Question Stem:Consider the equation \((2\mathrm{k} - 3)(2\mathrm{x} + \mathrm{m}) + 8 = -\frac{7}{5}\mathrm{x} - 2\), where k and m are...

1. INFER the condition for no solution

When does a linear equation in x have no solution? This happens when:

• The coefficients of x on both sides are equal, BUT
• The constant terms are different

This creates a contradiction like '5 = 7' which is impossible.

2. SIMPLIFY by expanding the left side

Starting with: \((2\mathrm{k} - 3)(2\mathrm{x} + \mathrm{m}) + 8 = -\frac{7}{5}\mathrm{x} - 2\)

Expand using distributive property:

• \((2\mathrm{k} - 3)(2\mathrm{x} + \mathrm{m}) = (2\mathrm{k} - 3)(2\mathrm{x}) + (2\mathrm{k} - 3)(\mathrm{m})\)
• \(= 2\mathrm{x}(2\mathrm{k} - 3) + \mathrm{m}(2\mathrm{k} - 3) = (4\mathrm{k} - 6)\mathrm{x} + (2\mathrm{k} - 3)\mathrm{m}\)

So our equation becomes:

\((4\mathrm{k} - 6)\mathrm{x} + (2\mathrm{k} - 3)\mathrm{m} + 8 = -\frac{7}{5}\mathrm{x} - 2\)

3. INFER which coefficients must be equal

For no solution, the x-coefficients must be equal:

\(4\mathrm{k} - 6 = -\frac{7}{5}\)

4. SIMPLIFY to solve for k

\(4\mathrm{k} - 6 = -\frac{7}{5}\)
\(4\mathrm{k} = -\frac{7}{5} + 6\)
\(4\mathrm{k} = -\frac{7}{5} + \frac{30}{5} = \frac{23}{5}\)
\(\mathrm{k} = \frac{23}{20}\)

5. APPLY CONSTRAINTS to verify the solution works

With \(\mathrm{k} = \frac{23}{20}\), we get \(2\mathrm{k} - 3 = \frac{23}{10} - \frac{30}{10} = -\frac{7}{10}\)

The constant terms are:

• Left side: \((-\frac{7}{10})\mathrm{m} + 8\)
• Right side: \(-2\)

For these to be equal: \((-\frac{7}{10})\mathrm{m} + 8 = -2\)

Solving: \(\mathrm{m} = \frac{100}{7} \approx 14.29\)

Since m must be a nonnegative integer, and \(\frac{100}{7}\) is not an integer, the constants cannot be equal. This confirms our equation has no solution.

Answer: \(\frac{23}{20}\)

Why Students Usually Falter on This Problem

Most Common Error Path:

Weak INFER skill: Not understanding what 'no solution' means for a linear equation

Students might think 'no solution' simply means solving for k in any way, missing that they need coefficients equal but constants different. They might try to solve the entire equation as if it has a solution, leading to confusion about what value of k to find.

This leads to confusion and random guessing among answer choices.

Second Most Common Error:

Poor SIMPLIFY execution: Making algebraic errors during expansion

Students might incorrectly expand \((2\mathrm{k} - 3)(2\mathrm{x} + \mathrm{m})\), getting the wrong coefficient for x. For example, getting \((2\mathrm{k} - 3)(2\mathrm{x}) = 4\mathrm{k}\mathrm{x} - 6\mathrm{x}\) instead of \((4\mathrm{k} - 6)\mathrm{x}\). This leads to the wrong equation for k.

This causes them to calculate an incorrect value of k that doesn't match any reasonable answer.

The Bottom Line:

This problem requires understanding the deeper concept of when equations have no solutions, not just mechanical algebra. Students need to INFER the mathematical meaning behind 'no solution' before they can set up the correct algebraic relationships.