Picture a safety zone around a construction site on a straight road. If you must stay more than seven units from its center, you may go far left or far right. You cannot stand between the two boundary positions. A greater than absolute value inequality makes this outside choice. Its answer has two separate stretches when the inside is a line with x in it. The word 'or' says either stretch is acceptable. You do not have to stand in both at once. In interval notation, the union symbol ∪ collects the two permitted stretches into one solution set. If the sign includes an equals bar, standing on the safety boundary is allowed too. Each outside stretch is a ray: it continues forever in one direction.
The V is above height 7 to the left of −4 and to the right of 3.
Reminder
Letters naming whole expressions. For |2x + 1| > 7, A means 2x + 1 and B means 7.
Union and endpoint notation. (−∞, −4) ∪ (3, ∞) accepts either outside ray. Parentheses leave −4 and 3 out.
Negative multiplication. −4x ≤ −7 becomes x ≥ 74 after dividing by −4.
Why it works. If B > 0 and A is farther than B from zero, A is either to the left of −B or to the right of B. That is A < −B or A > B. A value between those boundaries has distance at most B and cannot satisfy the strict outside condition. The two outside conditions cannot both hold for the same A because −B < B. Asking for 'and' would reject every input. The union collects all inputs accepted by either branch, which is why it matches 'or'.
RuleFor B > 0: |A| > B means A < −B or A > B; |A| ≥ B means A ≤ −B or A ≥ B. For an inside line with a nonzero number multiplying x, solve each alternative and join its two outside intervals with ∪.
The same idea, five ways
Say it
two x plus one is more than seven units from zero
Write it
x is less than negative four or greater than three.
In math
|2x + 1| > 7
2x + 1 < −7 or 2x + 1 > 7
x < −4 or x > 3
(−∞, −4) ∪ (3, ∞)
{x | x < −4 or x > 3}
{x : x < −4 or x > 3}
Like
You can clear a safety zone by moving beyond its left edge or beyond its right edge.
See itTwo rays show both permitted alternatives, with the boundary numbers excluded.
The same idea, other ways
As a safety zone
Beyond the left edge or beyond the right edge is safe. The middle is forbidden. These are alternatives, so use 'or'.
Either outside ray qualifies for |x| > 3.
As a V above a line
For |2x + 1| > 7, compare the V's height with the flat line at 7. Only the distant arms stand above that line.
The V is above height 7 to the left of −4 and to the right of 3.
With small numbers
For |x| > 3, x = 4 and x = −4 work because each has distance 4. x = 3 fails because its distance equals 3. x = 0 fails because its distance is 0.
Form (B > 0)
Means
Write
Picture
|A| > B
Farther than B: outside
A < −B or A > B
Two pieces, joined by ∪
|A| ≥ B
At least B away: outside
A ≤ −B or A ≥ B
Two pieces, boundaries included
.1Strict outside
Farther than the limit permits either distant side but rejects the boundary itself.
For B > 0, |A| > B gives A < −B or A > B.
Strict finite boundaries, actual end numbers, use hollow dots and parentheses.
Both outside rays belong, and hollow dots exclude −3 and 3.
Reminder
Strict signs. 4 > 3 passes, while 3 > 3 fails.
Worked exampleBeyond three units
Solve |x| > 3.
Both outside rays belong, and hollow dots exclude −3 and 3.
What it asks. Find all numbers farther than three units from zero.
Plan. Select the left and right rays and exclude the distance three endpoints.
The word 'farther' requires a greater distance.
Only one of the outside alternatives has to hold.
x < −3 or x > 3.The distance must exceed three on one side of zero.
Write (−∞, −3) ∪ (3, ∞).Either outside interval is allowed, so join them with ∪. The boundary distance equals 3 and fails the strict greater than comparison.
Counterexamples: what it is not, and the tempting wrong moves
✗ Not this: Write (−3, 3) for |x| > 3.
That interval contains 0, whose distance is smaller than 3.
✓ Instead: Select (−∞, −3) ∪ (3, ∞).
Tips and tricks
A strict outside answer uses two hollow boundary dots.
.2Inclusive outside
At least the limit allows the boundary as well as the more distant positions.
|A| ≥ B gives A ≤ −B or A ≥ B when B > 0.
