Question

Решить систему уравнений :2х2-у2=32,2х-у=8

Answer 1

$left { {{2 x^{2} - y^{2} =32} atop {2x-y=8}} right. \ \ left { {{y=2x-8} atop {(2 x^{2} -(2x-8)^{2}=32}} right. \ \ left { {{y=2x-8} atop {2 x^{2} -4 x^{2} +32x-64=32}} right. \ \ left { {{y=2x-8} atop {-2 x^{2} +32x-96=0}} right. \ \ 2 x^{2} -32x+96=0 \ x^{2} -16x+48=0 \ \ left { {{x_{1}+x_{2}=16} atop {x_{1}x_{2}=48}} right.$
$x_{1}=12 \ x_{2}=4$

Ответ: $12; 4$