<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] If you assume assume $2x+5>0$ and $x+1>0$ you end up with :$$3(2x+5)\leq 2(x+1)$$
$$3(2x+5)\leq 2(x+1)\Rightarrow 6x+15\leq2x+2\Rightarrow 4x\leq -13\Rightarrow x\leq??$$
Suppose we assume $x+1$ is positive, then we would end up with above conclusion (Why??)
Suppose we assume $2x+5$ is positive but $x+1 <0$ then???
Please do it by yourself :)
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