Beispiel Nr: 12
$\text{Gegeben:} \\ a1 \cdot x +b1 \cdot y =c1\\ a2 \cdot x +b2 \cdot y =c2 \\ \\ \text{Gesucht:} \\\text{x und y} \\ \\ \textbf{Gegeben:} \\ \\ \frac{2}{3}x -\frac{5}{7}y =\frac{2}{3}\\ 1x +1y = 10\frac{2}{3} \\ \\ \\ \\ \textbf{Rechnung:} \\\begin{array}{l|l} \begin{array}{l} I \qquad \frac{2}{3} x -\frac{5}{7} y =\frac{2}{3}\\ II \qquad 1 x +1 y = 10\frac{2}{3} \\ \text{I nach x auflösen}\\ \frac{2}{3} x -\frac{5}{7} y =\frac{2}{3} \\ \frac{2}{3} x -\frac{5}{7} y =\frac{2}{3} \qquad /+\frac{5}{7} y\\ \frac{2}{3} x =\frac{2}{3} +\frac{5}{7} y \qquad /:\frac{2}{3} \\ x =1 +1\frac{1}{14} y \\ \text{I in II}\\ 1 (1 +1\frac{1}{14} y ) + 1 y = 10\frac{2}{3} \\ 1 +1\frac{1}{14} y +1 y = 10\frac{2}{3} \qquad / -1 \\ +1\frac{1}{14} y +1 y = 10\frac{2}{3} -1 \\ 2\frac{1}{14} y = 9\frac{2}{3} \qquad /:2\frac{1}{14} \\ y = \frac{9\frac{2}{3}}{2\frac{1}{14}} \\ y=4\frac{2}{3} \\ x =1 +1\frac{1}{14} y \\ x =1 +1\frac{1}{14} \cdot 4\frac{2}{3} \\ x=6 \\ L=\{6/4\frac{2}{3}\} \end{array} & \begin{array}{l} I \qquad \frac{2}{3} x -\frac{5}{7} y =\frac{2}{3}\\ II \qquad 1 x +1 y = 10\frac{2}{3} \\ \text{I nach y auflösen}\\ \frac{2}{3} x -\frac{5}{7} y =\frac{2}{3} \\ \frac{2}{3} x -\frac{5}{7} y =\frac{2}{3} \qquad /-\frac{2}{3} x\\ -\frac{5}{7} y =\frac{2}{3} -\frac{2}{3}x \qquad /:\left(-\frac{5}{7}\right) \\ y =-\frac{14}{15} +\frac{14}{15}x \\ \text{I in II}\\ 1x + 1(-\frac{14}{15} +\frac{14}{15} x ) = 10\frac{2}{3} \\ -\frac{14}{15} +\frac{14}{15} x +1 x = 10\frac{2}{3} \qquad / -\left(-\frac{14}{15}\right) \\ +\frac{14}{15} x +1 x = 10\frac{2}{3} -\left(-\frac{14}{15}\right) \\ 1\frac{14}{15} x = 11\frac{3}{5} \qquad /:1\frac{14}{15} \\ x = \frac{11\frac{3}{5}}{1\frac{14}{15}} \\ x=6 \\ y =-\frac{14}{15} +\frac{14}{15} x \\ y =-\frac{14}{15} +\frac{14}{15} \cdot 6 \\ y=4\frac{2}{3} \\ L=\{6/4\frac{2}{3}\} \end{array} \end{array}$