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: $(-2.5;-1) \cup (-\infty;-3.25)$

Write $$\frac{3}{x+1}-\frac{2}{2x+5}=\frac{4x+13}{\left(x+1\right)\left(2x+5\right)}\leq0$$ There are
'special' values $-1,-2.5$ and $-3.25$, found if one of the factors in numerator or denominator is asked to equalize $0$.
Ask yourself the question:
What is the sign of this fraction if I fill in a value $x>-1$ or
$-2.5<x<-1$ or $-3.25<x<-2.5$ or $x\leq-3.25$? The answers to these questions  lead you to the final answer.
