|x − a| is the distance from a. "Less than" traps x between two fences; "greater than" throws it outside them.
The problem
|x − 3| < 5
"Within 5 of 3": step 5 both ways from 3 to get −2 < x < 8. B solves only one side; C answers the OPPOSITE question (> instead of <); D steps both ways from the wrong centre.
|x − 3| measures how far x sits from 3 on the number line — no cases, no formalism. Every absolute value inequality is a sentence about distance: |x − 3| < 5 says "x is within 5 of 3". Translate first; the algebra becomes geography.
|x − a| < b → a − b < x < a + b
one connected stretch of the line
|x − 3| < 5 → −2 < x < 8
3 minus 5, 3 plus 5
Worked example: solve |2x + 1| < 9. A coefficient inside the bars changes nothing about the shape — one sandwich, then undo the inside on all three parts.
−9 < 2x + 1 < 9
less-thAND: one sandwich
−10 < 2x < 8
− 1 from all three parts
−5 < x < 4
÷ 2 all three parts (positive, so no flip)
x = 0: |0 + 1| = 1 < 9 ✓
a point inside the band works
|x − a| > b → x < a − b or x > a + b
two separate pieces — the word "or" is compulsory
|x − 3| > 5 → x < −2 or x > 8
far from 3, in either direction
The mnemonic pair: less-thAND squeezes into one strip, greatOR splits into two rays. Writing "−2 > x > 8" — a between-answer for a greater-than question — is the classic mark-loser.
2|x| + 1 < 7
the bars are not alone yet
2|x| < 6
− 1 both sides
|x| < 3
÷ 2 both sides — now translate
−3 < x < 3
less-thAND: within 3 of 0
Drop the bars too early and you write −7 < 2x + 1 < 7, which gives −8 < 2x < 6 and −4 < x < 3 — lopsided, and wrong: x = −3.5 passes it, but 2(3.5) + 1 = 8 is not less than 7. Only |…| itself can be split into a sandwich.
| Statement | Solution |
|---|---|
| |x| < −2 | none — a distance is never negative |
| |x| > −2 | every real number — all distances beat −2 |
Every interval has a centre and a half-width. Write 2 ≤ x ≤ 10 as an absolute value inequality.
centre = 2 + 102 = 6
the midpoint of the fences
half-width = 10 − 22 = 4
the distance allowed
|x − 6| ≤ 4
between, ends included → ≤
Outside instead of between: x < −3 or x > 5 has fences −3 and 5, centre −3 + 52 = 1 and half-width 4, so it is |x − 1| > 4 — greater than, because x is far from the centre, and strict, because the fences are open.
A centre of −2 gives |x − (−2)| = |x + 2|. Writing |x − 2| for a centre at −2 moves the whole interval. Check by putting a fence back in: it must make the two sides equal.
Solve −3|x − 1| + 2 ≥ −10.
−3|x − 1| ≥ −12
− 2 both sides
|x − 1| ≤ 4
÷ (−3): the sign flips
−3 ≤ x ≤ 5
less-thAND; x = 0: −3 + 2 = −1 ≥ −10 ✓
Different: the flip happens while isolating the bars, and it turns a ≥ question into a “between” answer.
Keeping ≥ leaves |x − 1| ≥ 4 and the answer x ≤ −3 or x ≥ 5 — exactly the wrong region. Test x = 0: it satisfies the original, so 0 must be in the answer.
Solve |3 − 4x| > 5.
3 − 4x > 5 → −4x > 2 → x < −½
÷ (−4): flip
3 − 4x < −5 → −4x < −8 → x > 2
÷ (−4): flip
x < −½ or x > 2
x = 3: |−9| = 9 > 5 ✓; x = 0: 3 ✗
Different: with −4x inside, each branch needs its own flip — the “> 5” branch gives the left-hand ray, x < −½.
Solve |x + 4| ≤ 0, then |x + 4| > 0.
|x + 4| ≤ 0 → |x + 4| = 0
a distance cannot be below 0
x = −4 only
one point
|x + 4| > 0 → every x except −4
any distance but zero
Different: with 0 on the right the “strip” shrinks to a single point, and the “outside” is the whole line with one hole.
2 ≤ |x + 1| < 5 says x is at least 2 but less than 5 away from −1. Solve each half and keep the overlap.
|x + 1| < 5 → −6 < x < 4
less-thAND: one strip
|x + 1| ≥ 2 → x ≤ −3 or x ≥ 1
greatOR: two rays
−6 < x ≤ −3 or 1 ≤ x < 4
the strip with a hole cut from its middle
integers: −5, −4, −3, 1, 2, 3
six in all
For integers there is a quicker route: |x + 1| can only be 2, 3 or 4, so x + 1 = ±2, ±3, ±4 and x = 1, −3, 2, −4, 3, −5 — the same six.
Expect a solve (isolate the bars first), a count of integers, a number line or tolerance story to match with |x − c| ≤ r, or a bars-between-two-numbers item. Read every one as a distance from the centre — within means ≤ or <, beyond means ≥ or > — then test one point inside and one outside.
The problem
|2x − 1| ≥ 7
GreatOR — two pieces: 2x − 1 ≥ 7 gives x ≥ 4, and 2x − 1 ≤ −7 gives x ≤ −3: x ≤ −3 or x ≥ 4. A answers the between question; B forgets the negative branch entirely; D solves |2x + 1| ≥ 7, moving the 1 across with the wrong sign. Test x = 0: |−1| = 1, not ≥ 7 — so 0 must be excluded, which kills A on the spot.
