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Difficulty: HardQuestion Stem Simplification and Target Rephrasing

For all real numbers xx and yy such that xy≠0xy \neq 0 and ∣x∣≠∣y∣|x| \neq |y|, the Data Sufficiency Yes/No target question "Is x2y−xy2x3y−xy3>0\frac{x^2 y - x y^2}{x^3 y - x y^3} > 0?" is algebraically equivalent to asking "Is x+y>0x + y > 0?".

Answer: Answer

Answer

The statement is True because algebraic factorization and cancellation simplify the target inequality directly to asking whether x+y>0x + y > 0.
The statement is True. Factoring the given expression yields xy(x−y)xy(x−y)(x+y)\frac{xy(x - y)}{xy(x - y)(x + y)}, which simplifies to 1x+y\frac{1}{x + y}. The inequality 1x+y>0\frac{1}{x + y} > 0 is satisfied if and only if x+y>0x + y > 0.

Step-by-Step Solution

1
Factor the numerator of the rational expression in the target question.
x2y−xy2=xy(x−y)x^2 y - x y^2 = xy(x - y)
Extract the greatest common monomial factor xyxy.
2
Factor the denominator of the rational expression.
x3y−xy3=xy(x2−y2)=xy(x−y)(x+y)x^3 y - x y^3 = xy(x^2 - y^2) = xy(x - y)(x + y)
Extract xyxy and then apply the difference of squares formula x2−y2=(x−y)(x+y)x^2 - y^2 = (x - y)(x + y).
3
Cancel common factors between the numerator and denominator.
xy(x−y)xy(x−y)(x+y)=1x+y\frac{xy(x - y)}{xy(x - y)(x + y)} = \frac{1}{x + y}
Since xy≠0xy \neq 0 and ∣x∣≠∣y∣|x| \neq |y|, xy(x−y)≠0xy(x - y) \neq 0, allowing safe cancellation.
4
Rephrase the target inequality 1x+y>0\frac{1}{x + y} > 0.
1x+y>0  ⟺  x+y>0\frac{1}{x + y} > 0 \iff x + y > 0
A unit fraction is positive if and only if its denominator is strictly positive.

Key Concept

Target stem simplification via algebraic factoring and difference of squares
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Topics

Question Stem Simplification and Target Rephrasing

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