Compound Inequalities Through the Lens of Distance
An absolute value inequality converts into a compound inequality through one of two standard translations — one for "less than," one for "greater than." The less-than form produces a bounded interval, the greater-than form produces two unbounded rays, and the geometric interpretation as distance on the number line makes both conversions intuitive. Once the absolute value is removed, the resulting compound inequality is solved by whatever method its internal structure requires: linear techniques for linear expressions, sign charts for quadratic or polynomial expressions, and so on. The absolute value is the outer shell; the content inside determines the difficulty.
The number ∣x∣ is how far x is from 0 on the number line, whichever side it is on: ∣2∣=2 and ∣−2∣=2. Read that way, ∣x∣<3 asks for every point closer than 3 to zero. There are such points on both sides, and they form one unbroken stretch.
The bar is every point whose distance from 0 is under 3. The open dots at −3 and 3 are exactly 3 away, so they are left out.
The same condition can also be written as two ordinary inequalities.
Drop the picture and use the rule instead: ∣x∣<3 is the same as −3<x<3 — the number is above −3 and, at the same time, below 3. In general, with any expression f(x) inside the bars:
Two Fundamental Forms
−k<f(x)<k
Both conditions must hold simultaneously. No distances are involved, only two comparisons joined by AND.
Both descriptions give the same answer on every value.
Test the same values against each one: the distance reading in one column, the double inequality in the next.
x
distance |x|
closer than 3?
between −3 and 3?
−4
4
no
no
−3
3
no
no
−1
1
yes
yes
0
0
yes
yes
2
2
yes
yes
3
3
no
no
4
4
no
no
The two columns never disagree. Both accept −1, 0 and 2, both reject ±4, and both reject the borders ±3 for the same reason: a distance of exactly 3 is not less than 3.
The graph of ∣x∣−3 is below the axis exactly where ∣x∣<3: on (−3,3), the "yes" rows of the table, with the open circles at ±3. The same state is on the inequality visual explorer.
The two descriptions were never separate, because each side of zero gives one half of the chain.
On the right of zero the distance is x itself; on the left it is −x. So "closer than 3" turns into one ordinary inequality on each side.
side of 0
|x| is
|x| < 3 says
which means
x ≥ 0
x
x < 3
right part stops before 3
x < 0
−x
−x < 3
x > −3: left part stops after −3
both sides
−3 < x < 3
The right side contributes x<3, the left side contributes x>−3, and a point belongs if it satisfies its own side's condition. Put together that is −3<x<3: the double inequality is the distance picture, written one side at a time.
The definition
An absolute value inequality is an inequality in which the unknown appears inside absolute value bars, such as ∣x∣<3 or ∣2x−3∣≥5. With one absolute value and a constant it takes one of two forms, each with its own conversion:
An absolute value inequality is a statement about distance, and its answer is a set.
It is not a number and not a function. It is true for some values of x and false for others, and its answer collects the values that make it true. Less-than keeps the points near the centre, so its answer is one interval. Greater-than keeps the points far away, so its answer is two rays pointing apart.
One interval or two rays
With k = 3: |x| < 3 asks for points within 3 of zero, a single bounded interval from −3 to 3, written as the conjunction −3 < x < 3. |x| > 3 asks for points more than 3 from zero, two rays heading away from each other, written as the disjunction x < −3 or x > 3. Both boundaries are open because both inequalities are strict.
Less-than becomes AND, greater-than becomes OR: the shape of the answer follows.
Read the distance when the inside is simple; convert when it is not.
For ∣x∣<3 or ∣x−5∣<2 the distance picture gives the answer at a glance. For ∣2x−3∣<5 or ∣x2−4∣>1 there is no single centre to read from, and the conversion does the work. The less-than form becomes the chain above. The greater-than form becomes a disjunction — two separate inequalities:
Two Fundamental Forms
f(x)<−korf(x)>k
At least one condition must hold. The solution is the set of x values for which f(x) falls outside the interval [−k,k].
