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Difficulty: HardAlgebraic Equations and Systems in Data Sufficiency

If xx and yy are real numbers such that x+y≠0x + y \neq 0, what is the value of the product xyxy?

(1) x3+y3=28(x+y)x^3 + y^3 = 28(x + y)
(2) x2+y2=20−xyx^2 + y^2 = 20 - xy

  1. A
    Statement (1) ALONE is sufficient, but statement (2) alone is not sufficient.
  2. B
    Statement (2) ALONE is sufficient, but statement (1) alone is not sufficient.
  3. BOTH statements TOGETHER are sufficient, but NEITHER statement ALONE is sufficient.Answer
  4. D
    EACH statement ALONE is sufficient.
  5. E
    Statements (1) and (2) TOGETHER are NOT sufficient.

Answer

BOTH statements TOGETHER are sufficient, but NEITHER statement ALONE is sufficient.
Combining both statements provides two independent linear equations in terms of (x+y)2(x+y)^2 and xyxy. Subtracting Statement (1) from Statement (2) eliminates the squared sum term (x+y)2(x+y)^2 and yields a unique value of −4-4 for the product xyxy, with real solutions existing for xx and yy. Therefore, both statements together are sufficient.

Step-by-Step Solution

1
Analyze Statement (1) algebraically.
(x+y)2−3xy=28(x + y)^2 - 3xy = 28
Factor x3+y3x^3 + y^3 as (x+y)(x2−xy+y2)(x + y)(x^2 - xy + y^2). Since x+y≠0x + y \neq 0, divide both sides by (x+y)(x + y) to obtain x2−xy+y2=28x^2 - xy + y^2 = 28. Express x2+y2x^2 + y^2 as (x+y)2−2xy(x + y)^2 - 2xy, yielding (x+y)2−3xy=28(x + y)^2 - 3xy = 28. Because (x+y)2(x + y)^2 is unknown, xyxy can take multiple real values (e.g., if (x+y)2=16(x+y)^2 = 16, xy=−4xy = -4; if (x+y)2=36(x+y)^2 = 36, xy=8/3xy = 8/3). Thus, Statement (1) alone is insufficient.
2
Analyze Statement (2) algebraically.
(x+y)2−xy=20(x + y)^2 - xy = 20
Rearrange x2+y2=20−xyx^2 + y^2 = 20 - xy to x2+xy+y2=20x^2 + xy + y^2 = 20. Express x2+y2x^2 + y^2 as (x+y)2−2xy(x + y)^2 - 2xy, yielding (x+y)2−xy=20(x + y)^2 - xy = 20. Again, since (x+y)2(x + y)^2 is unknown, xyxy is not uniquely determined (e.g., if (x+y)2=16(x+y)^2 = 16, xy=−4xy = -4; if (x+y)2=4(x+y)^2 = 4, xy=−16xy = -16). Thus, Statement (2) alone is insufficient.
3
Combine both statements and solve the system of equations.
xy=−4xy = -4
Let u=(x+y)2u = (x + y)^2 and v=xyv = xy. Statement (1) gives u−3v=28u - 3v = 28 and Statement (2) gives u−v=20u - v = 20. Subtracting the first equation from the second yields (u−v)−(u−3v)=20−28  ⟹  2v=−8  ⟹  v=−4(u - v) - (u - 3v) = 20 - 28 \implies 2v = -8 \implies v = -4. Hence, xy=−4xy = -4 uniquely.
4
Verify existence of real solutions for xx and yy.
Real solutions exist because the discriminant is positive.
With u=16u = 16, we have x+y=4x + y = 4 or x+y=−4x + y = -4. For x+y=4x + y = 4 and xy=−4xy = -4, xx and yy are roots of t2−4t−4=0t^2 - 4t - 4 = 0, which has discriminant 16−4(1)(−4)=32>016 - 4(1)(-4) = 32 > 0. Thus real values of xx and yy exist.

Key Concept

System rephrasing in Data Sufficiency by substituting composite variables like (x+y)2(x+y)^2 and xyxy.
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Topics

Algebraic Equations and Systems in Data Sufficiency

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