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: 10
$\begin{array}{l}
\text{Gegeben:} \\
a1 \cdot x +b1 \cdot y =c1\\
a2 \cdot x +b2 \cdot y =c2 \\
\\ \text{Gesucht:} \\\text{x und y}
\\ \\ \textbf{Gegeben:} \\
\\
-1x +1y =3\\
\frac{1}{2}x -4y = 5 \\
\\
\\ \\ \textbf{Rechnung:} \\\begin{array}{l|l}
\begin{array}{l}
I \qquad -1 x +1 y =3\\
II \qquad \frac{1}{2} x -4 y = 5 \\
\text{I nach x auflösen}\\
-1 x +1 y =3 \\
-1 x +1 y =3 \qquad /-1 y\\
-1 x =3 -1 y \qquad /:\left(-1\right) \\
x =-3 +1 y \\
\text{I in II}\\
\frac{1}{2} (-3 +1 y ) + -4 y = 5 \\
-1\frac{1}{2} +\frac{1}{2} y -4 y = 5 \qquad / -\left(-1\frac{1}{2}\right) \\
+\frac{1}{2} y -4 y = 5 -\left(-1\frac{1}{2}\right) \\
-3\frac{1}{2} y = 6\frac{1}{2} \qquad /:\left(-3\frac{1}{2}\right) \\
y = \frac{6\frac{1}{2}}{-3\frac{1}{2}} \\
y=-1\frac{6}{7} \\
x =-3 +1 y \\
x =-3 +1 \cdot \left(-1\frac{6}{7}\right) \\
x=-4\frac{6}{7} \\
L=\{-4\frac{6}{7}/-1\frac{6}{7}\}
\end{array} &
\begin{array}{l}
I \qquad -1 x +1 y =3\\
II \qquad \frac{1}{2} x -4 y = 5 \\
\text{I nach y auflösen}\\
-1 x +1 y =3 \\
-1 x +1 y =3 \qquad /+1 x\\
1 y =3 +1x \qquad /:1 \\
y =3 +1x \\
\text{I in II}\\
\frac{1}{2}x + -4(3 +1 x ) = 5 \\
-12 -4 x -4 x = 5 \qquad / -\left(-12\right) \\
-4 x -4 x = 5 -\left(-12\right) \\
-3\frac{1}{2} x = 17 \qquad /:\left(-3\frac{1}{2}\right) \\
x = \frac{17}{-3\frac{1}{2}} \\
x=-4\frac{6}{7} \\
y =3 +1 x \\
y =3 +1 \cdot \left(-4\frac{6}{7}\right) \\
y=-1\frac{6}{7} \\
L=\{-4\frac{6}{7}/-1\frac{6}{7}\}
\end{array}
\end{array}
\end{array}$