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Difficulty: MediumAlgebraic Equations and Systems in Data Sufficiency

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

(1) x2−y2=15x^2 - y^2 = 15
(2) x2y−xy2=12x^2 y - x y^2 = 12

  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.
Evaluating each statement individually reveals that neither provides enough constraints on its own to solve for the sum of the variables. When both statements are combined, substituting the expression for the sum into the product equation yields a quartic polynomial in terms of the difference of the variables. Because the function is strictly increasing for positive values, it possesses exactly one positive real root. This single valid root uniquely determines the sum of the variables, making both statements together sufficient.

Step-by-Step Solution

1
Evaluate Statement (1) independently.
Statement (1) gives x2−y2=(x−y)(x+y)=15x^2 - y^2 = (x-y)(x+y) = 15. Multiple pairs of positive real numbers satisfy this (e.g., x=4,y=1x=4, y=1 gives x+y=5x+y=5; x=8,y=7x=8, y=7 gives x+y=15x+y=15). Thus, Statement (1) alone is NOT sufficient.
A single non-linear equation in two variables does not restrict x+yx+y to a unique value.
2
Evaluate Statement (2) independently.
Statement (2) gives x2y−xy2=xy(x−y)=12x^2 y - x y^2 = xy(x-y) = 12. Infinitely many positive real pairs satisfy this condition (e.g., x=4,y=1x=4, y=1 gives x+y=5x+y=5; x=3,y=1x=3, y=1 does not yield 12, but continuous choices of x,y>0x, y > 0 allow multiple sums). Thus, Statement (2) alone is NOT sufficient.
Knowing the product xy(x−y)=12xy(x-y) = 12 leaves the value of x+yx+y undetermined.
3
Combine Statement (1) and Statement (2).
Let u=x−yu = x-y and v=x+yv = x+y. Note that u>0u > 0 because x2−y2=15>0x^2 - y^2 = 15 > 0 and x,y>0x, y > 0.
From (1), v=15uv = \frac{15}{u}.
Also, xy=(x+y)2−(x−y)24=v2−u24xy = \frac{(x+y)^2 - (x-y)^2}{4} = \frac{v^2 - u^2}{4}.
Substituting into (2): (v2−u24)u=12  ⟹  u(v2−u2)=48\left(\frac{v^2 - u^2}{4}\right) u = 12 \implies u(v^2 - u^2) = 48.
Substitute v=15uv = \frac{15}{u} into the equation: u(225u2−u2)=48  ⟹  225u−u3=48  ⟹  u4+48u−225=0u\left(\frac{225}{u^2} - u^2\right) = 48 \implies \frac{225}{u} - u^3 = 48 \implies u^4 + 48u - 225 = 0.
Express the system in terms of the difference u=x−yu = x-y and sum v=x+yv = x+y to analyze uniqueness.
4
Determine the number of positive real roots for u4+48u−225=0u^4 + 48u - 225 = 0.
Let f(u)=u4+48u−225f(u) = u^4 + 48u - 225. For u>0u > 0, the derivative f′(u)=4u3+48>0f'(u) = 4u^3 + 48 > 0, meaning f(u)f(u) is strictly increasing for all positive uu. Since f(0)=−225<0f(0) = -225 < 0 and lim⁡u→∞f(u)=+∞\lim_{u \to \infty} f(u) = +\infty, there is exactly ONE positive real root for uu. Testing integer values shows f(3)=34+48(3)−225=81+144−225=0f(3) = 3^4 + 48(3) - 225 = 81 + 144 - 225 = 0, so u=3u = 3 uniquely. Substituting u=3u = 3 into v=15uv = \frac{15}{u} yields v=x+y=5v = x+y = 5 uniquely (with positive numbers x=4,y=1x = 4, y = 1).
A strictly increasing continuous function transitioning from negative to positive values has exactly one root.

Key Concept

Non-Linear Algebraic Systems and Root Uniqueness in Data Sufficiency
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Topics

Algebraic Equations and Systems in Data Sufficiency

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