Question:y = sqrt(x+10)y = x - 2Which ordered pair \(\mathrm{(x, y)}\) is a solution to the given system of equations?Answer...

1. INFER the solving strategy

• Since both equations equal y, we can set the right sides equal to each other:
  \(\sqrt{\mathrm{x+10}} = \mathrm{x - 2}\)
• This eliminates y and gives us one equation with one variable to solve

2. SIMPLIFY by eliminating the square root

• Square both sides to remove the radical:
  \((\sqrt{\mathrm{x+10}})^2 = (\mathrm{x - 2})^2\)
  \(\mathrm{x + 10 = x^2 - 4x + 4}\)
• Rearrange into standard quadratic form:
  \(\mathrm{0 = x^2 - 4x + 4 - x - 10}\)
  \(\mathrm{0 = x^2 - 5x - 6}\)

3. SIMPLIFY by solving the quadratic

• Factor the quadratic expression:
  \(\mathrm{0 = (x - 6)(x + 1)}\)
• This gives us: \(\mathrm{x = 6}\) or \(\mathrm{x = -1}\)

4. APPLY CONSTRAINTS by checking for extraneous solutions

• Since we squared both sides, we must verify both solutions work in the original system
• For \(\mathrm{x = 6}\):
  • From \(\mathrm{y = x - 2}\): \(\mathrm{y = 6 - 2 = 4}\)
  • Check in \(\mathrm{y = \sqrt{x+10}}\): \(\mathrm{4 = \sqrt{6+10} = \sqrt{16} = 4}\) ✓
  • Solution \(\mathrm{(6, 4)}\) is valid
• For \(\mathrm{x = -1}\):
  • From \(\mathrm{y = x - 2}\): \(\mathrm{y = -1 - 2 = -3}\)
  • Check in \(\mathrm{y = \sqrt{x+10}}\): \(\mathrm{-3 = \sqrt{-1+10} = \sqrt{9} = 3}\) ✗
  • This doesn't work since \(\mathrm{-3 ≠ 3}\), so \(\mathrm{x = -1}\) is extraneous

Answer: D. \(\mathrm{(6, 4)}\)

Why Students Usually Falter on This Problem

Most Common Error Path:

Weak APPLY CONSTRAINTS skill: Students find both \(\mathrm{x = 6}\) and \(\mathrm{x = -1}\) from the quadratic, but fail to check these solutions in the original radical equation. They see \(\mathrm{(-1, -3)}\) as an answer choice and select it without verification.

When \(\mathrm{x = -1}\), they get \(\mathrm{y = -3}\) from the linear equation, but don't realize that \(\mathrm{\sqrt{9} = 3}\), not -3. The square root function only outputs non-negative values.

This leads them to select Choice A \(\mathrm{(-1, -3)}\).

Second Most Common Error:

Poor INFER reasoning: Students don't recognize they can set the equations equal to each other. Instead, they try substituting each answer choice into both equations individually, which is inefficient and prone to calculation errors.

This approach can work but often leads to computational mistakes or confusion about which choice actually satisfies both equations simultaneously.

The Bottom Line:

The key challenge is remembering that squaring both sides of an equation can introduce solutions that don't work in the original problem. Always check your solutions when you've squared both sides!