Question:2/3x + 1/2y = 5/6 + 1/3x3/4y - 1/2 = kx + 1/4yIn the given system of equations, k is...
GMAT Algebra : (Alg) Questions
\(\frac{2}{3}\mathrm{x} + \frac{1}{2}\mathrm{y} = \frac{5}{6} + \frac{1}{3}\mathrm{x}\)
\(\frac{3}{4}\mathrm{y} - \frac{1}{2} = \mathrm{kx} + \frac{1}{4}\mathrm{y}\)
In the given system of equations, \(\mathrm{k}\) is a constant. If the system has no solution, what is the value of \(\mathrm{k}\)?
Express your answer as a fraction in lowest terms.
\(-3\)
\(-\frac{2}{3}\)
\(-\frac{1}{2}\)
\(-\frac{1}{3}\)
1. TRANSLATE the problem information
- Given: System of equations with parameter k
- Find: Value of k that makes system have no solution
- Key insight: "No solution" is a specific mathematical condition
2. SIMPLIFY both equations to standard form
- First equation: \(\frac{2}{3}\mathrm{x} + \frac{1}{2}\mathrm{y} = \frac{5}{6} + \frac{1}{3}\mathrm{x}\)
- Move x terms to left: \((\frac{2}{3} - \frac{1}{3})\mathrm{x} + \frac{1}{2}\mathrm{y} = \frac{5}{6}\)
- Result: \(\frac{1}{3}\mathrm{x} + \frac{1}{2}\mathrm{y} = \frac{5}{6}\)
- Second equation: \(\frac{3}{4}\mathrm{y} - \frac{1}{2} = \mathrm{kx} + \frac{1}{4}\mathrm{y}\)
- Move y terms to left: \((\frac{3}{4} - \frac{1}{4})\mathrm{y} = \mathrm{kx} + \frac{1}{2}\)
- Simplify: \(\frac{1}{2}\mathrm{y} = \mathrm{kx} + \frac{1}{2}\)
- Rearrange: \(-\mathrm{kx} + \frac{1}{2}\mathrm{y} = \frac{1}{2}\)
3. INFER the condition for no solution
- No solution occurs when equations represent parallel lines
- Parallel lines: same slope, different y-intercepts
- Need to convert to slope-intercept form: \(\mathrm{y} = \mathrm{mx} + \mathrm{b}\)
4. SIMPLIFY to slope-intercept form
- Equation 1: \(\frac{1}{3}\mathrm{x} + \frac{1}{2}\mathrm{y} = \frac{5}{6}\)
- Solve for y: \(\frac{1}{2}\mathrm{y} = -\frac{1}{3}\mathrm{x} + \frac{5}{6}\)
- Multiply by 2: \(\mathrm{y} = -\frac{2}{3}\mathrm{x} + \frac{5}{3}\)
- Equation 2: \(-\mathrm{kx} + \frac{1}{2}\mathrm{y} = \frac{1}{2}\)
- Solve for y: \(\frac{1}{2}\mathrm{y} = \mathrm{kx} + \frac{1}{2}\)
- Multiply by 2: \(\mathrm{y} = 2\mathrm{kx} + 1\)
5. APPLY CONSTRAINTS for parallel lines
- For same slope: \(-\frac{2}{3} = 2\mathrm{k}\)
- Solve: \(\mathrm{k} = -\frac{2}{3} \div 2 = -\frac{1}{3}\)
- Verify different y-intercepts: \(\frac{5}{3} \neq 1\) ✓
Answer: D) -1/3
Why Students Usually Falter on This Problem
Most Common Error Path:
Weak SIMPLIFY skill: Making fraction arithmetic errors when collecting like terms or converting between forms.
Students often struggle with operations like \((\frac{2}{3} - \frac{1}{3})\mathrm{x}\) or converting \(\frac{1}{2}\mathrm{y} = -\frac{1}{3}\mathrm{x} + \frac{5}{6}\) to slope-intercept form. A common mistake is getting \(\mathrm{y} = -\frac{1}{3}\mathrm{x} + \frac{5}{6}\) instead of \(\mathrm{y} = -\frac{2}{3}\mathrm{x} + \frac{5}{3}\), leading to the wrong slope comparison. This may lead them to select Choice B (-2/3) or get confused and guess.
Second Most Common Error:
Incomplete INFER reasoning: Understanding that no solution means something special, but not connecting it to the parallel lines condition.
Some students recognize they need equal slopes but forget to verify different y-intercepts, or they set the entire equations equal instead of just the slopes. This leads to confusion and guessing among the available choices.
The Bottom Line:
This problem combines algebraic manipulation with systems theory. Success requires both technical fraction skills and conceptual understanding of what "no solution" means geometrically.
\(-3\)
\(-\frac{2}{3}\)
\(-\frac{1}{2}\)
\(-\frac{1}{3}\)