Question

Rezolvati urmatorul sistem de ecuatii

×+y=-1
× . y = -6

Va rogg!

Answer 1

$\left\{\begin{matrix} x + y = - 1\\ x \times y = - 6\end{matrix}\right.$

$x + y = - 1$

$x = - 1 - y$

$x \times y = - 6$

$( - 1 - y) \times y = - 6$

$- y - {y}^{2} = - 6$

$- {y}^{2} - y + 6 = 0$

${y}^{2} + y - 6 = 0$

${y}^{2} + 3y - 2y - 6 = 0$

$y(y + 3) - 2(y + 3) = 0$

$(y + 3)(y - 2) = 0$

$y + 3 = 0 = > y _{1}= - 3$

$y - 2 = 0 = > y_{2} = 2$

$x=-1-y$

$1)x_{1}=-1-(-3)$

$x_{1}=-1+3$

$x_{1}=2$

$2)x_{2}=-1-2$

$x_{2}=-3$

$S=\left \{ (x_{1},y_{1}) ,(x_{2},y_{2})\right \}$

$S=\left \{ (2,-3) ,(-3,2)\right \}$

Answer 2

$\left \{ {{x+y=-1} \atop {x \cdot y=-6}} \right. \\ \\ \left \{ {{x=-1-y} \atop {(-1-y) \cdot y=-6}} \right. \\ \\ \left \{ {{x=-1-y} \atop {-y-y^{2}+6=0$
$- y^{2} -y+6=0$  inmultim cu (-1)

[tex] y^{2}+y-6=0 [/tex]
y²+y-6=0
Este o ecuatie de gradul II cu:
Δ=b²-4×a×c=1²-4×(-6)×1=1+24= 25⇒ Δ = 25,
are Δ pozitiv deci ecuatia are 2 solutii diferite
y₁=(-b+√Δ)/2a=(-1+5)/2=2
y₂=(-b-√Δ)/2a=(-1-5)/2=- 3

daca y₁=2  atunci x₁= -1-y= -1-2=-3       deci y₁= 2  si x₁ = - 3

daca y₂=-3   atunci x₂= -1-y = -1 + 3= 2 ⇒y₂= - 3   si  x₂= 2