Question

me ajuda ai pessoal urgente​

Answer 1

$\left \{ {{12x+6y=24} \atop {x^2+4y=4}} \right.$

$\left \{ {{24x+12y=48} \atop {3x^2+12y=12}} \right.$

$(24x+12y)-(3x^2+12y)=48-12$

$24x+12y-3x^2-12y=36$

$-3x^2+24x=36$

$-3x^2+24x-36=0$

$3x^2-24x+36=0$

$x^2-8x+12=0$

$\triangle=(-8)^2-4.1.12=64-48=16$

$x_1=\frac{8+\sqrt{16} }{2}=\frac{8+4}{2}=\frac{12}{2}=6$

$x_2=\frac{8-\sqrt{16} }{2}=\frac{8-4}{2}=\frac{4}{2}=2$

$(x_1)^2+4y_1=4$

$6^2+4y_1=4$

$36+4y_1=4$

$9+y_1=1$

$y_1=1-9$

$y_1=-8$

$(x_2)^2+4y_2=4$

$2^2+4y_2=4$

$4+4y_2=4$

$1+y_2=1$

$y_2=1-1$

$y_2=0$

Assim os seguintes pares ordenados definem o conjunto solução "V":

$V=\{(2,0);(6,-8)\}$