The equation \(9\mathrm{x} + 5 = \mathrm{a(x + b)}\), where a and b are constants, has no solutions. Which of...

1. SIMPLIFY the equation by expanding

• Given: \(9\mathrm{x} + 5 = \mathrm{a(x + b)}\)
• Expand the right side: \(9\mathrm{x} + 5 = \mathrm{ax} + \mathrm{ab}\)
• Now we can compare terms directly

2. INFER what "no solutions" means for linear equations

• For a linear equation \(\mathrm{ax} + \mathrm{b} = \mathrm{cx} + \mathrm{d}\) to have no solutions:
  • The x-coefficients must be equal: \(\mathrm{a} = \mathrm{c}\)
  • The constant terms must be different: \(\mathrm{b} ≠ \mathrm{d}\)
• This creates a contradiction (like \(0 = 5\)), making the equation impossible to solve

3. INFER the required conditions from our equation

• From \(9\mathrm{x} + 5 = \mathrm{ax} + \mathrm{ab}\):
• x-coefficients: 9 must equal a → Statement I is true: \(\mathrm{a} = 9\)
• Constant terms: 5 must NOT equal ab → Since \(\mathrm{a} = 9\), we need \(5 ≠ 9\mathrm{b}\)
• This means \(\mathrm{b} ≠ \frac{5}{9}\) → Statement III is true: \(\mathrm{b} ≠ \frac{5}{9}\)

4. INFER whether statement II is necessary

• Statement II claims \(\mathrm{b} = 5\)
• We know \(\mathrm{b} ≠ \frac{5}{9}\), but this doesn't force b to equal 5
• b could be 5, but it could also be any other value except \(\frac{5}{9}\)
• Statement II is not necessarily true

Answer: D. I and III only

Why Students Usually Falter on This Problem

Most Common Error Path:

Weak INFER skill: Misunderstanding what "no solutions" means for linear equations

Students might think "no solutions" means the equation can't be solved at all, rather than understanding it means the equation leads to a contradiction. They may randomly evaluate the statements without systematic analysis of the algebraic structure.

This leads to confusion and guessing among the answer choices.

Second Most Common Error:

Inadequate SIMPLIFY execution: Not properly expanding the right side

Students might try to work with \(9\mathrm{x} + 5 = \mathrm{a(x + b)}\) without expanding, making it difficult to compare coefficients systematically. This prevents them from identifying the necessary conditions clearly.

This may lead them to select Choice A (None) or Choice B (I only) because they can't establish the relationships definitively.

The Bottom Line:

Success requires understanding that linear equations with "no solutions" have a specific algebraic structure - equal variable coefficients but unequal constants create the necessary contradiction.