The Central Idea

Consider

\[ |x|=5. \]

Absolute value measures distance from zero, so the equation asks:

Which real numbers are 5 units from zero?

There are two:

\[ x=5 \]

and

\[ x=-5. \]

So

\[ |x|=5 \quad\Longleftrightarrow\quad x=5 \ \text{or}\ x=-5. \]

Absolute-value equations often create multiple branches because distance does not record direction.

Absolute Value as Distance

For a real number \(x\),

\[ |x| \]

is the distance from \(x\) to \(0\).

More generally,

\[ |x-a| \]

is the distance from \(x\) to \(a\).

For example,

\[ |x-3|=4 \]

asks for the points 4 units from \(3\).

Those points are

\[ 3+4=7 \]

and

\[ 3-4=-1. \]

Therefore,

\[ x=7 \]

or

\[ x=-1. \]

The Basic Two-Case Rule

If

\[ |A|=k \]

with

\[ k>0, \]

then

\[ A=k \]

or

\[ A=-k. \]

Symbolically,

\[ |A|=k \quad\Longleftrightarrow\quad A=k \ \text{or}\ A=-k, \qquad k>0. \]

This is the fundamental solving rule for absolute-value equations.

Why There Are Two Cases

If

\[ |A|=5, \]

then \(A\) could be \(5\) because

\[ |5|=5, \]

or \(A\) could be \(-5\) because

\[ |-5|=5. \]

Absolute value removes sign information, so solving must restore both sign possibilities.

When the Right Side Is Zero

If

\[ |A|=0, \]

then the only possibility is

\[ A=0. \]

For example,

\[ |x-7|=0 \]

gives

\[ x-7=0, \]

so

\[ x=7. \]

An absolute-value equation with right side zero has at most one real solution.

When the Right Side Is Negative

Absolute value is never negative.

Therefore, an equation such as

\[ |x+2|=-3 \]

has no real solution.

Its solution set is

\[ \varnothing. \]

This is often the fastest case to recognize: if an isolated absolute value equals a negative number, stop immediately.

Three Basic Outcomes

Equation Outcome
\(|A|=k,\ k>0\) Usually two branches: \(A=k\) or \(A=-k\)
\(|A|=0\) One branch: \(A=0\)
\(|A|=k,\ k<0\) No real solution

One-Step Absolute-Value Equation

Solve

\[ |x|=8. \]

Split into two cases:

\[ x=8 \]

or

\[ x=-8. \]

Therefore, the solution set is

\[ \{-8,8\}. \]

Absolute Value Around a Linear Expression

Solve

\[ |x-4|=7. \]

Case 1:

\[ x-4=7. \]

Then

\[ x=11. \]

Case 2:

\[ x-4=-7. \]

Then

\[ x=-3. \]

So

\[ \boxed{x=-3,\ 11}. \]

Interpretation on the Number Line

The equation

\[ |x-4|=7 \]

asks for points whose distance from \(4\) is \(7\).

Move 7 units right:

\[ 4+7=11. \]

Move 7 units left:

\[ 4-7=-3. \]

The algebraic two-case rule and the geometric distance interpretation describe the same relationship.

Isolate the Absolute Value First

Consider

\[ 2|x-1|=10. \]

Before splitting into cases, isolate the absolute-value expression.

Divide by \(2\):

\[ |x-1|=5. \]

Now split:

\[ x-1=5 \]

or

\[ x-1=-5. \]

So

\[ x=6 \]

or

\[ x=-4. \]

Multistep Absolute-Value Equation

Solve

\[ 3|2x+1|-4=11. \]

Add \(4\):

\[ 3|2x+1|=15. \]

Divide by \(3\):

\[ |2x+1|=5. \]

Now split into two cases.

Case 1:

\[ 2x+1=5, \]

so

\[ 2x=4, \] \[ x=2. \]

Case 2:

\[ 2x+1=-5, \]

so

\[ 2x=-6, \] \[ x=-3. \]

Therefore,

\[ \boxed{x=-3,\ 2}. \]

Do Not Split Too Early

Suppose

\[ 4|x+2|+3=19. \]

The absolute value is not yet isolated.

