Beispiel Nr: 17
$\text{Gegeben:} \\ a1 \cdot x +b1 \cdot y =c1\\ a2 \cdot x +b2 \cdot y =c2 \\ \\ \text{Gesucht:} \\\text{x und y} \\ \\ \textbf{Gegeben:} \\ \\ -\frac{1}{2} x +4 y =6\\ -2 x -8 y = 2 \\ \\ \\ \\ \textbf{Rechnung:} \\\begin{array}{l|l} \begin{array}{l} \\I \qquad -\frac{1}{2} x +4 y =6\\ II \qquad -2 x -8 y = 2 \\ I \qquad -\frac{1}{2} x +4 y =6 \qquad / \cdot\left(-2\right)\\ II \qquad -2 x -8 y = 2 \qquad / \cdot\frac{1}{2}\\ I \qquad 1 x -8 y =-12\\ II \qquad -1 x -4 y = 1 \\ \text{I + II}\\ I \qquad 1 x -1 x-8 y -4 y =-12 +1\\ -12 y = -11 \qquad /:\left(-12\right) \\ y = \frac{-11}{-12} \\ y=\frac{11}{12} \\ \text{y in I}\\ I \qquad -\frac{1}{2} x +4\cdot \frac{11}{12} =6 \\ -\frac{1}{2} x +3\frac{2}{3} =6 \qquad / -3\frac{2}{3} \\ -\frac{1}{2} x =6 -3\frac{2}{3} \\ -\frac{1}{2} x =2\frac{1}{3} \qquad / :\left(-\frac{1}{2}\right) \\ x = \frac{2\frac{1}{3}}{-\frac{1}{2}} \\ x=-4\frac{2}{3} \\ L=\{-4\frac{2}{3}/\frac{11}{12}\} \end{array} & \begin{array}{l} \\I \qquad -\frac{1}{2} x +4 y =6\\ II \qquad -2 x -8 y = 2 \\ I \qquad -\frac{1}{2} x +4 y =6 \qquad / \cdot\left(-2\right)\\ II \qquad -2 x -8 y = 2 \qquad / \cdot\left(-1\right)\\ I \qquad 1 x -8 y =-12\\ II \qquad 2 x +8 y = -2 \\ \text{I + II}\\ I \qquad 1 x +2 x-8 y +8 y =-12 -2\\ 3 x = -14 \qquad /:3 \\ x = \frac{-14}{3} \\ x=-4\frac{2}{3} \\ \text{x in I}\\ I \qquad -\frac{1}{2} \cdot \left(-4\frac{2}{3}\right) +4y =6 \\ 4 y +2\frac{1}{3} =6 \qquad / -2\frac{1}{3} \\ 4 y =6 -2\frac{1}{3} \\ 4 y =3\frac{2}{3} \qquad / :4 \\ y = \frac{3\frac{2}{3}}{4} \\ y=\frac{11}{12} \\ L=\{-4\frac{2}{3}/\frac{11}{12}\} \end{array} \end{array}$