48/(|x + 3| + 2) = 6 The given equation has two solutions. What is the positive difference between the...

1. TRANSLATE the problem information

• Given equation: \(\frac{48}{|\mathrm{x} + 3| + 2} = 6\)
• Find: The positive difference between the two solutions

2. INFER the solution approach

• The equation contains an absolute value in a rational expression
• Strategy: First isolate the absolute value term, then split into cases
• Key insight: We need to clear the denominator first before dealing with the absolute value

3. SIMPLIFY to isolate the absolute value term

• Multiply both sides by \((|\mathrm{x} + 3| + 2)\):
  \(48 = 6(|\mathrm{x} + 3| + 2)\)
• Divide both sides by 6:
  \(8 = |\mathrm{x} + 3| + 2\)
• Subtract 2 from both sides:
  \(6 = |\mathrm{x} + 3|\)

4. CONSIDER ALL CASES for the absolute value equation

• Since \(|\mathrm{x} + 3| = 6\), we have two cases:
  • Case 1: \(\mathrm{x} + 3 = 6\)
    Solving: \(\mathrm{x} = 3\)
  • Case 2: \(\mathrm{x} + 3 = -6\)
    Solving: \(\mathrm{x} = -9\)

5. SIMPLIFY to find the final answer

• The two solutions are \(\mathrm{x} = 3\) and \(\mathrm{x} = -9\)
• Positive difference = \(|3 - (-9)| = |3 + 9| = 12\)

Answer: 12

Why Students Usually Falter on This Problem

Most Common Error Path:

Weak INFER skill: Students try to work with the absolute value immediately without first isolating it, leading to complex nested equations they cannot solve systematically.

Instead of clearing the denominator first, they might attempt to split \(|\mathrm{x} + 3|\) into cases while it's still buried in the fraction, creating equations like:

• Case 1: \(\frac{48}{(\mathrm{x} + 3) + 2} = 6\)
• Case 2: \(\frac{48}{-(\mathrm{x} + 3) + 2} = 6\)

This creates much more complicated rational equations that are difficult to solve, leading to confusion and potential abandonment of systematic solution.

Second Most Common Error:

Poor CONSIDER ALL CASES execution: Students solve the absolute value equation correctly but only find one solution, missing the second case entirely.

They might correctly get to \(|\mathrm{x} + 3| = 6\), but then only solve \(\mathrm{x} + 3 = 6\) to get \(\mathrm{x} = 3\), forgetting that absolute value equations typically have two solutions. This leads them to believe there's only one solution, making the "positive difference" question confusing.

The Bottom Line:

This problem tests whether students can systematically work through a multi-step process: algebraic simplification followed by absolute value case analysis. The key is recognizing that you must isolate the absolute value term before splitting into cases.