Решите систему уравнений {2x-3y=0, 5x^2+2y=3

Решение

{2x-3y=0,

{5x^2+2y=3

{3y = 2x

{5x^2+2y=3

{y = (2/3) x

{5x² + 2 * (2/3) x - 3 = 0 умножим на 3

{y = 2/3 x

{15x² + 4x - 9 = 0

15x² + 4x - 9 = 0

D = 16 + 4*15*9 = 556

x₁ = ( - 4 - 2√139) / 30

x₁ = ( - 2 - √139) / 15

x₂ = ( - 4 + 2√139) / 30

x₂ = ( - 2 + √139) / 15

x₁ = ( - 2 - √139) / 15

y₁ = (2/3) * ( - 2 - √139) / 15

y₁ = ( - 4 - 2√139) / 45

x₂ = ( - 2 + √139) / 15

y₂ = (2/3) * ( - 2 + √139) / 15

y₂ = ( - 4 + 2√139) / 45

Ответ: x₁ = ( - 2 - √139) / 15; y₁ = ( - 4 - 2√139) / 45;

x₂ = ( - 2 + √139) / 15; y₂ = ( - 4 + 2√139) / 45.