Beispiel Nr: 13
$ \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}{2}x -2y =9\\ \frac{2}{5}x +\frac{1}{3}y = 5 \\ \\ \\ \\ \textbf{Rechnung:} \\\begin{array}{l|l} \begin{array}{l} I \qquad 1\frac{1}{2} x -2 y =9\\ II \qquad \frac{2}{5} x +\frac{1}{3} y = 5 \\ \text{I nach x auflösen}\\ 1\frac{1}{2} x -2 y =9 \\ 1\frac{1}{2} x -2 y =9 \qquad /+2 y\\ 1\frac{1}{2} x =9 +2 y \qquad /:1\frac{1}{2} \\ x =6 +1\frac{1}{3} y \\ \text{I in II}\\ \frac{2}{5} (6 +1\frac{1}{3} y ) + \frac{1}{3} y = 5 \\ 2\frac{2}{5} +\frac{8}{15} y +\frac{1}{3} y = 5 \qquad / -2\frac{2}{5} \\ +\frac{8}{15} y +\frac{1}{3} y = 5 -2\frac{2}{5} \\ \frac{13}{15} y = 2\frac{3}{5} \qquad /:\frac{13}{15} \\ y = \frac{2\frac{3}{5}}{\frac{13}{15}} \\ y=3 \\ x =6 +1\frac{1}{3} y \\ x =6 +1\frac{1}{3} \cdot 3 \\ x=10 \\ L=\{10/3\} \end{array} & \begin{array}{l} I \qquad 1\frac{1}{2} x -2 y =9\\ II \qquad \frac{2}{5} x +\frac{1}{3} y = 5 \\ \text{I nach y auflösen}\\ 1\frac{1}{2} x -2 y =9 \\ 1\frac{1}{2} x -2 y =9 \qquad /-1\frac{1}{2} x\\ -2 y =9 -1\frac{1}{2}x \qquad /:\left(-2\right) \\ y =-4\frac{1}{2} +\frac{3}{4}x \\ \text{I in II}\\ \frac{2}{5}x + \frac{1}{3}(-4\frac{1}{2} +\frac{3}{4} x ) = 5 \\ -1\frac{1}{2} +\frac{1}{4} x +\frac{1}{3} x = 5 \qquad / -\left(-1\frac{1}{2}\right) \\ +\frac{1}{4} x +\frac{1}{3} x = 5 -\left(-1\frac{1}{2}\right) \\ \frac{13}{20} x = 6\frac{1}{2} \qquad /:\frac{13}{20} \\ x = \frac{6\frac{1}{2}}{\frac{13}{20}} \\ x=10 \\ y =-4\frac{1}{2} +\frac{3}{4} x \\ y =-4\frac{1}{2} +\frac{3}{4} \cdot 10 \\ y=3 \\ L=\{10/3\} \end{array} \end{array} $