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Re: If x > y and x < 0, then which of the following must be true?
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02 Apr 2019, 05:38
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Top Contributor
Bunuel wrote:
If \(x > y\) and \(x < 0\), then which of the following must be true?
I. \(\frac{1}{x} < \frac{1}{y}\)
II. \(\frac{1}{x−1} < \frac{1}{y−1}\)
III. \(\frac{1}{x+1} < \frac{1}{y+1}\)
(A) I only (B) II only (C) III only (D) I and II only (E) I and III only
We're told that x is NEGATIVE, and we're told that y < x, so y is also NEGATIVE
Take each statement and manipulate it (by applying the rules of inequalities) to see which one(s) obey the restriction that y < x
I. \(\frac{1}{x} < \frac{1}{y}\)
Multiply both sides by x to get: \(1 > \frac{x}{y}\) [since we multiplied by a NEGATIVE value, we REVERSED the inequality symbol]
Multiply both sides by y to get: \(y < x\) [since we multiplied by a NEGATIVE value, we REVERSED the inequality symbol]
PERFECT! This obeys the restriction that y < x
Statement I is true. Check the answer choices . . . ELIMINATE B and C
II. \(\frac{1}{x−1} < \frac{1}{y−1}\) NOTE: Since x is NEGATIVE, we know that x-1 is NEGATIVE Likewise, since y is NEGATIVE, we know that y-1 is NEGATIVE
Multiply both sides by \(x-1\) to get: \(1 > \frac{x-1}{y−1}\) [since we multiplied by a NEGATIVE value, we REVERSED the inequality symbol]
Multiply both sides by \(y-1\) to get: \(y-1 < x-1\) [since we multiplied by a NEGATIVE value, we REVERSED the inequality symbol]
Add 1 to both sides to get: \(y < x\)
PERFECT! This obeys the restriction that y < x
Statement II is true. Check the remaining answer choices . . . ELIMINATE A and E
By the process of elimination, the correct answer is D