Question:3|2x + 1| - 12 = 15What is the sum of the solutions to the given equation?-10-5-149

1. SIMPLIFY to isolate the absolute value

• Start with: \(3|2\mathrm{x} + 1| - 12 = 15\)
• Add 12 to both sides: \(3|2\mathrm{x} + 1| = 27\)
• Divide by 3: \(|2\mathrm{x} + 1| = 9\)

2. CONSIDER ALL CASES for the absolute value equation

• When we have |expression| = positive number, we get two cases:
  • Case 1: expression = positive number
  • Case 2: expression = negative number
• So \(|2\mathrm{x} + 1| = 9\) gives us:
  • Case 1: \(2\mathrm{x} + 1 = 9\)
  • Case 2: \(2\mathrm{x} + 1 = -9\)

3. SIMPLIFY each case to find x-values

• Case 1: \(2\mathrm{x} + 1 = 9\)
  • \(2\mathrm{x} = 8\)
  • \(\mathrm{x} = 4\)
• Case 2: \(2\mathrm{x} + 1 = -9\)
  • \(2\mathrm{x} = -10\)
  • \(\mathrm{x} = -5\)

4. Verify both solutions work in the original equation

• For \(\mathrm{x} = 4\): \(3|2(4) + 1| - 12 = 3|9| - 12 = 27 - 12 = 15\) ✓
• For \(\mathrm{x} = -5\): \(3|2(-5) + 1| - 12 = 3|-9| - 12 = 27 - 12 = 15\) ✓

5. INFER what the question is asking for

• The question asks for "the sum of the solutions"
• Sum = \(4 + (-5) = -1\)

Answer: C) -1

Why Students Usually Falter on This Problem

Most Common Error Path:

Weak CONSIDER ALL CASES skill: Students solve only one case of the absolute value equation, typically just \(2\mathrm{x} + 1 = 9\), getting \(\mathrm{x} = 4\). They then select this as their final answer without considering the negative case.

This may lead them to select Choice D (4).

Second Most Common Error:

Poor INFER reasoning: Students correctly find both solutions (\(\mathrm{x} = 4\) and \(\mathrm{x} = -5\)) but misread the question. Instead of finding the sum, they might select one of the individual solutions or try to find some other combination.

This may lead them to select Choice B (-5) or Choice D (4).

The Bottom Line:

Absolute value equations require systematic consideration of both positive and negative cases. The key challenge is remembering that \(|\mathrm{A}| = \mathrm{k}\) always gives two solutions when \(\mathrm{k} \gt 0\), and then carefully reading what the question asks you to do with those solutions.