Absolute Value Equations: Case-Splitting Method

Definition of Absolute Value

The absolute value of a real number

x

, written

|x|

, is defined piecewise:

Definition of Absolute Value

|x| = \begin{cases} x & \text{if } x \geq 0 \\ -x & \text{if } x < 0 \end{cases}

The notation

-x

in the second case does not mean "a negative number." It means "the opposite of

x

." When

x

is already negative,

-x

is positive.

RULE

The absolute value function takes any input and returns a non-negative output:

|x| \geq 0

for every real

x

, with equality only when

x = 0

.

Geometrically,

|x|

is the distance from

x

to the origin on the number line. The number

-7

sits

7

units from zero, so

|-7| = 7

. The number

4

sits

4

units from zero, so

|4| = 4

. Distance is inherently non-negative, which is why absolute value never produces a negative result.

This non-negativity has an immediate consequence for equations: if

|f(x)| = k

and

k < 0

, the equation has no solution. No expression inside absolute value bars can produce a negative output. The sign of the right-hand side is the first thing to check before any case analysis begins.

The absolute value page develops this distance reading and the rules the bars obey, such as

|ab| = |a|\,|b|

and

\sqrt{x^2} = |x|

.

|x| + 5 = 2: no solution

The V-shaped curve |x| + 5 never drops below 5, so it never reaches the level 2: the equation, which says |x| = −3, asks a distance to be negative and has no solution. Checking the sign of the right-hand side settles this before any case split. Slide the level and watch solutions appear on the equation visual explorer.

A negative right-hand side means an empty solution set, whatever is inside the bars.

Absolute Value Notation

Notation

Absolute Value Notation

The bars as a bracket pair, the most overloaded glyph in mathematics, and the algebraic alias that lets the bars be squared away.

The

\pm

behind every case split — radical notation; the cases brace in the definition — piecewise functions.

|a + b|

The absolute value of the whole quantity a plus b

The bars are a bracket pair: everything between them is one block, evaluated first — like the radical's vinculum, an invisible parenthesis with a built-in operation. The mathematical keyboard inserts the pair without typing either bar.

CasesNested bars are legal but hard on the eye:

\left||x| - 1\right|

resolves inside-out. Print sizes the outer pair taller; handwriting has no such mercy, which is why worked solutions relabel the inner block instead.

Also written

\operatorname{abs}(x)

— the functional spelling, universal in programming, rare in print.

Do not confuseA distributable mark.

|a + b| \neq |a| + |b|

in general — the gap between them is the triangle inequality

|a+b| \leq |a| + |b|

, and treating the bars like a factor erases it.

|x|

·

|z|

·

|A|

·

|M|

The same two bars, four different jobs

The vertical pair is the most overloaded bracket in mathematics. Here it means distance to zero, per the Definition above — and only some of its other jobs keep that meaning.

Cases

|z|

on a complex number is the same idea — distance to the origin, plane instead of line.

\|v\|

doubles the bars for vector length. Both generalize this page's bars honestly.

Do not confuseThe unrelated jobs.

|A|

on a set is cardinality — a count;

|M|

on a matrix is the determinant — which can be negative: the one place the bars carry no non-negativity promise at all.

|x| = \sqrt{x^2}

The absolute value is the square root of the square

The bars have an algebraic alias: squaring erases the sign, the principal root keeps the result non-negative. One identity, two directions — it defines

|x|

without cases, and it lets equations shed their bars by squaring.

Cases

|A| = |B|

is equivalent to

A^2 = B^2

— the squaring route for Equations with Absolute Value on Both Sides below. Calculus leans on the alias too:

\sqrt{x^2}

differentiates where the piecewise form needs case-work.

Also written

\max(x, -x)

— the order-theoretic spelling, common in analysis.

Do not confuse

\sqrt{x^2} = x

, with the bars dropped. That identity is false for negative

x

— the whole story is told at the even-root identity, whose answer is exactly these bars.

All algebra symbols →↑ Equations

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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 equationsThis page

|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.

