|X| = k says the expression X sits at distance k from zero, and for a positive k there are exactly two places it can be. That is the whole method: isolate the bars, split into two equations, and substitute every candidate back into the original.
Core ideas
Everything follows from one reading of the bars, and one discipline about when the rule may be applied.
|X| is how far X lies from zero, with no direction attached. Foundations of Algebra defined it in two cases; here the distance reading is the useful one, because |X| = 5 then says that X is 5 away from 0, and there are exactly two such places.
|x − 3| = 5 becomes x − 3 = 5 or x − 3 = −5. Both get solved and both answers count. This is not a choice between two possibilities; it is two equations whose solution sets are combined.
The rule is stated for |X| = k, not for 3|X| + 2 = k. Isolating the absolute value is a step of the method, not a tidying-up, and splitting early produces numbers that fail the check.
The material
The two-case rule, where it comes from, and the situations in which the number of solutions is not two.
|a| is the distance from a to 0 on the number line. In cases, |a| = a when a ≥ 0, and |a| = −a when a < 0.
The second case is easy to misread. −a is not a negative number when a is itself negative: |−4| = −(−4) = 4. The minus sign there reverses a sign; it does not impose one.
For k > 0, the equation |X| = k is true exactly when X = k or X = −k.
For k = 0 it is true exactly when X = 0, a single case. For k < 0 it is never true, because a distance is never negative.
Read the theorem on the number line. The numbers at distance 5 from 0 are 5 and −5, and there are no others. X may be any expression — x − 3, 2x + 1, anything at all — and the rule applies to the whole of what the bars enclose, not to x.
Both candidates survive, and both belong in the answer: the solution set is { 5, −3 }. Reporting one of them is not a smaller answer, it is a wrong one, because the question asked which numbers make the statement true and two of them do.
The middle case repays a closer look. Splitting |x + 4| = −2 mechanically gives x + 4 = −2 and x + 4 = 2, so x = −6 and x = −2. Substituting: |−6 + 4| = 2 and |−2 + 4| = 2, and neither equals −2. The split produced two numbers and neither is a solution, which is exactly what the check exists to discover.
When the right side contains x, its sign is not known in advance, and the theorem — stated for a positive k — carries no guarantee that the two cases will deliver solutions. Split anyway, then substitute every candidate into the original equation. Here the check is not a safeguard; it is part of the method.
|2x − 1| = x + 3. Case one: 2x − 1 = x + 3, so x = 4; then the left side is |7| = 7 and the right side is 4 + 3 = 7. Case two: 2x − 1 = −(x + 3) = −x − 3, so 3x = −2 and x = −2/3; then the left side is |−7/3| = 7/3 and the right side is −2/3 + 3 = 7/3. Both check, and the solution set has two members.
A candidate thrown out by the check is called extraneous. It is not the mark of an arithmetic slip: the splitting step can produce equations with solutions the original never had, and only substitution finds out which. The same phenomenon returns in Rational and Radical Expressions, when both sides of an equation get squared.
Interactive
Two cases, and a check that throws one out
Method
Step 1 comes before everything else, and step 4 is not a formality.
Reach |X| = (something) with nothing multiplying the bars and nothing added to them. Until that is done there is no rule available to apply, and applying one anyway is the commonest way to get this topic wrong.
A negative number: no solutions, and you are finished. Zero: one case, X = 0. A positive number: two cases. An expression containing x: split, and rely on step 4 to sort the candidates out.
X = k and X = −k. Write the second with a bracket — X = −(k) — so that a right side with several terms gets every one of its signs flipped.
Not into a line halfway down your working, which has already had the bars removed. The splitting step is where extraneous values enter, and only the original statement can reject them.
|2x − 1| = x − 5. The right side contains x, so the check is what decides the answer.
Both candidates fail and the equation has no solutions. That is a complete answer, not an unfinished one, and writing no solutions is the correct final line.
The failure was predictable. |2x − 1| is never negative, so any solution would need x − 5 ≥ 0, that is x ≥ 5. Both candidates are below 5, so neither could have worked. Checking the sign of the right side first would have saved two substitutions — but it is the substitutions that establish the rejection, and the sign argument only predicts it.
Change the right side to x + 3 and the same two cases produce x = 4 and x = −2/3, and both survive. The algebra of the split is identical; everything that differs is in the check.
For a faded solve, isolate the bars in 2|x + 2| − 1 = 9; the supplied result is |x + 2| = 5. Write both equations, finish both candidates and substitute each into the original statement before using the lab.
Solve |2x + 3| = x without the lab. State the sign restriction on the right side, solve both case equations, and substitute both candidates into the original. If neither survives, write the empty solution set and identify why the case equations still produced numbers.
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