Question:y - x^2 = 2 - 4x3y + 6x = 15If x_1, y_1 and x_2, y_2 are the two solutions...

1. TRANSLATE the problem information

• Given information:
  • System: \(\mathrm{y - x^2 = 2 - 4x}\) and \(\mathrm{3y + 6x = 15}\)
  • We need \(\mathrm{x_1 + x_2}\) (sum of x-coordinates from both solutions)

2. INFER the solution strategy

• Since we want x-values and have one linear equation, use substitution
• The linear equation will be easier to solve for y, then substitute into the non-linear equation
• This should give us a quadratic equation in x with two solutions

3. SIMPLIFY the linear equation

• From \(\mathrm{3y + 6x = 15}\), divide everything by 3:
• \(\mathrm{y + 2x = 5}\)
• Therefore: \(\mathrm{y = 5 - 2x}\)

4. SIMPLIFY by substitution

• Substitute \(\mathrm{y = 5 - 2x}\) into the first equation:
• \(\mathrm{(5 - 2x) - x^2 = 2 - 4x}\)
• \(\mathrm{5 - 2x - x^2 = 2 - 4x}\)
• Rearrange to standard form:
• \(\mathrm{5 - 2x - x^2 - 2 + 4x = 0}\)
• \(\mathrm{3 + 2x - x^2 = 0}\)
• \(\mathrm{x^2 - 2x - 3 = 0}\)

5. SIMPLIFY to find the roots

• Factor: Look for two numbers that multiply to -3 and add to -2
• Those numbers are -3 and +1: \(\mathrm{(x - 3)(x + 1) = 0}\)
• So \(\mathrm{x = 3}\) or \(\mathrm{x = -1}\)

6. INFER the final answer

• The question asks for \(\mathrm{x_1 + x_2}\)
• \(\mathrm{x_1 + x_2 = 3 + (-1) = 2}\)

Answer: B. 2

Why Students Usually Falter on This Problem

Most Common Error Path:

Weak SIMPLIFY execution: Students make algebraic mistakes when substituting or combining like terms, particularly when handling the negative signs and \(\mathrm{x^2}\) terms.

For example, they might incorrectly simplify \(\mathrm{(5 - 2x) - x^2 = 2 - 4x}\) as \(\mathrm{5 - 2x - x^2 = 2 - 4x}\), then combine terms incorrectly to get \(\mathrm{x^2 + 2x - 3 = 0}\) instead of \(\mathrm{x^2 - 2x - 3 = 0}\). This leads to roots of \(\mathrm{x = 1}\) and \(\mathrm{x = -3}\), giving a sum of -2. This may lead them to select Choice A (-2).

Second Most Common Error:

Missing INFER insight: Students find the individual x-values correctly (\(\mathrm{x = 3}\) and \(\mathrm{x = -1}\)) but then select one of these values as their answer, forgetting that the question asks for the sum.

This may lead them to select Choice D (3) by choosing the positive root, or get confused when -1 isn't among the choices.

The Bottom Line:

This problem tests whether students can systematically work through substitution while maintaining algebraic accuracy and keeping track of what the question actually asks for.