Inequalities

(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

|x| = 3 has solutions x = −3 and x = 3

The level y = 3 cuts both arms of the V, so B > 0 gives two solutions; at B = 0 the level touches only the corner and there is one; below the axis the level misses the graph and there is none. The three structures of the section are three positions of one line. Slide the level through the three cases on the equation visual explorer.

Everything else on this page is this picture with a more complicated expression inside the bars.

The Basic Equation

The obvious move with |2x − 3| = 7 is to drop the bars and solve what is left.

Without the bars,

2x - 3 = 7

gives

2x = 10

and

x = 5

. Checking it:

|2 \cdot 5 - 3| = |7| = 7

. It works, so the job looks done.

But x = −2 works too — and dropping the bars never finds it.

At

x = -2

the inside is

-7

, and the bars turn it into

7

. The graph shows the problem: the V shape reaches height

7

in two places, and the shortcut found only one.

x	2x − 3	\|2x − 3\|	equals 7?
5	7	7	yes — found by dropping the bars
−2	−7	7	yes — missed

The loss was predictable: absolute value forgets the sign, so two numbers have absolute value 7.

Being "7 kilometres from home" does not say in which direction — east or west both qualify. In the same way

|p| = 7

allows

p = 7

and

p = -7

, and dropping the bars keeps only the first.

The equation

|A| = B

, where

A

is an expression and

B

is a constant, has three possible structures depending on the sign of

B

.

The definition

An absolute value equation has the unknown inside absolute value bars. For a positive constant

b

:

|p| = b

exactly when

p = b

or

p = -b

;

|p| = 0

exactly when

p = 0

; and

|p| = b

with

b < 0

has no solution.

When

B > 0

, the equation splits into two cases:

Absolute Value Equation

|p| = b \implies p = b \text{ or } p = -b \quad (b > 0)

Learn more about this formula: Absolute Value Equation →

Browse all algebra formulas →

So

|2x - 3| = 7

means

2x - 3 = 7

or

2x - 3 = -7

:

x = 5

or

x = -2

.

It is an equation that splits into cases — usually two, sometimes one or none.

Both must be solved, and both solutions (if they exist) belong to the solution set. The equation

|2x - 3| = 7

becomes

2x - 3 = 7

or

2x - 3 = -7

, giving

x = 5

or

x = -2

.

When

B = 0

, the equation reduces to a single case:

A = 0

. There is no split because zero is its own opposite. The equation

|4x + 1| = 0

gives

4x + 1 = 0

, so

x = -\frac{1}{4}

.

Sign of B	Resulting case(s)	Solution count
B > 0	A = B or A = −B	two (typically)
B = 0	A = 0	one
B < 0	(impossible — absolute value is never negative)	none — empty set

The sign of the right-hand side decides the split: two cases when positive, one when zero, none when negative.

Read backwards, the cases give a test to run first: look at the sign of the right-hand side.

When

B < 0

, no solution exists. The equation

|x + 3| = -2

asks for a non-negative quantity to equal

-2

, which is impossible. The solution set is empty, and no algebra is needed to determine this.

Pitfall: Splitting a negative right-hand side

Splitting

|x + 3| = -2

into

x + 3 = -2

or

x + 3 = 2

gives

x = -5

or

x = -1

, but both make

|x + 3| = 2

, not

-2

. The equation has no solution.

Checking the sign of the right-hand side before proceeding is not a shortcut — it is the logical first step. A negative right-hand side eliminates the equation immediately, and recognizing this saves the effort of setting up cases that cannot produce valid solutions.

When the inside is a longer expression, the same split applies to it as a whole, as the next section shows.

The rest of the page covers equations of the form |f(x)| = k, absolute values on both sides, several absolute value terms, and extraneous solutions.

Solving Equations of the Form |f(x)| = k

When the expression inside the absolute value is a function of

x

rather than

x

alone, the case-splitting method applies in the same way. The equation

|f(x)| = k

(with

k > 0

) generates two separate equations:

Solving Equations of the Form |f(x)| = k

f(x) = k \quad \text{and} \quad f(x) = -k

Each equation is solved independently using whatever method the form of

f(x)

requires. The union of both solution sets is the complete answer.

