Algebra-Lineares Gleichungssystem-Determinantenverfahren (2)

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Beispiel Nr: 11
$\begin{array}{l} D_h=\begin{array}{|cc|}a1\ & b1 \\ a2&b2 \\ \end{array}= a1 \cdot b2 -b1 \cdot a2 \\ D_x=\begin{array}{|cc|}c1\ & b1 \\ c2&b2 \\ \end{array}= c1 \cdot b2 -b1 \cdot c2 \\ D_y=\begin{array}{|cc|}a1\ & c1 \\ a2&c2 \\ \end{array}= a1 \cdot c2 -c1 \cdot a2\\ x=\frac{D_x}{D_h} \\ y=\frac{D_y}{D_h} \\ \\ \textbf{Gegeben:} \\ \\ 1\frac{1}{5} x -1\frac{1}{3} y =5\frac{1}{3}\\ 2\frac{1}{2} x -\frac{1}{4} y = 12\frac{3}{8} \\ \\ \\ \\ \textbf{Rechnung:} \\ D_h=\begin{array}{|cc|}1\frac{1}{5}\ & -1\frac{1}{3} \\ 2\frac{1}{2}&-\frac{1}{4} \\ \end{array}= 1\frac{1}{5} \cdot \left(-\frac{1}{4}\right) -\left(-1\frac{1}{3}\right) \cdot 2\frac{1}{2}=3\frac{1}{30} \\ D_x=\begin{array}{|cc|}5\frac{1}{3}\ & -1\frac{1}{3} \\ 12\frac{3}{8}&-\frac{1}{4} \\ \end{array}= 5\frac{1}{3} \cdot \left(-\frac{1}{4}\right) -\left(-1\frac{1}{3}\right) \cdot 12\frac{3}{8}=15\frac{1}{6} \\ D_y=\begin{array}{|cc|}1\frac{1}{5}\ & 5\frac{1}{3} \\ 2\frac{1}{2}&12\frac{3}{8} \\ \end{array}= 1\frac{1}{5} \cdot 12\frac{3}{8} -5\frac{1}{3} \cdot 2\frac{1}{2}=1\frac{31}{60} \\ \ x=\frac{15\frac{1}{6}}{3\frac{1}{30}} \\ x=5 \\ y=\frac{1\frac{31}{60}}{3\frac{1}{30}} \\ y=\frac{1}{2} \\ L=\{5/\frac{1}{2}\}\\ \, \end{array}$