Imagine a dog on a leash attached to a post on a straight path. If the leash reaches four units, the dog can be on either side, but it must stay between the two farthest reachable positions. An absolute value inequality describes a whole set of positions. A less than sign means closer than the allowed distance. A less than or equal sign permits the farthest positions too. A compound inequality joins two comparisons. For a between answer, both comparisons must be true for the same position. This is the word 'and'. The answer is one uninterrupted stretch of the number line, called an interval. Its endpoint symbols tell you whether the ends belong.
The allowed interval runs from 1 through 9, with both endpoints included.
Reminder
Letters naming whole expressions. In |2x − 1| < 5, A names 2x − 1 and B names 5. Keep that whole A together.
Changing all three parts. −4 ≤ x − 5 ≤ 4 becomes 1 ≤ x ≤ 9 after adding 5 everywhere.
Flipping after negative division. −2x < 6 becomes x > −3. A three part chain has two signs to flip.
Endpoint notation. 1 < x < 9 uses hollow dots and (1, 9); 1 ≤ x ≤ 9 uses filled dots and [1, 9].
Why it works. For B > 0, let A name the whole expression inside the bars. A distance less than B cannot reach B on the right or −B on the left. Both A > −B and A < B must hold, giving −B < A < B. Allowing equality includes the two boundary positions. Addition slides every quantity by the same amount and preserves order. Division by a negative turns the line around and reverses both comparisons, so the final chain may need reading from right to left.
RuleFor B > 0: |A| < B means −B < A < B; |A| ≤ B means −B ≤ A ≤ B. Both comparisons must hold: 'and'. A is everything inside the bars; B is the number on the other side.
The same idea, five ways
Say it
x is at most four units from five
Write it
x is between one and nine, including both ends.
In math
|x − 5| ≤ 4
−4 ≤ x − 5 ≤ 4
1 ≤ x ≤ 9
[1, 9]
{x | 1 ≤ x ≤ 9}
{x : 1 ≤ x ≤ 9}
Like
A four unit leash attached at position five reaches positions one through nine.
See itFilled dots include 1 and 9, and the whole connecting stretch belongs.
The same idea, other ways
As a leash
A post at 5 and a four-unit leash allow movement left to 1 and right to 9. You may stand anywhere in between; you cannot skip past either limit.
As two guards
−4 ≤ A says the inside value cannot go too far left. A ≤ 4 says it cannot go too far right. Both guards must approve the same A, so the connector is 'and'.
As a V under a line
The graph y = |x − 5| lies at or below y = 4 exactly between its two crossings. This turns a vertical comparison of outputs into a horizontal interval of inputs.
At or below the comparison line means inputs between the crossings.
Why the endpoints change
At x = 1 and x = 9 the distance is exactly 4. The condition |x − 5| ≤ 4 accepts equality. The condition |x − 5| < 4 rejects it. Interior inputs satisfy either condition.
Form (B > 0)
Means
Write
Picture
|A| < B
Closer than B: between
−B < A < B
One piece, open endpoints
|A| ≤ B
At most B away: between
−B ≤ A ≤ B
One piece, closed endpoints
.1Strict between
A shorter leash stops before either farthest position. A strict inequality uses < or > and leaves its boundary out. For a closer than question, both ends are excluded.
For B > 0, |A| < B means −B < A < B.
Strict endpoints use hollow dots and parentheses.
For example, |x| < 2 gives −2 < x < 2; the harder shifted example follows.
Hollow dots exclude 1 and 9 from the strict interval (1, 9).
Reminder
Strict comparison. 3 < 4 is true; 4 < 4 is false. Equal distance fails a strict comparison.
Worked exampleStrictly less than four from five
Solve |x − 5| < 4.
The strict comparison leaves both boundaries out.
What it asks. Find every input closer than four units to five, with the ends left out.
Plan. Use a strict between chain and undo the subtraction of five.
The middle expression is x − 5.
Distance exactly four does not pass a strict condition.
Write −4 < x − 5 < 4.The whole inside must stay farther right than −4 and farther left than 4.
Add 5 everywhere: −4 + 5 < x − 5 + 5 < 4 + 5, giving 1 < x < 9.This removes the subtraction and finds the allowed inputs without changing either strict sign.
