Algebra-Lineares Gleichungssystem-Einsetzverfahren (2)
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Beispiel Nr: 12
$\begin{array}{l}
\text{Gegeben:} \\
a1 \cdot x +b1 \cdot y =c1\\
a2 \cdot x +b2 \cdot y =c2 \\
\\ \text{Gesucht:} \\\text{x und y}
\\ \\ \textbf{Gegeben:} \\
\\
\frac{2}{3}x -\frac{5}{7}y =\frac{2}{3}\\
1x +1y = 10\frac{2}{3} \\
\\
\\ \\ \textbf{Rechnung:} \\\begin{array}{l|l}
\begin{array}{l}
I \qquad \frac{2}{3} x -\frac{5}{7} y =\frac{2}{3}\\
II \qquad 1 x +1 y = 10\frac{2}{3} \\
\text{I nach x auflösen}\\
\frac{2}{3} x -\frac{5}{7} y =\frac{2}{3} \\
\frac{2}{3} x -\frac{5}{7} y =\frac{2}{3} \qquad /+\frac{5}{7} y\\
\frac{2}{3} x =\frac{2}{3} +\frac{5}{7} y \qquad /:\frac{2}{3} \\
x =1 +1\frac{1}{14} y \\
\text{I in II}\\
1 (1 +1\frac{1}{14} y ) + 1 y = 10\frac{2}{3} \\
1 +1\frac{1}{14} y +1 y = 10\frac{2}{3} \qquad / -1 \\
+1\frac{1}{14} y +1 y = 10\frac{2}{3} -1 \\
2\frac{1}{14} y = 9\frac{2}{3} \qquad /:2\frac{1}{14} \\
y = \frac{9\frac{2}{3}}{2\frac{1}{14}} \\
y=4\frac{2}{3} \\
x =1 +1\frac{1}{14} y \\
x =1 +1\frac{1}{14} \cdot 4\frac{2}{3} \\
x=6 \\
L=\{6/4\frac{2}{3}\}
\end{array} &
\begin{array}{l}
I \qquad \frac{2}{3} x -\frac{5}{7} y =\frac{2}{3}\\
II \qquad 1 x +1 y = 10\frac{2}{3} \\
\text{I nach y auflösen}\\
\frac{2}{3} x -\frac{5}{7} y =\frac{2}{3} \\
\frac{2}{3} x -\frac{5}{7} y =\frac{2}{3} \qquad /-\frac{2}{3} x\\
-\frac{5}{7} y =\frac{2}{3} -\frac{2}{3}x \qquad /:\left(-\frac{5}{7}\right) \\
y =-\frac{14}{15} +\frac{14}{15}x \\
\text{I in II}\\
1x + 1(-\frac{14}{15} +\frac{14}{15} x ) = 10\frac{2}{3} \\
-\frac{14}{15} +\frac{14}{15} x +1 x = 10\frac{2}{3} \qquad / -\left(-\frac{14}{15}\right) \\
+\frac{14}{15} x +1 x = 10\frac{2}{3} -\left(-\frac{14}{15}\right) \\
1\frac{14}{15} x = 11\frac{3}{5} \qquad /:1\frac{14}{15} \\
x = \frac{11\frac{3}{5}}{1\frac{14}{15}} \\
x=6 \\
y =-\frac{14}{15} +\frac{14}{15} x \\
y =-\frac{14}{15} +\frac{14}{15} \cdot 6 \\
y=4\frac{2}{3} \\
L=\{6/4\frac{2}{3}\}
\end{array}
\end{array}
\end{array}$