The equation (kx + 15)/3 + (2x - 9)/5 = (7x + 50)/15 has no solution, where k is a...

1. TRANSLATE the problem information

• Given: The equation \(\frac{\mathrm{kx} + 15}{3} + \frac{2\mathrm{x} - 9}{5} = \frac{7\mathrm{x} + 50}{15}\) has no solution
• Find: The value of constant k
• This means we need to find k such that the equation becomes a contradiction

2. SIMPLIFY by clearing fractions

• Multiply the entire equation by the LCD of 15:
  \(15 \times [\frac{\mathrm{kx} + 15}{3} + \frac{2\mathrm{x} - 9}{5}] = 15 \times [\frac{7\mathrm{x} + 50}{15}]\)
• This eliminates all fractions: \(5(\mathrm{kx} + 15) + 3(2\mathrm{x} - 9) = 7\mathrm{x} + 50\)

3. SIMPLIFY by expanding and collecting terms

• Distribute: \(5\mathrm{kx} + 75 + 6\mathrm{x} - 27 = 7\mathrm{x} + 50\)
• Combine like terms: \((5\mathrm{k} + 6)\mathrm{x} + 48 = 7\mathrm{x} + 50\)

4. INFER the no-solution condition

• For a linear equation to have no solution, we need:
  • Coefficients of x to be equal (so x terms cancel)
  • Constants to be unequal (creating a contradiction)
• Set coefficients equal: \(5\mathrm{k} + 6 = 7\)
• Check constants: \(48 \neq 50\) ✓ (already different)

5. SIMPLIFY to solve for k

• From \(5\mathrm{k} + 6 = 7\): \(5\mathrm{k} = 1\)
• Therefore: \(\mathrm{k} = \frac{1}{5}\)

6. Verify the answer

• When \(\mathrm{k} = \frac{1}{5}\): \((5(\frac{1}{5}) + 6)\mathrm{x} + 48 = 7\mathrm{x} + 50\)
• This becomes: \(7\mathrm{x} + 48 = 7\mathrm{x} + 50\)
• Subtracting 7x: \(48 = 50\) (false - confirms no solution)

Answer: 1/5, 0.2

Why Students Usually Falter on This Problem

Most Common Error Path:

Weak SIMPLIFY execution: Students make algebraic errors when clearing fractions or expanding terms, leading to incorrect coefficient relationships.

For example, they might incorrectly expand \(5(\mathrm{kx} + 15) + 3(2\mathrm{x} - 9)\) as \(5\mathrm{kx} + 15 + 6\mathrm{x} - 27\), missing the multiplication of \(5 \times 15 = 75\). This gives wrong coefficients, leading them to set up an incorrect equation for k. Such errors result in wrong values that don't create the no-solution condition.

Second Most Common Error:

Insufficient INFER reasoning: Students don't recognize what "no solution" means mathematically and attempt to solve the original equation directly.

They might try to solve \(\frac{\mathrm{kx} + 15}{3} + \frac{2\mathrm{x} - 9}{5} = \frac{7\mathrm{x} + 50}{15}\) for x in terms of k, getting confused when they can't find a specific value. Without understanding that "no solution" requires equal coefficients but unequal constants, they get stuck and guess randomly.

The Bottom Line:

This problem requires both solid algebraic manipulation skills AND conceptual understanding of when linear equations have no solution. Students must recognize that the contradiction comes from equal variable coefficients with unequal constants.