Question:3|x - 1| = 12What is the positive solution to the given equation?

1. SIMPLIFY the equation to isolate the absolute value

• Given: \(3|\mathrm{x} - 1| = 12\)
• Divide both sides by 3: \(|\mathrm{x} - 1| = 4\)

2. INFER that absolute value creates two cases

• The equation \(|\mathrm{x} - 1| = 4\) means the expression \(\mathrm{x} - 1\) can equal either \(+4\) or \(-4\)
• This gives us two separate linear equations to solve

3. CONSIDER ALL CASES by solving both equations

• Case 1: \(\mathrm{x} - 1 = 4\)
  • Add 1 to both sides: \(\mathrm{x} = 5\)
• Case 2: \(\mathrm{x} - 1 = -4\)
  • Add 1 to both sides: \(\mathrm{x} = -3\)

4. APPLY CONSTRAINTS to select the final answer

• Both \(\mathrm{x} = 5\) and \(\mathrm{x} = -3\) are mathematically correct solutions
• Since the problem asks for the positive solution: \(\mathrm{x} = 5\)
• Verify: \(3|5 - 1| = 3|4| = 12\) ✓

Answer: 5

Why Students Usually Falter on This Problem

Most Common Error Path:

Weak INFER skill: Students treat \(|\mathrm{x} - 1| = 4\) like a regular linear equation and try to "cancel out" the absolute value bars without recognizing it creates two cases.

They might write something like: \(3(\mathrm{x} - 1) = 12\), so \(\mathrm{x} - 1 = 4\), therefore \(\mathrm{x} = 5\). While this happens to give the positive answer, they miss the complete solution process and would struggle with problems asking for all solutions or the negative solution.

Second Most Common Error:

Poor CONSIDER ALL CASES execution: Students correctly set up both cases but make arithmetic errors in one of them, or they find both solutions (\(\mathrm{x} = 5\) and \(\mathrm{x} = -3\)) but don't read carefully that the problem asks specifically for the positive solution.

This leads them to list both answers or guess between them, potentially selecting the wrong one if they misread the question.

The Bottom Line:

The key insight is recognizing that absolute value equations inherently have two solutions because absolute value represents distance, and two different numbers can be the same distance from a reference point. Students who master this concept can systematically solve any absolute value equation.