In the system of equations below, a and c are constants. 1/2x + 1/3y = 1/6 ax + y =...

1. INFER the key condition for infinite solutions

• When a system has infinitely many solutions, the two equations must be equivalent
• This means one equation is simply a multiple of the other equation
• All coefficients and constants must be proportional by the same factor

2. INFER which coefficients to compare first

• Look for the simplest coefficients to work with
• The y coefficients are: \(\frac{1}{3}\) (first equation) and \(1\) (second equation)
• Since these are easier to work with than mixed variables, start here

3. SIMPLIFY to find the proportionality factor

• Compare y coefficients: \(\frac{1}{3}\) in first equation, \(1\) in second equation
• What times \(\frac{1}{3}\) gives \(1\)?
• \(1 \div \frac{1}{3} = 1 \times 3 = 3\)
• So the second equation is 3 times the first equation

4. INFER what this means for the x coefficient

• If the second equation is 3 times the first equation, then ALL coefficients must follow this pattern
• x coefficient in first equation: \(\frac{1}{2}\)
• x coefficient in second equation: a
• Therefore: \(\mathrm{a} = 3 \times \frac{1}{2} = \frac{3}{2}\)

5. Verify the logic

• First equation × 3: \(3\left(\frac{1}{2}\right)\mathrm{x} + 3\left(\frac{1}{3}\right)\mathrm{y} = 3\left(\frac{1}{6}\right)\)
• This gives: \(\frac{3}{2}\mathrm{x} + 1\mathrm{y} = \frac{1}{2}\)
• This matches the form \(\mathrm{ax} + \mathrm{y} = \mathrm{c}\) when \(\mathrm{a} = \frac{3}{2}\)

Answer: D. \(\frac{3}{2}\)

Why Students Usually Falter on This Problem

Most Common Error Path:

Weak INFER skill: Not recognizing that infinite solutions requires equivalent equations

Many students try to solve the system algebraically (substitution or elimination) without realizing that infinite solutions is a special condition. They might attempt to solve for specific x and y values, getting confused when the algebra doesn't work out to unique solutions. This leads to confusion and guessing among the answer choices.

Second Most Common Error:

Poor SIMPLIFY execution: Making fraction arithmetic errors when finding the proportionality factor

Students might recognize the equivalent equations concept but make errors like:

• Thinking \(1 \div \frac{1}{3} = \frac{1}{3}\) instead of \(3\)
• Computing \(\mathrm{a} = \frac{1}{2} \div 3 = \frac{1}{6}\) instead of \(\mathrm{a} = 3 \times \frac{1}{2} = \frac{3}{2}\)

This may lead them to select Choice A (\(-\frac{1}{2}\)) or Choice C (\(\frac{1}{2}\)).

The Bottom Line:

This problem tests whether students understand that infinite solutions in systems occurs when equations are equivalent, not just when they're consistent. The key insight is recognizing proportional coefficients rather than trying to solve algebraically.