Algebra-Lineares Gleichungssystem-Determinantenverfahren (2)

Beispiel Nr: 12
$D_h=\begin{array}{|cc|}a1\ & b1 \\ a2&b2 \\ \end{array}= a1 \cdot b2 -b1 \cdot a2 \\ D_x=\begin{array}{|cc|}c1\ & b1 \\ c2&b2 \\ \end{array}= c1 \cdot b2 -b1 \cdot c2 \\ D_y=\begin{array}{|cc|}a1\ & c1 \\ a2&c2 \\ \end{array}= a1 \cdot c2 -c1 \cdot a2\\ x=\frac{D_x}{D_h} \\ y=\frac{D_y}{D_h} \\ \\ \textbf{Gegeben:} \\ \\ \frac{2}{3} x -\frac{5}{7} y =\frac{2}{3}\\ 1 x +1 y = 10\frac{2}{3} \\ \\ \\ \\ \textbf{Rechnung:} \\ D_h=\begin{array}{|cc|}\frac{2}{3}\ & -\frac{5}{7} \\ 1&1 \\ \end{array}= \frac{2}{3} \cdot 1 -\left(-\frac{5}{7}\right) \cdot 1=1\frac{8}{21} \\ D_x=\begin{array}{|cc|}\frac{2}{3}\ & -\frac{5}{7} \\ 10\frac{2}{3}&1 \\ \end{array}= \frac{2}{3} \cdot 1 -\left(-\frac{5}{7}\right) \cdot 10\frac{2}{3}=8\frac{2}{7} \\ D_y=\begin{array}{|cc|}\frac{2}{3}\ & \frac{2}{3} \\ 1&10\frac{2}{3} \\ \end{array}= \frac{2}{3} \cdot 10\frac{2}{3} -\frac{2}{3} \cdot 1=6\frac{4}{9} \\ \ x=\frac{8\frac{2}{7}}{1\frac{8}{21}} \\ x=6 \\ y=\frac{6\frac{4}{9}}{1\frac{8}{21}} \\ y=4\frac{2}{3} \\ L=\{6/4\frac{2}{3}\}\\ \,$