|2x - 6| = yy = -x + 3 How many solutions are there to the system of equations above?...

1. TRANSLATE the problem information

• Given system:
  • \(|2\mathrm{x} - 6| = \mathrm{y}\)
  • \(\mathrm{y} = -\mathrm{x} + 3\)
• We need to find how many solutions exist

2. INFER the solution approach

• Since we have an absolute value equation, we need to use case analysis
• The absolute value \(|2\mathrm{x} - 6|\) behaves differently depending on whether \((2\mathrm{x} - 6)\) is positive or negative
• We can use substitution since the second equation gives us y directly

3. CONSIDER ALL CASES for the absolute value

• Find the boundary: Set \(2\mathrm{x} - 6 = 0\), so \(\mathrm{x} = 3\)
• Case 1: When \(\mathrm{x} \geq 3\), then \(2\mathrm{x} - 6 \geq 0\), so \(|2\mathrm{x} - 6| = 2\mathrm{x} - 6\)
• Case 2: When \(\mathrm{x} \lt 3\), then \(2\mathrm{x} - 6 \lt 0\), so \(|2\mathrm{x} - 6| = -(2\mathrm{x} - 6) = -2\mathrm{x} + 6\)

4. SIMPLIFY Case 1 \((\mathrm{x} \geq 3)\)

• Equation becomes: \(2\mathrm{x} - 6 = \mathrm{y}\)
• Substitute \(\mathrm{y} = -\mathrm{x} + 3\): \(2\mathrm{x} - 6 = -\mathrm{x} + 3\)
• Solve: \(3\mathrm{x} = 9\), so \(\mathrm{x} = 3\)
• Find y: \(\mathrm{y} = -3 + 3 = 0\)
• Solution candidate: \((3, 0)\)

5. APPLY CONSTRAINTS for Case 1

• Check: Does \(\mathrm{x} = 3\) satisfy \(\mathrm{x} \geq 3\)? Yes (boundary case)
• This gives us one valid solution: \((3, 0)\)

6. SIMPLIFY Case 2 \((\mathrm{x} \lt 3)\)

• Equation becomes: \(-2\mathrm{x} + 6 = \mathrm{y}\)
• Substitute \(\mathrm{y} = -\mathrm{x} + 3\): \(-2\mathrm{x} + 6 = -\mathrm{x} + 3\)
• Solve: \(-\mathrm{x} = -3\), so \(\mathrm{x} = 3\)
• Solution candidate: \((3, 0)\)

7. APPLY CONSTRAINTS for Case 2

• Check: Does \(\mathrm{x} = 3\) satisfy \(\mathrm{x} \lt 3\)? No
• This case yields no valid solutions

8. Verify our solution

• Check \((3, 0)\) in original equations:
  • \(|2(3) - 6| = |0| = 0\) ✓
  • \(0 = -(3) + 3 = 0\) ✓

Answer: C (There is exactly 1 solution)

Why Students Usually Falter on This Problem

Most Common Error Path:

Weak CONSIDER ALL CASES skill: Students only consider one case of the absolute value, typically treating \(|2\mathrm{x} - 6|\) as simply \(2\mathrm{x} - 6\) without checking when this is valid.

They solve \(2\mathrm{x} - 6 = -\mathrm{x} + 3\) to get \(\mathrm{x} = 3\), find \(\mathrm{y} = 0\), and conclude there's one solution without realizing they need to verify this works for their assumed case. While they accidentally get the right answer, their reasoning is incomplete and they might miss this verification step on other problems.

This may lead them to select Choice C for the wrong reasons, or if they make calculation errors, to select other choices.

Second Most Common Error:

Poor APPLY CONSTRAINTS reasoning: Students find \(\mathrm{x} = 3\) in both cases but fail to check whether \(\mathrm{x} = 3\) actually satisfies the constraint \(\mathrm{x} \lt 3\) for Case 2.

They might think both cases give valid solutions and conclude there are 2 solutions total, leading them to select Choice B (There are exactly 2 solutions).

The Bottom Line:

Absolute value problems require systematic case analysis with careful attention to which solutions are actually valid for each case. Students must check their solutions against the original constraints, not just solve the resulting equations.