x^2 + y^2 = 502x + 3y = yIf \(\mathrm{(x, y)}\) is a solution to the system of equations above...

1. TRANSLATE the problem information

• Given system:
  • \(\mathrm{x^2 + y^2 = 50}\) (quadratic equation)
  • \(\mathrm{2x + 3y = y}\) (linear equation)
• Constraint: \(\mathrm{y \lt 0}\)
• Find: \(\mathrm{x - y}\)

2. INFER the solution strategy

• The linear equation looks complex but can be simplified
• Strategy: Simplify the linear equation first to find a relationship between x and y, then substitute into the quadratic equation

3. SIMPLIFY the linear equation

• Start with: \(\mathrm{2x + 3y = y}\)
• Subtract 3y from both sides: \(\mathrm{2x = y - 3y}\)
• Combine like terms: \(\mathrm{2x = -2y}\)
• Divide by 2: \(\mathrm{x = -y}\)

4. SIMPLIFY by substitution into the quadratic equation

• Substitute \(\mathrm{x = -y}\) into \(\mathrm{x^2 + y^2 = 50}\):
• \(\mathrm{(-y)^2 + y^2 = 50}\)
• \(\mathrm{y^2 + y^2 = 50}\)
• \(\mathrm{2y^2 = 50}\)
• \(\mathrm{y^2 = 25}\)
• \(\mathrm{y = ±5}\)

5. APPLY CONSTRAINTS to select the valid solution

• Since \(\mathrm{y \lt 0}\), we choose \(\mathrm{y = -5}\)
• Therefore: \(\mathrm{x = -y = -(-5) = 5}\)
• Solution: \(\mathrm{(x, y) = (5, -5)}\)

6. Calculate the final answer

• \(\mathrm{x - y = 5 - (-5) = 5 + 5 = 10}\)

Answer: (D) 10

Why Students Usually Falter on This Problem

Most Common Error Path:

Weak INFER skill: Students often try to solve the system by substitution or elimination without first recognizing that the linear equation \(\mathrm{2x + 3y = y}\) can be dramatically simplified.

They might attempt to solve for y from the first equation (getting \(\mathrm{y = ±\sqrt{50 - x^2}}\)) and substitute this complex expression into the second equation, leading to a much more difficult algebraic mess. This approach quickly becomes unmanageable and causes students to abandon systematic solution and guess.

Second Most Common Error:

Poor APPLY CONSTRAINTS reasoning: Students correctly find that \(\mathrm{y = ±5}\) but either forget to apply the constraint \(\mathrm{y \lt 0}\), or misapply it by choosing the positive value.

When they use \(\mathrm{y = 5}\) instead of \(\mathrm{y = -5}\), they get \(\mathrm{x = -5}\), leading to \(\mathrm{x - y = -5 - 5 = -10}\). This may lead them to select Choice (A) (-10).

The Bottom Line:

This problem rewards students who recognize that seemingly complex equations often have simple underlying relationships. The key insight is that strategic simplification of the linear equation first makes the entire system much more manageable.