Question:|2x - 6| = x + 1How many distinct real solutions are there to the given equation?

1. INFER the domain requirement

• Given equation: \(|2\mathrm{x} - 6| = \mathrm{x} + 1\)
• Key insight: Since absolute value is always non-negative, we need \(\mathrm{x} + 1 \geq 0\)
• Therefore: \(\mathrm{x} \geq -1\) (this will be our domain constraint)

2. CONSIDER ALL CASES based on the expression inside the absolute value

• We need to determine when \(2\mathrm{x} - 6\) is positive vs negative
• Critical point: \(2\mathrm{x} - 6 = 0\) → \(\mathrm{x} = 3\)
• Case 1: When \(\mathrm{x} \geq 3\), then \(2\mathrm{x} - 6 \geq 0\)
• Case 2: When \(\mathrm{x} \lt 3\), then \(2\mathrm{x} - 6 \lt 0\)

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

• Since \(2\mathrm{x} - 6 \geq 0\), we have \(|2\mathrm{x} - 6| = 2\mathrm{x} - 6\)
• Equation becomes: \(2\mathrm{x} - 6 = \mathrm{x} + 1\)
• Solving: \(\mathrm{x} = 7\)

4. APPLY CONSTRAINTS to check Case 1 solution

• Is \(7 \geq 3\)? Yes ✓
• Is \(7 \geq -1\)? Yes ✓
• Verification: \(|2(7) - 6| = |8| = 8\) and \(7 + 1 = 8\) ✓

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

• Since \(2\mathrm{x} - 6 \lt 0\), we have \(|2\mathrm{x} - 6| = -(2\mathrm{x} - 6) = -2\mathrm{x} + 6\)
• Equation becomes: \(-2\mathrm{x} + 6 = \mathrm{x} + 1\)
• Solving: \(5 = 3\mathrm{x}\), so \(\mathrm{x} = \frac{5}{3}\)

6. APPLY CONSTRAINTS to check Case 2 solution

• Is \(\frac{5}{3} \lt 3\)? Yes (\(\frac{5}{3} \approx 1.67\)) ✓
• Is \(\frac{5}{3} \geq -1\)? Yes ✓
• Verification: \(|2(\frac{5}{3}) - 6| = |-\frac{8}{3}| = \frac{8}{3}\) and \(\frac{5}{3} + 1 = \frac{8}{3}\) ✓

Answer: C (Exactly two solutions: \(\mathrm{x} = 7\) and \(\mathrm{x} = \frac{5}{3}\))

Why Students Usually Falter on This Problem

Most Common Error Path:

Weak CONSIDER ALL CASES skill: Students solve only one case, typically treating the absolute value as if the expression inside is always positive.

They might write: \(|2\mathrm{x} - 6| = 2\mathrm{x} - 6 = \mathrm{x} + 1\), getting only \(\mathrm{x} = 7\), and conclude there's exactly one solution. This leads them to select Choice B (Exactly one).

Second Most Common Error:

Poor APPLY CONSTRAINTS execution: Students find both solutions but forget to verify they satisfy the domain requirement \(\mathrm{x} \geq -1\), or they incorrectly apply the case conditions.

For example, they might accept \(\mathrm{x} = \frac{5}{3}\) as a solution to Case 1 (even though \(\frac{5}{3} \lt 3\)), leading to confusion about which solutions are valid. This causes them to get stuck and guess.

The Bottom Line:

Absolute value equations require systematic case analysis - you can't just "remove" the absolute value bars without considering when the expression inside is positive versus negative.