If (x+6)/3 = (x+6)/13, the value of x + 6 is between which of the following pairs of values?

1. SIMPLIFY the rational equation by eliminating fractions

• Given equation: \(\frac{x+6}{3} = \frac{x+6}{13}\)
• Strategy: Multiply both sides by the LCD of 3 and 13, which is 39
• This gives us: \(39 \times \left[\frac{x+6}{3}\right] = 39 \times \left[\frac{x+6}{13}\right]\)
• Simplifying: \(13(x+6) = 3(x+6)\)

2. SIMPLIFY further by collecting like terms

• Subtract \(3(x+6)\) from both sides: \(13(x+6) - 3(x+6) = 0\)
• Factor out the common term: \(10(x+6) = 0\)
• Divide both sides by 10: \(x+6 = 0\)

3. APPLY CONSTRAINTS to select the correct interval

• We found that \(x+6 = 0\)
• Check which interval contains 0:
  • Choice A: between -7 and -3? No, \(0 \gt -3\)
  • Choice B: between -2 and 2? Yes, \(-2 \lt 0 \lt 2\) ✓
  • Choice C: between 2 and 7? No, \(0 \lt 2\)
  • Choice D: between 8 and 13? No, \(0 \lt 8\)

Answer: B

Why Students Usually Falter on This Problem

Most Common Error Path:

Weak SIMPLIFY execution: Students make algebraic errors when multiplying both sides by 39, particularly in the distribution step. They might incorrectly get something like \(\frac{39(x+6)}{3} = 13(x) + 13(6)\) instead of \(13(x+6)\), or make arithmetic mistakes in the subsequent steps.

This leads to incorrect values for \(x+6\) (such as getting \(x+6 = 6\) or \(x+6 = -6\)) and may cause them to select Choice A (-7 and -3) or Choice C (2 and 7).

Second Most Common Error:

Missing APPLY CONSTRAINTS reasoning: Students solve correctly to get \(x+6 = 0\), but then forget to check which interval actually contains 0, or misinterpret what "between" means. They might think 0 needs to be strictly greater than both boundary values.

This causes confusion and may lead to guessing among the choices or incorrectly selecting Choice C (2 and 7).

The Bottom Line:

This problem tests whether students can systematically work through algebraic steps without computational errors and then apply logical reasoning to match their result to the given constraints. The key insight is recognizing that the unusual form of the equation (same expression on both sides with different denominators) leads to that expression equaling zero.