Let \(\mathrm{f(x) = 3(2x - 1) + 4}\) and \(\mathrm{g(x) = 5x + 1}\). How many real solutions does the...

1. SIMPLIFY the first function

• Given: \(\mathrm{f(x) = 3(2x - 1) + 4}\)
• Apply distributive property:
  • \(\mathrm{f(x) = 3(2x) + 3(-1) + 4}\)
  • \(\mathrm{f(x) = 6x - 3 + 4}\)
  • \(\mathrm{f(x) = 6x + 1}\)

2. INFER the solution strategy

• We need to find where \(\mathrm{f(x) = g(x)}\)
• This means: \(\mathrm{6x + 1 = 5x + 1}\)
• This creates a linear equation we can solve directly

3. SIMPLIFY the equation

• Start with: \(\mathrm{6x + 1 = 5x + 1}\)
• Subtract \(\mathrm{5x}\) from both sides: \(\mathrm{x + 1 = 1}\)
• Subtract 1 from both sides: \(\mathrm{x = 0}\)

4. INFER the final answer

• We found exactly one solution: \(\mathrm{x = 0}\)
• Since both functions are linear with different slopes (6 vs 5), they can only intersect once
• This confirms our algebraic result

Answer: A (Exactly one)

Why Students Usually Falter on This Problem

Most Common Error Path:

Weak SIMPLIFY execution: Students make arithmetic errors when expanding \(\mathrm{f(x) = 3(2x - 1) + 4}\)

Common mistakes include:

• Getting \(\mathrm{3(2x - 1) = 6x - 1}\) instead of \(\mathrm{6x - 3}\) (forgetting to multiply -1 by 3)
• Computing \(\mathrm{-3 + 4 = 7}\) instead of 1
• Writing the final form as \(\mathrm{6x + 7}\) instead of \(\mathrm{6x + 1}\)

When they set \(\mathrm{6x + 7 = 5x + 1}\), they get \(\mathrm{x = -6}\), leading them to still select Choice A (since there's still exactly one solution), but with wrong reasoning.

Second Most Common Error:

Poor INFER reasoning: Students don't recognize what "How many solutions does \(\mathrm{f(x) = g(x)}\) have?" is asking

Some students try to:

• Solve \(\mathrm{f(x) = 0}\) and \(\mathrm{g(x) = 0}\) separately
• Find where the functions equal specific values
• Get confused about what equation to set up

This leads to confusion and guessing among the answer choices.

The Bottom Line:

This problem tests whether students can accurately expand expressions and understand that finding solutions to \(\mathrm{f(x) = g(x)}\) means solving one equation with one unknown.