Algebra-Lineares Gleichungssystem-Additionsverfahren (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
Beispiel Nr: 09
$\begin{array}{l}
\text{Gegeben:} \\
a1 \cdot x +b1 \cdot y =c1\\
a2 \cdot x +b2 \cdot y =c2 \\
\\ \text{Gesucht:} \\\text{x und y}
\\ \\ \textbf{Gegeben:} \\
\\
-\frac{1}{2} x +1 y =2\\
\frac{1}{2} x -3 y = -3 \\
\\
\\ \\ \textbf{Rechnung:} \\\begin{array}{l|l}
\begin{array}{l}
\\I \qquad -\frac{1}{2} x +1 y =2\\
II \qquad \frac{1}{2} x -3 y = -3 \\
I \qquad -\frac{1}{2} x +1 y =2 \qquad / \cdot\left(-\frac{1}{2}\right)\\
II \qquad \frac{1}{2} x -3 y = -3 \qquad / \cdot\left(-\frac{1}{2}\right)\\
I \qquad \frac{1}{4} x -\frac{1}{2} y =-1\\
II \qquad -\frac{1}{4} x +1\frac{1}{2} y = 1\frac{1}{2} \\
\text{I + II}\\
I \qquad \frac{1}{4} x -\frac{1}{4} x-\frac{1}{2} y +1\frac{1}{2} y =-1 +1\frac{1}{2}\\
1 y = \frac{1}{2} \qquad /:1 \\
y = \frac{\frac{1}{2}}{1} \\
y=\frac{1}{2} \\
\text{y in I}\\
I \qquad -\frac{1}{2} x +1\cdot \frac{1}{2} =2 \\
-\frac{1}{2} x +\frac{1}{2} =2 \qquad / -\frac{1}{2} \\
-\frac{1}{2} x =2 -\frac{1}{2} \\
-\frac{1}{2} x =1\frac{1}{2} \qquad / :\left(-\frac{1}{2}\right) \\
x = \frac{1\frac{1}{2}}{-\frac{1}{2}} \\
x=-3 \\
L=\{-3/\frac{1}{2}\}
\end{array} &
\begin{array}{l}
\\I \qquad -\frac{1}{2} x +1 y =2\\
II \qquad \frac{1}{2} x -3 y = -3 \\
I \qquad -\frac{1}{2} x +1 y =2 \qquad / \cdot\left(-3\right)\\
II \qquad \frac{1}{2} x -3 y = -3 \qquad / \cdot\left(-1\right)\\
I \qquad 1\frac{1}{2} x -3 y =-6\\
II \qquad -\frac{1}{2} x +3 y = 3 \\
\text{I + II}\\
I \qquad 1\frac{1}{2} x -\frac{1}{2} x-3 y +3 y =-6 +3\\
1 x = -3 \qquad /:1 \\
x = \frac{-3}{1} \\
x=-3 \\
\text{x in I}\\
I \qquad -\frac{1}{2} \cdot \left(-3\right) +1y =2 \\
1 y +1\frac{1}{2} =2 \qquad / -1\frac{1}{2} \\
1 y =2 -1\frac{1}{2} \\
1 y =\frac{1}{2} \qquad / :1 \\
y = \frac{\frac{1}{2}}{1} \\
y=\frac{1}{2} \\
L=\{-3/\frac{1}{2}\} \end{array}
\end{array}
\end{array}$