y = x^2 \(\mathrm{2y + 6 = 2(x + 3)}\) If \(\mathrm{(x, y)}\) is a solution of the system of...

1. TRANSLATE the problem information

• Given system:
  • \(\mathrm{y = x^2}\)
  • \(\mathrm{2y + 6 = 2(x + 3)}\)
  • Constraint: \(\mathrm{x \gt 0}\)
• Find: the value of \(\mathrm{xy}\)

2. INFER the best solution approach

• Since y is already isolated in the first equation (\(\mathrm{y = x^2}\)), substitution is the most efficient method
• Substitute \(\mathrm{x^2}\) for y in the second equation to get one equation with only x

3. SIMPLIFY through substitution and algebraic manipulation

• Substitute: \(\mathrm{2(x^2) + 6 = 2(x + 3)}\)
• Apply distributive property: \(\mathrm{2x^2 + 6 = 2x + 6}\)
• Subtract 6 from both sides: \(\mathrm{2x^2 = 2x}\)
• Subtract 2x from both sides: \(\mathrm{2x^2 - 2x = 0}\)
• Factor out 2x: \(\mathrm{2x(x - 1) = 0}\)

4. INFER the solutions from factored form

• Using zero product property: either \(\mathrm{2x = 0}\) or \(\mathrm{(x - 1) = 0}\)
• This gives us \(\mathrm{x = 0}\) or \(\mathrm{x = 1}\)

5. APPLY CONSTRAINTS to select valid solution

• Since \(\mathrm{x \gt 0}\), we reject \(\mathrm{x = 0}\)
• Therefore \(\mathrm{x = 1}\) is our solution

6. SIMPLIFY to find y and calculate xy

• When \(\mathrm{x = 1}\): \(\mathrm{y = x^2 = (1)^2 = 1}\)
• Therefore: \(\mathrm{xy = (1)(1) = 1}\)

Answer: A. 1

Why Students Usually Falter on This Problem

Most Common Error Path:

Weak SIMPLIFY skill: Students make algebraic errors when distributing or combining like terms, leading to an incorrect quadratic equation. For example, they might incorrectly distribute \(\mathrm{2(x + 3)}\) as \(\mathrm{2x + 3}\) instead of \(\mathrm{2x + 6}\), or forget to subtract terms properly when isolating the quadratic. These algebraic missteps create wrong equations that yield incorrect solutions.

This leads to confusion and random answer selection among the given choices.

Second Most Common Error:

Missing APPLY CONSTRAINTS step: Students correctly solve to get \(\mathrm{x = 0}\) or \(\mathrm{x = 1}\), but forget to apply the constraint \(\mathrm{x \gt 0}\). They might calculate xy using \(\mathrm{x = 0}\), getting \(\mathrm{xy = (0)(0) = 0}\), which isn't among the answer choices. This causes them to get stuck and guess.

The Bottom Line:

Success on this problem requires solid algebraic manipulation skills combined with careful attention to the given constraints. Students who rush through the algebra or ignore the \(\mathrm{x \gt 0}\) condition will struggle to reach the correct answer.