nichtlineare Gleichungssysteme

442b – FW Algebra

Lösen Sie das System ohne Fallunterscheidung.

$\begin{align*} \left| \begin{array}{rcl} a(x-y) & = & xy \\ b(x+y) & = & xy \end{array} \right| \end{align*}$

Lösung

$\begin{align*} \begin{array}{c} (1) \\ (2) \end{array} & \left| \begin{array}{rcl} a(x-y) & = & xy \\ b(x+y) & = & xy \end{array} \right| \end{align*}$

Einsetzmethode: die Gleichung (1) nach $x$ auflösen

$\begin{eqnarray*} ax-ay-xy & = & 0 \\ (a-y)x & = & ay \\ x & = & \frac{ay}{a-y} ~~~(*) \end{eqnarray*}$

in (2) einsetzen:

$\begin{eqnarray*} b\left(\frac{ay}{a-y}+y\right) & = & \frac{ay^2}{a-y} \\ b\left(\frac{ay+ay-y^2}{a-y}\right) & = & \frac{ay^2}{a-y} ~~~\mid \cdot (a-y) \\ b(2ay-y^2) & = & ay^2 \\ 2aby-by^2 & = & ay^2 \\ ay^2+by^2-2aby & = & 0 \\ (a+b)y^2-2aby & = & 0 \\ y\left[\left(a+b\right)y-2ab\right] & = & 0 ~~~(**) \end{eqnarray*}$

Den ersten Faktor gleich Null setzen:

$\begin{equation*} y_1 = 0 \end{equation*}$

aus (*):

$\begin{equation*} x_1 = 0 \end{equation*}$

Den zweiten Faktor von (**) gleich Null setzen:

$\begin{equation*} y_2 = \frac{2ab}{a+b} \end{equation*}$

aus (*):

$\begin{eqnarray*} x_2 & = & \frac{\frac{2a^2b}{a+b}}{a-\frac{2ab}{a+b}} \\ x_2 & = & \frac{2a^2b}{a^2-ab} \\ x_2 & = & \frac{2ab}{a-b} \end{eqnarray*}$

$\begin{equation*} \mathbb{L}=\left\{\left(0,0\right),\left(\frac{2ab}{a-b},\frac{2ab}{a+b}\right)\right\} \end{equation*}$

439a – FW Algebra

$\begin{align*} \left| \begin{array}{rcl} s^2 - st & = & 20 \\ t^2 - st & = & -5 \end{array} \right| \end{align*}$

Lösung

$\begin{align*} \begin{array}{c} (1) \\ (2) \end{array} & \left| \begin{array}{rcl} s^2 - st & = & 20 \\ t^2 - st & = & -5 \end{array} \right| \begin{array}{c} (*) \\ ~ \end{array} \end{align*}$

$\begin{align*} \begin{array}{c} (1) \\ (2) \end{array} & \left| \begin{array}{rcl} s(s-t) & = & 20 \\ t(t-s) & = & -5 \end{array} \right| \begin{array}{c} \\ \cdot(-1) \end{array} \end{align*}$

$\begin{align*} \begin{array}{c} (1) \\ (2) \end{array} & \left| \begin{array}{rcl} s(s-t) & = & 20 \\ t(s-t) & = & 5 \end{array} \right| \end{align*}$

Die Gleichung (1) durch die Gleichung (2) teilen:

$\begin{eqnarray*} \frac{s}{t} & = & 4 \\ s & = & 4t ~~~(**) \end{eqnarray*}$

in (*) einsetzen:

$\begin{eqnarray*} 16t^2-4t^2 & = & 20 \\ 12t^2 & = & 20 \\ t^2 & = & \frac{5}{3} \\ t_{1,2} & = & \pm\sqrt{\frac{5}{3}} \end{eqnarray*}$

in (**) einsetzen:

$\begin{equation*} s_{1,2} = \pm 4\sqrt{\frac{5}{3}} \end{equation*}$

$\begin{equation*} \mathbb{L}=\left\{\left(4\sqrt{\frac{5}{3}},\sqrt{\frac{5}{3}}\right),\left(-4\sqrt{\frac{5}{3}},-\sqrt{\frac{5}{3}}\right)\right\} \end{equation*}$

438a – FW Algebra

$\begin{align*} \left| \begin{array}{rcl} 2x^2+xy-3y^2 & = & x + 12 \\ x^2-4xy+4y^2 & = & 0 \end{array} \right| \end{align*}$

Lösung

$\begin{align*} \begin{array}{c} (1) \\ (2) \end{array} & \left| \begin{array}{rcl} 2x^2+xy-3y^2 & = & x + 12 \\ x^2-4xy+4y^2 & = & 0 \end{array} \right| \end{align*}$

Die linke Seite der Gleichung (2) lässt sich mit Hilfe der 2. binomischen Formel faktorisieren:

$\begin{eqnarray*} (x-2y)^2 & = & 0 ~~~~\mid \sqrt{\cdot} \\ x-2y & = & 0 \\ x & = & 2y ~~~(*) \end{eqnarray*}$

in (1) einsetzen:

