Решите уравнение: а) \frac{5y-2}{12} - \frac{6-Зу}{9} = \frac{6y+4}{6}; б) 3x²– 2x = 0.

а) \frac{5y-2}{12} - \frac{6-3y}{9} = \frac{6y+4}{6} \frac{3(5y-2) - 4(6-3y)}{36} = \frac{6(6y+4)}{36} 15y - 6 - 24 + 12y = 36y + 24 27y - 30 = 36y + 24 27y - 36y = 24 + 30 -9y = 54 y = -6 б) 3x²– 2x = 0 x(3x - 2) = 0 x₁ = 0 3x - 2 = 0 3x = 2 x₂ = \frac{2}{3} Ответ: а) y = -6; б) x₁ = 0, x₂ = \frac{2}{3}