Inequalities: Solving, Graphing & Interval Notation

What an Inequality Is

You can already compare numbers and solve equations.

Comparing two known numbers gives a plain yes or no:

3 < 7

is true,

5 > 9

is false. Solving an equation gives a short list of numbers:

(x + 2)(x - 3) = 0

holds when one of the factors is zero, so

x = -2

or

x = 3

. On the graph those are the two places where the curve meets the axis.

The two dots are the two answers of

(x + 2)(x - 3) = 0

: the curve is at height

0

there and nowhere else.

An equation cannot say where an expression is negative.

Ask a different question about the same expression: for which

x

is

(x + 2)(x - 3)

less than

0

? The equation's two answers do not answer it. Trying values shows what they really are.

x	(x + 2)(x − 3)	negative?
−3	6	no
−2	0	no — zero: a border
−1	−4	yes
0	−6	yes
1	−6	yes
2	−4	yes
3	0	no — zero: a border
4	6	no

Every value between

-2

and

3

works and every value outside fails. The equation's answers are not the answer here; they are the borders of it. The answer is a whole stretch of the line, and an equation has no way to write a stretch down.

The curve dips below the axis exactly between

-2

and

3

— the "yes" rows of the table — and the blue bar on the axis marks that stretch. The open circles are the two "zero" rows. The same state is on the inequality visual explorer.

The definition

An inequality is a statement that compares two expressions using one of the symbols

<

,

>

,

\leq

,

\geq

.

Its counterpart with an equals sign: equations.

An inequality is a statement, and its answer is a set.

It is not a number and not a function. With known numbers it is simply true or false. With an unknown it is true for some values and false for others, and solving it means collecting every value that makes it true — its solution set. For

(x + 2)(x - 3) < 0

that set is the interval

(-2, 3)

: a stretch of the line, not a list of points.

The bar is the stretch of "yes" rows; the open dots at

-2

and

3

are the borders, left out because

0 < 0

is false.

The equation still does half the work: it finds the borders.

Solve the matching equation first. Its answers cut the line into pieces, and on each piece the expression keeps one sign, so one test value per piece decides the whole piece.

piece of the line	test value	(x + 2)(x − 3)	negative?
left of −2	−3	6	no
between −2 and 3	0	−6	yes
right of 3	4	6	no

Only the middle piece passes, which gives

(-2, 3)

again, now without trying every value. Adding or subtracting the same quantity on both sides also works exactly as it does for equations.

Two things are new: the symbol decides whether a border belongs, and a negative multiplier flips the symbol.

Change

<

to

\leq

and the borders join the answer, because

0 \leq 0

is true:

(x + 2)(x - 3) \leq 0

gives

[-2, 3]

. Each of the four symbols makes that choice its own way.

Symbol	Meaning	Strictness	Endpoint behavior
<	strictly less than	strict	excluded — open dot, parenthesis
>	strictly greater than	strict	excluded — open dot, parenthesis
≤	less than or equal to	non-strict	included — solid dot, bracket
≥	greater than or equal to	non-strict	included — solid dot, bracket

And unlike an equation, an inequality reverses when both sides are multiplied or divided by a negative number:

-(x + 2)(x - 3) > 0

describes the same stretch, with the symbol turned around.

The same stretch both times. With

<

the dots are open and the brackets round; with

\leq

the dots are filled and the brackets square.

The rest of the page builds on this: what a solution set is, how to write it in interval notation and draw it on a number line, the properties that keep an inequality equivalent, compound inequalities, sign analysis, and then each type in turn — linear, quadratic, polynomial, rational and absolute value.

Solutions and Solution Sets

A solution to an inequality is any value of the variable that makes the statement true. The inequality

x > 3

is satisfied by

x = 4

, by

x = 3.01

, by

x = 1000

, and by every other real number greater than

3

. Unlike most equations, which have finitely many solutions, inequalities typically have infinitely many.

The solution set collects all solutions into one object. For

x > 3

, the solution set is the entire portion of the number line to the right of

3

. For

x^2 < 9

, the solution set is all

x

between

-3

and

3

. These are continuous stretches, not isolated values.

Pitfall: Taking square roots as in an equation

Reducing

x^2 < 9

to

x < 3

keeps values like

x = -5

, where

x^2 = 25

. Both

-3

and

3

bound the set, so the answer is

-3 < x < 3

.

The inequality

x^2 < 9

is a condition on the square of

x

, not on

x

itself. Squaring gives the same result for a number and its opposite, so

-3

and

3

sit at the same height,

9

, and the boundary of the solution set has a point on each side of

0

.

Set-builder notation provides a formal way to write solution sets:

\{x \in \mathbb{R} : x > 3\}

reads "the set of all real numbers

x

such that

x

is greater than

3

." This notation works for any inequality, no matter how complex the condition. In practice, interval notation (covered in the next section) is more compact and more commonly used.

(x + 2)(x − 1)(x − 5) < 0: two stretches of the line

The curve dips below the axis twice, and every input in those stretches makes the expression negative: the solution set is not a list of values but the blue bars on the axis, everything left of −2 and everything between 1 and 5. The open circles mark boundaries that do not satisfy the strict inequality. Try other expressions on the inequality visual explorer.

