Solve [tex]\dfrac{5|x|}{x+3}>1[/tex]

[Guest]

Hi,
I feel that I'm missing something, when solving this equality, like there is a third case I should discuss.
So, what I'm missing?
Thank you.

First, I define the domain D:

[tex]D=(-\infty,-3)\cup(-3,+\infty)[/tex]

Case 1: if [tex]x > -3[/tex]

[tex]x>-3 \implies 5|x|>x+3[/tex]
[tex]5|x|>x+3 \implies 5x>x+3[/tex] or [tex]5x<-(x+3)[/tex]
[tex]5x>x+3 \implies x>\dfrac{3}{4}[/tex]
[tex]5x<-(x+3) \implies x<-\dfrac{1}{2}[/tex]
[tex]S_1=(-\infty,-3)\cup(-3,-\dfrac{1}{2})\cup(\dfrac{3}{4},\infty)[/tex]

Case 2: if [tex]x < -3[/tex]
[tex]x < -3 \implies 5|x|<x+3[/tex]
[tex]5|x|<x+3 \implies -(x+3)<5x[/tex] or [tex]5x < x+3[/tex]
[tex]-(x+3)<5x \implies x>-\dfrac{1}{2}[/tex]
[tex]5x < x+3 \implies x < \dfrac{3}{4}[/tex]
[tex]S_2 = (-\dfrac{1}{2},\dfrac{3}{4})[/tex]

[Guest]

You must have 3 cases not 2.
The second case have to be splitted on 2 cases. The first one [tex]0 < x < 3[/tex] and the second one [tex]0 \le x[/tex]
If [tex]x < 0[/tex] then [tex]|x| = -x[/tex]

The second way is to have 2 cases but they are [tex]x > 0[/tex] and [tex]x \le 0[/tex]
and to solve this way
[tex]\frac{5|x| - (x + 3)}{x + 3} > 0[/tex]

[Guest]

Could you give me precisely the three cases without developing them, if this is not too asking.
I'll work on them and submit my answer. I really want to learn from this problem.
Thank you.

[Guest]

The best way I can think of explaining it is that that there are actually 2 sets of 2 conditions to consider.

The first is whether [tex]x+3[/tex] is positive or negative, i.e. [tex]x>-3[/tex] or [tex]x<-3[/tex].
The second is whether [tex]|x|[/tex] is the same as [tex]x[/tex] or [tex]-x[/tex], i.e. [tex]x\geq 0[/tex] or [tex]x<0[/tex].

You could take either of these as case 1 and case 2 then think of the other condition as 2 additional subcases. For example in your attempt you took [tex]x>-3[/tex] or [tex]x<-3[/tex] as cases 1 and 2 explicitly then implicitly split them into 2 subcases using the condition [tex]x\geq 0[/tex] or [tex]x<0[/tex].

Another valid way to deal with the conditions is to take all combinations of the conditions which results in a total of four cases:
Case 1: [tex]x>-3[/tex] and [tex]x\geq 0[/tex]
Case 2: [tex]x>-3[/tex] and [tex]x<0[/tex]
Case 3: [tex]x<-3[/tex] and [tex]x\geq 0[/tex]
Case 4: [tex]x<-3[/tex] and [tex]x<0[/tex]

If you consider the cases carefully you'll notice that we can sometimes simplify the conditions, for example case 1 could be rewritten as just [tex]x\geq 0[/tex]. Also case 3 obviously has no solutions, as it impossible for a number to be smaller than -3 but still be non-negative. So really there are only 3 cases:
Case 1: [tex]x\geq 0[/tex]
Case 2: [tex]-3<x<0[/tex]
Case 3: [tex]x<-3[/tex]

Often people will just state these 3 cases, as the intermediate steps of writing the 2 sets of 2 conditions then splitting it into 4 cases is easy and obvious to them.

Any of the ways described are valid ways of splitting up the question into cases, it's up to you to decide which you prefer.

After looking at your attempt I feel I should give you one more piece of advice. When you tackle a case, you use the conditions of the case to simplify solving the inequality, but once you have the solution you need to remember that the solution is subject to the original condition and so you need to take the intersection of the condition and solution to get the the true solution.

For example in Case 2 of your attempt, the condition is [tex]x<-3[/tex], in the third last line you implicitly use the condition [tex]x<0[/tex], then get a "solution" of [tex]x>-1/2[/tex]. Therefore we can say the original inequality is satisfied when:
[tex]x<-3[/tex] and [tex]x<0[/tex] and [tex]x>-1/2[/tex]
However, no such [tex]x[/tex] satisfies these conditions, so this subcase has actually yeilded no solutions.

Hope this helped,

R. Baber.

[Guest]

I should also have mentioned that one "trick" for determining the three cases is to realise that from our 2 sets of 2 conditions we know that -3 is an important value which separates cases, as is 0. So by removing these two values from the real line we get three connected regions [tex](-\infty, -3)[/tex], [tex](-3,0)[/tex] and [tex](0,\infty)[/tex], which essentially become our 3 cases. Though now you have to be careful and think about what happens at the actual values of [tex]x=-3[/tex] and [tex]x=0[/tex], do you deal with them as separate cases or do you add them to one of the regions and if so which one. In our example [tex]x=-3[/tex] is not in the domain so is not a problem, and it is easy to deal with [tex]x=0[/tex] as a special case or with a bit of thought we see that it is fine to lump it with either [tex](-3,0)[/tex] or [tex](0,\infty)[/tex] to get [tex](-3,0][/tex] or [tex][0,\infty)[/tex].

Hope this helped,

R. Baber.

[Guest]

Thank you so much for your kindness and the time you dedicated to help me. Your answers clear confusion of my mind.