Solution.

Let \(\mathcal {L}\{x\} = X\) and \(\mathcal {L}\{y\} = Y\)

\begin {align*} \mathcal {L}\{x'\} - 2\mathcal {L}\{x\} + 3 \mathcal {L}\{y\} & = 0 \hspace {0.5cm}\cdots \cdots \hspace {0.5cm} (1)\\\\ \mathcal {L}\{y'\} - \mathcal {L}\{y\} + 2 \mathcal {L}\{x\} & = 0 \hspace {0.5cm}\cdots \cdots \hspace {0.5cm} (2) \end {align*}

\begin {align*} sX - X(0) - 2X + 3Y & = 0\\ sY - Y(0) - Y + 2X & =0 \end {align*}

\begin {align*} (s-2)X + 3Y = 8 \hspace {0.5cm}& \}\times (s-1)\\ 2X + (s-1)Y = 3 \hspace {0.5cm}& \}\times (-3) \end {align*}

\[\implies \hspace {1cm} \big [(s-1)(s-2)\big ] X = 8(s-1) - 9\]

\[\implies \hspace {1cm} \big [s^2 - 3s -4\big ]X = 8s - 17\]

\[X(s) = \frac {8s -17}{s^2 - 3s - 4}\]

\[\implies \hspace {1cm} X(s) = \frac {8s - 17}{(s-4)(s+1)}\]

\[X(s) = \frac {5}{s + 1} + \frac {3}{s - 4}\]

\[x(t) = 5e^{-t} + 3e^{4t}\]

\begin {align*} \implies \hspace {1cm} 3y & = 2x - x'\\\\ x'(t) & = -5e^{-t} + 12 e^{4t}\\\\\\ \therefore 3y & = 2 \big ( 5e^{-t} + 3e^{4t}\big ) - \big (-5e^{-t} + 12e^{4t}\big ) \end {align*}

\begin {align*} \implies \hspace {1cm} 3y & = 15e^{-t} - 6e^{4t}\\\\ \therefore y(t) & = 5e^{-t} - 2e^{4t}\\\\ \end {align*}