Algebra-Lineares Gleichungssystem-Determinantenverfahren (2)

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Beispiel Nr: 14
$\begin{array}{l} D_h=\begin{array}{|cc|}a1\ & b1 \\ a2&b2 \\ \end{array}= a1 \cdot b2 -b1 \cdot a2 \\ D_x=\begin{array}{|cc|}c1\ & b1 \\ c2&b2 \\ \end{array}= c1 \cdot b2 -b1 \cdot c2 \\ D_y=\begin{array}{|cc|}a1\ & c1 \\ a2&c2 \\ \end{array}= a1 \cdot c2 -c1 \cdot a2\\ x=\frac{D_x}{D_h} \\ y=\frac{D_y}{D_h} \\ \\ \textbf{Gegeben:} \\ \\ 2 x +3 y =4\\ \frac{1}{3} x -\frac{1}{5} y = 12 \\ \\ \\ \\ \textbf{Rechnung:} \\ D_h=\begin{array}{|cc|}2\ & 3 \\ \frac{1}{3}&-\frac{1}{5} \\ \end{array}= 2 \cdot \left(-\frac{1}{5}\right) -3 \cdot \frac{1}{3}=-1\frac{2}{5} \\ D_x=\begin{array}{|cc|}4\ & 3 \\ 12&-\frac{1}{5} \\ \end{array}= 4 \cdot \left(-\frac{1}{5}\right) -3 \cdot 12=-36\frac{4}{5} \\ D_y=\begin{array}{|cc|}2\ & 4 \\ \frac{1}{3}&12 \\ \end{array}= 2 \cdot 12 -4 \cdot \frac{1}{3}=22\frac{2}{3} \\ \ x=\frac{-36\frac{4}{5}}{-1\frac{2}{5}} \\ x=26\frac{2}{7} \\ y=\frac{22\frac{2}{3}}{-1\frac{2}{5}} \\ y=-16\frac{4}{21} \\ L=\{26\frac{2}{7}/-16\frac{4}{21}\}\\ \, \end{array}$