A solution set of an inequality is usually a union of intervals, not a finite list.

Interval Notation

Notation

Interval Notation

Brackets and parentheses that turn a condition on

x

into a set on the number line — the bracket shape says whether each endpoint belongs. All are catalogued among the algebra symbols.

The inequality signs

<

,

\leq

,

>

,

\geq

themselves come from linear inequality notation.

(a, b)

the open interval from a to b

All real numbers strictly between

a

and

b

: every

x

with

a < x < b

. A parenthesis leaves its endpoint out.

CasesHalf-open intervals mix the two ends:

[a, b)

includes

a

but excludes

b

;

(a, b]

excludes

a

but includes

b

.

[a, b]

the closed interval from a to b

All

x

with

a \leq x \leq b

— the endpoints included. A square bracket keeps its endpoint in.

(a, \infty)

·

(-\infty, a]

from a to infinity; from minus infinity to a

Intervals that run on without end:

(a, \infty)

is all

x > a

,

(-\infty, a]

is all

x \leq a

, and

(-\infty, \infty)

is the whole real line.

Do not confuseA bracket on infinity.

\infty

is not a number and is never paired with a square bracket —

[a, \infty]

is meaningless, because infinity cannot be reached or included.

\cup

union — together with

Joins disconnected pieces into one solution set:

x < -2

or

x > 5

is

(-\infty, -2) \cup (5, \infty)

.

All algebra symbols →↑ Algebra

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Notation · Cheat sheet

Polynomials, Equations and Inequalities: Notation at a Glance

The marks from the notation cards across the polynomial, equation and inequality pages: how each is read and what it means. Every group links to the full card on its page.

Polynomial roots

a_n x^n + \cdots + a_0

a-n x to the n, down to a-zero

Standard form; the subscript of each coefficient is its power of

x

.

\deg P = n

the degree of P is n

The highest power with a nonzero coefficient.

(x - r)^m

x minus r, to the multiplicity m

The root

r

counted

m

times.

Polynomial rules

P = (x - c)\,Q + R

quotient and remainder

Division as a product: quotient

Q(x)

, remainder

R = P(c)

.

P(c), \; P(-x)

P of c; P of minus x

Put in, not multiply: a value, and a new polynomial.

\pm\frac{p}{q}

plus or minus p over q

Rational root candidates:

p \mid a_0

,

q \mid a_n

.

r_1, \ldots, r_n

r sub one to r sub n

The roots, listed once per multiplicity.

Quadratic equations

ax^2 + bx + c = 0

standard form

Descending powers, everything on the left,

a \neq 0

.

\Delta = b^2 - 4ac

delta, the discriminant

Its sign tells how many real roots there are.

x_1, \; x_2

x one and x two

The two roots; the subscripts only count.

Absolute value equations

|x|

the absolute value of x

Distance of

x

from

0

; never negative.

|a + b|

absolute value of a plus b

The bars cover the whole quantity: add first, then take the size.

|x| = \sqrt{x^2}

the root of the square

A second definition, handy with radicals.

InequalitiesThis page

(a, b)

open interval from a to b

Endpoints left out.

[a, b]

closed interval from a to b

Endpoints included.

(a, \infty)

·

(-\infty, a]

rays

Infinity always takes a round bracket.

\cup

union

Joins two intervals into one solution set.

Linear inequalities

<

·

\leq

·

>

·

\geq

less than … greater than or equal to

The bar under the sign adds equality.

-3 < 2x + 1 \leq 7

a compound inequality

Both conditions at once: an "and".

(-\infty, 9]

from minus infinity to nine, inclusive

The solution written as an interval.

Same marks, different job

|x|, \ |z|, \ |A|

Absolute value of a number, modulus of a complex number, determinant of a matrix: the same two bars.

(a, b)

An open interval here; a point in the plane in coordinate geometry.

p

in

\frac{p}{q}

Any divisor of

a_0

, not necessarily a prime.

Polynomial, equation and inequality symbols Algebra

Polynomials, Equations and Inequalities: Notation at a Glance·/algebra/polynomials/rootsLearn Math Class

The same expression with ≤: the boundaries join the set

Changing < to ≤ keeps the same stretches but fills in the circles at −2, 1 and 5, because the expression equals 0 there and 0 ≤ 0 is true. In interval notation the parentheses become brackets: (−∞, −2] ∪ [1, 5]. The infinite end keeps its parenthesis, since ∞ is never a number to include. Switch operators on the inequality visual explorer.

Filled circle, bracket and the non-strict symbol always travel together.

Number Line Representation

A number line graph makes the solution set visible. The boundary points are marked, and the region where the inequality holds is shaded.

A solid dot at a boundary point signals inclusion — the point itself satisfies the inequality. This corresponds to

\leq

or

\geq

and to a square bracket in interval notation. An open dot signals exclusion — the boundary is approached but not reached. This corresponds to

<

or

>

and to a parenthesis.