COROLLARY
The non-strict versions ≤ and ≥ work identically, with the boundary values included rather than excluded.
The conversions are the same; only the inequality symbols change from strict to non-strict throughout. One edge the conversion hides: a distance is never negative, so when k is negative ∣f(x)∣<k has no solutions at all and ∣f(x)∣>k holds for every x.
The rest of the page applies these two conversions: the less-than and greater-than forms in detail, worked linear examples, expressions with squares inside the bars, absolute value on both sides, and the distance reading again as a geometric interpretation.
Solving the Less-Than Form
The inequality ∣f(x)∣<k converts to −k<f(x)<k. The value of k determines whether a solution exists before any algebra begins.
When k>0, the double inequality defines a bounded region. The two inequalities f(x)>−k and f(x)<k are solved simultaneously, and the solution is their intersection — the set of x values satisfying both.
When k=0, the inequality ∣f(x)∣<0 asks for the absolute value to be strictly negative, which is impossible. The solution set is empty. The non-strict version ∣f(x)∣≤0 has a solution only at the roots of f(x)=0, because zero is the only non-negative value that satisfies ≤0.
COROLLARY
When k<0, both ∣f(x)∣<k and ∣f(x)∣≤k have empty solution sets. Absolute value is never negative, so it can never be less than a negative number.
For linear f(x), the double inequality is a three-part chain solved by isolating x in the middle.
Example
The inequality ∣2x−3∣<5 becomes −5<2x−3<5. Adding 3 throughout: −2<2x<8. Dividing by 2: −1<x<4. The solution is the open interval (−1,4).
|x − 2| − 4 < 0: the interval (−2, 6)
The V has its corner at x = 2 and dips below the axis between its zeros at −2 and 6, so the less-than inequality holds on that bounded interval: the compound inequality −4 < x − 2 < 4 drawn out. Shifting the corner shifts the interval without changing its width. Move the corner and the level on the inequality visual explorer.
The centre of the interval is wherever the expression inside the bars vanishes; its width is 2k divided by the size of the coefficient of x.
Solving the Greater-Than Form
The inequality ∣f(x)∣>k converts to f(x)<−k or f(x)>k. Each part is solved independently, and the solution is the union of the two individual solution sets.
When k>0, each part contributes a separate region, and the union typically produces two rays — one extending to the left and one to the right.
When k=0, the inequality ∣f(x)∣>0 asks where f(x) is nonzero. The solution is all real numbers except the roots of f(x)=0. The non-strict version ∣f(x)∣≥0 is satisfied by every real number, since absolute value is always non-negative.
COROLLARY
When k<0, the inequality ∣f(x)∣>k is satisfied by every real number — absolute value is always at least zero, which exceeds any negative number.
The solution is (−∞,∞).
Example
For a linear example: ∣4x+1∣≥7 becomes 4x+1≤−7 or 4x+1≥7. From the first: 4x≤−8, so x≤−2. From the second: 4x≥6, so x≥23. The solution is (−∞,−2]∪[23,∞).
Pitfall: Writing the greater-than form as a chain
Writing ∣4x+1∣≥7 as −7≤4x+1≤7 gives −2≤x≤23, the points where ∣4x+1∣≤7. The greater-than form splits instead: x≤−2 or x≥23.
The two parts of the disjunction are solved one at a time, and each answer is a ray. Here both rays start 47 away from x=−41, the point where 4x+1=0, and run outward in opposite directions.
k value
|f(x)| < k
|f(x)| ≤ k
|f(x)| > k
|f(x)| ≥ k
k > 0
bounded open interval (−k < f < k)
bounded closed interval (−k ≤ f ≤ k)
two open rays (f < −k or f > k)
two closed rays (f ≤ −k or f ≥ k)
k = 0
∅ (empty)
roots of f(x) = 0 only
all reals except roots of f(x) = 0
(−∞, ∞) all reals
k < 0
∅
∅
(−∞, ∞)
(−∞, ∞)
Greater-than splits into two separate conditions joined by OR, and the answer is the union of what each one allows.
