y - x + 6 = 0x^2 + y^2 = 18The solution to the system of equations above is \(\mathrm{(x,...

1. TRANSLATE the problem information

• Given information:
  • Linear equation: \(\mathrm{y - x + 6 = 0}\)
  • Quadratic equation: \(\mathrm{x^2 + y^2 = 18}\)
  • Need to find the value of y

2. INFER the solution strategy

• Since we have one linear and one quadratic equation, substitution method works well
• The linear equation is easier to manipulate, so solve it for one variable first
• Then substitute that expression into the quadratic equation

3. SIMPLIFY by solving the linear equation for y

• From \(\mathrm{y - x + 6 = 0}\):
• \(\mathrm{y = x - 6}\)

4. SIMPLIFY by substituting into the quadratic equation

• Substitute \(\mathrm{y = x - 6}\) into \(\mathrm{x^2 + y^2 = 18}\):
• \(\mathrm{x^2 + (x - 6)^2 = 18}\)

5. SIMPLIFY by expanding and combining terms

• Expand \(\mathrm{(x - 6)^2}\): \(\mathrm{x^2 + (x^2 - 12x + 36) = 18}\)
• Combine like terms: \(\mathrm{2x^2 - 12x + 36 = 18}\)
• Move everything to one side: \(\mathrm{2x^2 - 12x + 18 = 0}\)
• Divide by 2: \(\mathrm{x^2 - 6x + 9 = 0}\)

6. INFER the factoring approach and SIMPLIFY

• Recognize this as a perfect square trinomial
• Factor: \(\mathrm{(x - 3)^2 = 0}\)
• Therefore: \(\mathrm{x = 3}\)

7. SIMPLIFY to find the final answer

• Substitute \(\mathrm{x = 3}\) back into \(\mathrm{y = x - 6}\):
• \(\mathrm{y = 3 - 6 = -3}\)

Answer: -3 (Choice B)

Why Students Usually Falter on This Problem

Most Common Error Path:

Weak SIMPLIFY skill: Sign errors when expanding \(\mathrm{(x - 6)^2}\)

Students often get the middle term wrong, writing \(\mathrm{(x - 6)^2 = x^2 + 12x + 36}\) instead of \(\mathrm{x^2 - 12x + 36}\). This leads to \(\mathrm{2x^2 + 12x + 18 = 0}\), which gives \(\mathrm{x^2 + 6x + 9 = 0}\), factoring as \(\mathrm{(x + 3)^2 = 0}\), so \(\mathrm{x = -3}\). Then \(\mathrm{y = -3 - 6 = -9}\). Since -9 isn't among the choices, this leads to confusion and guessing.

Second Most Common Error:

Incomplete solution process: Solving for x but forgetting the question asks for y

Students correctly find \(\mathrm{x = 3}\) but then select Choice D (3) without realizing they need to substitute back to find \(\mathrm{y = -3}\). This happens when students don't carefully track what the question is actually asking for.

The Bottom Line:

This problem requires careful algebraic manipulation and attention to what variable the question asks for. The key insight is using substitution to convert the system into a single quadratic equation, then methodically expanding and factoring.