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Bunuel
If 3x < 2y < 0, which of the following must be the greatest?

A. 2y - 3x
B. 3x - 2y
C. -(3x - 2y)
D. -(3x + 2y)
E. 0


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We notice that -(3x - 2y) = -3x + 2y = 2y - 3x; i.e. the expression in answer choice A is equivalent to the expression in answer choice C. Therefore, neither A nor C can be the correct answer, and we eliminate both.

Next, notice that 3x - 2y is negative (since 3x < 2y); therefore, B cannot be the correct answer, either (because it is less than 0, which is answer choice E).

We need to decide between D and E. Notice that both 3x and 2y are strictly less than zero, and so is their sum. Since 3x + 2y is strictly less than zero, -(3x + 2y) is strictly greater than zero; which means D is the greatest.

Alternate Solution:

We can let x = -3 and y = -2 and check each answer choice.

Choice A: 2y - 3x = -4 - (-9) = 5

Choice B: 3x - 2y = -9 - (-4) = -5. Eliminate B.

Choice C: -(3x - 2y) = -(-5) = 5 (This is Choice B, with the sign reversed.) Because this is equal to Choice A, we can eliminate both A and C.

Choice D: -(3x + 2y) = -(-9 + (-4)) = -(-13) = 13. We compare this to Choice E and see that 13 is greater than 0. Choice D is the correct answer.

Answer: D
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Easiest thing to do is plug-in values that fit the constitute of the stem.
x=-2 and y=-1 for example.
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B option is negative
E option is zero.
Other options are positive. So we can safely eliminate B and E.
Out of other three -(3x+2y) is the greatest because in this we add two negative values and put a negative sign to make the whole value positive.
But the other two option are less in value because it takes the difference of two negative values.

Therefore D is the Correct option
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Btw, A and C are effectively same, so that is also a clue that neither of them can be right.
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Deconstructing the Question

Given \(3x<2y<0\), both \(3x\) and \(2y\) are negative, and \(3x<2y\).

From this:
\(0<2y-3x\), so \(2y-3x>0\).

Step-by-step

A: \(2y-3x>0\)

B: \(3x-2y=-(2y-3x)<0\)

C: \(-(3x-2y)=2y-3x>0\)

D: \(-(3x+2y)\). Since \(3x<0\) and \(2y<0\), \(3x+2y<0\), so \(-(3x+2y)>0\)

Compare D and A:

\(D-A = (-3x-2y)-(2y-3x) = -4y\)

Since \(2y<0 \Rightarrow y<0\), we have \(-4y>0\), so \(D>A\).

Thus D is greater than A and C, and all are greater than 0.

Answer: D
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