Algebra-Lineares Gleichungssystem-Additionsverfahren (2)
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Beispiel Nr: 20
$\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:} \\
\\
-1\frac{4}{5} x +1\frac{1}{3} y =-1\\
-\frac{2}{3} x +\frac{1}{9} y = 9 \\
\\
\\ \\ \textbf{Rechnung:} \\\begin{array}{l|l}
\begin{array}{l}
\\I \qquad -1\frac{4}{5} x +1\frac{1}{3} y =-1\\
II \qquad -\frac{2}{3} x +\frac{1}{9} y = 9 \\
I \qquad -1\frac{4}{5} x +1\frac{1}{3} y =-1 \qquad / \cdot\left(-\frac{2}{3}\right)\\
II \qquad -\frac{2}{3} x +\frac{1}{9} y = 9 \qquad / \cdot1\frac{4}{5}\\
I \qquad 1\frac{1}{5} x -\frac{8}{9} y =\frac{2}{3}\\
II \qquad -1\frac{1}{5} x +\frac{1}{5} y = 16\frac{1}{5} \\
\text{I + II}\\
I \qquad 1\frac{1}{5} x -1\frac{1}{5} x-\frac{8}{9} y +\frac{1}{5} y =\frac{2}{3} +16\frac{1}{5}\\
-\frac{31}{45} y = 16\frac{13}{15} \qquad /:\left(-\frac{31}{45}\right) \\
y = \frac{16\frac{13}{15}}{-\frac{31}{45}} \\
y=-24\frac{15}{31} \\
\text{y in I}\\
I \qquad -1\frac{4}{5} x +1\frac{1}{3}\cdot \left(-24\frac{15}{31}\right) =-1 \\
-1\frac{4}{5} x -32\frac{20}{31} =-1 \qquad / +32\frac{20}{31} \\
-1\frac{4}{5} x =-1 +32\frac{20}{31} \\
-1\frac{4}{5} x =31\frac{20}{31} \qquad / :\left(-1\frac{4}{5}\right) \\
x = \frac{31\frac{20}{31}}{-1\frac{4}{5}} \\
x=-17\frac{18}{31} \\
L=\{-17\frac{18}{31}/-24\frac{15}{31}\}
\end{array} &
\begin{array}{l}
\\I \qquad -1\frac{4}{5} x +1\frac{1}{3} y =-1\\
II \qquad -\frac{2}{3} x +\frac{1}{9} y = 9 \\
I \qquad -1\frac{4}{5} x +1\frac{1}{3} y =-1 \qquad / \cdot\frac{1}{9}\\
II \qquad -\frac{2}{3} x +\frac{1}{9} y = 9 \qquad / \cdot\left(-1\frac{1}{3}\right)\\
I \qquad -\frac{1}{5} x +\frac{4}{27} y =-\frac{1}{9}\\
II \qquad \frac{8}{9} x -\frac{4}{27} y = -12 \\
\text{I + II}\\
I \qquad -\frac{1}{5} x +\frac{8}{9} x+\frac{4}{27} y -\frac{4}{27} y =-\frac{1}{9} -12\\
\frac{31}{45} x = -12\frac{1}{9} \qquad /:\frac{31}{45} \\
x = \frac{-12\frac{1}{9}}{\frac{31}{45}} \\
x=-17\frac{18}{31} \\
\text{x in I}\\
I \qquad -1\frac{4}{5} \cdot \left(-17\frac{18}{31}\right) +1\frac{1}{3}y =-1 \\
1\frac{1}{3} y +31\frac{20}{31} =-1 \qquad / -31\frac{20}{31} \\
1\frac{1}{3} y =-1 -31\frac{20}{31} \\
1\frac{1}{3} y =-32\frac{20}{31} \qquad / :1\frac{1}{3} \\
y = \frac{-32\frac{20}{31}}{1\frac{1}{3}} \\
y=-24\frac{15}{31} \\
L=\{-17\frac{18}{31}/-24\frac{15}{31}\} \end{array}
\end{array}
\end{array}$