Algebra-Lineares Gleichungssystem-Einsetzverfahren (2)
1
2
3
4
5
6
7
8
9
10
11
12
13
14
15
16
17
18
19
20
21
22
23
24
25
26
Beispiel Nr: 14
$\begin{array}{l}
\text{Gegeben:} \\
a1 \cdot x +b1 \cdot y =c1\\
a2 \cdot x +b2 \cdot y =c2 \\
\\ \text{Gesucht:} \\\text{x und y}
\\ \\ \textbf{Gegeben:} \\
\\
2x +3y =4\\
\frac{1}{3}x -\frac{1}{5}y = 12 \\
\\
\\ \\ \textbf{Rechnung:} \\\begin{array}{l|l}
\begin{array}{l}
I \qquad 2 x +3 y =4\\
II \qquad \frac{1}{3} x -\frac{1}{5} y = 12 \\
\text{I nach x auflösen}\\
2 x +3 y =4 \\
2 x +3 y =4 \qquad /-3 y\\
2 x =4 -3 y \qquad /:2 \\
x =2 -1\frac{1}{2} y \\
\text{I in II}\\
\frac{1}{3} (2 -1\frac{1}{2} y ) + -\frac{1}{5} y = 12 \\
\frac{2}{3} -\frac{1}{2} y -\frac{1}{5} y = 12 \qquad / -\frac{2}{3} \\
-\frac{1}{2} y -\frac{1}{5} y = 12 -\frac{2}{3} \\
-\frac{7}{10} y = 11\frac{1}{3} \qquad /:\left(-\frac{7}{10}\right) \\
y = \frac{11\frac{1}{3}}{-\frac{7}{10}} \\
y=-16\frac{4}{21} \\
x =2 -1\frac{1}{2} y \\
x =2 -1\frac{1}{2} \cdot \left(-16\frac{4}{21}\right) \\
x=26\frac{2}{7} \\
L=\{26\frac{2}{7}/-16\frac{4}{21}\}
\end{array} &
\begin{array}{l}
I \qquad 2 x +3 y =4\\
II \qquad \frac{1}{3} x -\frac{1}{5} y = 12 \\
\text{I nach y auflösen}\\
2 x +3 y =4 \\
2 x +3 y =4 \qquad /-2 x\\
3 y =4 -2x \qquad /:3 \\
y =1\frac{1}{3} -\frac{2}{3}x \\
\text{I in II}\\
\frac{1}{3}x + -\frac{1}{5}(1\frac{1}{3} -\frac{2}{3} x ) = 12 \\
-\frac{4}{15} +\frac{2}{15} x -\frac{1}{5} x = 12 \qquad / -\left(-\frac{4}{15}\right) \\
+\frac{2}{15} x -\frac{1}{5} x = 12 -\left(-\frac{4}{15}\right) \\
\frac{7}{15} x = 12\frac{4}{15} \qquad /:\frac{7}{15} \\
x = \frac{12\frac{4}{15}}{\frac{7}{15}} \\
x=26\frac{2}{7} \\
y =1\frac{1}{3} -\frac{2}{3} x \\
y =1\frac{1}{3} -\frac{2}{3} \cdot 26\frac{2}{7} \\
y=-16\frac{4}{21} \\
L=\{26\frac{2}{7}/-16\frac{4}{21}\}
\end{array}
\end{array}
\end{array}$