Question

SISTEMA DE EQUAÇAO SEGUNDO GRAU METODO SUBSTITUIÇAO

x - y = -6
3x + 2y = 0


b) x = 1 + y
x/y + y/x = 5/2

a letra b e fraçao a parte de baixo

Answer 1

Ola!!!!!


Questão a

$\left \{ {{x-y=-6} \atop {3x+2y=0}} \right. = \left \{ {{x=-6+y} \atop {3(-6+y)+2y=0}} \right. = \left \{ {{------} \atop {-18+3y+2y=0}} \right. = \left \{ {{---} \atop {5y=18}} \right. = \left \{ {{----} \atop {y= \frac{18}{5} }} \right.$
$= \left \{ {{x=-6+ \frac{18}{5} } \atop {-----}} \right. = \left \{ {{x= \frac{-30+18}{5} } \atop {----}} \right. \left \{ {{x= \frac{-12}{5} } \atop {y= \frac{18}{5} }} \right.$


Questão b
$\left \{ {{x=1+y} \atop { \frac{x}{y} + \frac{y}{x} = \frac{5}{2} }} \right.= \left \{ {{-----} \atop { \frac{ x^{2} + y^{2} }{xy} = \frac{5}{2} }} \right. = \left \{ {{-----} \atop {2 x^{2} +2 y^{2}=5xy }} \right. = \left \{ {{-----} \atop {2(1+y)^{2}+2y^{2} =5y(1+y)}} \right.$
$= \left \{ {{----} \atop {2(1+2y+ y^{2} )+2y^{2} =5y+5y^{2} }} \right. = \left \{ {{----} \atop {2+4y+2y^{2}+2y^{2}-5y-5y^{2} =0}} \right. = \left \{ {{---} \atop {-y^{2}-y+2 =0}} \right.$
$= \left \{ {{---} \atop {y^{2}+y-2 =0}} \right. = \left \{ {{---} \atop {(y+2)(y-1) =0}} \right.$
$y=-2$     $ou$     $y=1$


$x=1+y$
se y=-2
$x=1-2$
$x=-1$

se y=1
$x=1+y$
$x=1+1$
$x=2$