Write (1, 9).Parentheses say neither equality boundary is included.
Answer
1 < x < 9.
Interval: (1, 9).
Check x = 6 gives |1| = 1 < 4, while x = 1 and 9 each give 4 and fail the strict comparison.
Counterexamples: what it is not, and the tempting wrong moves
✗ Not this: The strict answer is [1, 9].
The endpoints give distance exactly 4 rather than less than 4.
✓ Instead: Write (1, 9), with both ends excluded.
Tips and tricks
A hollow dot means the boundary number is left out.
.2Inclusive between
A full length leash permits standing at either farthest position. An inclusive inequality uses an equals bar, as in ≤ or ≥ with the numbers around it. The endpoint is allowed along with the positions inside.
For B > 0, |A| ≤ B means −B ≤ A ≤ B.
Finite endpoints are actual numbers, rather than infinity. Include an allowed finite endpoint with a filled dot and a bracket.
For example, |x| ≤ 2 gives [−2, 2]; the score example uses the same idea around 80.
The score interval includes both endpoints, 60 and 100.
Reminder
Names for variables. In |S − 80| ≤ 20, S is a score, as R represented a resistance earlier.
Worked exampleScores near eighty
Scores within 20 points of 80 pass. Include the maximum distance. Write the condition and score interval.
The permitted scores fill [60, 100].
What it asks. Write every score no more than twenty points from eighty, including the limits.
Plan. Measure the score's distance from eighty, use a between chain, and add eighty to each part.
Use S for the score; a variable can have a letter other than x.
The word 'include' permits equality at the two ends.
Let S be the score and write |S − 80| ≤ 20.The bars measure the distance from the score to 80, and a passing distance can equal 20.
Write −20 ≤ S − 80 ≤ 20.The score's signed difference must fit between the two permitted distance limits.
Add 80 to every part: −20 + 80 ≤ S ≤ 20 + 80, giving 60 ≤ S ≤ 100.This restores the score itself and finds the lowest and highest passing scores.
Write [60, 100].The filled endpoint choice records equality at both passing limits.
Answer
|S − 80| ≤ 20
60 ≤ S ≤ 100
Interval: [60, 100]
Check Scores 60 and 100 both differ from 80 by twenty points; 80 differs by zero.
Counterexamples: what it is not, and the tempting wrong moves
✗ Not this: Exclude scores 60 and 100.
Both are exactly twenty points from 80, and the question explicitly allows that distance.
✓ Instead: Keep the filled endpoints and write [60, 100].
Tips and tricks
An equals bar means the end number counts.
.3Read the same solution in three notations
Think of writing the same street address in words, in a map label, and on an envelope. Inequality notation, interval notation, and set-builder notation describe the same allowed numbers with different symbols. A set is a collection of numbers. Set-builder notation describes that collection by a condition. The vertical bar or the colon inside the braces means 'such that'; it has a different job from a pair of absolute value bars.
{x | 1 ≤ x ≤ 9} reads 'the set of all real numbers x such that x is between 1 and 9, ends included'.
{x : 1 ≤ x ≤ 9} has the same meaning; the colon avoids confusing the divider with absolute value bars.
1 ≤ x ≤ 9, [1, 9], and {x : 1 ≤ x ≤ 9} all name the same collection.
One filled interval is the picture shared by all three notations.
Reminder
Set notation for individual answers. {4} contains one number. A condition in {x : 1 ≤ x ≤ 9} selects every number in a whole interval.
Worked examplePut a between answer into set-builder notation
Write the solution 1 ≤ x ≤ 9 in interval notation and set-builder notation.
Changing notation leaves every allowed input in the same place.
What it asks. Name the same allowed stretch using two other ways of writing it.
Plan. Keep the endpoints and their inclusion, then place the condition after 'such that'.
The interval runs from the smaller endpoint to the larger.
Both comparison signs allow equality.
Write [1, 9].Both end numbers belong, so both ends use brackets.
Write {x | 1 ≤ x ≤ 9}, or {x : 1 ≤ x ≤ 9}.The condition after the divider selects exactly the same real numbers as the interval.
