Question

решить систему уравнений​

Answer 1

Объяснение:

$\left \{ {{x-4y=3} \atop {x^2-28=21y}} \right. \ \ \ \ \left \{ {{x=4y+3} \atop {(4y+3)^2-21y-28=0}} \right.\ \ \ \ \ \left \{ {{x=4y+3} \atop {16y^2+24y+9-21y-28=0}} \right.\ \ \ \ \left \{ {{x=4y+3} \atop {16x^2+3y-28=0}} \right. .$

$16y^2+3y-19=0\\D=1225\ \ \ \ \ \sqrt{D}=35\\ y_1=-\frac{19}{16}\ \ \ \ \ x_1=4*(-\frac{19}{16})+3=-\frac{19}{4}+3=-4\frac{3}{4} +3=-1\frac{3}{4}=-\frac{7}{4}. \\y_2=1. \ \ \ \ \ \ \ \ \ \ \ x_2=4*1+3=4+3=7.$

Ответ: (-7/4;-19/16),   (7;1).