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: $(-2.5;-1) \cup (-\infty;-3.25)$
 A: Equivalently, you may use this way, if you want. We draw a table as follows. Each row is for a term in the nominator and in factored denominator. I mean $4x+13$;  $x+5/2$ and $x+1$. Then put the roots of these terms on the top of the table as you see from the small one to the large one. Now make some $+$ and $-$ in the boxes to identify the sign of the terms in that subintervals. Finally, find the sign on the main fraction in each column as you see and select the subinterval in which the fraction is negative. This is very similar to @lab’s approach.

A: 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 :)
A: 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 $<0$
A: 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.
A: 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 $0$:
$$
\textstyle\left(-\infty,-\frac{13}{4}\right]\cup\left(-\frac52,-1\right)
$$
A: I realize this has been answered but here is a way for not making silly mistakes
Starting with
$$\frac{4x+13}{(x+1)(2x+5)} \le 0$$
The correct way to simplify is to multiply by the square of the denominator. Since we will be multiplying by a quantity that can never be negative, the sense of inequality will not change. So
$$ (4x+13)(x+1)(2x+5) \le 0$$
Now find the roots and order them in ascending order as $$-3.25, -2.5, -1$$
So to the left of $-3.25$ all the terms in the product is negative.
Between $-3.25$ and $-2.5$ one term is positive, the other two are negative.
Between $-2.5$ and $-1$ two terms are positive, and one negative.
To the right of $-1$ all are positive.
So the final answer: $$(-2.5;-1) \cup (-\infty;-3.25)$$
Note: As someone already pointed out there is an error in your original algebra. You had $4x+17$ and it should be $4x+13$
A: Let $f(x)=4x+13 \text{and}  g(x)=(x+1)(2x+5)$
and $H(x)=\frac{f(x)}{g(x)}$
$$H(x)<=0 $$So
either g(x) or H(x) is -ve and another is +ve but
$g(x)>0$  or  $g(x)<0$ 
since if g(x)=0 than Denominator is zero then  H(x) become Undefined .
Now solve below inequality .
g(x)>0 and h(x)<0
g(x)<0  and h(x)>0
Represent both inequality on number line  and take  intersection of both parts and 
finally take union of two solution 
This will give you solution .
$(-2.5,-1) \cup (-\infty,-3.25]$