Answer
Interval: [1, 9]
Set-builder: {x | 1 ≤ x ≤ 9}
Equivalent colon form: {x : 1 ≤ x ≤ 9}
Check x = 5 passes every form, x = 1 and x = 9 are included in every form, and x = 10 fails every form.
Counterexamples: what it is not, and the tempting wrong moves
✗ Not this: Read the lone divider in {x | x ≥ 3} as absolute value.
Absolute value uses two bars enclosing an expression. Here one divider separates the member name x from its condition.
✓ Instead: Read it 'all x such that x is three or more'.
Tips and tricks
Use a colon when an absolute value condition already contains a pair of bars.
Strategy: step by step
Get the bars alone, then check that B, the number on the other side, is positive. Zero or negative B needs the final lesson's cases.
Write the three part inequality with −B on the left and B on the right. Keep equality if the question allows it.
Undo each inside operation on all three parts. Dividing by a negative reverses both signs.
Read the finished bounds from smaller to larger and write the interval.
A strict sign excludes the endpoint: use a hollow dot and a parenthesis. An inclusive sign allows equality: use a filled dot and a bracket. Check the center and both endpoints in the original question.
Strategy
Solve a between inequality
1
Are the bars alone?
YesInspect B, the number on the other side.
NoUndo outside addition or subtraction, then outside multiplication or division. Negative multiplication or division flips the comparison.
↓
2
Is B positive?
YesUse the between chain.
NoUse the final lesson's zero or negative B cases before writing any chain.
↓
3
Does the original sign allow equality?
YesKeep equality in both chain signs, and include the final boundaries.
NoKeep both chain signs strict, and exclude the final boundaries.
↓
4
Are you dividing by a negative?
YesReverse both signs and rewrite the resulting bounds in increasing order.
NoKeep both signs.
Get the bars alone, then check that B, the number on the other side, is positive. Zero or negative B needs the final lesson's cases.
Write the three part inequality with −B on the left and B on the right. Keep equality if the question allows it.
Undo each inside operation on all three parts. Dividing by a negative reverses both signs.
Read the finished bounds from smaller to larger and write the interval.
A strict sign excludes the endpoint: use a hollow dot and a parenthesis. An inclusive sign allows equality: use a filled dot and a bracket. Check the center and both endpoints in the original question.
Worked exampleA distance interval with included endpoints
Solve |x − 5| ≤ 4.
Every input in [1, 9] makes the distance at most 4.
What it asks. Find every number that is four or fewer units from five.
Plan. Write the between chain, add five to all three parts, and show whether the two ends count.
The center is five. The distance limit B is four.
Less distance requires both a left limit and a right limit.
The equals bar means the endpoints count.
Write −4 ≤ x − 5 ≤ 4.A signed difference whose distance is at most 4 must be at least −4 and at most 4.
Add 5 to all three parts: −4 + 5 ≤ x − 5 + 5 ≤ 4 + 5.This undoes the subtraction from x while sliding both comparison limits by the same amount.
Simplify: 1 ≤ x ≤ 9.−4 + 5 = 1, −5 + 5 = 0, and 4 + 5 = 9. This finds every input that passes both limits.
Write [1, 9].At either end the distance equals 4, and the stated inequality allows equality.
Answer
1 ≤ x ≤ 9.
Interval: [1, 9].
Check The endpoint distances are |1 − 5| = 4 and |9 − 5| = 4. The center 5 has distance 0; the outside input 10 has distance 5 and fails.
Ladder: from easy to exam-hard. Press Try it first on any rung to hide its steps and use them as hints.
Rung 1Rung 1: strictly near zero
Solve |x| < 2.
The strict interval excludes −2 and 2.
What it asks. Find all positions fewer than two units from zero.
Plan. Use the strict between rule directly.
The inside expression is x itself.
Distance exactly two is excluded.
−2 < x < 2.These are all positions strictly less than two units from zero.
Write (−2, 2).Strict comparisons exclude both finite endpoints.
Answer
(−2, 2).
Check |0| = 0 < 2 passes. |−2| = |2| = 2 fails, so both ends are excluded. |3| = 3 fails outside the interval.
Rung 2Rung 2: include shifted endpoints
Solve |x + 3| ≤ 2.
The allowed interval is [−5, −1].
What it asks. Find all positions at most two units from −3.
