Решите неравенство. Срочно

$\frac{2x-3}{x+1} \ \textgreater \ 1, \frac{2x-3}{x+1}-1\ \textgreater \ 0$
$\frac{2x-3-(x+1)}{x+1}\ \textgreater \ 0 , \frac{x-4}{x+1} \ \textgreater \ 0$
метод интервалов:
$\left \{ {{x-4=0} \atop {x+1 \neq 0}} \right. , \left \{ {{x=4} \atop {x \neq -1}} \right.$
+ - +
-------------(-1)-----------(4)-------------->x

x∈(-∞;-1)∪(4;∞)