Question

Difficulty: MediumPositive and Negative Number Properties

If xx and yy are non-zero real numbers such that xyx>2\frac{x - y}{x} > 2 and xy2<0x y^2 < 0, which of the following inequalities must be true?

  1. A
    x+y<0x + y < 0
  2. B
    xy>0x y > 0
  3. x+y>0x + y > 0Answer
  4. D
    xy>0x - y > 0
  5. E
    yx>1\frac{y}{x} > -1

Answer

x+y>0x + y > 0
From xy2<0x y^2 < 0, since y0y \neq 0, y2y^2 is strictly positive, which forces x<0x < 0. Next, expanding xyx>2\frac{x - y}{x} > 2 by multiplying both sides by negative xx reverses the inequality direction to xy<2xx - y < 2x. Subtracting xx from both sides gives y<x-y < x, which is equivalent to x+y>0x + y > 0. Thus, x+y>0x + y > 0 must always be true.

Step-by-Step Solution

1
Determine the sign of xx using xy2<0x y^2 < 0.
Since y0y \neq 0, y2>0y^2 > 0. For the product xy2x y^2 to be negative, xx must be negative (x<0x < 0).
A positive quantity multiplied by a negative quantity yields a negative product.
2
Simplify the inequality xyx>2\frac{x - y}{x} > 2 taking the sign of xx into account.
Multiplying both sides by xx (and reversing the inequality because x<0x < 0) gives xy<2xx - y < 2x.
Multiplying or dividing an inequality by a negative value reverses the inequality sign.
3
Rearrange the terms of xy<2xx - y < 2x to isolate the relationship between xx and yy.
Subtracting xx from both sides yields y<x-y < x, which rearranges to x+y>0x + y > 0 (or y>xy > -x).
Adding yy to both sides converts y<x-y < x directly into x+y>0x + y > 0.

Key Concept

Deducing signs and algebraic bounds in inequalities involving negative variables
Estimated Time:1m 30s
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