Algebra-Lineares Gleichungssystem-Einsetzverfahren (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 \\
\text{I nach x auflösen}\\
-1\frac{4}{5} x +1\frac{1}{3} y =-1 \\
-1\frac{4}{5} x +1\frac{1}{3} y =-1 \qquad /-1\frac{1}{3} y\\
-1\frac{4}{5} x =-1 -1\frac{1}{3} y \qquad /:\left(-1\frac{4}{5}\right) \\
x =\frac{5}{9} +\frac{20}{27} y \\
\text{I in II}\\
-\frac{2}{3} (\frac{5}{9} +\frac{20}{27} y ) + \frac{1}{9} y = 9 \\
-\frac{10}{27} -\frac{40}{81} y +\frac{1}{9} y = 9 \qquad / -\left(-\frac{10}{27}\right) \\
-\frac{40}{81} y +\frac{1}{9} y = 9 -\left(-\frac{10}{27}\right) \\
-\frac{31}{81} y = 9\frac{10}{27} \qquad /:\left(-\frac{31}{81}\right) \\
y = \frac{9\frac{10}{27}}{-\frac{31}{81}} \\
y=-24\frac{15}{31} \\
x =\frac{5}{9} +\frac{20}{27} y \\
x =\frac{5}{9} +\frac{20}{27} \cdot \left(-24\frac{15}{31}\right) \\
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 \\
\text{I nach y auflösen}\\
-1\frac{4}{5} x +1\frac{1}{3} y =-1 \\
-1\frac{4}{5} x +1\frac{1}{3} y =-1 \qquad /+1\frac{4}{5} x\\
1\frac{1}{3} y =-1 +1\frac{4}{5}x \qquad /:1\frac{1}{3} \\
y =-\frac{3}{4} +1\frac{7}{20}x \\
\text{I in II}\\
-\frac{2}{3}x + \frac{1}{9}(-\frac{3}{4} +1\frac{7}{20} x ) = 9 \\
-\frac{1}{12} +\frac{3}{20} x +\frac{1}{9} x = 9 \qquad / -\left(-\frac{1}{12}\right) \\
+\frac{3}{20} x +\frac{1}{9} x = 9 -\left(-\frac{1}{12}\right) \\
-\frac{31}{60} x = 9\frac{1}{12} \qquad /:\left(-\frac{31}{60}\right) \\
x = \frac{9\frac{1}{12}}{-\frac{31}{60}} \\
x=-17\frac{18}{31} \\
y =-\frac{3}{4} +1\frac{7}{20} x \\
y =-\frac{3}{4} +1\frac{7}{20} \cdot \left(-17\frac{18}{31}\right) \\
y=-24\frac{15}{31} \\
L=\{-17\frac{18}{31}/-24\frac{15}{31}\}
\end{array}
\end{array}
\end{array}$