For the inequality

x \geq -1

, the graph shows a solid dot at

-1

with shading extending to the right toward

\infty

. For

-2 < x \leq 5

, the graph shows an open dot at

-2

, a solid dot at

5

, and shading between them. For

x < -3

or

x > 4

, two separate shaded regions appear, with open dots at

-3

and

4

.

The number line is a direct translation of interval notation into geometry. Every interval

[a, b]

,

(a, b)

,

(-\infty, a)

, or union of intervals has a corresponding picture, and reading the picture back produces the interval notation.

Interval	Meaning	Endpoints	Number line
(a, b)	a < x < b	both excluded	open dots at a and b, shaded between
[a, b]	a ≤ x ≤ b	both included	solid dots at a and b, shaded between
[a, b)	a ≤ x < b	a included, b excluded	solid dot at a, open dot at b
(a, b]	a < x ≤ b	a excluded, b included	open dot at a, solid dot at b
(a, ∞)	x > a	a excluded; ∞ never bracketed	open dot at a, shading extends right
(−∞, a]	x ≤ a	a included; −∞ never bracketed	solid dot at a, shading extends left
(−∞, ∞)	all real numbers	no endpoints	entire line shaded

Properties of Inequalities

The rules for transforming inequalities parallel those for equations, with one critical difference.

Adding or subtracting the same quantity on both sides preserves the direction of the inequality. If

a < b

, then

a + c < b + c

for any real number

c

. This property works identically to its equation counterpart and causes no complications.

Multiplying or dividing both sides by a positive number also preserves the direction. If

a < b

and

c > 0

, then

ac < bc

. Again, no surprises.

Multiplying or dividing both sides by a negative number reverses the direction. If

a < b

and

c < 0

, then

ac > bc

. The inequality flips. This is the single rule that distinguishes inequality manipulation from equation manipulation. Forgetting the flip is the most common error in inequality solving.

Fact: Reciprocals flip the order too

From

2 < 3

follows

\frac{1}{2} > \frac{1}{3}

: taking reciprocals reverses an inequality whenever both sides have the same sign, because

\frac{1}{x}

falls as

x

grows on each side of

0

. With opposite signs it does not:

-1 < 2

and

-1 < \frac{1}{2}

.

The reason for the reversal is arithmetic:

3 < 5

, but multiplying both sides by

-1

gives

-3

and

-5

, and

-3 > -5

. Negation mirrors the number line, swapping the relative positions of two values.

Transitivity provides a chain rule: if

a < b

and

b < c

, then

a < c

. This property justifies writing compound inequalities like

1 < x < 7

as a single chain rather than two separate statements.

Squaring both sides is unreliable because squaring is not a monotone operation — it treats positive and negative inputs differently. The inequality

-5 < 2

is true, but

(-5)^2 = 25 > 4 = 2^2

. Squaring can reverse, preserve, or destroy an inequality depending on the signs of both sides.

Operation on both sides	Direction	Note
add or subtract any real number	preserved	always safe; behaves exactly like equations
multiply or divide by a positive constant	preserved	behaves like equations
multiply or divide by a negative constant	reversed	the defining rule: 3 < 5 becomes −3 > −5

Any step that keeps the order of numbers is safe; a step that reverses the order flips the symbol.

The moves above side by side, each marked by what it does to the direction of the inequality.

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inequalities

What each move does to the direction

Most moves keep the inequality as it is; a few flip it, and squaring cannot be trusted.

Add or subtractKeeps direction

a < b \Rightarrow a + c < b + c

for any real

c

, exactly as with equations

Multiply or divide by a positiveKeeps direction

a < b,\ c > 0 \Rightarrow ac < bc

c > 0

Multiply or divide by a negativeFlips direction

a < b,\ c < 0 \Rightarrow ac > bc

c < 0

the one rule that sets inequalities apart from equations

3 < 5

, but multiplying by

-1

gives

-3 > -5

: negation mirrors the number line.

Take reciprocalsFlips direction

2 < 3 \Rightarrow \frac{1}{2} > \frac{1}{3}

both sides of the same sign

With opposite signs it does not flip:

-1 < 2

and

-1 < \frac{1}{2}

.

TransitivityKeeps direction

a < b,\ b < c \Rightarrow a < c

why a chain like

1 < x < 7

is allowed

Square both sidesUnreliable

-5 < 2

but

25 > 4

squaring is not monotone: it can keep, flip or destroy the inequality

What each move does to the direction·/algebra/inequalitiesLearn Math Class

Every flip comes from the same place: the move reverses the order of the number line, as negation does and as

\frac{1}{x}

does on each side of

0

. Squaring has no single direction at all, which is why it is not a safe step when solving an inequality.

Compound Inequalities

A compound inequality combines two inequality conditions into one statement. The two types — conjunction and disjunction — produce very different solution sets.

A conjunction (AND) requires both conditions to hold simultaneously. The compound inequality

-3 < x

AND

x < 5

is typically written as a chain:

-3 < x < 5

. The solution is the intersection of the two individual solution sets — the overlap. Geometrically, it is the region where both shadings coincide on the number line. For

-3 < x < 5

, the solution is the open interval

(-3, 5)

.

