Let \(\mathrm{f(x) = 3(2x - 5) + k}\), where k is a constant. If \(\mathrm{f(x)}\) is equivalent to 6x -...

1. SIMPLIFY the given function

• Given: \(\mathrm{f(x) = 3(2x - 5) + k}\)
• Expand using distributive property:
  \(\mathrm{f(x) = 3(2x) + 3(-5) + k}\)
  \(\mathrm{f(x) = 6x - 15 + k}\)

2. INFER what equivalence means

• Since \(\mathrm{f(x)}\) is equivalent to \(\mathrm{6x - 7}\) for ALL real numbers \(\mathrm{x}\)
• This means: \(\mathrm{6x - 15 + k = 6x - 7}\) for every possible \(\mathrm{x}\) value
• For this to work, the coefficients of like terms must be identical

3. SIMPLIFY by equating coefficients

• Coefficients of \(\mathrm{x}\) terms: \(\mathrm{6 = 6}\) ✓ (already match)
• Constant terms: \(\mathrm{-15 + k = -7}\)
• Solve for \(\mathrm{k}\): \(\mathrm{k = -7 + 15 = 8}\)

Answer: C (8)

Why Students Usually Falter on This Problem

Most Common Error Path:

Weak SIMPLIFY execution: Students make sign errors when solving the equation \(\mathrm{-15 + k = -7}\). Instead of adding 15 to both sides to get \(\mathrm{k = -7 + 15 = 8}\), they might subtract 15 from -7, calculating \(\mathrm{k = -7 - 15 = -22}\).

This leads them to select Choice A (-22).

Second Most Common Error:

Missing conceptual knowledge about equivalent expressions: Students might not understand that "equivalent for all real numbers x" means the expressions must be identical in form. They may try substituting specific x values or attempt other unnecessary approaches, leading to confusion and guessing.

The Bottom Line:

This problem tests whether students understand that algebraic equivalence requires matching coefficients, not just equal values at specific points. The key insight is recognizing that after expanding, you simply match up like terms.