$\begin{array}{l} D_h=\begin{array}{|cc|}a1\ & b1 \\ a2&b2 \\ \end{array}= a1 \cdot b2 -b1 \cdot a2 \\ D_x=\begin{array}{|cc|}c1\ & b1 \\ c2&b2 \\ \end{array}= c1 \cdot b2 -b1 \cdot c2 \\ D_y=\begin{array}{|cc|}a1\ & c1 \\ a2&c2 \\ \end{array}= a1 \cdot c2 -c1 \cdot a2\\ x=\frac{D_x}{D_h} \\ y=\frac{D_y}{D_h} \\ \\ \textbf{Gegeben:} \\ \\ 1\frac{1}{2} x -2 y =9\\ \frac{2}{5} x +\frac{1}{3} y = 5 \\ \\ \\ \\ \textbf{Rechnung:} \\ D_h=\begin{array}{|cc|}1\frac{1}{2}\ & -2 \\ \frac{2}{5}&\frac{1}{3} \\ \end{array}= 1\frac{1}{2} \cdot \frac{1}{3} -\left(-2\right) \cdot \frac{2}{5}=1\frac{3}{10} \\ D_x=\begin{array}{|cc|}9\ & -2 \\ 5&\frac{1}{3} \\ \end{array}= 9 \cdot \frac{1}{3} -\left(-2\right) \cdot 5=13 \\ D_y=\begin{array}{|cc|}1\frac{1}{2}\ & 9 \\ \frac{2}{5}&5 \\ \end{array}= 1\frac{1}{2} \cdot 5 -9 \cdot \frac{2}{5}=3\frac{9}{10} \\ \ x=\frac{13}{1\frac{3}{10}} \\ x=10 \\ y=\frac{3\frac{9}{10}}{1\frac{3}{10}} \\ y=3 \\ L=\{10/3\}\\ \, \end{array}$