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2x + 3y = 5 and 4x + ky = 7. In the given system of equations, k is a...

GMAT Algebra : (Alg) Questions

Source: Prism
Algebra
Systems of 2 linear equations in 2 variables
HARD
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Notes
Post a Query

\(2\mathrm{x} + 3\mathrm{y} = 5\) and \(4\mathrm{x} + \mathrm{k}\mathrm{y} = 7\). In the given system of equations, \(\mathrm{k}\) is a constant. If the system has no solution, what is the value of \(\mathrm{k}\)?

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Solution

1. TRANSLATE the problem requirements

  • Given: System of equations \(\mathrm{2x + 3y = 5}\) and \(\mathrm{4x + ky = 7}\)
  • Find: Value of k that makes the system have no solution

2. INFER what "no solution" means mathematically

  • A system has no solution when the equations represent parallel lines
  • Parallel lines occur when coefficient ratios are equal BUT constant ratios are different
  • This means: (coefficient of x in eq1)/(coefficient of x in eq2) = (coefficient of y in eq1)/(coefficient of y in eq2)
  • BUT: (constant in eq1)/(constant in eq2) ≠ (coefficient ratio)

3. SIMPLIFY by setting up the coefficient proportion

  • For our system: \(\mathrm{\frac{2}{4} = \frac{3}{k}}\)
  • Simplify the left side: \(\mathrm{\frac{1}{2} = \frac{3}{k}}\)
  • Cross-multiply: \(\mathrm{k = 6}\)

4. APPLY CONSTRAINTS by verifying no solution condition

  • When \(\mathrm{k = 6}\): \(\mathrm{2x + 3y = 5}\) and \(\mathrm{4x + 6y = 7}\)
  • Divide second equation by 2: \(\mathrm{2x + 3y = 3.5}\)
  • Now we have: \(\mathrm{2x + 3y = 5}\) and \(\mathrm{2x + 3y = 3.5}\)
  • Since \(\mathrm{5 \neq 3.5}\), these are parallel lines with no intersection

Answer: D) 6




Why Students Usually Falter on This Problem

Most Common Error Path:

Weak INFER skill: Students don't recognize the specific conditions for no solution. They might think "no solution" just means the equations are different, or they might confuse it with infinite solutions (which occurs when equations are identical). This conceptual confusion leads to random guessing among the answer choices.

Second Most Common Error:

Inadequate SIMPLIFY execution: Students correctly identify that coefficients must be proportional but make algebraic errors when solving \(\mathrm{\frac{1}{2} = \frac{3}{k}}\). Common mistakes include getting \(\mathrm{k = \frac{3}{2}}\) or \(\mathrm{k = \frac{6}{2} = 3}\), which may lead them to select Choice A) 3.

The Bottom Line:

This problem tests whether students understand the geometric meaning of "no solution" (parallel lines) and can translate that into the algebraic condition of proportional coefficients with non-proportional constants. The key insight is recognizing that you need BOTH conditions to ensure no solution.

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