Дробно-рациональные неравенства

Решите задачу:

$3+\frac{4}{x+2}\ \textgreater \ \frac{3}{x}, \\ x \neq 0, \ x+2 \neq 0, \ x \neq -2,\\ 3+\frac{4}{x+2}-\frac{3}{x}\ \textgreater \ 0, \\ \frac{3x(x+2+4x-3(x+2)}{x(x+2)}\ \textgreater \ 0, \\ \frac{3x^2+7x-6}{x(x+2)}\ \textgreater \ 0, \\ 3x^2+7x-6=0, \\ D=121, \\ x_1=-3, \ x_2=\frac{2}{3}, \\ 3x^2+7x-6=3(x+3)(x-\frac{2}{3}), \\ \frac{3(x+3)(x-\frac{2}{3})}{x(x+2)}\ \textgreater \ 0, \\ (x+3)(x-\frac{2}{3})x(x+2)\ \textgreater \ 0, \\ \begin{array}{c|ccccccccc}x&&-3&&-2&&0&&\frac{2}{3}&\\\frac{3(x+3)(x-\frac{2}{3})}{x(x+2)}&+&0&-&\times&+&\times&-&0&+\end{array}$
$x\in(-\infty;-3)\cup(-2;0)\cup(\frac{2}{3};+\infty).$