When

f(x)

is linear, each case produces a linear equation with exactly one solution, so the original absolute value equation has exactly two solutions (assuming

k > 0

). The equation

|3x + 5| = 11

yields

3x + 5 = 11

(so

x = 2

) and

3x + 5 = -11

(so

x = -\frac{16}{3}

).

When

f(x)

is quadratic, each case is a quadratic equation that may contribute zero, one, or two solutions.

Example

The equation

|x^2 - 4| = 5

splits into

x^2 - 4 = 5

and

x^2 - 4 = -5

. The first gives

x^2 = 9

, so

x = \pm 3

. The second gives

x^2 = -1

, which has no real solutions. The complete solution set is

\{-3, 3\}

.

Pitfall: Dropping the negative case

Solving

|3x + 5| = 11

only as

3x + 5 = 11

gives

x = 2

and stops there. The case

3x + 5 = -11

adds

x = -\frac{16}{3}

, so the solution set is

\left\{-\frac{16}{3}, 2\right\}

.

The method extends to any expression inside the absolute value. The internal structure determines how hard each case is to solve, but the case-splitting framework is always the same.

To solve your own absolute value equation with each case shown, use the absolute value equation solver.

Two cases, one of them empty

The section's |x² − 4| = 5. The bars fold the dipping part of x² − 4 (dashed) up above the axis. The case x² − 4 = 5 meets the level at x = ±3; the case x² − 4 = −5 would need the dashed parabola to reach −5, which it never does, so it adds nothing. The solution set is {−3, 3}.

Each case is solved by its own method, and only real solutions join the set.

Equations with Absolute Value on Both Sides

When absolute value appears on both sides —

|f(x)| = |g(x)|

— the equation asserts that two expressions are equal in magnitude, possibly differing in sign. This decomposes into two cases:

Equations with Absolute Value on Both Sides

f(x) = g(x) \quad \text{or} \quad f(x) = -g(x)

Each case is solved separately, and the results are combined.

Example

The equation

|2x - 1| = |x + 4|

produces two equations. From

2x - 1 = x + 4

:

x = 5

. From

2x - 1 = -(x + 4)

:

2x - 1 = -x - 4

, so

3x = -3

and

x = -1

. Both candidates should be verified:

|2(5) - 1| = |9| = 9

and

|5 + 4| = |9| = 9

, confirming

x = 5

. Similarly,

|2(-1) - 1| = |-3| = 3

and

|-1 + 4| = |3| = 3

, confirming

x = -1

. The solution set is

\{-1, 5\}

.

Pitfall: Negating only the first term

Writing the second case of

|2x - 1| = |x + 4|

as

2x - 1 = -x + 4

gives

x = \frac{5}{3}

, where the sides are

\frac{7}{3}

and

\frac{17}{3}

. Negating the whole side,

-(x + 4) = -x - 4

, gives

x = -1

.

The second case is where the sign change enters. Its right-hand side is

-g(x)

, the opposite of the entire expression on the right, which for this equation is

-(x + 4) = -x - 4

. Both candidates are then confirmed by substitution, as above.

RULE

An alternative approach uses the algebraic identity

|A| = |B|

if and only if

A^2 = B^2

.

Squaring both sides removes all absolute value bars at once, producing the equation

(2x - 1)^2 = (x + 4)^2

. Expanding and simplifying yields a polynomial equation. This method is algebraically equivalent to case-splitting: both sides are non-negative, so

A^2 = B^2

holds exactly when

|A| = |B|

, and squaring introduces no extraneous solutions here. Checking the candidates in the original equation is still a good habit against arithmetic slips.

Two V-shapes, two crossings

The section's |2x − 1| = |x + 4|, each side drawn as its own V. They meet where the magnitudes agree: at x = 5 the insides are equal (9 and 9), and at x = −1 they are opposites (−3 and 3). These are exactly the two cases f(x) = g(x) and f(x) = −g(x).

Checking both candidates in the original equation confirms the picture.

Equations with Multiple Absolute Value Terms

Equations containing two or more absolute value terms — such as

|x - 1| + |x + 3| = 8

— cannot be handled by a single case split. Each absolute value expression changes its behavior at a critical point where the expression inside equals zero, and the equation must be analyzed separately on each interval defined by these critical points.

