NMMRUS 3 Loesung
Eierspeise
Wir nennen die Anzahl an Eieren, die der Koch nach Hause getragen hat und nach denen gefragt wird x. Ein Duzent Eier koster p Cent (ein Duzent = 12 Stück).
Failed to parse (MathML with SVG or PNG fallback (recommended for modern browsers and accessibility tools): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle (x-2) \cdot {p \over 12} = 12}
Das sind die Eier, die der Koch tatsächlich bezahlt hat (x-2) - zum ursprünglichen Preis p. p wird durch 12 dividiert, weil p der Preis für ein Duzent Eier ist.
Failed to parse (MathML with SVG or PNG fallback (recommended for modern browsers and accessibility tools): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle x \cdot {{p - 1} \over 12} = 12}
Das sind die Eier, für die der Koch den Preis zurückrechnet - wie erwähnt p-1.
Weil mir das "durch 12" nicht gefällt, werden beide Gleichungen mit 12 multipliziert und dann gleich gesetzt:
Failed to parse (MathML with SVG or PNG fallback (recommended for modern browsers and accessibility tools): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle (x-2) \cdot p = 144}
Failed to parse (MathML with SVG or PNG fallback (recommended for modern browsers and accessibility tools): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle (x-2) \cdot p = x \cdot (p-1)}
Die letzte Gleichung wird ausmultipliziert.
Failed to parse (MathML with SVG or PNG fallback (recommended for modern browsers and accessibility tools): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle xp - 2p = xp - x}
Auf beiden Seiten kann man xp subtrahieren.
Failed to parse (MathML with SVG or PNG fallback (recommended for modern browsers and accessibility tools): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle -2p = -x }
Failed to parse (MathML with SVG or PNG fallback (recommended for modern browsers and accessibility tools): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle 2p=x}
Die letze Erkenntnis kann man in eine der Gleichungen von vorher einsetzen (x=2p).
Failed to parse (MathML with SVG or PNG fallback (recommended for modern browsers and accessibility tools): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle (2p-2) \cdot p = 144}
Failed to parse (MathML with SVG or PNG fallback (recommended for modern browsers and accessibility tools): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle 2p^2-2p=144}
Failed to parse (MathML with SVG or PNG fallback (recommended for modern browsers and accessibility tools): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle p^2-p=72}
Failed to parse (MathML with SVG or PNG fallback (recommended for modern browsers and accessibility tools): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle p^2-p-72=0}
Jetzt heißt es Quadratische Gleichungen auflösen ;-)
Failed to parse (MathML with SVG or PNG fallback (recommended for modern browsers and accessibility tools): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle p_{1,2} = {1 \over 2 } \pm \sqrt{{1 \over 4} + 72}}
Failed to parse (MathML with SVG or PNG fallback (recommended for modern browsers and accessibility tools): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle p_{1,2} = {1 \over 2 } \pm \sqrt{{1 + 4\cdot72} \over 4}}
Failed to parse (MathML with SVG or PNG fallback (recommended for modern browsers and accessibility tools): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle p_{1,2} = {1 \over 2 } \pm {{\sqrt{1+288}} \over 2}}
Failed to parse (MathML with SVG or PNG fallback (recommended for modern browsers and accessibility tools): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle p_{1,2} = {1 \over 2 } \pm { 17 \over 2}}
Da der Preis eine positive Zahl ist (der Koch muss für die Eier Geld hergeben - er bekommt kein Geld dafür, dass er Eier "kauft") - kommt nur eine Lösung für p in Frage - jene mit dem + .
Failed to parse (MathML with SVG or PNG fallback (recommended for modern browsers and accessibility tools): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle p = {1 \over 2 } + { 17 \over 2}}
Failed to parse (MathML with SVG or PNG fallback (recommended for modern browsers and accessibility tools): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle p = {18 \over 2 }}
Failed to parse (MathML with SVG or PNG fallback (recommended for modern browsers and accessibility tools): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle p = 9}
Von vorher wissen wir, dass x=2p ist.
Failed to parse (MathML with SVG or PNG fallback (recommended for modern browsers and accessibility tools): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle x=2\cdot9=18}
Der Koch hat also 18 Eier heimgetragen. Der ursprüngliche Preis für ein Duzent Eier ist 9 Cent.