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# Umformen Gaußsche Flächenformel

## Tags: Umformen einer Gleichung

17:23 Uhr, 24.07.2020

Hallo,
ich hab ein kleines Problem mit einer Schulaufgabe:

Gegeben ist: $2F\phantom{\rule{0.12em}{0ex}},$ alle anderen Variabeln bis auf $y\phantom{\rule{0.12em}{0ex}}4$
Gesucht ist: $y\phantom{\rule{0.12em}{0ex}}4$

$2F\phantom{\rule{0.12em}{0ex}}=\left(x\phantom{\rule{0.12em}{0ex}}5-x\phantom{\rule{0.12em}{0ex}}2\right)\cdot y\phantom{\rule{0.12em}{0ex}}1+\left(x\phantom{\rule{0.12em}{0ex}}1-x\phantom{\rule{0.12em}{0ex}}3\right)\cdot y\phantom{\rule{0.12em}{0ex}}2+\left(x\phantom{\rule{0.12em}{0ex}}2-x\phantom{\rule{0.12em}{0ex}}4\right)\cdot y\phantom{\rule{0.12em}{0ex}}3+\left(x\phantom{\rule{0.12em}{0ex}}3-x\phantom{\rule{0.12em}{0ex}}5\right)\cdot y\phantom{\rule{0.12em}{0ex}}4+\left(x\phantom{\rule{0.12em}{0ex}}4-x\phantom{\rule{0.12em}{0ex}}1\right)\cdot y\phantom{\rule{0.12em}{0ex}}5$

Mein Ansatz:

$2F\phantom{\rule{0.12em}{0ex}}=\left(x\phantom{\rule{0.12em}{0ex}}5-x\phantom{\rule{0.12em}{0ex}}2\right)\cdot y\phantom{\rule{0.12em}{0ex}}1+\left(x\phantom{\rule{0.12em}{0ex}}1-x\phantom{\rule{0.12em}{0ex}}3\right)\cdot y\phantom{\rule{0.12em}{0ex}}2+\left(x\phantom{\rule{0.12em}{0ex}}2-x\phantom{\rule{0.12em}{0ex}}4\right)\cdot y\phantom{\rule{0.12em}{0ex}}3+\left(x\phantom{\rule{0.12em}{0ex}}3-x\phantom{\rule{0.12em}{0ex}}5\right)\cdot y\phantom{\rule{0.12em}{0ex}}4+\left(x\phantom{\rule{0.12em}{0ex}}4-x\phantom{\rule{0.12em}{0ex}}1\right)\cdot y\phantom{\rule{0.12em}{0ex}}5|:y\phantom{\rule{0.12em}{0ex}}4$

$2F\phantom{\rule{0.12em}{0ex}}:y\phantom{\rule{0.12em}{0ex}}4=\left(x\phantom{\rule{0.12em}{0ex}}5-x\phantom{\rule{0.12em}{0ex}}2\right)\cdot y\phantom{\rule{0.12em}{0ex}}1+\left(x\phantom{\rule{0.12em}{0ex}}1-x\phantom{\rule{0.12em}{0ex}}3\right)\cdot y\phantom{\rule{0.12em}{0ex}}2+\left(x\phantom{\rule{0.12em}{0ex}}2-x\phantom{\rule{0.12em}{0ex}}4\right)\cdot y\phantom{\rule{0.12em}{0ex}}3+\left(x\phantom{\rule{0.12em}{0ex}}3-x\phantom{\rule{0.12em}{0ex}}5\right)+\left(x\phantom{\rule{0.12em}{0ex}}4-x\phantom{\rule{0.12em}{0ex}}1\right)\cdot y\phantom{\rule{0.12em}{0ex}}5|:2F\phantom{\rule{0.12em}{0ex}}$

$y\phantom{\rule{0.12em}{0ex}}4=\left[\left(x\phantom{\rule{0.12em}{0ex}}5-x\phantom{\rule{0.12em}{0ex}}2\right)\cdot y\phantom{\rule{0.12em}{0ex}}1+\left(x\phantom{\rule{0.12em}{0ex}}1-x\phantom{\rule{0.12em}{0ex}}3\right)\cdot y\phantom{\rule{0.12em}{0ex}}2+\left(x\phantom{\rule{0.12em}{0ex}}2-x\phantom{\rule{0.12em}{0ex}}4\right)\cdot y\phantom{\rule{0.12em}{0ex}}3+\left(x\phantom{\rule{0.12em}{0ex}}3-x\phantom{\rule{0.12em}{0ex}}5\right)+\left(x\phantom{\rule{0.12em}{0ex}}4-x\phantom{\rule{0.12em}{0ex}}1\right)\cdot y\phantom{\rule{0.12em}{0ex}}5\right]:2F\phantom{\rule{0.12em}{0ex}}$

