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Difficulty: Very hardAlgebraic Equations and Systems in Data Sufficiency

If xx and yy are real numbers, what is the value of x+yx + y?

(1) x2y+xy2=30x^2 y + xy^2 = 30
(2) x2+y2=19−xyx^2 + y^2 = 19 - 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. C
    BOTH statements TOGETHER are sufficient, but NEITHER statement ALONE is sufficient.
  4. D
    EACH statement ALONE is sufficient.
  5. Statements (1) and (2) TOGETHER are NOT sufficient.Answer

Answer

Statements (1) and (2) TOGETHER are NOT sufficient to determine a unique value for x+yx + y.
Evaluating both statements together produces the polynomial relation (x+y)3−19(x+y)−30=0(x + y)^3 - 19(x + y) - 30 = 0. Factoring gives (x+y−5)(x+y+2)(x+y+3)=0(x + y - 5)(x + y + 2)(x + y + 3) = 0. Each of the three roots (55, −2-2, −3-3) produces real numbers xx and yy satisfying both original equations. Because x+yx + y can take three different values, the two statements together are not sufficient to determine a unique value.

Step-by-Step Solution

1
Analyze Statement (1) independently
Factor Statement (1): x2y+xy2=xy(x+y)=30x^2 y + xy^2 = xy(x + y) = 30.
If (x,y)=(2,3)(x, y) = (2, 3), then xy(x+y)=6(5)=30xy(x + y) = 6(5) = 30, giving x+y=5x + y = 5. If (x,y)=(1,5)(x, y) = (1, 5), then xy(x+y)=5(6)=30xy(x + y) = 5(6) = 30, giving x+y=6x + y = 6. Multiple values exist, so Statement (1) alone is insufficient.
2
Analyze Statement (2) independently
Rearrange Statement (2): x2+y2+xy=19x^2 + y^2 + xy = 19, which equals (x+y)2−xy=19(x + y)^2 - xy = 19.
If (x,y)=(3,2)(x, y) = (3, 2), then 32+22+6=193^2 + 2^2 + 6 = 19, giving x+y=5x + y = 5. If (x,y)=(−5,3)(x, y) = (-5, 3), then (−5)2+32+(−15)=19(-5)^2 + 3^2 + (-15) = 19, giving x+y=−2x + y = -2. Multiple values exist, so Statement (2) alone is insufficient.
3
Evaluate Statements (1) and (2) together
Express xyxy in terms of S=x+yS = x + y: from Statement (1), xy=30Sxy = \frac{30}{S}; from Statement (2), xy=S2−19xy = S^2 - 19.
Equating the two expressions for xyxy yields 30S=S2−19  ⟹  S3−19S−30=0\frac{30}{S} = S^2 - 19 \implies S^3 - 19S - 30 = 0.
4
Solve the cubic polynomial for S=x+yS = x + y and check for real roots (x,y)(x, y)
S3−19S−30=(S−5)(S+2)(S+3)=0S^3 - 19S - 30 = (S - 5)(S + 2)(S + 3) = 0, giving S=5S = 5, S=−2S = -2, or S=−3S = -3.
For S=5S = 5, xy=6xy = 6, giving real solutions {x,y}={2,3}\{x, y\} = \{2, 3\}. For S=−2S = -2, xy=−15xy = -15, giving real solutions {x,y}={−5,3}\{x, y\} = \{-5, 3\}. For S=−3S = -3, xy=−10xy = -10, giving real solutions {x,y}={−5,2}\{x, y\} = \{-5, 2\}. Since three distinct valid real sums exist, the combined statements are not sufficient.

Key Concept

Non-linear systems of equations may yield multiple distinct real solutions. Two equations in two variables do not guarantee a unique solution in Data Sufficiency unless the system is linear or constrained to a single root.
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Algebraic Equations and Systems in Data Sufficiency

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