\(\frac{1}{4}(\mathrm{x} + 5) - \frac{1}{3}(\mathrm{x} + 5) = -7\)What value of x is the solution to the given equation?

1. INFER the most efficient approach

• Looking at: \(\frac{1}{4}(\mathrm{x} + 5) - \frac{1}{3}(\mathrm{x} + 5) = -7\)
• Key insight: Both terms on the left have the same factor \((\mathrm{x} + 5)\)
• Strategy: Factor out \((\mathrm{x} + 5)\) first, then deal with the fractions

2. SIMPLIFY by factoring out the common term

• Factor out \((\mathrm{x} + 5)\): \((\frac{1}{4} - \frac{1}{3})(\mathrm{x} + 5) = -7\)
• Now I need to compute \(\frac{1}{4} - \frac{1}{3}\)

3. SIMPLIFY the fraction arithmetic

• Find common denominator for \(\frac{1}{4}\) and \(\frac{1}{3}\):
  • \(\frac{1}{4} = \frac{3}{12}\)
  • \(\frac{1}{3} = \frac{4}{12}\)
• Subtract: \(\frac{1}{4} - \frac{1}{3} = \frac{3}{12} - \frac{4}{12} = -\frac{1}{12}\)
• Equation becomes: \(-\frac{1}{12}(\mathrm{x} + 5) = -7\)

4. SIMPLIFY to solve for x

• Multiply both sides by -12: \(\mathrm{x} + 5 = (-7)(-12) = 84\)
• Subtract 5 from both sides: \(\mathrm{x} = 84 - 5 = 79\)

Answer: C. 79

Why Students Usually Falter on This Problem

Most Common Error Path:

Weak INFER skill: Students don't recognize the factoring opportunity and instead try to distribute each fraction individually, creating: \(\frac{1}{4}\mathrm{x} + \frac{5}{4} - \frac{1}{3}\mathrm{x} - \frac{5}{3} = -7\). This leads to a much more complex equation with multiple fractions to combine, causing computational confusion and likely errors.

This leads to confusion and guessing among the available choices.

Second Most Common Error:

Poor SIMPLIFY execution: Students correctly factor out \((\mathrm{x} + 5)\) but make an error computing \(\frac{1}{4} - \frac{1}{3}\), such as getting \(+\frac{1}{12}\) instead of \(-\frac{1}{12}\), or computing it as \(\frac{1}{7}\). With the wrong coefficient, they arrive at an incorrect value for x.

This may lead them to select Choice A (-12) or cause them to get stuck and guess.

The Bottom Line:

This problem rewards students who can spot the factoring pattern immediately. Without that insight, the algebra becomes unnecessarily complicated and error-prone.