Fact: A whole interval of solutions

Between

-3

and

1

the two distances always add up to

4

, the gap between the critical points. So

|x - 1| + |x + 3| = 4

is solved by every

x

in

[-3, 1]

: infinitely many solutions from two linear pieces.

For

|x - 1| + |x + 3| = 8

, the critical points are

x = 1

and

x = -3

. These divide the number line into three intervals:

x < -3

,

-3 \leq x \leq 1

, and

x > 1

. On each interval, every absolute value expression has a definite sign, so the piecewise definition removes the bars entirely.

On

x < -3

: both

x - 1 < 0

and

x + 3 < 0

, so the equation becomes

-(x - 1) + (-(x + 3)) = 8

, which simplifies to

-2x - 2 = 8

, giving

x = -5

. Since

-5 < -3

, this value lies in the interval and is valid.

On

-3 \leq x \leq 1

:

x - 1 \leq 0

and

x + 3 \geq 0

, so the equation becomes

-(x - 1) + (x + 3) = 8

, which simplifies to

4 = 8

. This is false, so no solution exists in this interval.

On

x > 1

: both expressions are positive, so the equation becomes

(x - 1) + (x + 3) = 8

, giving

2x + 2 = 8

and

x = 3

. Since

3 > 1

, this is valid.

Pitfall: Keeping a candidate outside its interval

Solving

|x - 1| + |x + 3| = 3

piece by piece gives

x = -\frac{5}{2}

on

x < -3

and

x = \frac{1}{2}

on

x > 1

. Neither lies in its interval, and both give a sum of

4

, so there is no solution.

The solution set is

\{-5, 3\}

. The interval method is systematic: identify critical points, solve on each interval using the appropriate piecewise form, and verify that each candidate belongs to the interval where it was derived.

Interval	\|x − 1\|	\|x + 3\|	Equation becomes	Result
x < −3	−(x − 1)	−(x + 3)	−2x − 2 = 8	x = −5 ✓
−3 ≤ x ≤ 1	−(x − 1)	(x + 3)	4 = 8	no solution ✗
x > 1	(x − 1)	(x + 3)	2x + 2 = 8	x = 3 ✓

Cut the number line wherever an expression inside the bars is zero, and on each piece the bars simply disappear.

Absolute Value and Extraneous Solutions

The case-splitting method, when applied correctly, does not generate extraneous solutions. Each case defines a condition under which the absolute value expression takes a specific sign, and solving within that condition produces only valid results — provided the solution is checked against the interval constraint.

Extraneous solutions enter through a different route: squaring.

Example

The equation

|x - 2| = 3x - 4

can be solved by case-splitting (the reliable method) or by squaring both sides to obtain

(x - 2)^2 = (3x - 4)^2

. Squaring eliminates the absolute value but is not reversible — it treats

|A| = B

and

|A| = -B

identically, and if

B

takes negative values for some

x

, those values may appear as solutions to the squared equation without satisfying the original.

For this equation, case-splitting gives two cases. From

x - 2 = 3x - 4

:

x = 1

. Check:

|1 - 2| = 1

and

3(1) - 4 = -1

. Since

1 \neq -1

, this candidate is extraneous. From

-(x - 2) = 3x - 4

:

-x + 2 = 3x - 4

, so

x = \frac{3}{2}

. Check:

|\frac{3}{2} - 2| = \frac{1}{2}

and

3(\frac{3}{2}) - 4 = \frac{1}{2}

. This one holds. The only solution is

x = \frac{3}{2}

.

The first candidate failed because the right-hand side

3x - 4

was negative at

x = 1

.

RULE

An absolute value equation

|A| = B

requires

B \geq 0

at the solution point.

This constraint is automatically enforced by careful case-splitting but easily missed when squaring.

Pitfall: Keeping a root where the right side is negative

Accepting both cases of

|x + 3| = 2x

keeps

x = -1

beside

x = 3

. At

x = -1

the right side is

2x = -2

while

|x + 3| = 2

, so the only solution is

x = 3

.

