Algebra-Lineares Gleichungssystem-Additionsverfahren (2)

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Beispiel Nr: 14
$\begin{array}{l} \text{Gegeben:} \\ a1 \cdot x +b1 \cdot y =c1\\ a2 \cdot x +b2 \cdot y =c2 \\ \\ \text{Gesucht:} \\\text{x und y} \\ \\ \textbf{Gegeben:} \\ \\ 2 x +3 y =4\\ \frac{1}{3} x -\frac{1}{5} y = 12 \\ \\ \\ \\ \textbf{Rechnung:} \\\begin{array}{l|l} \begin{array}{l} \\I \qquad 2 x +3 y =4\\ II \qquad \frac{1}{3} x -\frac{1}{5} y = 12 \\ I \qquad 2 x +3 y =4 \qquad / \cdot\frac{1}{3}\\ II \qquad \frac{1}{3} x -\frac{1}{5} y = 12 \qquad / \cdot\left(-2\right)\\ I \qquad \frac{2}{3} x +1 y =1\frac{1}{3}\\ II \qquad -\frac{2}{3} x +\frac{2}{5} y = -24 \\ \text{I + II}\\ I \qquad \frac{2}{3} x -\frac{2}{3} x+1 y +\frac{2}{5} y =1\frac{1}{3} -24\\ 1\frac{2}{5} y = -22\frac{2}{3} \qquad /:1\frac{2}{5} \\ y = \frac{-22\frac{2}{3}}{1\frac{2}{5}} \\ y=-16\frac{4}{21} \\ \text{y in I}\\ I \qquad 2 x +3\cdot \left(-16\frac{4}{21}\right) =4 \\ 2 x -48\frac{4}{7} =4 \qquad / +48\frac{4}{7} \\ 2 x =4 +48\frac{4}{7} \\ 2 x =52\frac{4}{7} \qquad / :2 \\ x = \frac{52\frac{4}{7}}{2} \\ x=26\frac{2}{7} \\ L=\{26\frac{2}{7}/-16\frac{4}{21}\} \end{array} & \begin{array}{l} \\I \qquad 2 x +3 y =4\\ II \qquad \frac{1}{3} x -\frac{1}{5} y = 12 \\ I \qquad 2 x +3 y =4 \qquad / \cdot\frac{1}{5}\\ II \qquad \frac{1}{3} x -\frac{1}{5} y = 12 \qquad / \cdot3\\ I \qquad \frac{2}{5} x +\frac{3}{5} y =\frac{4}{5}\\ II \qquad 1 x -\frac{3}{5} y = 36 \\ \text{I + II}\\ I \qquad \frac{2}{5} x +1 x+\frac{3}{5} y -\frac{3}{5} y =\frac{4}{5} +36\\ 1\frac{2}{5} x = 36\frac{4}{5} \qquad /:1\frac{2}{5} \\ x = \frac{36\frac{4}{5}}{1\frac{2}{5}} \\ x=26\frac{2}{7} \\ \text{x in I}\\ I \qquad 2 \cdot 26\frac{2}{7} +3y =4 \\ 3 y +52\frac{4}{7} =4 \qquad / -52\frac{4}{7} \\ 3 y =4 -52\frac{4}{7} \\ 3 y =-48\frac{4}{7} \qquad / :3 \\ y = \frac{-48\frac{4}{7}}{3} \\ y=-16\frac{4}{21} \\ L=\{26\frac{2}{7}/-16\frac{4}{21}\} \end{array} \end{array} \end{array}$