Answer
$ x_{1}=\frac{3s-2}{3(s^{2}-4)}$
$ x_{2}=\frac{2s-12}{5(s^{2}-4)}$
Work Step by Step
Writing the equation in matrix form $A\textbf{x}=\textbf{b} $
$\begin{bmatrix}3s&5\\12&5s\end{bmatrix} \,\,\begin{bmatrix}x_{1}\\x_{2}\end{bmatrix} =\begin{bmatrix}3\\2\end{bmatrix}$
To have a unique solution, $\textbf{A}$ must be irreversible. This happens if and only if $\textbf{det A} \ne0$
$\textbf{det }\begin{bmatrix}3s&5\\12&5s\end{bmatrix}=15s^{2}-60$
To find the limits on s:
$15s^{2}-60=0$
$15s^{2}=60$
$s^{2}=4$
$s=+2, or\, s=-2$
Therefore for a unique solution, $s\ne+2, or\,s\ne-2$
Compute
$A_{1}\textbf{(b)}=\begin{bmatrix}3&5\\2&5s\end{bmatrix} ,\, and\,, det(A_{1}\textbf{(b)})=15s-10$
$A_{2}\textbf{(b)}=\begin{bmatrix}3s&3\\12&2\end{bmatrix} ,\, and\,, det(A_{2}\textbf{(b)})=6s-36$
By using Cramer's rule, if $s\ne+2, or\,s\ne-2$, that is, A is Invertible then;
$ x_{1}=\frac{det(A_{1}(\textbf{b}))}{det(A_{})}=\frac{15s-10}{15(s^{2}-4)}=\frac{3s-2}{3(s^{2}-4)}$
$ x_{2}=\frac{det(A_{2}(\textbf{b}))}{det(A_{})}=\frac{6s-36}{15(s^{2}-4)}=\frac{2s-12}{5(s^{2}-4)}$
