Algebra-Lineares Gleichungssystem-Additionsverfahren (2)

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Beispiel Nr: 20
$\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:} \\ \\ -1\frac{4}{5} x +1\frac{1}{3} y =-1\\ -\frac{2}{3} x +\frac{1}{9} y = 9 \\ \\ \\ \\ \textbf{Rechnung:} \\\begin{array}{l|l} \begin{array}{l} \\I \qquad -1\frac{4}{5} x +1\frac{1}{3} y =-1\\ II \qquad -\frac{2}{3} x +\frac{1}{9} y = 9 \\ I \qquad -1\frac{4}{5} x +1\frac{1}{3} y =-1 \qquad / \cdot\left(-\frac{2}{3}\right)\\ II \qquad -\frac{2}{3} x +\frac{1}{9} y = 9 \qquad / \cdot1\frac{4}{5}\\ I \qquad 1\frac{1}{5} x -\frac{8}{9} y =\frac{2}{3}\\ II \qquad -1\frac{1}{5} x +\frac{1}{5} y = 16\frac{1}{5} \\ \text{I + II}\\ I \qquad 1\frac{1}{5} x -1\frac{1}{5} x-\frac{8}{9} y +\frac{1}{5} y =\frac{2}{3} +16\frac{1}{5}\\ -\frac{31}{45} y = 16\frac{13}{15} \qquad /:\left(-\frac{31}{45}\right) \\ y = \frac{16\frac{13}{15}}{-\frac{31}{45}} \\ y=-24\frac{15}{31} \\ \text{y in I}\\ I \qquad -1\frac{4}{5} x +1\frac{1}{3}\cdot \left(-24\frac{15}{31}\right) =-1 \\ -1\frac{4}{5} x -32\frac{20}{31} =-1 \qquad / +32\frac{20}{31} \\ -1\frac{4}{5} x =-1 +32\frac{20}{31} \\ -1\frac{4}{5} x =31\frac{20}{31} \qquad / :\left(-1\frac{4}{5}\right) \\ x = \frac{31\frac{20}{31}}{-1\frac{4}{5}} \\ x=-17\frac{18}{31} \\ L=\{-17\frac{18}{31}/-24\frac{15}{31}\} \end{array} & \begin{array}{l} \\I \qquad -1\frac{4}{5} x +1\frac{1}{3} y =-1\\ II \qquad -\frac{2}{3} x +\frac{1}{9} y = 9 \\ I \qquad -1\frac{4}{5} x +1\frac{1}{3} y =-1 \qquad / \cdot\frac{1}{9}\\ II \qquad -\frac{2}{3} x +\frac{1}{9} y = 9 \qquad / \cdot\left(-1\frac{1}{3}\right)\\ I \qquad -\frac{1}{5} x +\frac{4}{27} y =-\frac{1}{9}\\ II \qquad \frac{8}{9} x -\frac{4}{27} y = -12 \\ \text{I + II}\\ I \qquad -\frac{1}{5} x +\frac{8}{9} x+\frac{4}{27} y -\frac{4}{27} y =-\frac{1}{9} -12\\ \frac{31}{45} x = -12\frac{1}{9} \qquad /:\frac{31}{45} \\ x = \frac{-12\frac{1}{9}}{\frac{31}{45}} \\ x=-17\frac{18}{31} \\ \text{x in I}\\ I \qquad -1\frac{4}{5} \cdot \left(-17\frac{18}{31}\right) +1\frac{1}{3}y =-1 \\ 1\frac{1}{3} y +31\frac{20}{31} =-1 \qquad / -31\frac{20}{31} \\ 1\frac{1}{3} y =-1 -31\frac{20}{31} \\ 1\frac{1}{3} y =-32\frac{20}{31} \qquad / :1\frac{1}{3} \\ y = \frac{-32\frac{20}{31}}{1\frac{1}{3}} \\ y=-24\frac{15}{31} \\ L=\{-17\frac{18}{31}/-24\frac{15}{31}\} \end{array} \end{array} \end{array}$