Algebra-Lineares Gleichungssystem-Einsetzverfahren (2)

Beispiel Nr: 11
$\text{Gegeben:} \\ a1 \cdot x +b1 \cdot y =c1\\ a2 \cdot x +b2 \cdot y =c2 \\ \\ \text{Gesucht:} \\\text{x und y} \\ \\ \textbf{Gegeben:} \\ \\ 1\frac{1}{5}x -1\frac{1}{3}y =5\frac{1}{3}\\ 2\frac{1}{2}x -\frac{1}{4}y = 12\frac{3}{8} \\ \\ \\ \\ \textbf{Rechnung:} \\\begin{array}{l|l} \begin{array}{l} I \qquad 1\frac{1}{5} x -1\frac{1}{3} y =5\frac{1}{3}\\ II \qquad 2\frac{1}{2} x -\frac{1}{4} y = 12\frac{3}{8} \\ \text{I nach x auflösen}\\ 1\frac{1}{5} x -1\frac{1}{3} y =5\frac{1}{3} \\ 1\frac{1}{5} x -1\frac{1}{3} y =5\frac{1}{3} \qquad /+1\frac{1}{3} y\\ 1\frac{1}{5} x =5\frac{1}{3} +1\frac{1}{3} y \qquad /:1\frac{1}{5} \\ x =4\frac{4}{9} +1\frac{1}{9} y \\ \text{I in II}\\ 2\frac{1}{2} (4\frac{4}{9} +1\frac{1}{9} y ) + -\frac{1}{4} y = 12\frac{3}{8} \\ 11\frac{1}{9} +2\frac{7}{9} y -\frac{1}{4} y = 12\frac{3}{8} \qquad / -11\frac{1}{9} \\ +2\frac{7}{9} y -\frac{1}{4} y = 12\frac{3}{8} -11\frac{1}{9} \\ 2\frac{19}{36} y = 1\frac{19}{72} \qquad /:2\frac{19}{36} \\ y = \frac{1\frac{19}{72}}{2\frac{19}{36}} \\ y=\frac{1}{2} \\ x =4\frac{4}{9} +1\frac{1}{9} y \\ x =4\frac{4}{9} +1\frac{1}{9} \cdot \frac{1}{2} \\ x=5 \\ L=\{5/\frac{1}{2}\} \end{array} & \begin{array}{l} I \qquad 1\frac{1}{5} x -1\frac{1}{3} y =5\frac{1}{3}\\ II \qquad 2\frac{1}{2} x -\frac{1}{4} y = 12\frac{3}{8} \\ \text{I nach y auflösen}\\ 1\frac{1}{5} x -1\frac{1}{3} y =5\frac{1}{3} \\ 1\frac{1}{5} x -1\frac{1}{3} y =5\frac{1}{3} \qquad /-1\frac{1}{5} x\\ -1\frac{1}{3} y =5\frac{1}{3} -1\frac{1}{5}x \qquad /:\left(-1\frac{1}{3}\right) \\ y =-4 +\frac{9}{10}x \\ \text{I in II}\\ 2\frac{1}{2}x + -\frac{1}{4}(-4 +\frac{9}{10} x ) = 12\frac{3}{8} \\ 1 -\frac{9}{40} x -\frac{1}{4} x = 12\frac{3}{8} \qquad / -1 \\ -\frac{9}{40} x -\frac{1}{4} x = 12\frac{3}{8} -1 \\ 2\frac{11}{40} x = 11\frac{3}{8} \qquad /:2\frac{11}{40} \\ x = \frac{11\frac{3}{8}}{2\frac{11}{40}} \\ x=5 \\ y =-4 +\frac{9}{10} x \\ y =-4 +\frac{9}{10} \cdot 5 \\ y=\frac{1}{2} \\ L=\{5/\frac{1}{2}\} \end{array} \end{array}$