The graph of the equation ax + ky = 6 is a line in the xy-plane, where a and k...
GMAT Algebra : (Alg) Questions
The graph of the equation \(\mathrm{ax + ky = 6}\) is a line in the xy-plane, where a and k are constants. If the line contains the points \(\mathrm{(-2, -6)}\) and \(\mathrm{(0, -3)}\), what is the value of k?
\(\mathrm{-2}\)
\(\mathrm{-1}\)
\(\mathrm{2}\)
\(\mathrm{3}\)
1. TRANSLATE the problem information
- Given information:
- Equation: \(\mathrm{ax + ky = 6}\) represents a line
- Two points on the line: \((-2, -6)\) and \((0, -3)\)
- What this tells us: Both points must satisfy the equation
2. INFER the most efficient approach
- Notice that point \((0, -3)\) has x-coordinate 0
- This means when we substitute this point, the ax term disappears
- This gives us a direct path to find k
3. TRANSLATE the point condition into an equation
Substitute \((0, -3)\) into \(\mathrm{ax + ky = 6}\):
- \(\mathrm{a(0) + k(-3) = 6}\)
- \(\mathrm{0 - 3k = 6}\)
- \(\mathrm{-3k = 6}\)
4. SIMPLIFY to find k
- Divide both sides by -3: \(\mathrm{k = 6/(-3) = -2}\)
5. Verify with the second point (optional but recommended)
Using \((-2, -6)\) and \(\mathrm{k = -2}\):
- \(\mathrm{a(-2) + (-2)(-6) = 6}\)
- \(\mathrm{-2a + 12 = 6}\)
- \(\mathrm{a = 3}\)
Check: \(\mathrm{3(-2) + (-2)(-6) = -6 + 12 = 6}\) ✓
Answer: A. -2
Why Students Usually Falter on This Problem
Most Common Error Path:
Weak INFER skill: Students don't recognize the strategic advantage of using point \((0, -3)\) first. Instead, they set up a system of equations with both points simultaneously, creating unnecessary complexity with two unknowns. This leads to more opportunities for algebraic errors and may cause them to get confused during the solving process, leading to wrong answer selection or abandoning systematic solution and guessing.
Second Most Common Error:
Poor SIMPLIFY execution: Students correctly substitute \((0, -3)\) but make sign errors when dealing with \(\mathrm{-3k = 6}\). They might forget the negative sign on k or incorrectly divide by -3, getting \(\mathrm{k = 2}\) instead of \(\mathrm{k = -2}\). This leads them to select Choice C (2).
The Bottom Line:
This problem rewards strategic thinking - recognizing that one point provides a much cleaner path than trying to solve a full system. Students who miss this insight often get bogged down in unnecessary complexity.
\(\mathrm{-2}\)
\(\mathrm{-1}\)
\(\mathrm{2}\)
\(\mathrm{3}\)