\(2(\mathrm{kx} - \mathrm{n}) = -\frac{28}{15}\mathrm{x} - \frac{36}{15}\)In the given equation, k and n are constants and n gt 1. The...

1. TRANSLATE the problem information

• Given equation: \(2(\mathrm{kx} - \mathrm{n}) = -\frac{28}{15}\mathrm{x} - \frac{36}{15}\)
• Constants k and n where \(\mathrm{n} \gt 1\)
• The equation has no solution
• Need to find: value of k

2. INFER what 'no solution' means

• A linear equation has no solution when you get a false statement like \(3 = 5\)
• This happens when both sides have the same coefficient for x, but different constant terms
• Think of it as two parallel lines that never intersect

3. SIMPLIFY the equation to standard form

• Expand left side: \(2(\mathrm{kx} - \mathrm{n}) = 2\mathrm{kx} - 2\mathrm{n}\)
• Equation becomes: \(2\mathrm{kx} - 2\mathrm{n} = -\frac{28}{15}\mathrm{x} - \frac{36}{15}\)
• Now we can clearly see coefficients and constants on each side

4. INFER and apply the no-solution condition

• For no solution: coefficients of x must be equal, constants must be different
• Coefficient of x on left: \(2\mathrm{k}\)
• Coefficient of x on right: \(-\frac{28}{15}\)
• Set them equal: \(2\mathrm{k} = -\frac{28}{15}\)

5. SIMPLIFY to find k

• Divide both sides by 2: \(\mathrm{k} = -\frac{28}{15} \div 2\)
• \(\mathrm{k} = -\frac{28}{15} \times \frac{1}{2} = -\frac{28}{30} = -\frac{14}{15}\)
• As a decimal: \(\mathrm{k} = -0.9333...\)

Answer: \(\mathrm{k} = -\frac{14}{15}\) (or -0.933 or -.9333)

Why Students Usually Falter on This Problem

Most Common Error Path:

Weak INFER skill: Students don't connect 'no solution' to the mathematical condition about coefficients and constants.

Many students try to solve the equation normally, getting confused when they can't find a specific x-value. They might attempt to substitute values or manipulate the equation without understanding that 'no solution' is a special condition requiring equal coefficients but unequal constants.

This leads to confusion and guessing.

Second Most Common Error:

Poor SIMPLIFY execution: Students make algebraic errors when expanding or rearranging.

Common mistakes include forgetting to distribute the 2 to both terms in the parentheses, or errors when dividing fractions. These calculation errors lead to wrong values of k.

This may lead them to select an incorrect numerical answer.

The Bottom Line:

This problem tests understanding of when linear equations fail to have solutions, not just how to solve them. The key insight is recognizing that 'no solution' translates to a specific mathematical relationship between coefficients and constants.