How many solutions does the equation \(12(\mathrm{x} - 3) = -3(\mathrm{x} + 12)\) have?

1. INFER the problem type and approach

• This is a linear equation with variables on both sides
• Strategy: Use distributive property, then collect like terms to solve for x
• The number of solutions depends on what we get after simplifying

2. SIMPLIFY by distributing on both sides

• Left side: \(\mathrm{12(x - 3) = 12x - 36}\)
• Right side: \(\mathrm{-3(x + 12) = -3x - 36}\)
• New equation: \(\mathrm{12x - 36 = -3x - 36}\)

3. SIMPLIFY by collecting like terms

• Add 3x to both sides: \(\mathrm{12x + 3x - 36 = -3x + 3x - 36}\)
• This gives us: \(\mathrm{15x - 36 = -36}\)
• Add 36 to both sides: \(\mathrm{15x - 36 + 36 = -36 + 36}\)
• Result: \(\mathrm{15x = 0}\)

4. SIMPLIFY to find the final solution

• Divide both sides by 15: \(\mathrm{x = 0}\)
• INFER the conclusion: Since we found exactly one value (\(\mathrm{x = 0}\)), the equation has exactly one solution

Answer: A. Exactly one

Why Students Usually Falter on This Problem

Most Common Error Path:

Weak SIMPLIFY execution: Students make distribution or sign errors during algebraic manipulation

For example, they might incorrectly distribute \(\mathrm{-3(x + 12)}\) as \(\mathrm{-3x + 12}\) instead of \(\mathrm{-3x - 36}\), or make errors when adding/subtracting terms from both sides. These calculation errors can lead to getting equations like \(\mathrm{0 = 0}\) (infinitely many solutions) or something like \(\mathrm{5 = 0}\) (no solutions), causing them to select the wrong answer choice.

Second Most Common Error:

Conceptual confusion about solution types: Students correctly solve to get \(\mathrm{x = 0}\) but misinterpret what this means

Some students think that because \(\mathrm{x = 0}\) is "nothing," the equation has no solutions, leading them to select Choice D (Zero). Others might confuse \(\mathrm{x = 0}\) with the case where you get \(\mathrm{0 = 0}\), leading them to think there are infinitely many solutions and select Choice C (Infinitely many).

The Bottom Line:

Success on this problem requires both careful algebraic manipulation and understanding that finding a specific value (even \(\mathrm{x = 0}\)) means the equation has exactly one solution.