Корень x^2+6y^2=2+x 3x+4y=23 ето система

Решим систему уравнений методом подстановки:

{√ (x^2 + 6y^2) = 2 + x;

{3x + 4y = 23; {2 + x ≥ 0;

{x^2 + 6y^2 = (2 + x) ^2;

{3x + 4y = 23; {x ≥ - 2;

{x^2 + 6y^2 = 4 + 4x + x^2;

{3x = 23 - 4y; {x ≥ - 2;

{6y^2 = 4x + 4;

{x = (23 - 4y) / 3; {x ≥ - 2;

{3y^2 = 2 * (23 - 4y) / 3 + 2;

{x = (23 - 4y) / 3; {x ≥ - 2;

{9y^2 = 2 * (23 - 4y) + 6;

{x = (23 - 4y) / 3; {x ≥ - 2;

{9y^2 = 46 - 8y + 6;

{x = (23 - 4y) / 3; {x ≥ - 2;

{9y^2 + 8y - 52 = 0;

{x = (23 - 4y) / 3; D/4 = 4^2 + 9 * 52 = 16 + 468 = 484 = 22^2; y = (-4 ± 22) / 9;

1) y = (-4 - 22) / 9 = - 26/9;

x = (23 - 4y) / 3 = (23 + 4 * 26/9) / 3 = (23 + 104/9) / 3 = (207 + 104) / 27 = 311/27;

2) y = (-4 + 22) / 9 = 18/9 = 2;

x = (23 - 4y) / 3 = (23 - 4 * 2) / 3 = (23 - 8) / 3 = 15/3 = 5.

Ответ: (311/27; - 26/9), (5; 2).