Сапрей
Мудрец
(13882)
1 месяц назад
a)
{ x + 2y = 1 ------------------> x = 1 - 2y
{ xy = -1 ----------------------> (1 - 2y) * y = -1
=>
y - 2y^2 = -1
2y^2 - y - 1 = 0
=>
y1 = -1;
y2 = 2
x1 = 1 - 2y1 = 1 - 2*(-1) = 3;
x2 = 1 - 2y2 = 1 - 2*2 = -3
б)
{ x^2 + xy = 6
{ x - y = 4
Из второго уравнения х = у + 4
Подставим в первое
(y+4)^2 + (y+4)*y = 6
y^2 + 8y + 16 + y^2 + 4y - 6 = 0
2y^2 + 12y + 10 = 0
y^2 + 6y + 5 = 0
y1 = -1; y2 = -5;
x1 = y1 + 4 = -1+4 = 3
x2 = y2 + 4 = -5+4 = -1