|x - 3| = 2x - 6y = x + 1Which ordered pair \((\mathrm{x}, \mathrm{y})\) is the solution to the...

1. TRANSLATE the problem information

• Given system:
  • \(\mathrm{|x - 3| = 2x - 6}\)
  • \(\mathrm{y = x + 1}\)
• Need to find the ordered pair \(\mathrm{(x, y)}\) that satisfies both equations

2. INFER the solution approach

• The second equation is already solved for y in terms of x
• Focus on solving the absolute value equation first
• Once we find x, we can easily substitute to find y

3. CONSIDER ALL CASES for the absolute value equation

The expression inside the absolute value is \(\mathrm{(x - 3)}\), so we need two cases:

Case 1: When \(\mathrm{x - 3 \geq 0}\) (meaning \(\mathrm{x \geq 3}\))

• The absolute value becomes: \(\mathrm{|x - 3| = x - 3}\)
• Our equation becomes: \(\mathrm{x - 3 = 2x - 6}\)

4. SIMPLIFY Case 1

\(\mathrm{x - 3 = 2x - 6}\)
\(\mathrm{-3 + 6 = 2x - x}\)
\(\mathrm{3 = x}\)

5. APPLY CONSTRAINTS to check Case 1

• We found \(\mathrm{x = 3}\)
• Does this satisfy \(\mathrm{x \geq 3}\)? Yes, exactly
• Let's verify in the original equation: \(\mathrm{|3 - 3| = 0}\) and \(\mathrm{2(3) - 6 = 0}\) ✓

Case 2: When \(\mathrm{x - 3 \lt 0}\) (meaning \(\mathrm{x \lt 3}\))

• The absolute value becomes: \(\mathrm{|x - 3| = -(x - 3) = -x + 3}\)
• Our equation becomes: \(\mathrm{-x + 3 = 2x - 6}\)

6. SIMPLIFY Case 2

\(\mathrm{-x + 3 = 2x - 6}\)
\(\mathrm{3 + 6 = 2x + x}\)
\(\mathrm{9 = 3x}\)
\(\mathrm{x = 3}\)

7. APPLY CONSTRAINTS to check Case 2

• We found \(\mathrm{x = 3}\)
• Does this satisfy \(\mathrm{x \lt 3}\)? No! This contradicts our case condition
• Therefore, Case 2 provides no valid solution

8. Find the y-coordinate

• From Case 1, we have \(\mathrm{x = 3}\)
• Using \(\mathrm{y = x + 1}\): \(\mathrm{y = 3 + 1 = 4}\)

Answer: (B) (3, 4)

Why Students Usually Falter on This Problem

Most Common Error Path:

CONSIDER ALL CASES: Students often only work with one case for the absolute value equation, typically treating \(\mathrm{|x - 3|}\) as simply \(\mathrm{(x - 3)}\) without considering when this might be negative.

Working only with Case 1 might happen to give the right answer here, but if they worked only with Case 2, they might accept \(\mathrm{x = 3}\) without checking the constraint \(\mathrm{x \lt 3}\). This fundamental misunderstanding of absolute value can lead them to accept invalid solutions or miss valid ones on similar problems.

Second Most Common Error:

APPLY CONSTRAINTS: Even students who set up both cases correctly sometimes forget to verify that their solutions actually satisfy the case conditions.

In this problem, finding \(\mathrm{x = 3}\) from Case 2 and not checking that it contradicts \(\mathrm{x \lt 3}\) could lead to thinking there are two solutions. This confusion about which solution(s) to accept may cause them to get stuck and guess.

The Bottom Line:

Absolute value equations require systematic case analysis - you must consider all possibilities and verify that solutions are valid for their respective cases. Students who rush through or skip the constraint checking often make critical errors.