A disjunction (OR) requires at least one condition to hold. The compound inequality

x < -1

OR

x > 4

has a solution set that is the union of two rays:

(-\infty, -1) \cup (4, \infty)

. Any value satisfying either condition qualifies.

Example

Three-part chains like

2 \leq 3x - 1 < 8

are conjunctions in disguise. They are solved by applying operations to all three parts at once: add

1

to get

3 \leq 3x < 9

, then divide by

3

to get

1 \leq x < 3

. The solution is the half-open interval

[1, 3)

.

The distinction between AND and OR is structural, not cosmetic. AND narrows the solution set (intersection), OR expands it (union). Confusing the two changes the answer entirely.

Type	Logical condition	Set operation	Solution shape	Example
Conjunction (AND)	both must hold	intersection	bounded interval	−3 < x < 5 → (−3, 5)
Disjunction (OR)	at least one holds	union	two disjoint rays	x < −1 or x > 4 → (−∞, −1) ∪ (4, ∞)

Equivalent Inequalities

Two inequalities are equivalent when they have the same solution set. The inequalities

2x > 6

and

x > 3

are equivalent — every value satisfying one satisfies the other. Solving an inequality means transforming it, step by step, into a simpler equivalent inequality whose solution set is transparent.

RULE

The operations that guarantee equivalence are: adding or subtracting any real number on both sides, and multiplying or dividing both sides by a positive constant.

These are reversible operations that preserve both the solution set and the direction of the inequality.

Operations that risk breaking equivalence include multiplying or dividing by an expression whose sign depends on the variable. The inequality

\frac{x}{x - 1} > 0

cannot be cleared by multiplying both sides by

(x - 1)

without knowing whether

x - 1

is positive or negative — the direction of the inequality would depend on the answer. This is why sign analysis, rather than algebraic clearing, is the standard method for non-linear inequalities.

Squaring both sides is another non-equivalence-preserving operation. It may enlarge the solution set (introducing extraneous solutions) or collapse it, depending on the signs involved. Whenever a non-reversible operation is used, the result must be verified against the original inequality.

Sign Analysis

For inequalities involving products, quotients, or higher-degree expressions, direct algebraic isolation of the variable is either impossible or unreliable. The method that handles all these cases uniformly is sign analysis.

The idea is straightforward. Every polynomial or rational expression has a finite number of critical points — values where it equals zero or is undefined. Between consecutive critical points, the expression cannot change sign because an expression can change sign only at a zero or at a point where it is undefined. Testing a single value in each interval determines the sign of the entire expression throughout that interval.

The sign chart organizes this process. The critical points are marked on a number line, dividing it into intervals. For each factor in the expression, the sign across every interval is recorded in a row. The final row multiplies the signs together to give the sign of the overall expression in each interval. The solution to the inequality consists of the intervals where the sign matches the one demanded.

For polynomial inequalities, the critical points are the real roots of the polynomial. For rational inequalities, they also include the values where the denominator is zero — these points are always excluded from the solution set regardless of the inequality symbol.

Root multiplicity adds a refinement. At a critical point where a factor has odd multiplicity, the sign of that factor changes as

x

crosses the point. At even multiplicity, the sign stays the same — the expression touches zero but does not cross. Tracking multiplicity at each critical point determines whether the sign chart shows a change or a bounce at each boundary.

Endpoint inclusion is the final step. For non-strict inequalities (

\leq

or

\geq

), roots of the numerator are included because the expression equals zero there and zero satisfies the condition. Denominator zeros are never included because the expression is undefined. For strict inequalities (

<

or

>

), all critical points are excluded.

Critical point	Sign behavior	Strict (<, >)	Non-strict (≤, ≥)
numerator zero, odd multiplicity	sign changes (crossing)	✗	✓
numerator zero, even multiplicity	sign stays (tangent / bounce)	✗	✓
denominator zero	discontinuity; expression undefined	✗	✗

(x + 2)(x − 1)(x − 5) < 0: the sign chart

The three roots cut the axis into four intervals and the chart records the sign of the product on each: negative, positive, negative, positive from left to right. The inequality holds exactly on the shaded intervals where the curve is below the axis. Change an operator or a root and watch the shading redistribute on the inequality visual explorer.

Sign analysis replaces algebra on the intervals with a single test point per interval, or none at all when the factors are known.

Linear Inequalities

A linear inequality in one variable takes the form

ax + b < 0

(or

>

,

\leq

,

\geq

) with

a \neq 0

. The solution is always a single ray — a half-line extending from one boundary point toward positive or negative infinity.

The solving procedure is identical to that of a linear equation: isolate the variable using addition and multiplication. The only complication arises when dividing by a negative coefficient, which reverses the inequality direction. The inequality

-3x \geq 12

becomes

x \leq -4

after dividing by

-3

, not

x \geq -4

.

Compound linear inequalities — chains like

-1 < 2x + 3 \leq 9

or disjunctions like

x < -5

or

x > 2

— are handled by solving each part and combining the results through intersection (AND) or union (OR). The boundary point in every case is the solution of the corresponding linear equation.

