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