The constraint matters whenever the right-hand side contains

x

, because then its sign changes along the number line. Each case produces a candidate, and the right-hand side is evaluated at that candidate before it joins the solution set.

Where squaring goes wrong

The section's |x − 2| = 3x − 4. The V and the line meet once, at x = 3/2. Squaring both sides also produces x = 1, but on the dashed line x = 1 the two sides are 1 and −1: equal in square, not in value. Because the right side is negative there, x = 1 is extraneous.

Case-splitting with an interval check avoids the extraneous root that squaring creates.

Geometric Interpretation

Absolute value measures distance on the number line, and this interpretation converts many equations into geometric statements that can be read directly.

The equation

|x - a| = d

asks: which points on the number line are exactly

d

units from

a

? There are two such points, one on each side:

x = a + d

and

x = a - d

. The solutions are symmetric about

a

, and

a

is their midpoint. The equation

|x - 5| = 3

has solutions

x = 8

and

x = 2

, both exactly

3

units from

5

.

The equation

|x - a| = |x - b|

asks: which point is equidistant from

a

and

b

?

THEOREM

On the number line, exactly one point lies at equal distance from two distinct fixed points — the midpoint

x = \frac{a + b}{2}

.

The equation

|x - 1| = |x - 7|

has the single solution

x = 4

, the midpoint of

1

and

7

.

The equation

|x - a| + |x - b| = d

asks: which points have the property that their total distance to

a

and

b

equals

d

? The minimum possible value of

|x - a| + |x - b|

is

|a - b|

(achieved by any

x

between

a

and

b

). If

d < |a - b|

, no solution exists. If

d = |a - b|

, every point in the closed interval

[a, b]

is a solution. If

d > |a - b|

, exactly two solutions exist, one beyond each endpoint.

These geometric readings do not replace the algebraic method, but they provide a check on the number and location of solutions before any computation begins.

Equation form	Geometric question	Solution
\|x − a\| = d	which points are exactly d units from a?	x = a + d and x = a − d
\|x − a\| = \|x − b\|	which point is equidistant from a and b?	x = (a + b) ⁄ 2 (midpoint)
\|x − a\| + \|x − b\| = d, d < \|a − b\|	total distance less than the gap?	none (impossible)
\|x − a\| + \|x − b\| = d, d = \|a − b\|	total distance equals the gap?	every x in [a, b]
\|x − a\| + \|x − b\| = d, d > \|a − b\|	total distance greater than the gap?	two points, one outside each endpoint

An absolute value equation is a question about distances, and the number line often answers it before any algebra.

Summary of Absolute Value Equation Forms

The forms covered above all fit a small set of patterns. The table below collects them with the recommended method for each and the typical number of solutions to expect. The constraint

|A| \geq 0

is the unifying logic: every method either uses it directly (case-splitting) or has to verify it after the fact (squaring).

Equation form	Method	Number of solutions
\|A\| = k, k > 0	split into A = k and A = −k; solve each	two (typically)
\|A\| = 0	solve A = 0	one
\|A\| = k, k < 0	none — read off immediately	none (empty set)
\|f(x)\| = k, k > 0	split into f(x) = k and f(x) = −k; solve each by appropriate method	depends on f
\|f(x)\| = \|g(x)\|	f = g or f = −g (or square both sides; verify)	depends on f, g
\|f(x)\| = g(x)	split into two cases; require g(x) ≥ 0 at each candidate	verify carefully — squaring risks extraneous solutions
Multiple \|…\| terms (sum/difference)	interval method: find critical points, solve on each interval using piecewise definition	varies; verify each candidate lies in its interval

Common Pitfalls

Five mistakes account for most wrong answers to absolute value equations. 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.

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1Splitting a negative right-hand side

The right-hand side is a negative constant, but the equation is split into two cases as if it were positive, and the two results are reported as solutions.

Start

|x + 3| = -2

Wrong turn

Error here

\implies x + 3 = -2 \text{ or } x + 3 = 2

\implies x = -5 \text{ or } x = -1

Right path

An absolute value is never negative, so

|x + 3|

cannot equal

-2

for any

x

. The equation has no solution.

