textbf{b} $
$\begin{bmatrix}1&2s\\3&6s\end{bmatrix} \,\,\begin{bmatrix}x_{1}\\x_{2}\end{bmatrix} =\begin{bmatrix}-1\\4\end{bmatrix}$
To have a unique solution, $\textbf{A}$ must be invertible. This happens if and only if $\textbf{det A} \ne0$
$\textbf{det }\begin{bmatrix}1&2s\\3&6s\end{bmatrix}=6s^{2}-6s$
To find the limits on s:
$6s^{2}-6s\ne0$
$6s(s-1)\ne0$
$s\ne0\,and\,s\ne1$
Therefore for a unique solution, $s\ne0, or\,s\ne1$
Compute $A_{i}(\textbf{b})$ by replacing the ith column by vector $\begin{bmatrix}-1\\4\end{bmatrix}$
$A_{1}\textbf{(b)}=\begin{bmatrix}-1&2s\\4&6s\end{bmatrix} ,\, and\,, det(A_{1}\textbf{(b)})=-6s-8s$
$A_{2}\textbf{(b)}=\begin{bmatrix}s&-1\\3&4\end{bmatrix} ,\, and\,, det(A_{2}\textbf{(b)})=4s+3$
By using Cramer's rule, if $s\ne0, and\,s\ne1$, that is, A is Invertible then;
$ x_{1}=\frac{det(A_{1}(\textbf{b}))}{det(A_{})}=\frac{-6s-8s}{6s(s-1)}=\frac{-7s}{3s(s-1)}$
$ x_{2}=\frac{det(A_{2}(\textbf{b}))}{det(A_{})}=\frac{4s+3}{6s(s-1)}$
