Solutions for $\frac{3}{x+1}\le\frac{2}{2x+5}$ Im in search of the solutions for:
$$\frac{3}{x+1}\le\frac{2}{2x+5}$$
So first i tried to combine the two sites:
$$\frac{6x + 15 - 2x + 2}{2x^2 +7x + 5}\le{0}$$
$$\frac{4x + 17}{2x^2 +7x +5}\le{0}$$
My problem is that now i have two solutions for the denominator and i dont know how to continue:
$2x^2+7x+5 = -1 \text{ and } -2.5$
The solution should be: $(-2.5;-1) \cup (-\infty;-3.25)$

HINT:
$$\frac{4x+13}{2x^2+7x+5}=0\implies 4x+13=0$$
$$\frac{4x+13}{2x^2+7x+5}<0 \iff (4x+13)(2x^2+7x+5)<0$$
$$\iff(4x+13)(2x+5)(x+1)<0\iff\left(x+\frac{13}4\right)\left(x+\frac52\right)(x+1)<0$$
So, we need odd number of factor(s) to be $<0$