Danke für die Hilfe

17:31 Uhr, 24.07.2020

$2F\phantom{\rule{0.12em}{0ex}}=\left({x\phantom{\rule{0.12em}{0ex}}}_{5}-{x\phantom{\rule{0.12em}{0ex}}}_{2}\right)\cdot {y\phantom{\rule{0.12em}{0ex}}}_{1}+\left({x\phantom{\rule{0.12em}{0ex}}}_{1}-{x\phantom{\rule{0.12em}{0ex}}}_{3}\right)\cdot {y\phantom{\rule{0.12em}{0ex}}}_{2}+\left({x\phantom{\rule{0.12em}{0ex}}}_{2}-{x\phantom{\rule{0.12em}{0ex}}}_{4}\right)\cdot {y\phantom{\rule{0.12em}{0ex}}}_{3}+\left({x\phantom{\rule{0.12em}{0ex}}}_{3}-{x\phantom{\rule{0.12em}{0ex}}}_{5}\right)\cdot {y\phantom{\rule{0.12em}{0ex}}}_{4}+\left({x\phantom{\rule{0.12em}{0ex}}}_{4}-{x\phantom{\rule{0.12em}{0ex}}}_{1}\right)\cdot {y\phantom{\rule{0.12em}{0ex}}}_{5}$

$\left({x\phantom{\rule{0.12em}{0ex}}}_{3}-{x\phantom{\rule{0.12em}{0ex}}}_{5}\right)\cdot {y\phantom{\rule{0.12em}{0ex}}}_{4}=2F\phantom{\rule{0.12em}{0ex}}-\left({x\phantom{\rule{0.12em}{0ex}}}_{5}-{x\phantom{\rule{0.12em}{0ex}}}_{2}\right)\cdot {y\phantom{\rule{0.12em}{0ex}}}_{1}-\left({x\phantom{\rule{0.12em}{0ex}}}_{1}-{x\phantom{\rule{0.12em}{0ex}}}_{3}\right)\cdot {y\phantom{\rule{0.12em}{0ex}}}_{2}-\left({x\phantom{\rule{0.12em}{0ex}}}_{2}-{x\phantom{\rule{0.12em}{0ex}}}_{4}\right)\cdot {y\phantom{\rule{0.12em}{0ex}}}_{3}-\left({x\phantom{\rule{0.12em}{0ex}}}_{4}-{x\phantom{\rule{0.12em}{0ex}}}_{1}\right)\cdot {y\phantom{\rule{0.12em}{0ex}}}_{5}$

${y\phantom{\rule{0.12em}{0ex}}}_{4}=\frac{2F\phantom{\rule{0.12em}{0ex}}-\left({x\phantom{\rule{0.12em}{0ex}}}_{5}-{x\phantom{\rule{0.12em}{0ex}}}_{2}\right)\cdot {y\phantom{\rule{0.12em}{0ex}}}_{1}-\left({x\phantom{\rule{0.12em}{0ex}}}_{1}-{x\phantom{\rule{0.12em}{0ex}}}_{3}\right)\cdot {y\phantom{\rule{0.12em}{0ex}}}_{2}-\left({x\phantom{\rule{0.12em}{0ex}}}_{2}-{x\phantom{\rule{0.12em}{0ex}}}_{4}\right)\cdot {y\phantom{\rule{0.12em}{0ex}}}_{3}-\left({x\phantom{\rule{0.12em}{0ex}}}_{4}-{x\phantom{\rule{0.12em}{0ex}}}_{1}\right)\cdot {y\phantom{\rule{0.12em}{0ex}}}_{5}}{{x\phantom{\rule{0.12em}{0ex}}}_{3}-{x\phantom{\rule{0.12em}{0ex}}}_{5}}$

mfG

Atlantik

17:37 Uhr, 24.07.2020

Dein Fehler ist, dass du nur $2F\phantom{\rule{0.12em}{0ex}}$ durch ${y\phantom{\rule{0.12em}{0ex}}}_{4}$ geteilt hast.

mfG

Atlantik

17:47 Uhr, 24.07.2020

Hallo,
vielen Dank für deine Mühe!

Warum nimmst du im zweiten Schritt den Gesamten Ausdruch(x3-x5)y4 "mit"?

