x + y = 12 y = x^2 If \(\mathrm{(x, y)}\) is a solution to the system of equations above,...

1. TRANSLATE the problem information

• Given system:
  • \(\mathrm{x + y = 12}\) (linear equation)
  • \(\mathrm{y = x²}\) (quadratic equation)
• Need to find possible value of x from the given choices

2. INFER the solution strategy

• Since the second equation already expresses y in terms of x, substitution is the most efficient approach
• Replace y in the first equation with \(\mathrm{x²}\) from the second equation

3. SIMPLIFY through substitution

• Substitute \(\mathrm{y = x²}\) into \(\mathrm{x + y = 12}\):
  \(\mathrm{x + x² = 12}\)
• Rearrange to standard quadratic form:
  \(\mathrm{x² + x - 12 = 0}\)

4. SIMPLIFY by factoring the quadratic

• Need two numbers that multiply to -12 and add to +1
• Those numbers are +4 and -3 (since \(\mathrm{4 × (-3) = -12}\) and \(\mathrm{4 + (-3) = 1}\))
• Factor: \(\mathrm{x² + x - 12 = (x + 4)(x - 3) = 0}\)

5. SIMPLIFY using zero product property

• If \(\mathrm{(x + 4)(x - 3) = 0}\), then either:
  • \(\mathrm{x + 4 = 0}\), so \(\mathrm{x = -4}\)
  • \(\mathrm{x - 3 = 0}\), so \(\mathrm{x = 3}\)

6. APPLY CONSTRAINTS to select from answer choices

• Solutions are \(\mathrm{x = -4}\) and \(\mathrm{x = 3}\)
• Looking at choices: A. 0, B. 1, C. 2, D. 3
• Only \(\mathrm{x = 3}\) appears in the choices

Answer: D. 3

Why Students Usually Falter on This Problem

Most Common Error Path:

Weak INFER skill: Students may not recognize that substitution is the key strategy, instead trying to solve the system by elimination or getting confused about how to handle the mixed linear-quadratic system.

Some students might attempt to substitute \(\mathrm{x = 12 - y}\) into the second equation, leading to \(\mathrm{y = (12 - y)²}\), which creates a more complex quadratic in y. While this approach can work, it's less efficient and more prone to algebraic errors. This confusion about strategy can lead to abandoning systematic solution and guessing.

Second Most Common Error:

Poor SIMPLIFY execution: Students may correctly set up \(\mathrm{x² + x - 12 = 0}\) but struggle with factoring, either making sign errors or not recognizing the factor pairs for -12.

For example, they might incorrectly factor as \(\mathrm{(x + 3)(x - 4) = 0}\), leading to solutions \(\mathrm{x = -3}\) and \(\mathrm{x = 4}\). Since 4 isn't among the choices, this creates confusion and may lead them to select Choice A (0) or Choice C (2) as seemingly reasonable guesses.

The Bottom Line:

This problem tests whether students can efficiently combine substitution with quadratic factoring. The key insight is recognizing that the second equation being "\(\mathrm{y = x²}\)" makes substitution the natural choice, then executing clean algebraic manipulation to reach the factored form.