Worked Examples with Linear Expressions
When the expression inside the absolute value is linear, the compound inequality after conversion is also linear, and the solution follows from the methods on the linear inequalities page.
Solve ∣3x−7∣<2. Convert: −2<3x−7<2. Add 7: 5<3x<9. Divide by 3: 35<x<3. The solution is (35,3).
Solve ∣1−2x∣≥5. Convert: 1−2x≤−5 or 1−2x≥5. From the first: −2x≤−6, so x≥3 (direction flips when dividing by −2). From the second: −2x≥4, so x≤−2. The solution is (−∞,−2]∪[3,∞).
Solve ∣5x+3∣<−1. No conversion needed — absolute value is never negative, so it is certainly never less than −1. The solution set is empty.
Solve ∣x−4∣>0. Convert: x−4<0 or x−4>0, meaning x<4 or x>4. This is every real number except x=4. The solution is (−∞,4)∪(4,∞), or equivalently R∖{4}.
Pitfall: Forgetting to flip when dividing by a negative
Dividing −2x≤−6 and −2x≥4 by −2 without flipping gives x≤3 or x≥−2, which is every real number. Flipping both signs gives (−∞,−2]∪[3,∞).
In every linear case with k>0, the less-than form produces a single bounded interval, and the greater-than form produces two rays. The center of symmetry is the root of f(x)=0, and for f(x)=ax+b the radius is ∣a∣k.
Absolute Value Inequalities with Quadratic or Other Expressions
When the expression inside the absolute value is not linear, the compound inequality after conversion requires more powerful methods — typically the sign chart.
Example
Solve ∣x2−4∣<5. Convert to −5<x2−4<5. This is two simultaneous inequalities: x2−4>−5 and x2−4<5.
From x2−4>−5: x2>−1. Since x2≥0 for all real x, this holds everywhere. It imposes no restriction.
From x2−4<5: x2<9, which gives −3<x<3.
The intersection of "all real numbers" and "−3<x<3" is simply (−3,3).
Solve ∣x2−7x+10∣≥2. Convert to x2−7x+10≤−2 or x2−7x+10≥2.
From the first: x2−7x+12≤0. Factor: (x−3)(x−4)≤0. This is a quadratic inequality with roots 3 and 4, positive leading coefficient. The expression is non-positive between the roots: [3,4].
From the second: x2−7x+8≥0. The discriminant is 49−32=17>0. The roots are 27±17, approximately 1.44 and 5.56. With positive leading coefficient, the expression is non-negative outside the roots: (−∞,27−17]∪[27+17,∞).
Pitfall: Solving only the positive part
Solving only x2−7x+10≥2 gives the two outer rays and loses [3,4], where the inside is −2 or less: at x=3.5, ∣−2.25∣=2.25≥2. The part x2−7x+10≤−2 supplies that piece.
The solution is the union of both parts: (−∞,27−17]∪[3,4]∪[27+17,∞).
The absolute value conversion reduces the problem to standard inequality types. The difficulty is determined by whatever sits inside the bars, not by the absolute value itself.
Three pieces from one inequality
The section's |x² − 7x + 10| ≥ 2. The bars fold the dip of the parabola between 2 and 5 upward, and the folded hump just reaches above 2 between x = 3 and x = 4. Together with the two outer arms, the graph is on or above the level on three pieces: x ≤ (7 − √17)/2 ≈ 1.44, 3 ≤ x ≤ 4, and x ≥ (7 + √17)/2 ≈ 5.56.
With a non-linear inside, each half of the conversion can contribute its own pieces.
Inequalities with Absolute Value on Both Sides
When absolute value appears on both sides — ∣f(x)∣<∣g(x)∣ or ∣f(x)∣>∣g(x)∣ — the standard less-than/greater-than conversion does not apply directly because the right-hand side is not a constant. Two alternative approaches handle these cases.