To solve a specific inequality, the linear inequality solver tracks every direction flip and gives the answer in interval notation.

Why dividing by a negative flips the sign

The section's −3x ≥ 12. The line y = −3x falls as x grows, so it sits on or above the level 12 only for x at or left of −4. Dividing both sides by −3 must therefore turn ≥ into ≤: the answer is x ≤ −4, not x ≥ −4.

Multiplying by a negative reverses order because it mirrors the number line.

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Quadratic Inequalities

A quadratic inequality has the form

ax^2 + bx + c > 0

(or

<

,

\leq

,

\geq

) with

a \neq 0

. The solution depends on the roots of the corresponding quadratic equation

ax^2 + bx + c = 0

and on the sign of the leading coefficient

a

.

The discriminant

\Delta = b^2 - 4ac

determines the structure before any intervals are tested. When

\Delta > 0

, two distinct real roots divide the line into three intervals. When

\Delta = 0

, a single repeated root creates two intervals. When

\Delta < 0

, no real roots exist and the expression keeps the same sign everywhere — the sign of

a

alone decides whether the inequality is always true or never true.

Pitfall: Reading no roots as no solution

For

x^2 + 1 > 0

,

\Delta = -4 < 0

, and stopping there gives "no solution". But

a = 1 > 0

, so

x^2 + 1

is positive everywhere: the solution is

(-\infty, \infty)

.

The parabola

y = ax^2 + bx + c

provides a geometric shortcut. The inequality

ax^2 + bx + c > 0

asks where the parabola lies above the

x

-axis, and

ax^2 + bx + c < 0

asks where it lies below. The opening direction (upward if

a > 0

, downward if

a < 0

) together with the root locations makes the answer visible.

To solve a specific inequality, the quadratic inequality solver finds the roots, tests each interval and gives the solution in interval notation.

(x + 2)(x − 3) < 0: between the roots

The parabola dips below the axis only between its two roots, so the solution is the single interval (−2, 3); with a negative discriminant the parabola would never dip and the solution would be empty or everything. The roots are the boundaries, the leading coefficient the orientation. Flip the sign of the leading coefficient on the inequality visual explorer.

The three discriminant cases are treated in full on the quadratic inequalities page.

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Polynomial Inequalities

A polynomial inequality of degree three or higher — such as

x^3 - 4x > 0

or

(x + 2)(x - 1)^2(x - 5) \leq 0

— is solved by the sign chart method introduced in the sign analysis section above.

The real roots of the polynomial are found first, partitioning the number line into intervals. The sign of the polynomial is determined in each interval by testing a single point or by tracking the signs of the individual factors. The intervals where the sign matches the inequality form the solution.

Root multiplicity plays a decisive role.

RULE

At a root of odd multiplicity the polynomial changes sign, so the boundary separates a positive interval from a negative one. At a root of even multiplicity the polynomial keeps its sign, merely touching zero without crossing — the boundary is a tangent point rather than a crossing.

Pitfall: Switching the sign at every root

Alternating signs for

(x + 2)(x - 1)^2(x - 5) \leq 0

gives

(-\infty, -2] \cup [1, 5]

, which keeps

x = -3

, where the product is

128

. The double root at

1

keeps the sign, so the answer is

[-2, 5]

.

This distinction determines the shape of the solution set near each root.

A sign chart is usually filled in from one end of the number line to the other, one critical point at a time. At each root the exponent of its factor is read off the factored form, and that exponent decides what is written in the next interval.

(x + 3)(x − 2)² < 0: a double root does not change the sign

At x = 2 the curve touches the axis and turns back without crossing, so the sign is the same on both sides of that root, while at the simple root x = −3 it crosses and the sign flips. The solution is the interval left of −3 alone. Watch which roots flip the chart on the inequality visual explorer.

Multiplicity is the one fact the sign chart cannot do without.

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Rational Inequalities

A rational inequality involves at least one fraction with the variable in the denominator. The inequality

\frac{x - 2}{x + 1} > 0

asks where a rational expression is positive — a question that the sign chart answers directly.

The critical points include both the zeros of the numerator (where the expression equals zero) and the zeros of the denominator (where the expression is undefined). Both types appear on the sign chart and create intervals for testing, but they are treated differently at the endpoint step: numerator zeros may be included in the solution for non-strict inequalities, while denominator zeros are always excluded because the expression does not exist there.

A common and dangerous shortcut is cross-multiplying to clear the denominator. Because the denominator can change sign across the number line (at its real zeros), the direction of the inequality would need to flip in some intervals but not others — making a single multiplication invalid. The sign chart avoids this entirely by analyzing the factored form directly, never multiplying through.

(x − 1)/(x + 2) < 0: two kinds of critical point

The numerator zero at x = 1 and the pole at x = −2 both cut the axis, and the expression is negative only between them; the pole is drawn as an asymptote and can never belong to the solution, while the numerator zero joins it only for a non-strict inequality. Toggle strictness and watch the endpoints open and close on the inequality visual explorer.

Rational inequalities differ from polynomial ones only at the poles, and the poles are always excluded.

