Algebra-Lineares Gleichungssystem-Additionsverfahren (2)

Beispiel Nr: 10
$\text{Gegeben:} \\ a1 \cdot x +b1 \cdot y =c1\\ a2 \cdot x +b2 \cdot y =c2 \\ \\ \text{Gesucht:} \\\text{x und y} \\ \\ \textbf{Gegeben:} \\ \\ -1 x +1 y =3\\ \frac{1}{2} x -4 y = 5 \\ \\ \\ \\ \textbf{Rechnung:} \\\begin{array}{l|l} \begin{array}{l} \\I \qquad -1 x +1 y =3\\ II \qquad \frac{1}{2} x -4 y = 5 \\ I \qquad -1 x +1 y =3 \qquad / \cdot\left(-\frac{1}{2}\right)\\ II \qquad \frac{1}{2} x -4 y = 5 \qquad / \cdot\left(-1\right)\\ I \qquad \frac{1}{2} x -\frac{1}{2} y =-1\frac{1}{2}\\ II \qquad -\frac{1}{2} x +4 y = -5 \\ \text{I + II}\\ I \qquad \frac{1}{2} x -\frac{1}{2} x-\frac{1}{2} y +4 y =-1\frac{1}{2} -5\\ 3\frac{1}{2} y = -6\frac{1}{2} \qquad /:3\frac{1}{2} \\ y = \frac{-6\frac{1}{2}}{3\frac{1}{2}} \\ y=-1\frac{6}{7} \\ \text{y in I}\\ I \qquad -1 x +1\cdot \left(-1\frac{6}{7}\right) =3 \\ -1 x -1\frac{6}{7} =3 \qquad / +1\frac{6}{7} \\ -1 x =3 +1\frac{6}{7} \\ -1 x =4\frac{6}{7} \qquad / :\left(-1\right) \\ x = \frac{4\frac{6}{7}}{-1} \\ x=-4\frac{6}{7} \\ L=\{-4\frac{6}{7}/-1\frac{6}{7}\} \end{array} & \begin{array}{l} \\I \qquad -1 x +1 y =3\\ II \qquad \frac{1}{2} x -4 y = 5 \\ I \qquad -1 x +1 y =3 \qquad / \cdot\left(-4\right)\\ II \qquad \frac{1}{2} x -4 y = 5 \qquad / \cdot\left(-1\right)\\ I \qquad 4 x -4 y =-12\\ II \qquad -\frac{1}{2} x +4 y = -5 \\ \text{I + II}\\ I \qquad 4 x -\frac{1}{2} x-4 y +4 y =-12 -5\\ 3\frac{1}{2} x = -17 \qquad /:3\frac{1}{2} \\ x = \frac{-17}{3\frac{1}{2}} \\ x=-4\frac{6}{7} \\ \text{x in I}\\ I \qquad -1 \cdot \left(-4\frac{6}{7}\right) +1y =3 \\ 1 y +4\frac{6}{7} =3 \qquad / -4\frac{6}{7} \\ 1 y =3 -4\frac{6}{7} \\ 1 y =-1\frac{6}{7} \qquad / :1 \\ y = \frac{-1\frac{6}{7}}{1} \\ y=-1\frac{6}{7} \\ L=\{-4\frac{6}{7}/-1\frac{6}{7}\} \end{array} \end{array}$