<s>[INST] 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: $(-2.5;-1) \cup (-\infty;-3.25)$
 [/INST] The first problem is that the difference of the two sides is
$$
\begin{align}
\frac{6x+15-2x\color{#C00000}{-}2}{(x+1)(2x+5)}=\frac{4x+\color{#C00000}{13}}{(x+1)(2x+5)}\le0
\end{align}
$$
So there are three points to consider: $-\frac{13}{4}$, $-\frac52$, and $-1$.
To the left of all three points, all three terms, $4x+13$, $x+1$, and $2x+5$, are negative.
Between $-\frac{13}{4}$ and $-\frac52$, only two terms are negative.
Between $-\frac52$ and $-1$, only one term is negative.
To the right of all three points, none of the terms are negative.
Note that all the finite endpoints should be closed, not open, where it doesn't divide by $0$:
$$
\textstyle\left(-\infty,-\frac{13}{4}\right]\cup\left(-\frac52,-1\right)
$$
</s>