Why it happensThe split into

p = b

or

p = -b

becomes a reflex, and it is applied before anyone looks at the condition

b > 0

.

How to checkSubstitute the results.

x = -5

gives

|-2| = 2

, and

x = -1

gives

|2| = 2

; neither equals

-2

.

2Dropping the negative case

Only the case with the positive right-hand side is solved, so one of the two solutions is missing.

Start

|3x + 5| = 11

Wrong turn

Error here

\implies 3x + 5 = 11 \implies x = 2

Right path

Both cases are solved.

3x + 5 = 11 \implies x = 2

3x + 5 = -11 \implies 3x = -16 \implies x = -\frac{16}{3}

The solution set is

\left\{-\frac{16}{3}, 2\right\}

.

Why it happensRemoving the bars looks like removing a pair of brackets, and a linear equation is expected to have one answer.

How to checkThe inside

3x + 5

can be

11

or

-11

, and both have absolute value

11

. Substituting

x = -\frac{16}{3}

gives

|-16 + 5| = |-11| = 11

.

3Negating only the first term

In the second case of

|f(x)| = |g(x)|

, the minus sign is attached to the first term of

g(x)

instead of to the whole expression.

Start

|2x - 1| = |x + 4|

Wrong turn

Error here

\implies 2x - 1 = -x + 4

\implies 3x = 5 \implies x = \frac{5}{3}

Right path

The whole expression is negated.

2x - 1 = -(x + 4) = -x - 4

\implies 3x = -3 \implies x = -1

Together with

x = 5

from the first case, the solution set is

\{-1, 5\}

.

Why it happensThe brackets around

x + 4

are dropped when the minus sign is written, so the sign reaches only the

x

.

How to checkSubstitute.

x = \frac{5}{3}

gives

\frac{7}{3}

on the left and

\frac{17}{3}

on the right, which differ.

x = -1

gives

3

on both sides.

4Keeping a candidate outside its interval

Each interval between the critical points gives its own equation, and the solution of that equation is kept without checking that it lies in the interval.

Start

|x - 1| + |x + 3| = 3

x < -3: \quad -2x - 2 = 3 \implies x = -\frac{5}{2}

-3 \leq x \leq 1: \quad 4 = 3, \text{ no solution}

x > 1: \quad 2x + 2 = 3 \implies x = \frac{1}{2}

Wrong turn

Error here

Each interval is solved and its result is kept. Both values are reported as solutions.

Right path

-\frac{5}{2}

is not less than

-3

, and

\frac{1}{2}

is not greater than

1

, so both are rejected. The equation has no solution.

Why it happensEach piece is a plain linear equation whose answer looks final, and the interval it came from is written at the top of the working and forgotten by the end.

How to checkSubstitute into the original. Both values give

|x - 1| + |x + 3| = 4

, not

3

. The distance reading agrees: the total distance from

x

to

1

and to

-3

is never less than

4

.

5Keeping a root where the right side is negative

The right-hand side contains

x

, both cases are solved, and both results are kept without checking the sign of the right-hand side.

Start

|x + 3| = 2x

\implies x + 3 = 2x \text{ or } x + 3 = -2x

Wrong turn

Error here

\implies x = 3 \text{ or } x = -1

Right path

|A| = B

requires

B \geq 0

. At

x = 3

the right side is

6

and

|3 + 3| = 6

, so

x = 3

holds. At

x = -1

the right side is

-2

, so

x = -1

is extraneous.

The only solution is

x = 3

.

Why it happensWith a positive constant on the right, both cases always survive, and that habit carries over to right-hand sides that change sign.

How to checkSubstitute each candidate into the original equation.

x = -1

gives

|2| = 2

on the left and

-2

on the right.

Common pitfalls in absolute value equations·/algebra/equations/absolute-valueLearn Math Class

Substituting every answer back into the original equation catches four of the five slips, because each of them reports a value that does not satisfy it. Dropping the negative case is different: the answer it gives is correct, and the error is the solution that is missing. So two checks belong at the end of every absolute value equation: each answer satisfies the original equation, and every case that could give an answer has been solved.