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Absolute Value Inequalities

An absolute value inequality translates into a compound inequality through two standard conversions, one for each direction.

RULE

The "less than" form

|f(x)| < k

(with

k > 0

) says that

f(x)

lies within

k

units of zero. This converts to the conjunction

-k < f(x) < k

, a bounded compound inequality whose solution, for linear

f

, is an interval.

The inequality

|2x - 3| < 5

becomes

-5 < 2x - 3 < 5

, which solves to

-1 < x < 4

.

RULE

The "greater than" form

|f(x)| > k

says that

f(x)

lies more than

k

units from zero. This converts to the disjunction

f(x) < -k

or

f(x) > k

, which for linear

f

produces two rays as the solution.

The inequality

|x + 1| \geq 4

becomes

x + 1 \leq -4

or

x + 1 \geq 4

, giving

x \leq -5

or

x \geq 3

.

Pitfall: Using AND for greater than

Converting

|x + 1| \geq 4

to

-4 \leq x + 1 \leq 4

gives

-5 \leq x \leq 3

, which keeps

x = 0

, where

|0 + 1| = 1

. Greater than means OR:

x \leq -5

or

x \geq 3

.

The geometric reading — absolute value as distance on the number line — provides immediate intuition. The inequality

|x - a| < d

describes all points within distance

d

of

a

, forming the interval

(a - d, a + d)

. The inequality

|x - a| > d

describes all points farther than

d

from

a

, forming two rays.

Form	Geometric meaning	Converts to	Solution shape	Example
\|f(x)\| < k (k > 0)	f(x) within k of zero	−k < f(x) < k (AND)	bounded interval	\|2x − 3\| < 5 → (−1, 4)
\|f(x)\| > k (k > 0)	f(x) more than k from zero	f(x) < −k or f(x) > k (OR)	two disjoint rays	\|x + 1\| ≥ 4 → (−∞, −5] ∪ [3, ∞)

|x| − 3 < 0: the V dips below the axis on (−3, 3)

The V-shaped graph is below the axis exactly between its two zeros, so the less-than form gives one bounded interval; the greater-than form would give the two outer rays instead. That is the compound inequality −3 < x < 3 as a picture. Switch the operator on the inequality visual explorer.

The two conversions are the two ways a level line can sit against a V.

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Inequalities Beyond Algebra

Inequalities appear throughout mathematics, and several types fall outside the purely algebraic family because they involve transcendental functions.

Exponential inequalities contain the variable in the exponent. The inequality

2^x > 8

asks where an exponential function exceeds a threshold. When the base exceeds

1

, the exponential is increasing, so the inequality direction is preserved when converting to the exponent:

2^x > 2^3

gives

x > 3

. When the base is between

0

and

1

, the function is decreasing and the direction reverses.

Logarithmic inequalities contain the variable inside a logarithm. The inequality

\log_2(x - 1) < 3

converts to

x - 1 < 2^3

when the base exceeds

1

, but domain restrictions must be enforced: the argument of the logarithm must be positive, adding the constraint

x > 1

.

Trigonometric inequalities involve the variable inside trigonometric functions. The periodicity of sine, cosine, and tangent typically produces infinitely many disjoint solution intervals, one per period. These are treated in the trigonometry section.

Summary of Algebraic Inequality Types

The five algebraic inequality types covered above all produce solution sets shaped as intervals or unions of intervals, but they differ in how those intervals are found. The table below collects each type with its canonical form, the typical solution shape, and the primary solving technique.

Type	Canonical form	Solution shape	Primary technique
Linear	ax + b < 0 (a ≠ 0)	single ray	isolate the variable; flip direction when dividing by a negative
Quadratic	ax² + bx + c < 0	interval, two rays, all reals, or empty	discriminant Δ + sign chart, or read off the parabola
Polynomial	P(x) < 0, deg ≥ 3	union of intervals	sign chart on the factored form, tracking root multiplicity
Rational	P(x) ⁄ Q(x) < 0	union of intervals, excluding zeros of Q	sign chart on the factored form; never cross-multiply
Absolute value	\|f(x)\| < k or \|f(x)\| > k	bounded interval, or two rays	convert to a compound inequality (AND for <, OR for >)

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Before you read

Five families of algebraic inequalities

Each family has its own method and its own trap. The answer is always a set of numbers, usually written as intervals.

Algebraic inequalities5 families

Linear

ax + b < 0

Solutionone ray

flip the sign when dividing by a negative

Quadratic

ax^2 + bx + c < 0

Methoddiscriminant plus a sign chart or the parabola

Solutionan interval, two rays, all reals or empty

Polynomialdegree

\geq 3

Methodsign chart on the factored form

Crossingroot multiplicity decides

Rational

P/Q < 0

Critical pointszeros of numerator and denominator

denominator zeros always excluded; never cross-multiply

Absolute value

|f| < k

,

|f| > k

|f| < k

AND: one interval

|f| > k

OR: two rays

familya trap to watch

Takeaway

Signs decide everything

Linear inequalities flip when divided by a negative; the higher families are solved with a sign chart around their critical points.

