Question:x = y^2 - 3y + 3x - y = 2If \(\mathrm{(x_1, y_1)}\) and \(\mathrm{(x_2, y_2)}\) are the two solutions...

1. TRANSLATE the problem information

• Given system:
  • \(\mathrm{x = y^2 - 3y + 3}\) (quadratic equation)
  • \(\mathrm{x - y = 2}\) (linear equation)
• We need: \(\mathrm{y_1 + y_2}\) (sum of the y-coordinates of both solutions)

2. INFER the most efficient approach

• Since we only need the sum \(\mathrm{y_1 + y_2}\), we don't need to find individual values
• Strategy: Use substitution to get one quadratic in y, then apply Vieta's formula

3. TRANSLATE the linear equation for substitution

• From \(\mathrm{x - y = 2}\): \(\mathrm{x = y + 2}\)
• This expresses x in terms of y

4. SIMPLIFY by substituting into the quadratic equation

• Replace x in the first equation: \(\mathrm{y + 2 = y^2 - 3y + 3}\)
• Rearrange to standard form:
  • \(\mathrm{y + 2 = y^2 - 3y + 3}\)
  • \(\mathrm{0 = y^2 - 3y + 3 - y - 2}\)
  • \(\mathrm{0 = y^2 - 4y + 1}\)

5. INFER the direct path to the answer

• We have quadratic \(\mathrm{y^2 - 4y + 1 = 0}\) in standard form \(\mathrm{ay^2 + by + c = 0}\)
• Here: \(\mathrm{a = 1}\), \(\mathrm{b = -4}\), \(\mathrm{c = 1}\)
• By Vieta's formula: \(\mathrm{y_1 + y_2 = -b/a = -(-4)/1 = 4}\)

Answer: C

Why Students Usually Falter on This Problem

Most Common Error Path:

Weak INFER skill: Students attempt to solve the quadratic completely to find \(\mathrm{y_1}\) and \(\mathrm{y_2}\) individually, then add them.

Using the quadratic formula on \(\mathrm{y^2 - 4y + 1 = 0}\) gives messy expressions involving \(\mathrm{\sqrt{12}}\), leading to calculation errors or time wastage. They might get confused with the arithmetic and select a wrong answer or abandon the systematic approach.

This leads to confusion and potentially guessing among the answer choices.

Second Most Common Error:

Poor SIMPLIFY execution: Students make algebraic errors when rearranging \(\mathrm{y + 2 = y^2 - 3y + 3}\).

Common mistakes include:

• Forgetting to move all terms to one side
• Sign errors when rearranging
• Getting \(\mathrm{y^2 - 2y - 1 = 0}\) instead of \(\mathrm{y^2 - 4y + 1 = 0}\)

This leads to applying Vieta's formula to the wrong quadratic, potentially selecting Choice A (2) if they get \(\mathrm{y^2 - 2y + c = 0}\).

The Bottom Line:

This problem rewards recognizing that you don't need individual solutions - just their sum. Students who jump straight to the quadratic formula miss the elegant shortcut that Vieta's formulas provide.