10-е задание помогите решить

Решите задачу:

$\frac{6x - 3(x + y)}{(x - y)(x + y)} = \frac{6x - 3x - 3y}{(x - y)(x + y)} = \frac{3x - 3y}{(x + y)(x - y)}$
$\frac{3(x - y)}{(x + y)(x - y)} = \frac{3}{x + y}$
$3 \div \frac{( - 2)}{3} + \frac{3}{4 } = \\ - 3 \times \frac{3}{2} = - \frac{9}{2 } \\$
$- \frac{9}{2} + \frac{3}{4} = - \frac{18}{4} + \frac{3}{4} = - \frac{15}{4} = \\ = - 3 \frac{3}{4}$