Question

Помогите решить систему уравнений методом подстановки
3y+x=11; y²+2x=14

Answer 1

$3y + x = 11 \ {y}^{2} + 2x = 14 \ \ x = 11 - 3y \ {y}^{2} + 2(11 - 3y) = 14 \ {y}^{2} + 22 - 6y = 14 \ {y}^{2} - 6y = - 8 \ \ y = frac{6 + sqrt{36 - 4 times 1 times 8} }{2} \ y = frac{6 - sqrt{36 - 4 times 1 times 8} }{2} \ y1 = 4 \ y2 = 2 \ x1 = 11 - 3 times 4 \ x1 = - 1 \ x2 = 11 - 3 times 2 \ x2 = 5$