Algebra-Lineares Gleichungssystem-Additionsverfahren (2)
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Beispiel Nr: 14
$\begin{array}{l}
\text{Gegeben:} \\
a1 \cdot x +b1 \cdot y =c1\\
a2 \cdot x +b2 \cdot y =c2 \\
\\ \text{Gesucht:} \\\text{x und y}
\\ \\ \textbf{Gegeben:} \\
\\
2 x +3 y =4\\
\frac{1}{3} x -\frac{1}{5} y = 12 \\
\\
\\ \\ \textbf{Rechnung:} \\\begin{array}{l|l}
\begin{array}{l}
\\I \qquad 2 x +3 y =4\\
II \qquad \frac{1}{3} x -\frac{1}{5} y = 12 \\
I \qquad 2 x +3 y =4 \qquad / \cdot\frac{1}{3}\\
II \qquad \frac{1}{3} x -\frac{1}{5} y = 12 \qquad / \cdot\left(-2\right)\\
I \qquad \frac{2}{3} x +1 y =1\frac{1}{3}\\
II \qquad -\frac{2}{3} x +\frac{2}{5} y = -24 \\
\text{I + II}\\
I \qquad \frac{2}{3} x -\frac{2}{3} x+1 y +\frac{2}{5} y =1\frac{1}{3} -24\\
1\frac{2}{5} y = -22\frac{2}{3} \qquad /:1\frac{2}{5} \\
y = \frac{-22\frac{2}{3}}{1\frac{2}{5}} \\
y=-16\frac{4}{21} \\
\text{y in I}\\
I \qquad 2 x +3\cdot \left(-16\frac{4}{21}\right) =4 \\
2 x -48\frac{4}{7} =4 \qquad / +48\frac{4}{7} \\
2 x =4 +48\frac{4}{7} \\
2 x =52\frac{4}{7} \qquad / :2 \\
x = \frac{52\frac{4}{7}}{2} \\
x=26\frac{2}{7} \\
L=\{26\frac{2}{7}/-16\frac{4}{21}\}
\end{array} &
\begin{array}{l}
\\I \qquad 2 x +3 y =4\\
II \qquad \frac{1}{3} x -\frac{1}{5} y = 12 \\
I \qquad 2 x +3 y =4 \qquad / \cdot\frac{1}{5}\\
II \qquad \frac{1}{3} x -\frac{1}{5} y = 12 \qquad / \cdot3\\
I \qquad \frac{2}{5} x +\frac{3}{5} y =\frac{4}{5}\\
II \qquad 1 x -\frac{3}{5} y = 36 \\
\text{I + II}\\
I \qquad \frac{2}{5} x +1 x+\frac{3}{5} y -\frac{3}{5} y =\frac{4}{5} +36\\
1\frac{2}{5} x = 36\frac{4}{5} \qquad /:1\frac{2}{5} \\
x = \frac{36\frac{4}{5}}{1\frac{2}{5}} \\
x=26\frac{2}{7} \\
\text{x in I}\\
I \qquad 2 \cdot 26\frac{2}{7} +3y =4 \\
3 y +52\frac{4}{7} =4 \qquad / -52\frac{4}{7} \\
3 y =4 -52\frac{4}{7} \\
3 y =-48\frac{4}{7} \qquad / :3 \\
y = \frac{-48\frac{4}{7}}{3} \\
y=-16\frac{4}{21} \\
L=\{26\frac{2}{7}/-16\frac{4}{21}\} \end{array}
\end{array}
\end{array}$