Let y be a real number. Consider the equation |2y - 6| = 2|y - 3|. How many real solutions...

1. TRANSLATE the problem information

• Given equation: \(|2y - 6| = 2|y - 3|\)
• Need to find: How many real solutions exist

2. INFER the strategic approach

• Notice that the left side can be factored: \(2y - 6 = 2(y - 3)\)
• This suggests we can use the absolute value property \(|ab| = |a||b|\)
• Key insight: If we can make both sides identical, we have an identity

3. SIMPLIFY by factoring and applying absolute value properties

• Factor the left side: \(|2y - 6| = |2(y - 3)|\)
• Apply the property \(|ab| = |a||b|\): \(|2(y - 3)| = |2| \cdot |y - 3| = 2|y - 3|\)
• The equation becomes: \(2|y - 3| = 2|y - 3|\)

This is an identity! It's true for every real value of y.

4. CONSIDER ALL CASES to verify our conclusion

Let's double-check by examining what happens in different regions:

Case 1: When \(y \geq 3\)

• \(|y - 3| = y - 3\) (since \(y - 3 \geq 0\))
• \(|2y - 6| = 2y - 6\) (since \(2y - 6 \geq 0\) when \(y \geq 3\))
• Left side: \(2y - 6\)
• Right side: \(2(y - 3) = 2y - 6\)
• Equal! ✓

Case 2: When \(y \lt 3\)

• \(|y - 3| = -(y - 3) = 3 - y\) (since \(y - 3 \lt 0\))
• \(|2y - 6| = -(2y - 6) = 6 - 2y\) (since \(2y - 6 \lt 0\) when \(y \lt 3\))
• Left side: \(6 - 2y\)
• Right side: \(2(3 - y) = 6 - 2y\)
• Equal! ✓

Since the equation holds for all \(y \geq 3\) AND all \(y \lt 3\), it holds for all real numbers.

Answer: D (Infinitely many)

Why Students Usually Falter on This Problem

Most Common Error Path:

Weak INFER skill: Students don't recognize the factoring opportunity and instead try to solve by setting up cases immediately: "When \(y \geq 3\), then \(|2y - 6| = 2y - 6\)..." This leads to solving two separate equations without realizing they can simplify first. They might find solutions like \(y = 3\) or specific values, missing that ALL real numbers work. This may lead them to select Choice B (1) or Choice C (2).

Second Most Common Error:

Incomplete CONSIDER ALL CASES execution: Students correctly factor and get \(2|y - 3| = 2|y - 3|\), but then think "this means \(y - 3 = 0\), so \(y = 3\)" without recognizing this is an identity. They conclude there's only one solution where the expressions inside the absolute value bars equal zero. This leads them to select Choice B (1).

The Bottom Line:

This problem tests whether students can recognize when algebraic manipulation reveals an identity equation, rather than getting bogged down in case-by-case solving. The key insight is that factoring first dramatically simplifies the problem.