In algebra, the real numbers x and y satisfy x = 1 − y and x^2 = 2 − y^2 simultaneously. Using these two equations, determine the exact value of the product x·y.

Introduction / Context:
This question involves solving a system of equations that contains both linear and quadratic relationships between x and y. Once x and y are related through two equations, the task is to find the value of the product x·y without necessarily computing each variable individually in decimal form. This style of problem is common in algebra based aptitude tests, where the focus is on symbolic manipulation rather than heavy numerical computation.

Given Data / Assumptions:
- x and y are real numbers.
- They satisfy x = 1 − y (a linear relation) and x^2 = 2 − y^2 (a quadratic relation).
- We are asked to find the exact value of x·y.
- Standard algebraic rules and methods for solving simultaneous equations apply.

Concept / Approach:
We use substitution. Since x = 1 − y, we can substitute this expression for x in the second equation x^2 = 2 − y^2. After simplification, we obtain an equation solely in terms of y, which we can then solve. Once y is known, x follows from x = 1 − y. Finally, we compute the product x·y. Because the equations are symmetric in a certain way, the product will turn out to be the same for both solution pairs, leading to a unique value for x·y.

Step-by-Step Solution:
1) From x = 1 − y, substitute into x^2 = 2 − y^2 to get (1 − y)^2 = 2 − y^2. 2) Expand the left side: (1 − y)^2 = 1 − 2y + y^2. 3) Set up the equation: 1 − 2y + y^2 = 2 − y^2. 4) Bring all terms to one side: 1 − 2y + y^2 − 2 + y^2 = 0, which simplifies to 2y^2 − 2y − 1 = 0. 5) Solve the quadratic 2y^2 − 2y − 1 = 0 using the quadratic formula, giving y = (2 ± √(4 + 8)) / 4 = (2 ± √12) / 4 = (2 ± 2√3) / 4 = (1 ± √3) / 2. 6) Use x = 1 − y. For y = (1 + √3) / 2, x = (1 − √3) / 2. For y = (1 − √3) / 2, x = (1 + √3) / 2. 7) In both cases, the product x·y = ((1 + √3)/2)·((1 − √3)/2) = (1 − 3) / 4 = −2 / 4 = −1/2.

Verification / Alternative check:
We can substitute one pair, for example x = (1 + √3)/2 and y = (1 − √3)/2, back into the original equations. For the linear relation, x = 1 − y becomes (1 + √3)/2 = 1 − (1 − √3)/2, which simplifies correctly. For the quadratic relation, x^2 = 2 − y^2, computing each side shows equality. Since both solution pairs lead to the same product x·y, the final answer −1/2 is consistent and unique for the product.

Why Other Options Are Wrong:
Option B (1), Option C (−1), Option D (2), and Option E (1/2) correspond to guessing the product or confusing it with sums like x + y or differences like x − y. None of these values match the exact product obtained from the correct solution of the system. Only Option A reflects the properly computed value of x·y.

Common Pitfalls:
Some learners try to solve for x and y numerically too early, leading to approximate decimal values and rounding errors instead of an exact product. Others may expand (1 − y)^2 incorrectly, or forget to move all terms to one side of the quadratic equation. The symmetry between the two solutions might also be overlooked, leading to confusion about which product to choose. Working symbolically and carefully checking algebraic steps prevents these errors.

Final Answer:
Solving the system and multiplying the resulting values of x and y shows that x·y = −1/2.