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Zorluk: OrtaAlgebraic Equations and Systems in Data Sufficiency

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

(1) x3+y3=28x^3 + y^3 = 28
(2) x2−xy+y2=7x^2 - xy + y^2 = 7

  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.Cevap
  4. D
    EACH statement ALONE is sufficient.
  5. E
    Statements (1) and (2) TOGETHER are NOT sufficient.

Cevap

BOTH statements TOGETHER are sufficient, but NEITHER statement ALONE is sufficient.
The sum of cubes factors as x3+y3=(x+y)(x2−xy+y2)x^3 + y^3 = (x + y)(x^2 - xy + y^2). Neither statement alone provides enough information to determine x+yx + y, but combining both statements gives 28=(x+y)(7)28 = (x + y)(7), which uniquely determines x+y=4x + y = 4.

Adım Adım Çözüm

1
Rephrase the question target using the sum of cubes identity.
Recall x3+y3=(x+y)(x2−xy+y2)x^3 + y^3 = (x + y)(x^2 - xy + y^2). Thus, if x2−xy+y2≠0x^2 - xy + y^2 \neq 0, then x+y=x3+y3x2−xy+y2x + y = \frac{x^3 + y^3}{x^2 - xy + y^2}.
Factoring allows connecting the expression in Statement (1) directly to the expression in Statement (2).
2
Evaluate Statement (1) alone.
Statement (1) gives x3+y3=28x^3 + y^3 = 28. For example, if x=3x=3 and y=−1y=-1, x3+y3=27−1=26≠28x^3+y^3 = 27 - 1 = 26 \neq 28; if x=283x=\sqrt[3]{28} and y=0y=0, then x+y=283x+y=\sqrt[3]{28}. If x=3,y=1x=3, y=1, x3+y3=28x^3+y^3=28 and x+y=4x+y=4. Multiple values are possible.
A single cubic equation in two variables does not fix the sum x+yx+y.
3
Evaluate Statement (2) alone.
Statement (2) gives x2−xy+y2=7x^2 - xy + y^2 = 7. Multiple pairs such as (x,y)=(3,2)(x,y) = (3,2) give 9−6+4=79-6+4=7 (where x+y=5x+y=5) and (x,y)=(3,1)(x,y) = (3,1) give 9−3+1=79-3+1=7 (where x+y=4x+y=4).
A quadratic equation in two variables permits multiple sums for x+yx+y.
4
Evaluate both statements together.
Substitute Statement (1) and Statement (2) into the identity: 28=(x+y)(7)  ⟹  x+y=428 = (x + y)(7) \implies x + y = 4.
Dividing x3+y3x^3 + y^3 by x2−xy+y2x^2 - xy + y^2 yields a single unique value of 44 for x+yx + y.

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Algebraic Equations and Systems in Data Sufficiency

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