RULE
The squaring method exploits the fact that ∣A∣<∣B∣ if and only if A2<B2 (since both sides are non-negative).
The inequality ∣f(x)∣<∣g(x)∣ becomes [f(x)]2<[g(x)]2, which rearranges to [g(x)]2−[f(x)]2>0. Factoring the difference of squares:
Inequalities with Absolute Value on Both Sides
(g(x)−f(x))(g(x)+f(x))>0
This is a polynomial inequality solvable by sign chart. The critical points are the roots of g(x)−f(x)=0 and g(x)+f(x)=0.
Example
For ∣2x−1∣<∣x+3∣, square both sides: (2x−1)2<(x+3)2. Rearrange: (x+3)2−(2x−1)2>0. Factor: ((x+3)−(2x−1))((x+3)+(2x−1))>0, which gives (4−x)(3x+2)>0. The roots are x=4 and x=−32. The sign chart yields the solution (−32,4).
Pitfall: Dropping the bars on both sides
Dropping the bars in ∣2x−1∣<∣x+3∣ gives 2x−1<x+3, so x<4. That keeps x=−1, where ∣−3∣=3 is not less than ∣2∣=2. Squaring gives (−32,4).
The case-splitting method is the alternative: consider the signs of f(x) and g(x) on each interval defined by their roots, remove the absolute value bars using the piecewise definition, and solve the resulting inequality on each interval. This approach is more laborious but avoids squaring and works for non-polynomial expressions as well.
Method
Procedure
Resulting problem
Note
Squaring
replace |A| compared to |B| with A² compared to B²; rearrange and factor as a difference of squares
polynomial inequality (g − f)(g + f) compared to 0; solve by sign chart
cleanest when f and g are polynomial; less natural for transcendental expressions
Case-splitting
find intervals where f(x) and g(x) have constant signs; remove |·| using the piecewise definition on each interval; solve on each
several inequalities, one per sign-region; combine the partial solutions
more laborious but works for non-polynomial expressions
Bars on both sides can be removed by squaring, which is safe because both sides are sizes and never negative.
Geometric Interpretation
Absolute value measures distance on the number line, and this interpretation turns every absolute value inequality into a geometric statement about proximity or separation. The reading of ∣a−b∣ as the distance between a and b is set out on the absolute value page.
Fact: Calculus is written in this language
A function f is continuous at a when every ε>0 has a δ>0 such that ∣x−a∣<δ forces ∣f(x)−f(a)∣<ε. Both conditions are distance inequalities of exactly this kind: an interval around a centre.
The inequality ∣x−a∣<d asks: which points lie within distance d of a? The answer is the open interval (a−d,a+d), centered at a with radius d. The inequality ∣x−5∣<3 describes all points within 3 units of 5: the interval (2,8).
The inequality ∣x−a∣>d asks: which points lie farther than distance d from a? The answer is two rays: (−∞,a−d)∪(a+d,∞). The inequality ∣x−5∣>3 describes all points more than 3 units from 5: (−∞,2)∪(8,∞).
The inequality ∣x−a∣<∣x−b∣ asks: which points are closer to a than to b? On the number line, the set of points closer to a is everything on a's side of the midpoint m=2a+b. If a<b, the solution is (−∞,m). This follows from the squaring method: ∣x−a∣<∣x−b∣ becomes (x−a)2<(x−b)2, which simplifies to a linear inequality whose solution is the ray ending at the midpoint.
Pitfall: Reading the centre with the wrong sign
Reading ∣x+2∣<3 as distance from 2 gives (−1,5). Since ∣x+2∣=∣x−(−2)∣, the centre is −2 and the answer is (−5,1).
These geometric readings provide instant answers for simple absolute value inequalities and serve as sanity checks for more complex ones. If the algebraic solution to ∣x−5∣<3 produces something other than a symmetric interval centered at 5 with radius 3, the algebra contains an error.
