Distance from zero. Two numbers are always that far away — one each side.
Solve for x
|2x − 5| = 9
The expression inside can be 9 or −9. From 2x − 5 = 9, x = 7. From 2x − 5 = −9, 2x = −4 and x = −2. Check both: |14 − 5| = 9 ✓ and |−4 − 5| = |−9| = 9 ✓ A takes only the positive case, which is the standard omission. Two cases, every time.
|a| is the distance from a to 0, so it is never negative.
|8| = 8
8 is 8 units from 0
|−8| = 8
−8 is also 8 units from 0: drop the sign
|0| = 0
the only number at distance zero
A horizontal line above the vertex cuts the V twice. That is the picture behind "two cases". A line below the vertex misses entirely — which is why some equations have no solution.
|A| = c becomes A = c or A = −c, for c ≥ 0.
|2x − 5| = 9
two cases
2x − 5 = 9 → x = 7
case 1
2x − 5 = −9 → x = −2
case 2
|x + 4| = −3 has no solution at all, because an absolute value is a distance and distances are never negative. Spotting that saves splitting into two impossible cases.
|A| < c becomes −c < A < c.
|x − 3| < 5
within 5 of 3
−5 < x − 3 < 5
one compound statement
−2 < x < 8
+3 throughout
Read it literally: "the distance from x to 3 is less than 5" describes everything within 5 units of 3, which is exactly −2 to 8.
|A| > c becomes A < −c or A > c.
|x + 1| ≥ 4
at least 4 from −1
x + 1 ≤ −4 or x + 1 ≥ 4
two separate branches
x ≤ −5 or x ≥ 3
not a single band
That says a number is both at most −4 and at least 4, which nothing satisfies. A "greater than" absolute value inequality must be written as two statements joined by or.
The same split; these are the three places students stall.
The bars are not alone yet: 2|x + 1| − 3 = 9
2|x + 1| = 12
add 3
|x + 1| = 6
divide by 2 — now split
x + 1 = 6 or x + 1 = −6
two cases
x = 5 or x = −7
check: 2(6) − 3 = 9 ✓
Different: isolate the bars before splitting. Splitting 2|x + 1| − 3 = ±9 straight away gives wrong values.
Equal to zero: |4x − 10| = 0
4x − 10 = 0
+0 and −0 are the same case
x = 52
exactly one solution
Different: only a distance of zero gives one answer — the bottom tip of the V.
A negative x inside: |5 − 2x| < 3
−3 < 5 − 2x < 3
less than: a band
−8 < −2x < −2
−5 throughout
4 > x > 1
÷ (−2) flips both signs
1 < x < 4
test 2: |1| = 1 < 3 ✓
Different: dividing by a negative reverses both inequality signs; rewrite it smallest first.
The mnemonic is "less thAND, greatOR" — less than gives "and", greater than gives "or". Clumsy, but it survives an exam.
Expect a "solve" item with two answers to choose between, an inequality to match with a band or two rays, and a quick "no solution" item with a negative on the right. The fastest route is to isolate the bars, read the sign (= two cases, < a band, > two rays) and solve. The 10-second check: substitute one answer back, or test x = 0 to see whether the middle belongs.
Solve for x
|3x − 6| > 9
"Greater than" gives two branches: 3x − 6 < −9 or 3x − 6 > 9. The first gives 3x < −3, so x < −1; the second gives 3x > 15, so x > 5. The answer is x < −1 or x > 5. B is the "less than" answer — a band instead of two rays. Test x = 0: |−6| = 6, which is not greater than 9, so 0 must be excluded ✓
Hint: greater than means "or", so expect two separate pieces. Test x = 0 afterwards to confirm the middle is excluded.