Five families of algebraic inequalities·/algebra/inequalitiesLearn Math Class

Common Pitfalls

Six mistakes cause most wrong answers in the inequality types on this page. Four of them are marked where they happen in the sections above; the other two are named in the text of their sections. Here each is worked through in full, with the wrong and the right working side by side.

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1Taking square roots as in an equation

An inequality of the form

x^2 < k

is solved by taking the square root of both sides, as for the equation

x^2 = k

, and only the positive root is kept.

Start

x^2 < 9

Wrong turn

Error here

\implies x < 3

Right path

Move everything to one side and factor.

x^2 - 9 < 0

\implies (x + 3)(x - 3) < 0

The product is negative only between its roots, so

-3 < x < 3

.

Why it happensFor the equation

x^2 = 9

the step gives

x = \pm 3

, and the inequality looks like the same problem with a different symbol. Writing

x < \pm 3

does not help, because it has no clear meaning as a set.

How to checkTest a value from the wrong answer outside the right one.

x = -5

is less than

3

, but

(-5)^2 = 25

, which is not less than

9

. A value from the right answer,

x = 0

, gives

0 < 9

.

2Forgetting the flip

Both sides are divided by a negative number, and the inequality symbol is kept as it was.

Start

-3x \geq 12

Wrong turn

Error here

\implies x \geq -4

Right path

Dividing by

-3

reverses the direction.

\implies x \leq -4

Why it happensThe step is copied from equation solving, where dividing by a negative number changes nothing about the

=

sign.

How to checkTest a value from the wrong answer.

x = 0

satisfies

x \geq -4

, but

-3 \cdot 0 = 0

, which is not at least

12

. A value from the right answer,

x = -5

, gives

15 \geq 12

.

3Reading no roots as no solution

The discriminant of a quadratic inequality is negative, and the conclusion "no real roots" is turned into "no solution".

Start

x^2 + 1 > 0, \quad \Delta = 0^2 - 4 \cdot 1 \cdot 1 = -4 < 0

Wrong turn

Error here

\implies \text{no solution}

Right path

With no real roots the expression never changes sign, and the leading coefficient

a = 1 > 0

makes the parabola open upward, so it lies above the

x

-axis everywhere. The solution is every real number,

(-\infty, \infty)

. The opposite inequality

x^2 + 1 < 0

is the one with no solution.

Why it happensFor the equation

x^2 + 1 = 0

, a negative discriminant does mean there is no real solution, and that habit carries over.

How to checkTest any single value.

x = 0

gives

1 > 0

, so the solution set is not empty.

4Switching the sign at every root

The sign chart is filled in by changing the sign at each critical point, without looking at the multiplicity of the factors.

Wrong turn

Error here

For

(x + 2)(x - 1)^2(x - 5) \leq 0

, starting with

+

to the right of

5

and alternating gives

-

on

(1, 5)

,

+

on

(-2, 1)

and

-

to the left of

-2

, so the answer is taken as

(-\infty, -2] \cup [1, 5]

.

Right path

The factor

(x - 1)^2

has even multiplicity and is never negative, so the sign does not change at

x = 1

. The signs are

+

to the left of

-2

,

-

on

(-2, 1)

,

-

on

(1, 5)

and

+

to the right of

5

. With the zeros

-2

,

1

and

5

included, the solution is

[-2, 5]

.

Why it happensFor products of distinct linear factors the signs really do alternate, so alternation becomes a reflex.

How to checkTest a value in each interval.

x = -3

gives

(-1)(16)(-8) = 128

, which is not

\leq 0

;

x = 0

gives

(2)(1)(-5) = -10 \leq 0

.

5Cross-multiplying

The denominator of a rational inequality is cleared by multiplying both sides by it, as if it were a positive number.

Start

\frac{x - 2}{x + 1} > 0

Wrong turn

Error here

\implies x - 2 > 0

\implies x > 2

Right path

The critical points are

x = 2

(numerator zero) and

x = -1

(denominator zero). The quotient is positive to the left of

-1

, negative between

-1

and

2

, and positive to the right of

2

. The solution is

(-\infty, -1) \cup (2, \infty)

.

Why it happensMultiplying by the denominator is the standard first step for rational equations. For an inequality it is valid only where the denominator is positive; where it is negative the direction would have to flip.

How to checkTest a value left of the pole.

x = -2

gives

\frac{-4}{-1} = 4 > 0

, so it belongs to the solution, but the wrong working lost it.

6Using AND for greater than

A "greater than" absolute value inequality is converted with the bounded chain that belongs to the "less than" form.

Start

|x + 1| \geq 4

Wrong turn

Error here

\implies -4 \leq x + 1 \leq 4

\implies -5 \leq x \leq 3

Right path

|x + 1| \geq 4

says

x + 1

is at least

4

units from zero, which is a disjunction.

x + 1 \leq -4 \text{ or } x + 1 \geq 4

\implies x \leq -5 \text{ or } x \geq 3

Why it happensThe chain

-k < f(x) < k

is learned first and is the familiar shape. A related slip writes the OR case as a single chain,

-4 \geq x + 1 \geq 4

, which no number satisfies.