Absolute Value Equations FAQ

Why do vertical bars mean different things in different places?

+

They are the most overloaded bracket in mathematics. Some uses keep this meaning:

|z|

for a complex number is still distance to the origin, and

\|v\|

doubles the bars for vector length. Others do not:

|A|

on a set is cardinality, a count, and

|M|

on a matrix is the determinant, which can be negative.Read more →

Can you get rid of absolute value bars by squaring?

+

Yes, safely when both sides carry bars. Because

|x| = \sqrt{x^2}

, the equation

|A| = |B|

is equivalent to

A^2 = B^2

, which turns a case analysis into one polynomial equation. With bars on one side only, squaring can add roots:

|x - 2| = 3x - 4

squared gives

x = 1

and

x = \frac{3}{2}

, but at

x = 1

the right side is

-1

, so only

x = \frac{3}{2}

solves the original.Read more →

Can an absolute value equation have infinitely many solutions?

+

Yes. For any

x

between

-3

and

1

, the distances from

x

to

1

and to

-3

add up to

4

, the gap between the two points, so

|x - 1| + |x + 3| = 4

holds for every

x

in

[-3, 1]

. In general

|x - a| + |x - b| = d

has no solution when

d < |a - b|

, the whole interval between

a

and

b

when

d = |a - b|

, and two solutions when

d > |a - b|

.Read more →

Why does $|x - 1| = |x - 7|$ have only one solution?

+

Because one of its two cases collapses. The case

x - 1 = x - 7

reduces to

-1 = -7

, which is false, so it gives nothing; the case

x - 1 = -(x - 7)

gives

2x = 8

and

x = 4

. Read as distance, the equation asks for the point equally far from

1

and

7

, and on the number line the only such point is their midpoint,

4

.Read more →

Why does $|x^2 - 4| = 5$ have only two solutions?

+

Because only one of its two cases has real roots. The case

x^2 - 4 = 5

gives

x^2 = 9

, so

x = \pm 3

; the case

x^2 - 4 = -5

needs

x^2 = -1

, which no real number satisfies. With a quadratic inside, each case can contribute zero, one or two solutions, so the total can be anything from none to four.Read more →

Equations with Absolute Value

Distance, Cases, and the Geometry of the Number Line

Key Terms

Core Concept

Equation Solving

Formulas Used on This Page

Definition of Absolute Value

Absolute Value Notation

Absolute Value Notation

The Basic Equation

Solving Equations of the Form |f(x)| = k

Equations with Absolute Value on Both Sides

Equations with Multiple Absolute Value Terms

Absolute Value and Extraneous Solutions

Geometric Interpretation

Summary of Absolute Value Equation Forms

Common Pitfalls

Absolute Value Equations FAQ

Why do vertical bars mean different things in different places?

Can you get rid of absolute value bars by squaring?

Can an absolute value equation have infinitely many solutions?

Why does $|x - 1| = |x - 7|$ have only one solution?

Why does $|x^2 - 4| = 5$ have only two solutions?

Equations with Absolute Value

Distance, Cases, and the Geometry of the Number Line

Key Terms

Core Concept

Equation Solving

Formulas Used on This Page

Definition of Absolute Value

Absolute Value Notation

Absolute Value Notation

The Basic Equation

Solving Equations of the Form |f(x)| = k

Equations with Absolute Value on Both Sides

Equations with Multiple Absolute Value Terms

Absolute Value and Extraneous Solutions

Geometric Interpretation

Summary of Absolute Value Equation Forms

Common Pitfalls

Absolute Value Equations FAQ

Why do vertical bars mean different things in different places?

Can you get rid of absolute value bars by squaring?

Can an absolute value equation have infinitely many solutions?

Why does ∣x−1∣=∣x−7∣|x - 1| = |x - 7|∣x−1∣=∣x−7∣ have only one solution?

Why does ∣x2−4∣=5|x^2 - 4| = 5∣x2−4∣=5 have only two solutions?

Why does $|x - 1| = |x - 7|$ have only one solution?

Why does $|x^2 - 4| = 5$ have only two solutions?