Beispiel Nr: 09
$ 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:} \\ \\ -\frac{1}{2} x +1 y =2\\ \frac{1}{2} x -3 y = -3 \\ \\ \\ \\ \textbf{Rechnung:} \\ D_h=\begin{array}{|cc|}-\frac{1}{2}\ & 1 \\ \frac{1}{2}&-3 \\ \end{array}= -\frac{1}{2} \cdot \left(-3\right) -1 \cdot \frac{1}{2}=1 \\ D_x=\begin{array}{|cc|}2\ & 1 \\ -3&-3 \\ \end{array}= 2 \cdot \left(-3\right) -1 \cdot \left(-3\right)=-3 \\ D_y=\begin{array}{|cc|}-\frac{1}{2}\ & 2 \\ \frac{1}{2}&-3 \\ \end{array}= -\frac{1}{2} \cdot \left(-3\right) -2 \cdot \frac{1}{2}=\frac{1}{2} \\ \ x=\frac{-3}{1} \\ x=-3 \\ y=\frac{\frac{1}{2}}{1} \\ y=\frac{1}{2} \\ L=\{-3/\frac{1}{2}\}\\ \, $