How can i solve these? I have done them but I am unsure if I did them right.How could i solve this?
$\frac{2x^2-x-19}{x^2+3x+2}>1$
$-1<\frac{2x+3}{x-1}<1$

For two see See Solve this inequality equation with 3 sides?
For one,
$$\dfrac{2x^2-x-19}{x^2+3x+2}<1\iff\dfrac{2x^2-x-19}{x^2+3x+2}-1<0$$
$$\iff\dfrac{x^2-4x-21}{x^2+3x+2}<0$$
$$\iff\dfrac{(x-7)(x+3)}{(x+2)(x+1)}<0$$
$$\iff0>(x-7)(x+3)(x+2)(x+1)=(x-7)\{x-(-1)\}\{x-(-2)\}\{x-(-3)\}$$
So, we need odd number of negative multiplicands
$-1<x<7\implies $ only $x-7<0$
and $-3<x<-2\implies$ only $x+3>0$
