Решите систему уравнений У-2Х=2 5Х^2-У=1

У=2+2х 5х^2-у=1 5х^2-(2+2х)=1 5х^2-2-2х=1 5х^(-2х)=1 -10х2=1 10х2=1 Х1=√1 х2=-√1 У1=4 у=0

[latex] \left \{ {{y-2x=2,} \atop {5x^2-y=1;}} \right. \ \left \{ {{y=2x+2,} \atop {5x^2-(2x+2)=1;}} \right. \ \left \{ {{y=2x+2,} \atop {5x^2-2x-2-1=0;}} \right. \ \left \{ {{y=2x+2,} \atop {5x^2-2x-3=0;}} \right. \\ 5x^2-2x-3=0, \\ D_{/4}=(-1)^2-5\cdot(-3)=1+15=16, \\ x=\frac{1\pm4}{5}, \\ x_1=-0,6, \ x_2=1; \\ \left \{ {{y=2x+2,} \atop { \left [ {{x=-0,6,} \atop {x=1;}} \right. }} \right. \ \left [ {{ \left \{ {{x=-0,6,} \atop {y=0,8;}} \right. } \atop { \left \{ {{x=1,} \atop {y=4.}} \right. }} \right. [/latex]