How to checkTest a value from the wrong answer.

x = 0

gives

|0 + 1| = 1

, which is not at least

4

. A value from the right answer,

x = 4

, gives

|4 + 1| = 5 \geq 4

.

Common pitfalls with inequalities·/algebra/inequalitiesLearn Math Class

Every check above is the same move: pick one number and substitute it into the original inequality. A value from the wrong answer that fails, or a value from the right answer that the wrong working left out, proves the working wrong. One test value per interval between the critical points is enough, since the sign of the expression cannot change inside an interval.

Inequalities FAQ

How are the solutions of an equation related to the solutions of the matching inequality?

+

They are its borders, not its answer. The roots of

(x + 2)(x - 3) = 0

are

-2

and

3

. They cut the line into three pieces, and on each piece the expression keeps one sign, so one test value per piece decides the whole piece. For

(x + 2)(x - 3) < 0

only the middle piece passes, giving

(-2, 3)

. The borders join the answer only for

\leq

or

\geq

:

(x + 2)(x - 3) \leq 0

gives

[-2, 3]

.Read more →

Does taking reciprocals reverse an inequality?

+

Yes, when both sides have the same sign:

2 < 3

gives

\frac{1}{2} > \frac{1}{3}

, and

-3 < -2

gives

-\frac{1}{3} > -\frac{1}{2}

. The reason is that

\frac{1}{x}

falls as

x

grows on each side of

0

, and a step that reverses the order of numbers flips the symbol, just as multiplying by a negative does. With opposite signs there is no flip:

-1 < 2

and

-1 < \frac{1}{2}

, because a negative number stays below a positive one.Read more →

How do you choose a method for solving an inequality?

+

By its type. A linear inequality is solved like an equation, flipping the sign when dividing by a negative, and gives one ray. Quadratic, polynomial and rational inequalities are solved with a sign chart on the factored form: for a quadratic the discriminant settles the shape first, for a polynomial root multiplicity decides where the sign changes, and for a rational inequality the denominator zeros join the chart but never the answer. An absolute value inequality is first converted to a compound one: AND for

<

, OR for

>

.Read more →

Can you multiply both sides of an inequality by an expression containing $x$ ?

+

Not safely, unless its sign is known. Multiplying by a positive keeps the direction and multiplying by a negative flips it, so an expression that is positive for some

x

and negative for others would need both at once. Clearing

\frac{x}{x - 1} > 0

by multiplying by

x - 1

as if it were positive gives

x > 0

, which keeps

x = \frac{1}{2}

, where the fraction is

-1

, and loses

x = -1

, where it is

\frac{1}{2}

. A sign chart gives the right answer,

(-\infty, 0) \cup (1, \infty)

.Read more →

Are exponential and logarithmic inequalities solved with a sign chart?

+

Usually not. A single exponential or logarithm is always increasing or always decreasing, so the inequality passes straight to the exponent or the argument:

2^x > 8

is

2^x > 2^3

, so

x > 3

. A base above

1

keeps the direction and a base between

0

and

1

reverses it. A logarithm also adds a domain condition:

\log_2(x - 1) < 3

needs

x - 1 < 8

and

x - 1 > 0

, so

1 < x < 9

. A sign chart comes back only if what is left is a polynomial or rational inequality.Read more →

Inequalities

When Equality Is Not the Question

Key Terms

Core Concepts

Methods

Inequality Types

What an Inequality Is

Solutions and Solution Sets

Interval Notation

Interval Notation

Number Line Representation

Properties of Inequalities

What each move does to the direction

Compound Inequalities

Equivalent Inequalities

Sign Analysis

Linear Inequalities

Quadratic Inequalities

Polynomial Inequalities

Rational Inequalities

Absolute Value Inequalities

Inequalities Beyond Algebra

Summary of Algebraic Inequality Types

Common Pitfalls

Inequalities FAQ

How are the solutions of an equation related to the solutions of the matching inequality?

Does taking reciprocals reverse an inequality?

How do you choose a method for solving an inequality?

Can you multiply both sides of an inequality by an expression containing $x$ ?

Are exponential and logarithmic inequalities solved with a sign chart?

Inequalities

When Equality Is Not the Question

Key Terms

Core Concepts

Methods

Inequality Types

What an Inequality Is

Solutions and Solution Sets

Interval Notation

Interval Notation

Number Line Representation

Properties of Inequalities

What each move does to the direction

Compound Inequalities

Equivalent Inequalities

Sign Analysis

Linear Inequalities

Quadratic Inequalities

Polynomial Inequalities

Rational Inequalities

Absolute Value Inequalities

Inequalities Beyond Algebra

Summary of Algebraic Inequality Types

Common Pitfalls

Inequalities FAQ

How are the solutions of an equation related to the solutions of the matching inequality?

Does taking reciprocals reverse an inequality?

How do you choose a method for solving an inequality?

Can you multiply both sides of an inequality by an expression containing xxx?

Are exponential and logarithmic inequalities solved with a sign chart?

Can you multiply both sides of an inequality by an expression containing $x$ ?