Решить систему уравнений x^3+x^2y=12: x+y=3

{x³+x²y=12 x² * (x+y) = 12 x²*3=12 |:3 x²=4 x₁=2 x₂=-2 ⇒

{x+y=3 y=3-x y₁=3-2=1 y₂=3 - (-2) = 3+2=5.

Ответ: x₁=2 y₁=1 x₂=-2 y₂=5.

{x³ + x²y = 12

{x + y = 3

{y = 3-x

{x³ + x²y = 12

x³ + x²· (3-x) = 12

x³ + 3x² - x³ = 12

3x² = 12

x² = 12 : 3

x² = 4

x₁ = - √4 = - 2

x₂ = √4 = 2

x₁ = - 2 = > y₁ = 3 - (-2) = 3+2 = 5

x₂ = 2 = > y₂ = 3-2 = 1

Ответ: (-2; 5); (2; 1)