Absolute value as distance on the line
|x − 5| < 3 collects the points less than 3 away from 5: the interval (2, 8) centred at 5. |x − 1| < |x − 7| compares two distances: the points nearer to 1 than to 7 are everything left of their midpoint 4, the open ray x < 4. No algebra is needed once each side is read as a distance.
Reading |x − a| as distance from a turns many of these inequalities into a picture on the number line.
Summary of Absolute Value Inequality Forms
Every absolute value inequality is solved in two stages: first remove the absolute value bars by applying the appropriate conversion, then solve the resulting compound inequality using whatever method its internal structure demands. The table below collects the cases covered above, the conversion that removes the absolute value, and the kind of inequality that remains.
Five mistakes account for most wrong answers to absolute value inequalities. Each one is marked where it happens in the sections above; here each is worked through in full, with the wrong and the right working side by side.
The greater-than form is converted with the same double inequality as the less-than form, so the answer is the interval between the boundaries instead of the two rays outside them.
Start
∣4x+1∣≥7
Wrong turn
Error here
⟹−7≤4x+1≤7
⟹−2≤x≤23
Right path
Greater-than becomes a disjunction.
4x+1≤−7 or 4x+1≥7
⟹x≤−2 or x≥23
The solution is (−∞,−2]∪[23,∞).
Why it happensThe chain −k<f(x)<k is usually learned first and is the more familiar shape, so it gets used for both forms.
How to checkTest a point from the wrong answer. x=0 gives ∣1∣=1, which is not ≥7. A point from the right answer, x=2, gives ∣9∣=9≥7.
2Forgetting to flip when dividing by a negative
In ∣1−2x∣≥5, both parts end with a division by −2, and the inequality signs are kept as they were.
Start
1−2x≤−5
Wrong turn
Error here
⟹−2x≤−6
⟹x≤3
1−2x≥5
⟹−2x≥4
⟹x≥−2
Together, x≤3 or x≥−2 covers every real number.
Right path
Dividing by −2 reverses each sign.
−2x≤−6⟹x≥3
−2x≥4⟹x≤−2
The solution is (−∞,−2]∪[3,∞).
Why it happensThe sign rule belongs to linear inequalities, and it slips once attention is on the absolute value conversion.
How to checkA greater-than inequality with k>0 never holds for every x. Here x=0 gives ∣1∣=1, which is not ≥5, while x=3 gives ∣−5∣=5≥5.
3Solving only the positive part
For ∣x2−7x+10∣≥2, only x2−7x+10≥2 is solved, and the values where the inside is −2 or less are lost.
Start
x2−7x+8≥0
Wrong turn
Error here
⟹x≤27−17 or x≥27+17
Right path
The second part x2−7x+10≤−2 gives (x−3)(x−4)≤0, so 3≤x≤4. The solution is (−∞,27−17]∪[3,4]∪[27+17,∞).
Why it happensWith a linear inside, the negative part always gives a ray and is hard to overlook. With a quadratic inside, it can be a short bounded piece in the middle that looks as if it adds nothing.
How to checkTest a point between the two outer rays. x=3.5 gives 12.25−24.5+10=−2.25, and ∣−2.25∣=2.25≥2, so it belongs to the answer.
4Dropping the bars on both sides
With absolute value on both sides, the bars are removed and the inner expressions are compared directly.
Start
∣2x−1∣<∣x+3∣
Wrong turn
Error here
⟹2x−1<x+3
⟹x<4
Right path
Squaring keeps the comparison valid, because both sides are non-negative.
(x+3)2−(2x−1)2>0
⟹(4−x)(3x+2)>0
⟹−32<x<4
Why it happensRemoving matching bars from both sides looks like cancelling the same operation, but ∣A∣<∣B∣ compares sizes and says nothing about the signs of A and B.
How to checkTest a point from the wrong answer. x=−1 gives ∣−3∣=3 on the left and ∣2∣=2 on the right, and 3<2 is false.
