Recipe

Solve an absolute value equation or inequality

Use when: an absolute value expression sits somewhere in an equation or inequality

Goal: every valid solution to an equation, or the complete solution set to an inequality, with impossible cases ruled out before any needless case-splitting

Taught in Absolute value and inequalities

Quick recipe

Isolate the absolute value
Check whether the isolated value is even possible
Split by the comparison symbol
Solve each resulting piece
Assemble the answer and check it

1
Get the absolute value expression alone on one side, undoing anything added, subtracted, or multiplied outside the bars first.

Example

$3|2x+1| - 4 = 11$ : add 4, then divide by 3, to get $|2x+1| = 5$ .

Why: the bars can't be split into cases until the expression inside them is standing alone. An operation left outside would otherwise land on only one of the two resulting branches, corrupting both.
2
Look at what the absolute value is being compared to before doing any case work.

Example

$|x+1| = -3$ stops here: no solution, since a distance can never equal $-3$ .
- compared to a negative number, with $=$ , $<$ , or $\le$ — no solution, since a distance can't be negative.
- compared to a negative number, with $>$ or $\ge$ — every real number works, since a distance is always at least 0 and clears any negative target automatically.
- compared to zero or a positive number — continue to the next step and actually split into cases.
Why: splitting a "distance equals a negative number" claim into two equations wastes effort solving branches that were never going anywhere. Reading the sign first catches it in one glance.
3
How the split works depends on whether it's an equation, a less-than, or a greater-than.

Example

$|2x+1| = 5$ splits into $2x+1 = 5$ or $2x+1 = -5$ .
- equation, $|u| = a$ — two separate equations, $u = a$ or $u = -a$ .
- less-than, $|u| < a$ or $|u| \le a$ — one "and" compound: $-a < u < a$ (matching $\le$ at both ends for the non-strict version).
- greater-than, $|u| > a$ or $|u| \ge a$ — two separate pieces joined by "or": $u < -a$ or $u > a$ (matching $\ge$ at both ends for the non-strict version).
Why: distance from zero is symmetric. Landing exactly $a$ away means $+a$ or $-a$ ; landing closer than $a$ away sandwiches you between them; landing farther away puts you past one end or the other.
4
Solve every branch from the split with ordinary linear-equation or linear-inequality moves.

Example

$2x + 1 = 5 \Rightarrow x = 2$ ; and $2x + 1 = -5 \Rightarrow x = -3$ .

Why: once split, every branch is a plain linear problem, with no absolute-value machinery left to apply.
5
Combine the branches the way the split from Step 3 requires. An "and" split stays one interval. An "or" split, or an equation's two branches, stays two separate pieces: joined with $\cup$ for an inequality, or just listed together for an equation. Check every candidate in the ORIGINAL equation.

Example

The equation case gives $\{-3, 2\}$ . Check: $|2(2)+1| = 5$ ✓ and $|2(-3)+1| = 5$ ✓.

Why: the split only preserves the original problem's meaning if the pieces are combined the right way. Unlike squaring a radical equation, splitting an absolute value never manufactures an extraneous branch on its own, so this check confirms arithmetic, not structure.

Common mistakes

Splitting into cases before isolating the absolute value.
Trying to case-split something like $|u| = -3$ instead of recognizing it has no solution.
Using "and" (sandwich) logic on a greater-than inequality, or "or" (split) logic on a less-than one.
Forgetting that $|u| \ge 0$ makes a claim like " $|u| > -5$ " true for every real number, with nothing left to solve.