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

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You shouldn't use $\color{red};$ to separate between endpoints on an interval, but rather use $\color{red},$. –  Git Gud Dec 26 '13 at 11:13
Thanks! The ; used my professor! –  user2724695 Dec 26 '13 at 11:14
@drhab: $-3.25$ is correct, but $-3.25$ should be included in the solution, not excluded. –  robjohn Dec 26 '13 at 14:58
@ user2724695 Insted of 17 there should be 13 . –  john Dec 26 '13 at 15:14
It is better to choose for $(x+1)(2x+5)$ as denominator. It saves you the struggling with brackets and brightens the view on the fraction if you are looking for the sign (+ or -) of it. –  drhab Dec 26 '13 at 15:27

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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So in the end i have to make 4 cases right? Thanks –  user2724695 Dec 26 '13 at 11:22
Yes.... But I can assure you don't have to do all... you have to do at most three... :) I have already said this in the answer.... I would be happy if you say "I have to consider only two" :) –  Praphulla Koushik Dec 26 '13 at 11:24
Accept?? Have you done that completely? –  Praphulla Koushik Dec 26 '13 at 11:35
Yes! Thanks i had the same idea from the beginning but it seemed more easily to short it to one site! And yes i tried all 4 cases! Im not so good in imagine number areas! –  user2724695 Dec 26 '13 at 11:41
you tried all four cases??? I said not to do so :O I guess you did not understand how simple this could/should be.... –  Praphulla Koushik Dec 26 '13 at 11:45

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.

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Same way I use with my students. –  Adi Dani Dec 26 '13 at 13:35
It should be $4x+13$. You took over a mistake of the OP (just like I did) –  drhab Dec 26 '13 at 15:22
@drhab: Ohhh! Thanks. –  Babak S. Dec 26 '13 at 15:24
You overlooked the $5$-the cell in the first column. Still $4x+17$ there :-). –  drhab Dec 26 '13 at 15:30

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.

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Thanks for warning me. :-) –  Babak S. Dec 26 '13 at 15:28

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 $<0$

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It should be $4x+13$. You took over a mistake of the OP (just like I did) –  drhab Dec 26 '13 at 15:21
@drhab, thanks once again :) –  lab bhattacharjee Dec 26 '13 at 16:34

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 $0$: $$\textstyle\left(-\infty,-\frac{13}{4}\right]\cup\left(-\frac52,-1\right)$$

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I do not get your last sentence. –  Carsten Schultz Dec 26 '13 at 12:00
@CarstenSchultz: the intervals should be closed: e.g. $\left(-\infty,-\frac{13}{4}\right]\cup\left[-\frac52,-1\right]$. However, at $-\frac52$ and $-1$ we would be dividing by $0$. –  robjohn Dec 26 '13 at 14:48
Thank you for correcting me. I allready warned the others (lab and B.S.) –  drhab Dec 26 '13 at 15:23

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 $4x+13$

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Shouldn't $-3.25$ be included in the solution? –  robjohn Dec 26 '13 at 14:55
You are right! I just copied his answer without thinking! –  user44197 Dec 26 '13 at 15:20
Well, there is a way for not making silly mistakes :-). –  drhab Dec 26 '13 at 15:33
I am blushing and it is not a pretty sight :-) –  user44197 Dec 26 '13 at 15:34
I know the feeling... –  drhab Dec 26 '13 at 15:36

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 .

$(-2.5,-1) \cup (-\infty,-3.25]$

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Why is $-3.25$ excluded? –  robjohn Dec 26 '13 at 19:40
@robjohn I corrected now . Thanks for pointing mistake . –  john Dec 27 '13 at 18:04