Решите пожалуйста №3) СРОЧНО!!!

Решите задачу:

$\frac{3y^{2} - 12 }{2y ^{2} - 15y + 18 } \times \frac{6 - y}{y + 2} + \frac{y}{3 - 2y} = \\ = \frac{3(y ^{2} - 4)}{2y^{2} - 3y - 12y + 18 } \times \frac{ - (y - 6)}{y + 2} + \frac{y}{ - (2y - 3)} = \\ = \frac{3(y - 2)(y + 2)}{y(2y - 3) - 6(2y - 3)} \times \frac{ - (y - 6)}{y + 2} - \frac{y}{2y - 3} = \\ = \frac{3(y - 2)(y + 2)}{(2y - 3)(y - 6)} \times \frac{ - (y - 6)}{y + 2} - \frac{y}{2y - 3} = \\ = \frac{3(y - 2)}{2y - 3} \times( - 1) + \frac{y}{2y - 3} = \\ = - \frac{3y - 6 + y}{2y - 3} = - \frac{4y - 6}{2y - 3} = \\ = - \frac{2(2y - 3)}{2y - 3} = - 2$