Algebra-Lineares Gleichungssystem-Additionsverfahren (2)
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Beispiel Nr: 23
$\begin{array}{l}
\text{Gegeben:} \\
a1 \cdot x +b1 \cdot y =c1\\
a2 \cdot x +b2 \cdot y =c2 \\
\\ \text{Gesucht:} \\\text{x und y}
\\ \\ \textbf{Gegeben:} \\
\\
2 x +2 y =1\frac{7}{10}\\
3 x +6 y = 3 \\
\\
\\ \\ \textbf{Rechnung:} \\\begin{array}{l|l}
\begin{array}{l}
\\I \qquad 2 x +2 y =1\frac{7}{10}\\
II \qquad 3 x +6 y = 3 \\
I \qquad 2 x +2 y =1\frac{7}{10} \qquad / \cdot3\\
II \qquad 3 x +6 y = 3 \qquad / \cdot\left(-2\right)\\
I \qquad 6 x +6 y =5\frac{1}{10}\\
II \qquad -6 x -12 y = -6 \\
\text{I + II}\\
I \qquad 6 x -6 x+6 y -12 y =5\frac{1}{10} -6\\
-6 y = -\frac{9}{10} \qquad /:\left(-6\right) \\
y = \frac{-\frac{9}{10}}{-6} \\
y=\frac{3}{20} \\
\text{y in I}\\
I \qquad 2 x +2\cdot \frac{3}{20} =1\frac{7}{10} \\
2 x +\frac{3}{10} =1\frac{7}{10} \qquad / -\frac{3}{10} \\
2 x =1\frac{7}{10} -\frac{3}{10} \\
2 x =1\frac{2}{5} \qquad / :2 \\
x = \frac{1\frac{2}{5}}{2} \\
x=\frac{7}{10} \\
L=\{\frac{7}{10}/\frac{3}{20}\}
\end{array} &
\begin{array}{l}
\\I \qquad 2 x +2 y =1\frac{7}{10}\\
II \qquad 3 x +6 y = 3 \\
I \qquad 2 x +2 y =1\frac{7}{10} \qquad / \cdot3\\
II \qquad 3 x +6 y = 3 \qquad / \cdot\left(-1\right)\\
I \qquad 6 x +6 y =5\frac{1}{10}\\
II \qquad -3 x -6 y = -3 \\
\text{I + II}\\
I \qquad 6 x -3 x+6 y -6 y =5\frac{1}{10} -3\\
3 x = 2\frac{1}{10} \qquad /:3 \\
x = \frac{2\frac{1}{10}}{3} \\
x=\frac{7}{10} \\
\text{x in I}\\
I \qquad 2 \cdot \frac{7}{10} +2y =1\frac{7}{10} \\
2 y +1\frac{2}{5} =1\frac{7}{10} \qquad / -1\frac{2}{5} \\
2 y =1\frac{7}{10} -1\frac{2}{5} \\
2 y =\frac{3}{10} \qquad / :2 \\
y = \frac{\frac{3}{10}}{2} \\
y=\frac{3}{20} \\
L=\{\frac{7}{10}/\frac{3}{20}\} \end{array}
\end{array}
\end{array}$