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Is 1/(x - y) < y - x

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Is 1/(x - y) < y - x

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$$Is\ \ \ \ \frac{1}{x-y} < y-x\ .$$ $$(1)\ \ \ \frac{1}{x} < \frac{1}{y}$$ $$(2)\ \ \ 2x = 3y$$ The OA is the option C.

Experts, can you help me here? How can I show that each statement alone is not sufficient? Thanks.

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M7MBA wrote:
$$Is\ \ \ \ \frac{1}{x-y} < y-x\ .$$ $$(1)\ \ \ \frac{1}{x} < \frac{1}{y}$$ $$(2)\ \ \ 2x = 3y$$ The OA is the option C.

Experts, can you help me here? How can I show that each statement alone is not sufficient? Thanks.
Let's rephrase the question.

If x > y, then x - y > 0 and y -x < 0. The left side of the inequality would be larger than the right side.
If x < y, then x - y < 0, and y - x > 0. Now the left side of the inequality would be smaller than the right side - this is exactly what we're asked.

Thus, our rephrased question: Is x < y?

1) 1/x < 1/y

Case 1: x = 2 and y = 1. x is not less than y, so the answer is NO
Case 2: x = -1 and y = 1. x is less than y, so the answer is YES. Not Sufficient.

2) 2x = 3y
x = (3/2)y

Case 1: y = 1 and x = 3/2. x < y? NO
Case 2: y = -1 and x = -3/2. x < y? YES. Not Sufficient.

Together: Substitute x = (3/2)y into 1/x < 1/y to give us

1/[(3/2)y] < 1/y
(2/3) * 1/y < 1/y

This will only be true if y is positive. (If y were negative, 1/y would also be negative, and multiplying that value by 2/3 would make the result less negative.)
If y is positive, then and x = (3/2)y, x will be a larger positive number. If we know that x > y, the answer to the question is a definitive NO, and thus the statements together are sufficient to answer the question. The answer is C

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