5Reading the centre with the wrong sign
The inequality ∣x+2∣<3 is read as "within 3 of 2", taking the number inside the bars as the centre.
Wrong turn
Error here
Centre 2, radius 3, so the answer is (−1,5).
Right path
∣x+2∣=∣x−(−2)∣ is the distance from x to −2. The answer is (−2−3,−2+3)=(−5,1).
Why it happensThe pattern ∣x−a∣ has a minus sign, and when the bars hold a plus sign the number is copied as it appears.
How to checkConvert algebraically. −3<x+2<3 gives −5<x<1. Or test x=4 from the wrong answer: ∣6∣=6, which is not less than 3.
Four of the five slips happen as the bars come off. A greater-than form splits into two separate rays, the part u≤−k counts as much as u≥k, dividing by a negative reverses the sign, and bars on both sides come off only by squaring. The fifth reads ∣x+2∣ as a distance from 2 instead of from −2. Testing one point from the wrong answer, or one point it leaves out, against the original inequality catches each of them.
Absolute Value Inequalities FAQ
What happens to ∣f(x)∣<k and ∣f(x)∣>k when k is negative?
+
No algebra is needed, because an absolute value is never negative. Then ∣f(x)∣<k and ∣f(x)∣≤k have no solution at all: ∣5x+3∣<−1 is empty. And ∣f(x)∣>k and ∣f(x)∣≥k hold for every real x, since a value that is at least 0 always exceeds a negative number. Converting anyway gives nonsense: ∣5x+3∣<−1 would become 1<5x+3<−1, a chain no number satisfies.Read more →
Why does ∣x−4∣>0 leave out only x=4?
+
Because an absolute value is positive everywhere except where the inside is zero. So ∣f(x)∣>0 simply asks where f(x)=0: here x<4 or x>4, which is (−∞,4)∪(4,∞). The non-strict versions go to the extremes: ∣f(x)∣≥0 holds for every x, and ∣f(x)∣≤0 holds only at the roots of f(x)=0, since 0 is the only absolute value that is not positive.Read more →
How do you find the centre and width of the solution to ∣ax+b∣<k?
+
For k>0, the centre is where the inside is zero, x=−ab, and the interval reaches ∣a∣k to each side, so its width is ∣a∣2k. For ∣2x−3∣<5 the centre is 23 and the half-width 25, which gives (−1,4), the same as solving −5<2x−3<5. The greater-than form uses the same two numbers: for ∣4x+1∣≥7 the rays start 47 either side of −41, at −2 and 23.Read more →
How can an absolute value inequality have a solution in three pieces?
+
When the inside is not linear, each half of the conversion can give its own pieces. ∣x2−7x+10∣≥2 splits into x2−7x+10≥2, which gives two outer rays ending near 1.44 and 5.56, and x2−7x+10≤−2, which is (x−3)(x−4)≤0 and gives [3,4]. The bars fold the dip of the parabola upward, and the folded part rises above 2 between 3 and 4, so the answer has three separate pieces.Read more →
Why can you square both sides of ∣2x−1∣<∣x+3∣ when squaring usually breaks an inequality?
+
Because neither side is ever negative. Squaring keeps the order of non-negative numbers: if 0≤A<B, then A2<B2. It fails only when signs are mixed, as in −5<2 but 25>4. So ∣2x−1∣<∣x+3∣ becomes (x+3)2−(2x−1)2>0, which factors as (4−x)(3x+2)>0 and gives (−32,4). Dropping the bars instead gives x<4, which wrongly keeps x=−1.Read more →
What does ∣x−a∣<δ mean in calculus?
+
That x is less than δ away from a, so it lies in the open interval (a−δ,a+δ) centred at a, just as ∣x−5∣<3 is (2,8). The definition of continuity is written this way: f is continuous at a when every ε>0 has a δ>0 such that ∣x−a∣<δ forces ∣f(x)−f(a)∣<ε. Both conditions are less-than absolute value inequalities, each an interval around a centre.Read more →
