Bonjour, aidez moi svp trouvez moi la résolution de ce système d'équation [tex] \\ y - 1 = x {}^{2} - 2x \\ y - 3x = - 3[/tex] ​

Bonjour,

y - 1 = x² - 2x

y - 3x = - 3

y - 1 = x² - 2x

y = 3x - 3

3x - 3 - 1 = x² - 2x

y = 3x - 3

3x - 4 = x² - 2x

y = 3x - 3

x² - 2x - 3x + 4 = 0

y = 3x - 3

x² - 5x + 4 = 0

∆ = b² - 4ac = 25 - 4 × 1 × 4 = 25 - 16 = 9 > 0

x1 = (-b - √∆)/2a = (5 - 3)/2 = 2/2 = 1

x2 = (-b + √∆)/2 = (5 + 3)/2 = 8/2 = 4

S = { 1 ; 4 }

y = 3 × 1 - 3 = 3 - 3 = 0

ou y = 3 × 4 - 3 = 12 - 3 = 9

Le système admet deux couples de solutions :

x = 1 et y = 0 ou x = 4 et y = 9

Réponse :

Bonsoir,

Explications étape par étape

Voici une autre méthode:

[tex]\left\{\begin{array}{ccc}y-1&=&x^2-2x\\y-3x&=&-3\end{array}\right.\\\\\\\left\{\begin{array}{ccc}y&=&x^2-2x+1\\y&=&3x-3\end{array}\right.\\\\\\\left\{\begin{array}{ccc}y&=&(x-1)^2\\y&=&3(x-1)\end{array}\right.\\\\\\\left\{\begin{array}{ccc}(x-1)^2&=&3(x-1)\\y&=&3(x-1)\end{array}\right.\\\\\\\left\{\begin{array}{ccc}(x-1)*[x-1-3]&=&0\\y&=&3(x-1)\end{array}\right.\\\\\left\{\begin{array}{ccc}(x-1)*(x-4)&=&0\\y&=&3(x-1)\end{array}\right.\\[/tex]

[tex]\left\{\begin{array}{ccc}x&=&1\\y&=&0\end{array}\right.\\\\ou\\\\\left\{\begin{array}{ccc}x&=&4\\y&=&9\end{array}\right.\\[/tex]