9. Absolute Value and Piecewise Traps

Absolute value measures distance, so it creates two directions. Before splitting an equation into cases, isolate the absolute-value expression. If the isolated value is negative, stop: a distance cannot be negative.

Example 1 - two valid branches:
\[ |3x-5|=16. \]
Write two equations:
\[ 3x-5=16 \quad\text{or}\quad 3x-5=-16. \]
The first gives \(x=7\); the second gives \(x=-\frac{11}{3}\). Check both in the original equation.

Example 2 - impossible target:
\[ 2|x+4|+7=3. \]
Subtract 7 and divide by 2:
\[ |x+4|=-2. \]
Therefore there is no solution. Splitting into cases before isolating would waste time and often creates fake answers.

For inequalities, connect absolute value to distance:

• \(|x-a|inside the interval: \(a-r
• \(|x-a|>r\) means outside the interval: \(xa+r\).

Example 3:
\[ |2x-1|\le 7. \]
Use a compound inequality:
\[ -7\le 2x-1\le 7. \]
Add 1 and divide by 2:
\[ -3\le x\le 4. \]

Piecewise functions use the same boundary discipline. At a boundary such as \(x=2\), choose the branch whose condition contains the equality sign. An open condition like \(x<2\) does not own the point \(x=2\).

Common trap: treating \(|u|=k\) as only \(u=k\), or using "and" for an outside inequality that requires "or."

Advanced mastery lab

Learning targets

• isolate an absolute-value expression before creating cases
• translate inside and outside distance conditions correctly
• select the correct piecewise branch at a boundary

Visual model

How to read this SVG. Identify the center, then move the stated distance in both directions. For inequalities, inside distance uses an intersection of bounds, while outside distance uses two separate rays.

Decision protocol

1. Isolate the absolute-value expression.
2. Stop if it equals a negative number.
3. For an equation, create two cases; for an inequality, choose inside or outside form.
4. Check endpoints, branches, and every candidate in the original statement.

Decision table

Timed transfer challenge: Isolate before branching

Target time. 4 minutes, with both branches checked.

Problem. Solve \(2|3x-1|-5=13\).

Solution.

1. Add 5 and divide by 2: \(|3x-1|=9\).
2. Create cases \(3x-1=9\) and \(3x-1=-9\).
3. The cases give \(x=10/3\) and \(x=-8/3\).
4. Both candidates make the absolute value equal 9.

Answer. \(x=\frac{10}{3}\) or \(x=-\frac{8}{3}\)

Second independent check

Retrieval and reflection

Explain and versus or in absolute-value inequalities using distance language.

Mastery standard. Complete the work accurately, explain the decisive step, and verify the answer using the original condition or an independent representation.