$\begin{array}{l} \text{Gegeben:} \\ a1 \cdot x +b1 \cdot y =c1\\ a2 \cdot x +b2 \cdot y =c2 \\ \\ \text{Gesucht:} \\\text{x und y} \\ \\ \textbf{Gegeben:} \\ \\ 1 x +1 y =10\\ 1 x -1 y = 4 \\ \\ \\ \\ \textbf{Rechnung:} \\\begin{array}{l|l} \begin{array}{l} \\I \qquad 1 x +1 y =10\\ II \qquad 1 x -1 y = 4 \\ I \qquad 1 x +1 y =10 \qquad / \cdot1\\ II \qquad 1 x -1 y = 4 \qquad / \cdot\left(-1\right)\\ I \qquad 1 x +1 y =10\\ II \qquad -1 x +1 y = -4 \\ \text{I + II}\\ I \qquad 1 x -1 x+1 y +1 y =10 -4\\ 2 y = 6 \qquad /:2 \\ y = \frac{6}{2} \\ y=3 \\ \text{y in I}\\ I \qquad 1 x +1\cdot 3 =10 \\ 1 x +3 =10 \qquad / -3 \\ 1 x =10 -3 \\ 1 x =7 \qquad / :1 \\ x = \frac{7}{1} \\ x=7 \\ L=\{7/3\} \end{array} & \begin{array}{l} \\I \qquad 1 x +1 y =10\\ II \qquad 1 x -1 y = 4 \\ I \qquad 1 x +1 y =10 \qquad / \cdot\left(-1\right)\\ II \qquad 1 x -1 y = 4 \qquad / \cdot\left(-1\right)\\ I \qquad -1 x -1 y =-10\\ II \qquad -1 x +1 y = -4 \\ \text{I + II}\\ I \qquad -1 x -1 x-1 y +1 y =-10 -4\\ -2 x = -14 \qquad /:\left(-2\right) \\ x = \frac{-14}{-2} \\ x=7 \\ \text{x in I}\\ I \qquad 1 \cdot 7 +1y =10 \\ 1 y +7 =10 \qquad / -7 \\ 1 y =10 -7 \\ 1 y =3 \qquad / :1 \\ y = \frac{3}{1} \\ y=3 \\ L=\{7/3\} \end{array} \end{array} \end{array}$