The expression 84/(6x + 42) is equivalent to 14/(x + k), where k is a constant. What is the value...

1. TRANSLATE the problem information

• Given: \(\frac{84}{6\mathrm{x} + 42}\) is equivalent to \(\frac{14}{\mathrm{x} + \mathrm{k}}\)
• Find: The value of constant k
• This means we need to simplify the left side until it matches the form of the right side

2. INFER the solution strategy

• The key insight: To make \(\frac{84}{6\mathrm{x} + 42}\) look like \(\frac{14}{\mathrm{x} + \mathrm{k}}\), we need to factor and simplify
• Notice the numerator 84 and target numerator 14 suggest we'll divide by some factor
• The denominator \(6\mathrm{x} + 42\) likely has a common factor we can cancel

3. SIMPLIFY by factoring the denominator

• Look for the greatest common factor in \(6\mathrm{x} + 42\)
• Both terms are divisible by 6: \(6\mathrm{x} + 42 = 6(\mathrm{x} + 7)\)
• Now our expression is: \(\frac{84}{6(\mathrm{x} + 7)}\)

4. SIMPLIFY by canceling common factors

• We can separate this as: \(\frac{84}{6} \times \frac{1}{\mathrm{x} + 7}\)
• Calculate: \(84 \div 6 = 14\)
• Result: \(\frac{14}{\mathrm{x} + 7}\)

5. INFER the final answer by comparison

• Our simplified form: \(\frac{14}{\mathrm{x} + 7}\)
• Target form: \(\frac{14}{\mathrm{x} + \mathrm{k}}\)
• Since these are equivalent: \(\mathrm{x} + 7 = \mathrm{x} + \mathrm{k}\)
• Therefore: \(\mathrm{k} = 7\)

Answer: B (7)

Why Students Usually Falter on This Problem

Most Common Error Path:

Weak SIMPLIFY skill: Students struggle with factoring \(6\mathrm{x} + 42\) correctly or make arithmetic errors when dividing 84 by 6.

For example, they might factor as \(2(3\mathrm{x} + 21)\) instead of \(6(\mathrm{x} + 7)\), leading to a more complex simplification that doesn't clearly show \(\mathrm{k} = 7\). Or they might incorrectly calculate \(84 \div 6\) as something other than 14. This confusion often leads to guessing among the answer choices.

Second Most Common Error:

Poor INFER reasoning: Students don't recognize that they need to factor and simplify the original expression to match the target form.

Instead, they might try to work backwards from \(\frac{14}{\mathrm{x} + \mathrm{k}}\) or attempt to solve algebraically by cross-multiplying, creating unnecessary complexity. This approach typically leads them to select Choice A (6) because 6 appears prominently in the original denominator.

The Bottom Line:

This problem tests whether students can systematically simplify algebraic fractions by factoring and canceling. The key is recognizing that algebraic equivalence often requires strategic simplification, not just mechanical manipulation.