Infinity is a direction without an endpoint, so it always uses a parenthesis.
Both outside rays belong, and filled dots include −5 and 5.
Reminder
Infinity. ∞ means continuing without an end number. It always takes a parenthesis, even when the finite end is included.
Worked exampleInclude both outside boundaries
Solve |x| ≥ 5.
Both outside rays belong, and filled dots include −5 and 5.
What it asks. Find every number five or more units from zero.
Plan. Select both outside rays and include the two distance five boundaries.
At least includes equality.
Infinity never receives a bracket.
x ≤ −5 or x ≥ 5.Both distances equal to five and larger than five are allowed.
Write (−∞, −5] ∪ [5, ∞).Finite endpoints are included; infinity cannot be included as a number.
Answer
(−∞, −5] ∪ [5, ∞).
Check |−5| = |5| = 5 passes because equality is allowed. |−6| = |6| = 6 also passes. |0| = 0 is smaller than 5 and fails.
Counterexamples: what it is not, and the tempting wrong moves
✗ Not this: Use (−∞, −5) ∪ (5, ∞) for |x| ≥ 5.
That drops the two endpoint numbers even though distance exactly 5 is allowed.
✓ Instead: Use (−∞, −5] ∪ [5, ∞), with brackets at −5 and 5.
Tips and tricks
At least means a filled dot at each outside boundary.
Strategy: step by step
Get the bars alone and inspect B, the number on the other side. Use this method when B is positive.
Write the left outside inequality and the right outside inequality, connected by 'or'. Each alternative is a branch.
Solve each branch separately. Multiplying or dividing by a negative flips that branch's comparison sign.
Put the resulting intervals in order from left to right and join them with ∪.
Infinity is a direction, not an actual endpoint, so it always uses a parenthesis. At each finite boundary, an actual number, use a bracket only when equality passes the original comparison.
Strategy
Solve an outside inequality
1
Are the bars alone and is B positive?
YesWrite two outside alternatives joined by 'or'.
NoIsolate first and inspect B. If B is zero or negative, use the final lesson's cases.
↓
2
Does the sign include equality?
YesUse inclusive signs on both branches and test the finite boundaries.
NoUse strict signs on both branches and exclude the finite boundaries.
↓
3
Does solving a branch require negative multiplication or division?
YesFlip that branch's sign immediately.
NoKeep its sign.
Get the bars alone and inspect B, the number on the other side. Use this method when B is positive.
Write the left outside inequality and the right outside inequality, connected by 'or'. Each alternative is a branch.
Solve each branch separately. Multiplying or dividing by a negative flips that branch's comparison sign.
Put the resulting intervals in order from left to right and join them with ∪.
Infinity is a direction, not an actual endpoint, so it always uses a parenthesis. At each finite boundary, an actual number, use a bracket only when equality passes the original comparison.
Worked exampleTwo outside solution intervals
Solve |2x + 1| > 7.
Both outside rays belong, while −4 and 3 are excluded.
What it asks. Find every input for which 2x plus one is more than seven units from zero.
Plan. Write two outside alternatives, solve each, and join the two allowed intervals.
The right side B is positive seven.
Use 'or', because either outside direction works.
The divisor two is positive, so each comparison sign stays.
Write 2x + 1 < −7 or 2x + 1 > 7.An inside value more than seven units from zero must lie beyond one of the two signed boundaries.
Left branch: subtract 1 to get 2x < −8.This removes the inside addition without changing order.
Divide by 2: x < −4.The positive divisor leaves the sign unchanged and isolates every input passing the left alternative.
Right branch: subtract 1 to get 2x > 6.The same subtraction removes the inside addition in the other alternative.
Divide by 2: x > 3.This finds every input passing the right alternative.
Write (−∞, −4) ∪ (3, ∞).Either alternative qualifies, and both finite boundaries fail the strict comparison.
Answer
x < −4 or x > 3
Interval: (−∞, −4) ∪ (3, ∞)
Set-builder: {x : x < −4 or x > 3}
Check x = −5 gives |−9| = 9 > 7; x = 4 gives |9| = 9 > 7. The middle input 0 gives 1 and fails. Boundary inputs −4 and 3 give exactly 7 and fail.
