Which expression is equivalent to \(\frac{6\mathrm{y}(\mathrm{y}-3) + 9(3-\mathrm{y})}{4\mathrm{y}-12}\), where y gt 3?

1. TRANSLATE the problem information

• Given expression: \(\frac{6y(y-3) + 9(3-y)}{4y-12}\) where \(y \gt 3\)
• Goal: Simplify to find equivalent expression

2. INFER the key relationship in the numerator

• Look at the two terms: \(6y(y-3)\) and \(9(3-y)\)
• Key insight: Notice that \((3-y)\) is the opposite of \((y-3)\)
• This means: \((3-y) = -(y-3)\)

3. SIMPLIFY the numerator using this relationship

• Rewrite: \(6y(y-3) + 9(3-y) = 6y(y-3) + 9(-(y-3))\)
• Distribute: \(= 6y(y-3) - 9(y-3)\)
• Factor out \((y-3)\): \(= (y-3)(6y-9)\)

4. SIMPLIFY the denominator

• Factor: \(4y - 12 = 4(y-3)\)

5. SIMPLIFY the complete rational expression

• Substitute factored forms: \(\frac{(y-3)(6y-9)}{4(y-3)}\)

6. APPLY CONSTRAINTS to justify cancellation

• Since \(y \gt 3\), we know \((y-3) \neq 0\)
• This allows us to safely cancel \((y-3)\) from numerator and denominator
• Result: \(\frac{6y-9}{4}\)

Answer: B. \(\frac{6y-9}{4}\)

Why Students Usually Falter on This Problem

Most Common Error Path:

Weak INFER skill: Not recognizing that \((3-y) = -(y-3)\)

Students often try to work with \(6y(y-3) + 9(3-y)\) directly without seeing the relationship between the two binomials. They might expand everything:

• \(6y(y-3) = 6y^2 - 18y\)
• \(9(3-y) = 27 - 9y\)
• Getting \(6y^2 - 18y + 27 - 9y = 6y^2 - 27y + 27\)

This leads to a complex numerator that doesn't factor nicely with the denominator, causing confusion and likely guessing.

Second Most Common Error:

Poor SIMPLIFY execution: Arithmetic errors during factoring

Even if students recognize the \((3-y) = -(y-3)\) relationship, they might make sign errors:

• Writing \(6y(y-3) + 9(-(y-3))\) as \(6y(y-3) + 9(y-3)\) instead of \(6y(y-3) - 9(y-3)\)
• This gives \((y-3)(6y+9)\) instead of \((y-3)(6y-9)\)
• Leading to final answer \(\frac{6y+9}{4}\)

This may lead them to select Choice C \(\left(\frac{6y+9}{4}\right)\).

The Bottom Line:

The key challenge is recognizing equivalent forms of binomial expressions and using that insight to create factorable terms. Without this recognition, the algebraic manipulation becomes unnecessarily complex.