2x+y=7 y=x^2-1 система

2x+y=7

y=x^2-1

подстановка

2x+x^2 - 1 = 7

x^2 + 2x - 8 = 0

D = d^2 - 4ac = 2^2 - 4 * 1 * (-8) = 4+32 = 36 = 6^2

x1 = (-b + (D) ^1/2) / 2a = (-2 + 6) / 2 = 2

x2 = (-b - (D) ^1/2) / 2a = (-2 - 6) / 2 = - 4

y1 = 2^2 - 1 = 4-1 = 3;

y2 = - 4^2 - 1 = 16-1=15;

проверка

(x = 2, y = 3) :

2*2+3=7 = > 7=7

3 = 2^2 - 1 = > 3=3

(x = - 4, y = 15)

2 * (-4) + 15=7 = > - 8+15 = 7 = > 7=7

15 = (-4) ^2 - 1 = > 15 = 16-1 = > 15=15