18:04 Uhr, 24.07.2020

Wie meinst du das mit gesamten Ausdruck mitnehmen?

mfG

Atlantik

18:08 Uhr, 24.07.2020

1. (x3−x5)⋅y4=2F−(x5−x2)⋅y1−(x1−x3)⋅y2−(x2−x4)⋅y3−(x4−x1)⋅y5

2. $y\phantom{\rule{0.12em}{0ex}}4=$ 2F−(x5−x2)⋅y1−(x1−x3)⋅y2−(x2−x4)⋅y3−(x4−x1)⋅y5 / x3−x5

Wie bist du vom ersten Schritt zum zweiten gekommen? Könntest du das noch etwas genauer erklären?

Warum ändern sich die Vorzeichen

18:20 Uhr, 24.07.2020

Ich wähle mal Buchstaben:

$2F\phantom{\rule{0.12em}{0ex}}=a\phantom{\rule{0.12em}{0ex}}\cdot b\phantom{\rule{0.12em}{0ex}}+c\phantom{\rule{0.12em}{0ex}}\cdot d\phantom{\rule{0.12em}{0ex}}+e\phantom{\rule{0.12em}{0ex}}\cdot f\phantom{\rule{0.12em}{0ex}}+g\phantom{\rule{0.12em}{0ex}}\cdot h\phantom{\rule{0.12em}{0ex}}+i\phantom{\rule{0.12em}{0ex}}\cdot j\phantom{\rule{0.12em}{0ex}}$

Die Gleichung soll nun nach $h\phantom{\rule{0.12em}{0ex}}$ aufgelöst werden.

Ich tausche nun die Seiten: ( Den Schritt habe ich oben im Kopf gemacht)

$a\phantom{\rule{0.12em}{0ex}}\cdot b\phantom{\rule{0.12em}{0ex}}+c\phantom{\rule{0.12em}{0ex}}\cdot d\phantom{\rule{0.12em}{0ex}}+e\phantom{\rule{0.12em}{0ex}}\cdot f\phantom{\rule{0.12em}{0ex}}+g\phantom{\rule{0.12em}{0ex}}\cdot h\phantom{\rule{0.12em}{0ex}}+i\phantom{\rule{0.12em}{0ex}}\cdot j\phantom{\rule{0.12em}{0ex}}=2F\phantom{\rule{0.12em}{0ex}}$

Nun ist doch $g\phantom{\rule{0.12em}{0ex}}\cdot h\phantom{\rule{0.12em}{0ex}}=2F\phantom{\rule{0.12em}{0ex}}-\left(a\phantom{\rule{0.12em}{0ex}}\cdot b\phantom{\rule{0.12em}{0ex}}+c\phantom{\rule{0.12em}{0ex}}\cdot d\phantom{\rule{0.12em}{0ex}}+e\phantom{\rule{0.12em}{0ex}}\cdot f\phantom{\rule{0.12em}{0ex}}+i\phantom{\rule{0.12em}{0ex}}\cdot j\phantom{\rule{0.12em}{0ex}}\right)$

$g\phantom{\rule{0.12em}{0ex}}\cdot h\phantom{\rule{0.12em}{0ex}}=2F\phantom{\rule{0.12em}{0ex}}-a\phantom{\rule{0.12em}{0ex}}\cdot b\phantom{\rule{0.12em}{0ex}}-c\phantom{\rule{0.12em}{0ex}}\cdot d\phantom{\rule{0.12em}{0ex}}-e\phantom{\rule{0.12em}{0ex}}\cdot f\phantom{\rule{0.12em}{0ex}}-i\phantom{\rule{0.12em}{0ex}}\cdot j\phantom{\rule{0.12em}{0ex}}|:g\phantom{\rule{0.12em}{0ex}}$

$h\phantom{\rule{0.12em}{0ex}}=\frac{2F\phantom{\rule{0.12em}{0ex}}-a\phantom{\rule{0.12em}{0ex}}\cdot b\phantom{\rule{0.12em}{0ex}}-c\phantom{\rule{0.12em}{0ex}}\cdot d\phantom{\rule{0.12em}{0ex}}-e\phantom{\rule{0.12em}{0ex}}\cdot f\phantom{\rule{0.12em}{0ex}}-i\phantom{\rule{0.12em}{0ex}}\cdot j\phantom{\rule{0.12em}{0ex}}}{g\phantom{\rule{0.12em}{0ex}}}$

mfG

Atlantik

18:23 Uhr, 24.07.2020

Ahhhh!

Natürlich, vielen Dank für die Erklärung!

Schönes Wochenende!