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: outside around zero
Solve |x| > 2.
The strict outside condition allows both rays past −2 and 2.
What it asks. Find every position farther than two units from zero.
Plan. Use the strict outside rule directly.
Keep both directions.
The two distance two boundaries do not count.
x < −2 or x > 2.Distances greater than two lie on the two outside rays.
What it asks. Find all numbers at least three units from −2.
Plan. Use inclusive outside alternatives and subtract two in each.
x + 2 measures from −2.
Equality includes both boundaries.
x + 2 ≤ −3 or x + 2 ≥ 3.The distance includes the boundary and both farther sides.
Subtract 2: x ≤ −5 or x ≥ 1.Subtracting 2 from both sides of each branch undoes the + 2 inside the bars.
Answer
(−∞, −5] ∪ [1, ∞).
Check At −5 and 1 the distance from −2 is 3; at the center −2 it is 0 and fails.
Rung 3Rung 3: isolate an outside scale
Solve 2|3x − 2| − 1 > 7.
The two allowed rays exclude the exact boundaries −23 and 2.
What it asks. Find inputs for which the whole expression is greater than seven.
Plan. Undo outside operations, use two outside alternatives, and solve each.
Undo minus one before times two.
All divisors in this problem are positive.
Add 1: 2|3x − 2| > 8.This removes the outside subtraction and reveals the scaled distance.
Divide by 2: |3x − 2| > 4.The bars now stand alone; positive division preserves the comparison.
Write 3x − 2 < −4 or 3x − 2 > 4.A distance greater than 4 requires either outside alternative.
Add 2 in both alternatives: 3x < −2 or 3x > 6.This removes the inside subtraction.
Divide each by 3: x < −23 or x > 2.Positive division isolates the allowed inputs in both branches without flipping either sign.
Answer
(−∞, −23) ∪ (2, ∞).
Check Both finite boundaries give original output 7 and fail. At x = −1 the output is 9, and at x = 3 it is 13, so both outside pieces work.
Rung 4Rung 4: an inside negative reverses each branch
Solve |2 − 4x| ≥ 5.
Both rays include the exact boundary numbers −34 and 74.
What it asks. Find inputs making 2 minus four x at least five units from zero.
Plan. Write inclusive outside alternatives, then undo two and divide by negative four.
Dividing by −4 flips each branch's comparison.
Put the left ray first in the final union.
2 − 4x ≤ −5 or 2 − 4x ≥ 5.The inclusive outside condition uses two alternatives.
Subtract 2: −4x ≤ −7 or −4x ≥ 3.Both branches need the same addition undone.
Divide by −4: x ≥ 74 or x ≤ −34.Negative division reverses each branch's order.
Write the left piece first.The smaller boundary belongs at the left in interval notation.
Answer
(−∞, −34 ] ∪ [ 74, ∞).
Check At x = −34, 2 − 4x = 2 + 3 = 5. At x = 74, 2 − 4x = 2 − 7 = −5. Both give distance 5 and pass. At x = 0 the distance is 2 and fails. At x = −1 and x = 2, the distances are each 6 and pass.
Counterexamples: what it is not, and the tempting wrong moves
✗ Not this: The answer to |2x + 1| > 7 is x < −4 and x > 3.
No number is left of −4 and right of 3 at the same time. 'And' would reject both permitted rays.
✓ Instead: Write x < −4 or x > 3, then (−∞, −4) ∪ (3, ∞).
✗ Not this: Write a single chain −4 > x > 3.
That chain requires the same impossible two simultaneous comparisons.
✓ Instead: Keep two separate alternatives joined by 'or'.
✗ Not this: Include the boundary x = 3 in the strict answer.
At 3, |2 × 3 + 1| = 7. Equality fails a greater than condition.
✓ Instead: Use the open endpoint (3, ∞).
Tips and tricks
Memory device: greatOR. Greater distance accepts a far left position or a far right position.
Draw both allowed rays before writing the union. This catches a missing branch.
Test one input on each ray and one in the rejected middle.
Trap. Using 'and' for an outside condition. No x can be both less than −4 and greater than 3. Use 'or', and use ∪ to keep both permitted pieces.
