Absolute value is a distance.
|−5| and |5| are both 5 — distance has no sign.
Which is why every absolute value question produces two cases: something can sit the same distance from zero on either side.
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Absolute Value Equations and Inequalities
Distance from zero — so every question splits into two.
Solve for x
| 2x − 3 | = 7
An absolute value of 7 means the inside is either 7 or −7. From 2x − 3 = 7 you get x = 5; from 2x − 3 = −7 you get 2x = −4, so x = −2. D is the classic miss — solving only the positive case and stopping.
Equations: split into two.
2x − 3 = 7
x = 5
2x − 3 = −7
x = −2
Isolate the bars first
For 3|x − 1| + 4 = 19, take off the 4 and divide by 3 before splitting. Splitting too early scatters the constants into the wrong places.
And |x| = −3 has no solution at all. A distance can never be negative.
The inside can be a quadratic. |x² − 10| = 6 splits into x² − 10 = 6 or x² − 10 = −6, so x² = 16 or x² = 4, and x = ±4 or x = ±2 — four solutions, each case giving two.
Less than: a sandwich.
| x − 4 | < 3 means x is within 3 of the number 4.
−3 < x − 4 < 3
one inequality, three parts
1 < x < 7
+ 4 everywhere
One connected stretch, centred on the 4.
Greater than: two separate pieces.
| x + 2 | > 5 means x is more than 5 away from −2, in either direction.
x + 2 > 5
x > 3
x + 2 < −5
x < −7
Never write −5 > x + 2 > 5
That sandwich says x is below −5 and above 5 at once, which nothing satisfies. A “greater than” answer needs the word or, on two separate lines.
The pattern, and the impossible cases.
| Form | Becomes | Shape |
|---|---|---|
| |u| = a | u = a or u = −a | two points |
| |u| = 0 | u = 0 only | one point |
| |u| < a | −a < u < a | one stretch |
| |u| > a | u > a or u < −a | two pieces |
| |u| < negative | nothing works | empty |
| |u| > negative | every real number | the whole line |
“Less than” keeps you close to the centre. “Greater than” pushes you away from it.
A variable on the other side: check the sign.
When the right side contains x, it might turn out negative — and an absolute value can never equal a negative. Solve |x − 6| = 2x.
x − 6 = 2x → x = −6
first case
x − 6 = −2x → 3x = 6 → x = 2
second case
x = −6: right side 2(−6) = −12 < 0 ✗
|−12| = 12 ≠ −12
x = 2: |−4| = 4 and 2(2) = 4 ✓
keep
Both cases are candidates, not answers
With a number on the right, both cases always work. With x on the right, substitute each candidate back: the right side must come out ≥ 0. Sometimes both survive, sometimes only one.
Variations: a negative number in front of the bars, a single solution, a whole ray of solutions.
Solve 5 − 2|x + 1| = −7.
−2|x + 1| = −12
− 5 both sides
|x + 1| = 6
÷ (−2); an equation, so nothing flips
x + 1 = 6 or x + 1 = −6 → x = 5 or x = −7
x = 5: 5 − 12 = −7 ✓
Different: the bars carry a minus in front; isolate them fully before deciding anything — the −7 on the right is not a “no solution” signal.
Stopping at the −7
Seeing a negative on the right and answering “no solution” is wrong here. The rule applies only once the bars stand alone: |x + 1| = 6 is positive.
Solve |3x − 6| + 4 = 4.
|3x − 6| = 0
isolate
3x − 6 = 0 → x = 2 only
+0 and −0 are the same case
Different: a distance of zero means one point, so there is one answer, not two.
Solve |x − 1| = x − 1.
|u| = u exactly when u ≥ 0
a number equal to its own distance
x − 1 ≥ 0 → every x ≥ 1
x = 3: 2 = 2 ✓; x = 0: 1 ≠ −1 ✗
Different: the two cases are x − 1 = x − 1 (always true) and x − 1 = −(x − 1) (x = 1), and the sign check keeps a whole ray, not a single point.
Bars on both sides.
|A| = |B| means A and B are equal or opposite: A = B or A = −B. Solve |2x + 1| = |x − 4|.
2x + 1 = x − 4 → x = −5
equal
2x + 1 = −(x − 4) → 3x = 3 → x = 1
opposite: bracket the whole of B
x = −5: |−9| = |−9| ✓; x = 1: |3| = |−3| ✓
both work
Shortcut when both insides are x minus a number: |x − 1| = |x + 7| asks for the point equally far from 1 and −7, which is their midpoint, 1 + (−7)2 = −3. The “equal” case x − 1 = x + 7 has no solution, so −3 is the only answer.
On the exam
Absolute value items test the two cases — isolate the bars first — and then the traps: a negative right side (no solution), x on the right (check each root), bars on both sides, or “how many solutions”. Ten-second check: put each choice into the original; bars make substitution quick.
Check: test a value from each region.
x = 5
|10 − 3| = 7
✓
x = −2
|−4 − 3| = 7
✓
x = 5 in the sandwich
|5 − 4| = 1 < 3
inside ✓
x = 0 in the last one
|2| = 2, not > 5
correctly excluded ✓
Now you try
Practice problem
Solve for x
| 3x + 1 | = 10
Split it in two. 3x + 1 = 10 gives 3x = 9, so x = 3. 3x + 1 = −10 gives 3x = −11, so x = −113. D just negates the first root — but the second case is 3x = −11, not 3x = −9, so the roots are not opposites. C drops the minus from −113, and A solves 3x − 1 = ±10, moving the 1 across with the wrong sign.