This is a very intelligent back-and-forth going on, and I would like to contribute.
First of all, let me say, it's clear to me that both
pemdas and
neelgandham are rather intelligent individuals, and they both have my respect.
OK, back to the beginning.
Prompt: Is x>2a?
A straightforward inequality question.
Statement #1:
3x+2y>2a
I agree with
neelgandham's analysis at the top of this post, that for different values of x, y, and a, we can satisfy this inequality and still have either x>2a or x<2a.
As to
pemdas' assertion: 3x+2y>2a consists of (x>2a) and (2x+2y>0). That's simply not true. We can't even divide equations up like that, and inequalities are trickier than equations.
If 3x + 2y = 2a, we can't assume that x = 2a and 2x + 2y = zero. That
could be true if y < 0 and x = -y, but from Statement #1 alone, we have no way of knowing whether y is positive or negative, and no way of knowing any relationship between x and y. Remember, when we are in DS, and we are analyzing Statement #1, we have to completely ignore statement #2 at the start, and just ask whether Statement #1, by itself, is sufficient or not.
Here, Statement #1 is
insufficient. Again, the simple choice of numbers demonstrate this amply.
Statement #2: x-2y>2a
Again, I agree with
neelgandham's analysis at the top of this post, that for different values of x, y, and a, we can satisfy this inequality and still have either x>2a or x<2a.
Again, I caution against
pemdas' approach.
Given x-2y>2a, it
could be true that (x>2a) and (2y<0). That's absolutely a possibility, just not the only possibility. It could true that either one of them could be reversed for certain choices.
a) x = 5, y = -1, a = 2 ---> consistent with (x>2a) and (2y<0)
b) x = 10, y = 1, a = 2 ---> consistent with (x>2a) and (2y>0)
c) x = 3, y = -7, a = 2 ---> consistent with (x<2a) and (2y<0)
All three of those are possibilities, and none of them are eliminated by statement #2, so by itself, statement #2 does not provide information that allows us to give a definitive answer to the question. Statement #2 is
insufficient.
Now, the grand finale:
Statements #1 & #2 combined
Here, I will present two approaches.
In the first, following
neelgandham, I will add the two inequalities to get 4x > 4a, which simplifies to x > a. This is a really subtle point about systems of equations (and, by extension, systems of inequalities). We have three variables and two inequalities. If we eliminate one variable, and reduce to one inequality with two variables, that reduction will be unique. In other words, there is no alternate way to combine the two equations and eliminate y without producing an inequality that is in all ways equivalent to x > a. And, of course, if all we can deduce is x > a, we can't know whether or not x is greater than 2a. Combined, statements are
insufficient.
The second approach involves picking numbers that satisfy both inequalities.
Triplet #1: x = 5, y = 1, a = 2
Triplet #2: x = 4, y = -2, a = 3
Both are fully consistent with both inequalities, and in the first x>2a, and in the second, x<2a. Therefore, even when the two inequalities are combined, we cannot determine whether x is greater than or less than 2a. Combined, the statements are
insufficient. Answer =
E. This is in agreement with
neelgandham's answer.
With all due respect to
pemdas, I would caution you: on DS, we have to be hyper-vigilant about what the statements actually imply, that is, what must be true, vs. what
could be true given other conditions. We have to be hyper-vigilant about bringing no further assumptions into the problem, because further assumptions could be enough to make sufficient a statement that is properly insufficient. For example, when
pemdas said:
"to test whether two statements do not contradict each other we will combine to obtain |x|>|y| given y<0<x"
First of all, you can't assume y<0<x --- that's exactly the sort of extra assumption that, added to the statement, could change them from insufficient to sufficient. Adding extra outside assumptions is exactly what we
don't want to do on DS.
Also, part of answering DS is, as it were, keeping the full window of possibilities open for each statement. That means, not imposing additional limiting assumptions. Your test for consistency demonstrated the consistency of the limited assumptions you made ---- certain better than having inconsistent assumptions! --- but really the point is: don't make
any limiting assumptions. The whole point of DS is to see: given Statement #1, what kinds of things are possible? Does Statement #1 narrow down the field enough to be able to give a definitive answer to the question? If I impose extra limits, then those limits might point me to a definitive answer that is not the necessary conclusion of the statement itself.
In a way, I believe
pemdas is taking a more "find the answer" approach to DS, which is often the approach of someone naturally talented in math. When your goal is to find answer, it can be helpful to make simplifying assumptions if they wind up leading to an answer. The trouble is: "find the answer" is
not the goal on DS, and in fact, it can be an impediment. The DS questions are all about:
could you find the answer? It is possible, given only this statement, to determine a unique answer? In "finding the answer", you want to do everything you can to winnow the field down to fewer and fewer possibilities. In "could you find the answer?", you want the field of possibilities to remain as wide as possible, imposing absolutely no restraints except those absolutely necessitated by the Statements. The whole game is: is the field of possibilities wide enough that I can give both answers to the question? That's what you are
trying[u/] to do, and only abandon that when the statements themselves, by their own ineluctable implications, by complete necessity dictate only one way to answer the prompt.
You may also find help this blog I wrote:
https://magoosh.com/gmat/2012/introducti ... fficiency/
Does all that make sense? Please let me know if you have any questions about what I've said.
Mike 
