Algebra-Lineares Gleichungssystem-Additionsverfahren (2)

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Beispiel Nr: 12
$\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:} \\ \\ \frac{2}{3} x -\frac{5}{7} y =\frac{2}{3}\\ 1 x +1 y = 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} \\ I \qquad \frac{2}{3} x -\frac{5}{7} y =\frac{2}{3} \qquad / \cdot1\\ II \qquad 1 x +1 y = 10\frac{2}{3} \qquad / \cdot\left(-\frac{2}{3}\right)\\ I \qquad \frac{2}{3} x -\frac{5}{7} y =\frac{2}{3}\\ II \qquad -\frac{2}{3} x -\frac{2}{3} y = -7\frac{1}{9} \\ \text{I + II}\\ I \qquad \frac{2}{3} x -\frac{2}{3} x-\frac{5}{7} y -\frac{2}{3} y =\frac{2}{3} -7\frac{1}{9}\\ -1\frac{8}{21} y = -6\frac{4}{9} \qquad /:\left(-1\frac{8}{21}\right) \\ y = \frac{-6\frac{4}{9}}{-1\frac{8}{21}} \\ y=4\frac{2}{3} \\ \text{y in I}\\ I \qquad \frac{2}{3} x -\frac{5}{7}\cdot 4\frac{2}{3} =\frac{2}{3} \\ \frac{2}{3} x -3\frac{1}{3} =\frac{2}{3} \qquad / +3\frac{1}{3} \\ \frac{2}{3} x =\frac{2}{3} +3\frac{1}{3} \\ \frac{2}{3} x =4 \qquad / :\frac{2}{3} \\ x = \frac{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} \\ I \qquad \frac{2}{3} x -\frac{5}{7} y =\frac{2}{3} \qquad / \cdot\left(-1\right)\\ II \qquad 1 x +1 y = 10\frac{2}{3} \qquad / \cdot\left(-\frac{5}{7}\right)\\ I \qquad -\frac{2}{3} x +\frac{5}{7} y =-\frac{2}{3}\\ II \qquad -\frac{5}{7} x -\frac{5}{7} y = -7\frac{13}{21} \\ \text{I + II}\\ I \qquad -\frac{2}{3} x -\frac{5}{7} x+\frac{5}{7} y -\frac{5}{7} y =-\frac{2}{3} -7\frac{13}{21}\\ -1\frac{8}{21} x = -8\frac{2}{7} \qquad /:\left(-1\frac{8}{21}\right) \\ x = \frac{-8\frac{2}{7}}{-1\frac{8}{21}} \\ x=6 \\ \text{x in I}\\ I \qquad \frac{2}{3} \cdot 6 -\frac{5}{7}y =\frac{2}{3} \\ -\frac{5}{7} y +4 =\frac{2}{3} \qquad / -4 \\ -\frac{5}{7} y =\frac{2}{3} -4 \\ -\frac{5}{7} y =-3\frac{1}{3} \qquad / :\left(-\frac{5}{7}\right) \\ y = \frac{-3\frac{1}{3}}{-\frac{5}{7}} \\ y=4\frac{2}{3} \\ L=\{6/4\frac{2}{3}\} \end{array} \end{array} \end{array}$