x^2 = y\(\mathrm{(x + 3)^2 = (y + 3)^2}\)If \(\mathrm{(x, y)}\) is a solution to the system of equations above...

1. TRANSLATE the problem information

• Given system:
  • \(\mathrm{x^2 = y}\)
  • \(\mathrm{(x + 3)^2 = (y + 3)^2}\)
  • \(\mathrm{x \gt 0}\)
• We need to find: \(\mathrm{xy}\)

2. INFER the approach

• The first equation directly gives us y in terms of x
• The second equation involves squared expressions on both sides
• Key insight: When we have \(\mathrm{A^2 = B^2}\), this means \(\mathrm{A = ±B}\) (two cases)

3. CONSIDER ALL CASES from the second equation

Taking the square root of \(\mathrm{(x + 3)^2 = (y + 3)^2}\):

Case 1: \(\mathrm{x + 3 = y + 3}\)

This SIMPLIFIES to: \(\mathrm{x = y}\)

Case 2: \(\mathrm{x + 3 = -(y + 3)}\)

This SIMPLIFIES to: \(\mathrm{x + 3 = -y - 3}\), so \(\mathrm{y = -x - 6}\)

4. SIMPLIFY Case 1: \(\mathrm{x = y}\)

• Substitute \(\mathrm{y = x}\) into the first equation \(\mathrm{y = x^2}\):

\(\mathrm{x = x^2}\)

• Rearrange: \(\mathrm{x^2 - x = 0}\)
• Factor: \(\mathrm{x(x - 1) = 0}\)
• Solutions: \(\mathrm{x = 0}\) or \(\mathrm{x = 1}\)

5. APPLY CONSTRAINTS to Case 1

Since \(\mathrm{x \gt 0}\), we reject \(\mathrm{x = 0}\) and keep \(\mathrm{x = 1}\).

When \(\mathrm{x = 1}\), then \(\mathrm{y = 1^2 = 1}\).

So \(\mathrm{xy = 1 × 1 = 1}\).

6. SIMPLIFY Case 2: \(\mathrm{y = -x - 6}\)

• Substitute into \(\mathrm{y = x^2}\):

\(\mathrm{x^2 = -x - 6}\)

• Rearrange: \(\mathrm{x^2 + x + 6 = 0}\)
• Check the discriminant: \(\mathrm{b^2 - 4ac = 1^2 - 4(1)(6) = -23}\)
• Since the discriminant is negative, Case 2 has no real solutions.

7. INFER the final answer

Only Case 1 provides a valid solution with \(\mathrm{x \gt 0}\).

Answer: B (1)

Why Students Usually Falter on This Problem

Most Common Error Path:

Weak CONSIDER ALL CASES skill: Students see \(\mathrm{(x + 3)^2 = (y + 3)^2}\) and immediately conclude that \(\mathrm{x + 3 = y + 3}\), missing the negative case entirely.

This incomplete analysis leads them to find \(\mathrm{x = 1}\), \(\mathrm{y = 1}\) correctly, but they miss checking whether other solutions exist. Fortunately, this still leads them to select Choice B (1), which happens to be correct.

Second Most Common Error:

Poor SIMPLIFY execution: Students correctly identify both cases but make algebraic errors when solving the quadratics, particularly in Case 2.

For example, they might incorrectly solve \(\mathrm{x^2 + x + 6 = 0}\) and think real solutions exist, leading to confusion about which solution to use. This causes them to get stuck and guess randomly.

The Bottom Line:

This problem tests whether students recognize that squared equations create multiple cases. Even students who miss the second case can get lucky, but the complete solution requires systematic case analysis and careful constraint application.