Algebra-Lineares Gleichungssystem-Einsetzverfahren (2)

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