Math  /  Algebra

QuestionRésoudre dans R2\mathrm{R}^{2}, le système d'équations suivant : {xy=43200(x2)(y+300)=43200\left\{\begin{array}{l}x y=43200 \\ (x-2)(y+300)=43200\end{array}\right.

Studdy Solution
Vérifier les solutions dans le système original:
Pour (x,y)=(18,2400) (x, y) = (18, 2400) :
xy=18×2400=43200 xy = 18 \times 2400 = 43200 (x2)(y+300)=(182)(2400+300)=16×2700=43200 (x-2)(y+300) = (18-2)(2400+300) = 16 \times 2700 = 43200
Pour (x,y)=(16,2700) (x, y) = (-16, -2700) :
xy=16×2700=43200 xy = -16 \times -2700 = 43200 (x2)(y+300)=(162)(2700+300)=18×2400=43200 (x-2)(y+300) = (-16-2)(-2700+300) = -18 \times -2400 = 43200
Les deux solutions vérifient le système.
Les solutions sont (x,y)=(18,2400) (x, y) = (18, 2400) et (x,y)=(16,2700) (x, y) = (-16, -2700) .

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