Algebra-Lineares Gleichungssystem-Einsetzverfahren (2)

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Beispiel Nr: 14
$\begin{array}{l} \text{Gegeben:} \\ a1 \cdot x +b1 \cdot y =c1\\ a2 \cdot x +b2 \cdot y =c2 \\ \\ \text{Gesucht:} \\\text{x und y} \\ \\ \textbf{Gegeben:} \\ \\ 2x +3y =4\\ \frac{1}{3}x -\frac{1}{5}y = 12 \\ \\ \\ \\ \textbf{Rechnung:} \\\begin{array}{l|l} \begin{array}{l} I \qquad 2 x +3 y =4\\ II \qquad \frac{1}{3} x -\frac{1}{5} y = 12 \\ \text{I nach x auflösen}\\ 2 x +3 y =4 \\ 2 x +3 y =4 \qquad /-3 y\\ 2 x =4 -3 y \qquad /:2 \\ x =2 -1\frac{1}{2} y \\ \text{I in II}\\ \frac{1}{3} (2 -1\frac{1}{2} y ) + -\frac{1}{5} y = 12 \\ \frac{2}{3} -\frac{1}{2} y -\frac{1}{5} y = 12 \qquad / -\frac{2}{3} \\ -\frac{1}{2} y -\frac{1}{5} y = 12 -\frac{2}{3} \\ -\frac{7}{10} y = 11\frac{1}{3} \qquad /:\left(-\frac{7}{10}\right) \\ y = \frac{11\frac{1}{3}}{-\frac{7}{10}} \\ y=-16\frac{4}{21} \\ x =2 -1\frac{1}{2} y \\ x =2 -1\frac{1}{2} \cdot \left(-16\frac{4}{21}\right) \\ x=26\frac{2}{7} \\ L=\{26\frac{2}{7}/-16\frac{4}{21}\} \end{array} & \begin{array}{l} I \qquad 2 x +3 y =4\\ II \qquad \frac{1}{3} x -\frac{1}{5} y = 12 \\ \text{I nach y auflösen}\\ 2 x +3 y =4 \\ 2 x +3 y =4 \qquad /-2 x\\ 3 y =4 -2x \qquad /:3 \\ y =1\frac{1}{3} -\frac{2}{3}x \\ \text{I in II}\\ \frac{1}{3}x + -\frac{1}{5}(1\frac{1}{3} -\frac{2}{3} x ) = 12 \\ -\frac{4}{15} +\frac{2}{15} x -\frac{1}{5} x = 12 \qquad / -\left(-\frac{4}{15}\right) \\ +\frac{2}{15} x -\frac{1}{5} x = 12 -\left(-\frac{4}{15}\right) \\ \frac{7}{15} x = 12\frac{4}{15} \qquad /:\frac{7}{15} \\ x = \frac{12\frac{4}{15}}{\frac{7}{15}} \\ x=26\frac{2}{7} \\ y =1\frac{1}{3} -\frac{2}{3} x \\ y =1\frac{1}{3} -\frac{2}{3} \cdot 26\frac{2}{7} \\ y=-16\frac{4}{21} \\ L=\{26\frac{2}{7}/-16\frac{4}{21}\} \end{array} \end{array} \end{array}$