Algebra-Lineares Gleichungssystem-Einsetzverfahren (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 \\ \text{I nach x auflösen}\\ -1\frac{4}{5} x +1\frac{1}{3} y =-1 \\ -1\frac{4}{5} x +1\frac{1}{3} y =-1 \qquad /-1\frac{1}{3} y\\ -1\frac{4}{5} x =-1 -1\frac{1}{3} y \qquad /:\left(-1\frac{4}{5}\right) \\ x =\frac{5}{9} +\frac{20}{27} y \\ \text{I in II}\\ -\frac{2}{3} (\frac{5}{9} +\frac{20}{27} y ) + \frac{1}{9} y = 9 \\ -\frac{10}{27} -\frac{40}{81} y +\frac{1}{9} y = 9 \qquad / -\left(-\frac{10}{27}\right) \\ -\frac{40}{81} y +\frac{1}{9} y = 9 -\left(-\frac{10}{27}\right) \\ -\frac{31}{81} y = 9\frac{10}{27} \qquad /:\left(-\frac{31}{81}\right) \\ y = \frac{9\frac{10}{27}}{-\frac{31}{81}} \\ y=-24\frac{15}{31} \\ x =\frac{5}{9} +\frac{20}{27} y \\ x =\frac{5}{9} +\frac{20}{27} \cdot \left(-24\frac{15}{31}\right) \\ 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 \\ \text{I nach y auflösen}\\ -1\frac{4}{5} x +1\frac{1}{3} y =-1 \\ -1\frac{4}{5} x +1\frac{1}{3} y =-1 \qquad /+1\frac{4}{5} x\\ 1\frac{1}{3} y =-1 +1\frac{4}{5}x \qquad /:1\frac{1}{3} \\ y =-\frac{3}{4} +1\frac{7}{20}x \\ \text{I in II}\\ -\frac{2}{3}x + \frac{1}{9}(-\frac{3}{4} +1\frac{7}{20} x ) = 9 \\ -\frac{1}{12} +\frac{3}{20} x +\frac{1}{9} x = 9 \qquad / -\left(-\frac{1}{12}\right) \\ +\frac{3}{20} x +\frac{1}{9} x = 9 -\left(-\frac{1}{12}\right) \\ -\frac{31}{60} x = 9\frac{1}{12} \qquad /:\left(-\frac{31}{60}\right) \\ x = \frac{9\frac{1}{12}}{-\frac{31}{60}} \\ x=-17\frac{18}{31} \\ y =-\frac{3}{4} +1\frac{7}{20} x \\ y =-\frac{3}{4} +1\frac{7}{20} \cdot \left(-17\frac{18}{31}\right) \\ y=-24\frac{15}{31} \\ L=\{-17\frac{18}{31}/-24\frac{15}{31}\} \end{array} \end{array} \end{array}$