Plan. Use an inclusive between chain and subtract three from every part.
x + 3 is x − (−3), so the center is −3.
The ends count.
−2 ≤ x + 3 ≤ 2.The limit is positive and equality is allowed, so the inside lies from −2 through 2.
Subtract 3 everywhere: −5 ≤ x ≤ −1.Subtracting 3 from all three parts isolates x, so these bounds describe the allowed input positions instead of the inside difference.
Answer
[−5, −1].
Check The center −3 gives distance 0; endpoints −5 and −1 give distance 2 and are included.
Rung 3Rung 3: a scaled inside expression
Solve |2x − 1| < 5.
Only the inputs strictly between −2 and 3 pass.
What it asks. Find every input making the distance of 2x − 1 from zero smaller than five.
Plan. Use between, then undo subtracting one and multiplying by two.
Treat 2x − 1 as one inside expression.
A positive divisor keeps both signs.
−5 < 2x − 1 < 5.Strictly less distance means a strict between inequality.
Add 1: −4 < 2x < 6.Undo the subtraction on all three parts.
Divide by 2: −2 < x < 3.Dividing every part by the positive 2 preserves order and isolates x, giving the allowed input interval.
Answer
(−2, 3).
Check x = 0 gives |−1| = 1 < 5; endpoints −2 and 3 each give 5 and fail.
Rung 4Rung 4: isolate, then reverse order
Solve 2|1 − 3x| + 1 ≤ 9.
The endpoints −1 and 53 are both included; the fraction is exact.
What it asks. Find inputs making the whole expression at most nine.
Plan. Undo the operations outside the bars, then solve the between chain. The negative inside multiplier flips both signs.
Subtract one first, then divide by two.
Divide every part by −3 and reverse both signs.
Write the final bounds from smaller to larger.
Subtract 1: 2|1 − 3x| ≤ 8.This removes the outside addition and keeps the comparison.
Divide by 2: |1 − 3x| ≤ 4.This leaves the whole absolute value alone, and the positive divisor preserves order.
Write −4 ≤ 1 − 3x ≤ 4.Distance at most 4 gives two simultaneous inside limits.
Subtract 1 everywhere: −5 ≤ −3x ≤ 3.This undoes the inside addition while changing every part equally.
Divide every part by −3: 53 ≥ x ≥ −1.The negative divisor reverses both signs. −5 ÷ (−3) = 53 and 3 ÷ (−3) = −1.
Read in increasing order: −1 ≤ x ≤ 53.This identifies every allowed input in the order used by interval notation.
Answer
[−1, 53 ].
Check At x = −1, 1 − 3(−1) = 4. At x = 53, 1 − 3 × 53 = 1 − 5 = −4. Both give original value 2 × 4 + 1 = 9 and pass. At x = 0 the original value is 3, also at most 9. At x = 2 it is 2|−5| + 1 = 11 and fails.
Counterexamples: what it is not, and the tempting wrong moves
✗ Not this: |x − 5| ≤ 4 means only x − 5 ≤ 4.
That gives x ≤ 9 and accepts x = −100, whose distance from 5 is |−100 − 5| = 105, far larger than 4.
✓ Instead: Keep both limits: −4 ≤ x − 5 ≤ 4, so 1 ≤ x ≤ 9.
✗ Not this: Adding 5 to −4 ≤ x − 5 ≤ 4 gives −4 ≤ x ≤ 9.
Changing only two of the three amounts changes which numbers pass. The left bound needs the same addition.
✓ Instead: Add 5 everywhere: −4 + 5 ≤ x ≤ 4 + 5, so 1 ≤ x ≤ 9.
✗ Not this: |x − 5| < 4 includes x = 1.
|1 − 5| = 4 equals the limit, while the strict sign requires a smaller distance.
✓ Instead: Use (1, 9) for the strict sign and [1, 9] when equality is included.
Tips and tricks
Memory device: less thAND. A smaller distance stays between two limits, and both limits must hold.
Write the same arithmetic underneath all three parts. This prevents leaving one endpoint unmoved.
Check an endpoint in the original inequality before choosing its bracket or parenthesis.
Trap. Changing only the middle of a three-part inequality, or treating < and ≤ alike. Every algebra move acts on all three parts, and the original sign decides whether the boundary belongs.
