8x + y = -11 2x^2 = y + 341 The graphs of the equations in the given system of...

1. TRANSLATE the problem information

• Given information:
  • System: \(\mathrm{8x + y = -11}\) and \(\mathrm{2x^2 = y + 341}\)
  • Graphs intersect at point \(\mathrm{(x, y)}\)
• What this tells us: The intersection point is the solution to both equations simultaneously

2. INFER the solution strategy

• Since we have one linear and one quadratic equation, substitution is the most efficient method
• We'll solve the linear equation for y, then substitute into the quadratic equation

3. SIMPLIFY to isolate y from the linear equation

• From \(\mathrm{8x + y = -11}\)
• Subtract 8x from both sides: \(\mathrm{y = -11 - 8x}\)

4. SIMPLIFY by substituting into the quadratic equation

• Replace y in \(\mathrm{2x^2 = y + 341}\) with \(\mathrm{(-11 - 8x)}\):
• \(\mathrm{2x^2 = (-11 - 8x) + 341}\)
• \(\mathrm{2x^2 = -11 - 8x + 341}\)
• \(\mathrm{2x^2 = 330 - 8x}\)

5. SIMPLIFY to standard quadratic form

• Move all terms to one side: \(\mathrm{2x^2 + 8x - 330 = 0}\)
• Divide everything by 2: \(\mathrm{x^2 + 4x - 165 = 0}\)

6. SIMPLIFY by factoring the quadratic

• Need two numbers that multiply to -165 and add to 4
• Since \(\mathrm{15 \times (-11) = -165}\) and \(\mathrm{15 + (-11) = 4}\):
• \(\mathrm{(x + 15)(x - 11) = 0}\)

7. APPLY CONSTRAINTS using zero-product property

• If \(\mathrm{(x + 15)(x - 11) = 0}\), then either:
  • \(\mathrm{x + 15 = 0}\), so \(\mathrm{x = -15}\), OR
  • \(\mathrm{x - 11 = 0}\), so \(\mathrm{x = 11}\)
• Both values are mathematically valid solutions

8. INFER the final answer from available choices

• Looking at the answer choices, only \(\mathrm{x = -15}\) is listed
• Therefore, the answer is A. -15

Answer: A. -15

Why Students Usually Falter on This Problem

Most Common Error Path:

Weak SIMPLIFY execution: Students make algebraic errors during the substitution and simplification process.

The most frequent mistake occurs when substituting \(\mathrm{y = -11 - 8x}\) into the second equation. Students might write \(\mathrm{2x^2 = -11 - 8x + 341}\) but then incorrectly combine the constants, getting \(\mathrm{2x^2 = -8x + 330}\) instead of \(\mathrm{2x^2 = -8x + 330}\). This leads to the wrong quadratic equation and incorrect factors. Alternatively, they might make sign errors when rearranging to standard form, or struggle with factoring the resulting quadratic.

This may lead them to select Choice B (-11) or guess randomly among the remaining choices.

Second Most Common Error:

Incomplete INFER reasoning: Students correctly solve for the quadratic equation but fail to recognize that both \(\mathrm{x = -15}\) and \(\mathrm{x = 11}\) are valid mathematical solutions.

Some students find one solution (often \(\mathrm{x = 11}\) first, since it's positive) and stop there, not realizing they need to check which value appears in the answer choices. Others might incorrectly think that only one solution should exist for the system.

This causes them to get stuck and potentially guess among choices that seem reasonable.

The Bottom Line:

This problem challenges students to execute a multi-step algebraic process accurately while managing both positive and negative solutions. The key insight is recognizing that substitution creates a quadratic with two solutions, but the answer choices guide the final selection.