|X| < k says X is within k of zero, which is one interval and an and. |X| > k says X is further than k from zero, which is two rays and an or. The connector is not a convention to memorise; it is what the distance reading gives.
Core ideas
Two rules, one picture they both come from, and a set of right-hand sides that settle the question before any splitting.
|X| < 5 says the distance from X to 0 is under 5, so X lies between −5 and 5. |X| > 5 says the distance is over 5, so X lies outside that band. The two solution sets are complementary except at the two boundary points, where the distance is exactly 5 and neither strict statement holds.
A band is one region described by two conditions holding together: an and. Two rays are separate regions, and a number in either one qualifies: an or. Deriving the connector from the picture takes a second and is safer than recalling which word goes with which relation.
A distance is never negative and can be zero. |X| < −1 has no solutions, |X| > −1 has every real number, and |X| ≤ 0 holds only where X = 0. None of these needs a case split, and none of them is a special rule — each is read off what an absolute value is.
The material
The two rules, the picture each comes from, and the right-hand sides that end the question before it starts.
For k > 0, |X| < k is true exactly when −k < X < k.
With ≤ throughout: |X| ≤ k is true exactly when −k ≤ X ≤ k. The three-part form is legitimate here because the connector genuinely is and.
For k > 0, |X| > k is true exactly when X < −k or X > k.
With ≥: X ≤ −k or X ≥ k. There is no three-part form for this one, and writing one is the standard error of the topic.
Both are read straight off the number line. The numbers less than 5 away from 0 are those strictly between −5 and 5; the numbers more than 5 away are those below −5 together with those above 5. X may be any expression, and the rule applies to the whole of what the bars enclose.
Turn the relation round and the same two numbers reappear on the outside. |x − 2| > 5 gives x − 2 < −5 or x − 2 > 5, that is x < −3 or x > 7. The two solution sets do not overlap and do not quite cover the line: at x = −3 and at x = 7 the distance is exactly 5, so both strict statements fail.
The divisor there was 2, a positive number, so the relation stood. Had the statement been −2|x − 1| > −6, dividing by −2 would reverse it to |x − 1| < 3 — a band rather than two rays, and a different answer entirely. “Linear Inequalities”'s rule applies to the step that isolates the bars, before any case split exists.
None of these four needs a split, and none of them is a rule to memorise. Each is read off the two facts that an absolute value is never negative and is zero only where the inside is zero. The two theorems above were stated for k > 0, and reading the right-hand side first is how you stay inside the range they were stated for.
|x − 2| < 5 gives −3 < x < 7, and |x − 2| > 5 gives x < −3 or x > 7. Between them they account for every real number except −3 and 7. Replace both relations by ≤ and ≥ and the two answers instead overlap at exactly those two points, which is where the distance is exactly 5.
Interactive
Less than gives a band, greater than gives two rays
Method
The first two steps decide which rule you are about to use, or whether you need one at all.
Reach |X| < k or |X| > k with the bars alone on one side. Any multiplication or division you perform here obeys “Linear Inequalities”: a negative multiplier reverses the relation, and it does so before any case split exists.
Negative with < or ≤: no solutions. Negative with > or ≥: every real number. At zero, < gives none, ≤ gives X = 0, > gives everywhere except X = 0, and ≥ gives every real number. With a positive right side, go on to step 3.
|X| < k becomes −k < X < k, one statement, connector and. |X| > k becomes X < −k or X > k, two statements, connector or. Keep ≤ and ≥ if that is what you started with.
Each part is an ordinary linear inequality with its own reversal check. Finish by testing a number from inside the answer and one from outside against the original statement, with the bars still in it.
|3 − 2x| ≥ 7. The relation is a greater-than, so the connector is or, and both cases divide by a negative number.
Two reversals, one in each case, and they have nothing to do with the or that joined the cases. The connector came from the ≥ outside the bars; the reversals came from the −2 inside them. Mixing the two up is how this exercise usually goes wrong.
Both boundary numbers are in the answer because the relation is ≥ rather than >: at x = −2 the inside is 7 and at x = 5 it is −7, both of absolute value 7. Change the statement to |3 − 2x| > 7 and the answer becomes x < −2 or x > 5, with the same two numbers now excluded.
For a faded band, isolate 3|x − 2| − 1 ≤ 8; the supplied result is |x − 2| ≤ 3. Write the three-part inequality, solve all three parts, and test both endpoints plus one outside value before using the lab.
Solve −2|x + 1| ≤ −6 without the lab. Record the reversal while isolating the bars, derive the connector from the distance reading, solve both pieces and test both boundaries plus one excluded value in the original. Then classify all four statements |x| < 0, |x| ≤ 0, |x| > 0, and |x| ≥ 0.
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