$\begin{eqnarray*} 2\cdot 4y^2+2y^2-3y^2 & = & 2y+12 \\ 7y^2-2y-12 & = & 0 \\ y_{1,2} & = & \frac{2\pm\sqrt{4+336}}{14} \\ y_{1,2} & = & \frac{2\pm\sqrt{340}}{14} \\ y_{1,2} & = & \frac{2\pm2\sqrt{85}}{14} \\ y_{1,2} & = & \frac{1\pm\sqrt{85}}{7} \end{eqnarray*}$

in (*):

$\begin{equation*} x_{1,2} = \frac{2\pm2\sqrt{85}}{7} \end{equation*}$

$\begin{equation*} \mathbb{L}=\left\{\left(\frac{2+2\sqrt{85}}{7},\frac{1+\sqrt{85}}{7}\right),\left(\frac{2-2\sqrt{85}}{7},\frac{1-\sqrt{85}}{7}\right)\right\} \end{equation*}$

435c – FW Algebra

$\begin{align*} \left| \begin{array}{rcl} \sqrt{a-5} + \sqrt{b+2} & = & 5 \\ \sqrt{a+b} & = & 4 \end{array} \right| \end{align*}$

Lösung

$\begin{align*} \begin{array}{c} (1) \\ (2) \end{array} & \left| \begin{array}{rcl} \sqrt{a-5} + \sqrt{b+2} & = & 5 \\ \sqrt{a+b} & = & 4 \end{array} \right| \begin{array}{c} ^{\wedge}2 \\ ^{\wedge}2 \end{array} \end{align*}$

$\begin{align*} \begin{array}{c} (1) \\ (2) \end{array} & \left| \begin{array}{rcl} a-5 + 2\sqrt{(a-5)(b+2)} + b + 2 & = & 25 \\ a + b & = & 16 \end{array} \right| \begin{array}{c} \\ (*) \end{array} \end{align*}$

(2) in (1) einsetzen:

$\begin{eqnarray*} 16 - 5 + 2\sqrt{(a-5)(b+2)} + 2 & = & 25 \\ 2\sqrt{(a-5)(b+2)} & = & 12 \\ \sqrt{(a-5)(b+2)} & = & 6 ~~~\mid ~^{\wedge}2 \\ (a-5)(b+2) & = & 36 \end{eqnarray*}$

Zusammen mit (*) ergibt das

$\begin{align*} \begin{array}{c} (1) \\ (2) \end{array} & \left| \begin{array}{rcl} (a-5)(b+2) & = & 36 \\ a + b & = & 16 \end{array} \right| \begin{array}{c} \\ (**) \end{array} \end{align*}$

Einsetzmethode: (2) nach $a$ auflösen und in (1) einsetzen

$\begin{eqnarray*} (16-b-5)(b+2) & = & 36 \\ (11-b)(b+2) & = & 36 \\ 11b+22-b^2-2b & = & 36 \\ b^2-9b+14 & = & 0 \\ (b-2)(b-7) & = & 0 \\ b_1 & = & 2 \\ b_2 & = & 7 \end{eqnarray*}$

in (**) einsetzen und nach $a$ auflösen:

$\begin{eqnarray*} a_1 & = & 14 \\ a_2 & = & 9 \end{eqnarray*}$

$\begin{equation*} \mathbb{L}=\left\{\left(14,2\right),\left(9,7\right)\right\} \end{equation*}$

434e – FW Algebra

$\begin{align*} \left| \begin{array}{rcl} s^2-st-t^2 & = & 19 \\ s-t & = & 7 \end{array} \right| \end{align*}$

Lösung

$\begin{align*} \begin{array}{c} (1) \\ (2) \end{array} & \left| \begin{array}{rcl} s^2-st-t^2 & = & 19 \\ s-t & = & 7 \end{array} \right| \begin{array}{c} \\ (*) \end{array} \end{align*}$

$\begin{align*} \begin{array}{c} (1) \\ (2) \end{array} & \left| \begin{array}{rcl} s(s-t)-t^2 & = & 19 \\ s-t & = & 7 \end{array} \right| \end{align*}$

(2) in (1):

$\begin{align*} \begin{array}{c} (1) \\ (2) \end{array} & \left| \begin{array}{rcl} s\cdot 7 - t^2 & = & 19 \\ s-t & = & 7 \end{array} \right| \end{align*}$

Einsetzmethode: (2) nach $s$ auflösen und in (1) einsetzen

$\begin{eqnarray*} (7+t)\cdot 7 - t^2 & = & 19 \\ 49 + 7t - t^2 - 19 & = & 0 \\ t^2 - 7t - 30 & = & 0 \\ t_{1,2} & = & \frac{7\pm\sqrt{49+120}}{2} \\ t_1 & = & \frac{7+13}{2} = 10 \\ t_2 & = & \frac{7-13}{2} = -3 \end{eqnarray*}$

in (*):

$\begin{eqnarray*} s_1 & = & 17 \\ s_2 & = & 4 \end{eqnarray*}$

$\begin{equation*} \mathbb{L}=\left\{\left(17,10\right),\left(4,-3\right)\right\} \end{equation*}$