The system of equations 2x - 3y = 4 kx + 5/2y = m has no solution, where m gt...

1. INFER the key condition for no solution

• Given information:
  • System: \(2x - 3y = 4\) and \(kx + \frac{5}{2}y = m\)
  • The system has no solution
  • \(m \gt 10\)

• What this tells us: For a system to have no solution, the coefficients of x and y must be proportional, but the constant terms cannot maintain that same proportion.

2. TRANSLATE the proportional coefficient condition

• For no solution, we need:
  coefficient of x in equation 1 / coefficient of x in equation 2 = coefficient of y in equation 1 / coefficient of y in equation 2

• This gives us: \(\frac{2}{k} = \frac{-3}{\frac{5}{2}}\)

3. SIMPLIFY the right side of the equation

• \(\frac{-3}{\frac{5}{2}} = -3 \times \frac{2}{5} = \frac{-6}{5}\)
• So our equation becomes: \(\frac{2}{k} = \frac{-6}{5}\)

4. SIMPLIFY to solve for k

• Cross multiply: \(2 \times 5 = k \times (-6)\)
• \(10 = -6k\)
• \(k = \frac{10}{-6} = \frac{-10}{6} = \frac{-5}{3}\)

Answer: (B) -5/3

Why Students Usually Falter on This Problem

Most Common Error Path:

Weak INFER skill: Students don't recognize the specific condition needed for "no solution" and instead try to solve the system directly or look for when it has infinitely many solutions.

Without understanding that no solution requires proportional coefficients but non-proportional constants, they may try to set the entire equations proportional (including constants), leading them to work with \(\frac{2}{k} = \frac{-3}{\frac{5}{2}} = \frac{4}{m}\). This creates confusion about which relationships to use and may lead to guessing.

Second Most Common Error:

Poor SIMPLIFY execution: Students correctly set up \(\frac{2}{k} = \frac{-3}{\frac{5}{2}}\) but make errors when dividing by the fraction.

Common mistakes include forgetting to flip and multiply when dividing by 5/2, or sign errors with the negative. For example, they might get \(\frac{-3}{\frac{5}{2}} = \frac{-3}{5} \times 2 = \frac{-6}{5}\), but then incorrectly solve \(\frac{2}{k} = \frac{-6}{5}\) as \(k = 2 \times \frac{-6}{5}\) instead of \(k = \frac{2}{\frac{-6}{5}}\). This may lead them to select Choice (A) \(\frac{-10}{3}\) or other incorrect values.

The Bottom Line:

This problem tests both conceptual understanding of when linear systems have no solution and technical skill with fraction arithmetic. Success requires recognizing the specific proportionality condition AND executing the fraction calculations correctly.