So, you have good reasoning up to the point where you solve for both scenarios:
X < -3/5
AND
X > -3
Now here's where you are getting it a bit twisted: BOTH CONDITIONS MUST BE MET, NOT ONLY ONE.
So, values of X go from -3 to -3/5 (not including). Or in mathematical terms: X = (-3,-3/5)

Please see the image attached.
I hope this helps!

P.S: I have no idea why I can't quote the comment of einalemjs...

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Please see my response above...

Why is it that both scenarios must be true? I thought when you take an absolute value, you are either reversing the negative or leaving it as it was... but it can't be both negative and positive at once. Meaning, while it's useful to know both possible scenarios, it's not actually possible that they both occur simultaneously, since it can only be one OR the other. So what is the reasoning behind that?

It's not like finding the roots of a quadratic or accounting for positive/negative roots, where both are correct. It's a way of accounting for uncertainty in absolute value, but in this case it is either one case or the other, because the starting value could not have been both cases at once.

Given all of this, answer choice D makes the most sense, since it's possible that the |x| is less than, equal to, or greater than 4.

I really don't see how this is faulty reasoning. I hope someone can explain!

EDiT: I think I understand now, from doing more questions. It seems there's nothing wrong with my reasoning--the variables can't be both cases at once. But we're not precisely talking about the sign of the variable, but rather the output of the absolute value sign. Further, since we're dealing with inequalities, the values can satisfy both conditions. In the context of absolute value, since the resulting quantity will always be positive for the same inputs, the variable must satisfy the inequalities of both positive and negative values.

Thanks! Resolved.
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\(|3+3x| < -2x\)

We replace what is inside the modulus sign with the positive and the negative of what is inside, as both expressions can satisfy the inequality.

\(3 + 3x < -2x\)

\(-(3+3x)< -2x\)

We solve for \(x\), in each inequality.

The first inequality

\(3+3x < -2x\)

\(3+3x+2x<0\)

\(3+5x < 0\)

\(5x<-3\)

\(x<-3/5\)

The second inequality

\(-(3+3x) <-2x\)

\(-3-3x < -2x\)

\(-3-3x+2x<0\)

\(-3-x<0\)

\(-3<x\)

\(x> -3\)

Therefore, \(x\) lies between \(-3\) and \(-3/5\). Then \(|x|\) lies between \(3/5\) and \(3\), both of which are less than \(4\). Therefore, Quantity B is always greater.

Here (3+3x)<-2x is considered when we assume the value of x>0.

But the answer of this inequality is x<-3/5 which does not satisfy the condition of x>0.

Can anyone please clear this doubt of mine.

Posted from my mobile device

In this scenario you consider that, essentially, x < a or x > -a

our x is what we do have inside the module