Algebra-Lineares Gleichungssystem-Additionsverfahren (2)

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Beispiel Nr: 23
$\begin{array}{l} \text{Gegeben:} \\ a1 \cdot x +b1 \cdot y =c1\\ a2 \cdot x +b2 \cdot y =c2 \\ \\ \text{Gesucht:} \\\text{x und y} \\ \\ \textbf{Gegeben:} \\ \\ 2 x +2 y =1\frac{7}{10}\\ 3 x +6 y = 3 \\ \\ \\ \\ \textbf{Rechnung:} \\\begin{array}{l|l} \begin{array}{l} \\I \qquad 2 x +2 y =1\frac{7}{10}\\ II \qquad 3 x +6 y = 3 \\ I \qquad 2 x +2 y =1\frac{7}{10} \qquad / \cdot3\\ II \qquad 3 x +6 y = 3 \qquad / \cdot\left(-2\right)\\ I \qquad 6 x +6 y =5\frac{1}{10}\\ II \qquad -6 x -12 y = -6 \\ \text{I + II}\\ I \qquad 6 x -6 x+6 y -12 y =5\frac{1}{10} -6\\ -6 y = -\frac{9}{10} \qquad /:\left(-6\right) \\ y = \frac{-\frac{9}{10}}{-6} \\ y=\frac{3}{20} \\ \text{y in I}\\ I \qquad 2 x +2\cdot \frac{3}{20} =1\frac{7}{10} \\ 2 x +\frac{3}{10} =1\frac{7}{10} \qquad / -\frac{3}{10} \\ 2 x =1\frac{7}{10} -\frac{3}{10} \\ 2 x =1\frac{2}{5} \qquad / :2 \\ x = \frac{1\frac{2}{5}}{2} \\ x=\frac{7}{10} \\ L=\{\frac{7}{10}/\frac{3}{20}\} \end{array} & \begin{array}{l} \\I \qquad 2 x +2 y =1\frac{7}{10}\\ II \qquad 3 x +6 y = 3 \\ I \qquad 2 x +2 y =1\frac{7}{10} \qquad / \cdot3\\ II \qquad 3 x +6 y = 3 \qquad / \cdot\left(-1\right)\\ I \qquad 6 x +6 y =5\frac{1}{10}\\ II \qquad -3 x -6 y = -3 \\ \text{I + II}\\ I \qquad 6 x -3 x+6 y -6 y =5\frac{1}{10} -3\\ 3 x = 2\frac{1}{10} \qquad /:3 \\ x = \frac{2\frac{1}{10}}{3} \\ x=\frac{7}{10} \\ \text{x in I}\\ I \qquad 2 \cdot \frac{7}{10} +2y =1\frac{7}{10} \\ 2 y +1\frac{2}{5} =1\frac{7}{10} \qquad / -1\frac{2}{5} \\ 2 y =1\frac{7}{10} -1\frac{2}{5} \\ 2 y =\frac{3}{10} \qquad / :2 \\ y = \frac{\frac{3}{10}}{2} \\ y=\frac{3}{20} \\ L=\{\frac{7}{10}/\frac{3}{20}\} \end{array} \end{array} \end{array}$