$\text{Gegeben:} \\ a1 \cdot x +b1 \cdot y =c1\\ a2 \cdot x +b2 \cdot y =c2 \\ \\ \text{Gesucht:} \\\text{x und y} \\ \\ \textbf{Gegeben:} \\ \\ 1\frac{1}{2} x -2 y =9\\ \frac{2}{5} x +\frac{1}{3} y = 5 \\ \\ \\ \\ \textbf{Rechnung:} \\\begin{array}{l|l} \begin{array}{l} \\I \qquad 1\frac{1}{2} x -2 y =9\\ II \qquad \frac{2}{5} x +\frac{1}{3} y = 5 \\ I \qquad 1\frac{1}{2} x -2 y =9 \qquad / \cdot\frac{2}{5}\\ II \qquad \frac{2}{5} x +\frac{1}{3} y = 5 \qquad / \cdot\left(-1\frac{1}{2}\right)\\ I \qquad \frac{3}{5} x -\frac{4}{5} y =3\frac{3}{5}\\ II \qquad -\frac{3}{5} x -\frac{1}{2} y = -7\frac{1}{2} \\ \text{I + II}\\ I \qquad \frac{3}{5} x -\frac{3}{5} x-\frac{4}{5} y -\frac{1}{2} y =3\frac{3}{5} -7\frac{1}{2}\\ -1\frac{3}{10} y = -3\frac{9}{10} \qquad /:\left(-1\frac{3}{10}\right) \\ y = \frac{-3\frac{9}{10}}{-1\frac{3}{10}} \\ y=3 \\ \text{y in I}\\ I \qquad 1\frac{1}{2} x -2\cdot 3 =9 \\ 1\frac{1}{2} x -6 =9 \qquad / +6 \\ 1\frac{1}{2} x =9 +6 \\ 1\frac{1}{2} x =15 \qquad / :1\frac{1}{2} \\ x = \frac{15}{1\frac{1}{2}} \\ x=10 \\ L=\{10/3\} \end{array} & \begin{array}{l} \\I \qquad 1\frac{1}{2} x -2 y =9\\ II \qquad \frac{2}{5} x +\frac{1}{3} y = 5 \\ I \qquad 1\frac{1}{2} x -2 y =9 \qquad / \cdot\left(-\frac{1}{3}\right)\\ II \qquad \frac{2}{5} x +\frac{1}{3} y = 5 \qquad / \cdot\left(-2\right)\\ I \qquad -\frac{1}{2} x +\frac{2}{3} y =-3\\ II \qquad -\frac{4}{5} x -\frac{2}{3} y = -10 \\ \text{I + II}\\ I \qquad -\frac{1}{2} x -\frac{4}{5} x+\frac{2}{3} y -\frac{2}{3} y =-3 -10\\ -1\frac{3}{10} x = -13 \qquad /:\left(-1\frac{3}{10}\right) \\ x = \frac{-13}{-1\frac{3}{10}} \\ x=10 \\ \text{x in I}\\ I \qquad 1\frac{1}{2} \cdot 10 -2y =9 \\ -2 y +15 =9 \qquad / -15 \\ -2 y =9 -15 \\ -2 y =-6 \qquad / :\left(-2\right) \\ y = \frac{-6}{-2} \\ y=3 \\ L=\{10/3\} \end{array} \end{array}$