First subtract \(3\):

\[ 4|x+2|=16. \]

Then divide by \(4\):

\[ |x+2|=4. \]

Only now split into the two cases.

This keeps the case structure simple and reduces unnecessary work.

Example with a Negative Coefficient Outside

Solve

\[ -2|x-3|=-10. \]

Divide by \(-2\):

\[ |x-3|=5. \]

Then

\[ x-3=5 \]

or

\[ x-3=-5. \]

So

\[ x=8 \]

or

\[ x=-2. \]

Detect No-Solution Cases Early

Consider

\[ 2|x+1|+7=3. \]

Subtract \(7\):

\[ 2|x+1|=-4. \]

Divide by \(2\):

\[ |x+1|=-2. \]

This is impossible over the real numbers.

Therefore,

\[ \boxed{\varnothing}. \]

A Shortcut Using Minimum Values

Because

\[ |x+1|\ge0, \]

we know

\[ 2|x+1|+7\ge7. \]

So the expression can never equal \(3\).

Recognizing the range of an absolute-value expression can sometimes classify an equation before any detailed algebra is needed.

Absolute Value with Variables on Both Sides

Consider

\[ |x-2|=x. \]

Because the left side is nonnegative, any solution must satisfy

\[ x\ge0. \]

Now split into cases:

Case 1:

\[ x-2=x. \]

This gives

\[ -2=0, \]

so there is no solution from this branch.

Case 2:

\[ x-2=-x. \]

Then

\[ 2x=2, \] \[ x=1. \]

Check:

\[ |1-2|=1. \]

Therefore,

\[ \boxed{x=1}. \]

Why Verification Matters

When absolute value is isolated and the two-case rule is applied correctly, the branches are logically equivalent to the original equation.

Even so, checking remains useful because it can catch:

Verification becomes especially important when absolute value appears alongside fractions, radicals, or other restricted operations.

Piecewise Definition of Absolute Value

Absolute value can be defined piecewise:

\[ |u|= \begin{cases} u, & u\ge0,\\ -u, & u<0. \end{cases} \]

This explains why solving absolute-value equations naturally involves cases.

For example,

\[ |x-3| \]

means

\[ x-3 \]

when \(x\ge3\), and

\[ -(x-3) \]

when \(x<3\).

Solving by Piecewise Reasoning

Solve

\[ |x-3|=5. \]

If

\[ x\ge3, \]

then

\[ x-3=5, \]

so

\[ x=8. \]

This satisfies \(x\ge3\).

If

\[ x<3, \]

then

\[ -(x-3)=5. \]

So

\[ -x+3=5, \] \[ x=-2. \]

This satisfies \(x<3\).

The piecewise method produces the same two solutions as the basic two-case rule.

Absolute Value as a Function

The graph of

\[ y=|x| \]

is a V-shape with vertex at

\[ (0,0). \]

Solving

\[ |x|=5 \]

means finding where the graph

\[ y=|x| \]

intersects the horizontal line

\[ y=5. \]

There are two intersection points:

\[ (-5,5) \]

and

\[ (5,5). \]

The algebraic two-solution pattern has a direct graphical interpretation.

Shifted Absolute-Value Graphs

The graph of

\[ y=|x-h| \]

has vertex at

\[ (h,0). \]

So the equation

\[ |x-h|=k \]

with \(k>0\) asks for the two points at horizontal distance \(k\) from \(h\):

\[ x=h-k \]

and

\[ x=h+k. \]

This gives the compact solution rule

\[ |x-h|=k \quad\Longleftrightarrow\quad x=h\pm k. \]

Equations with Absolute Value on Both Sides

Consider

\[ |x-1|=|x+3|. \]

This asks for points that are equally distant from \(1\) and \(-3\).

Geometrically, the answer is the midpoint:

\[ x=-1. \]

Algebraically, square both sides:

\[ (x-1)^2=(x+3)^2. \]

Expand:

\[ x^2-2x+1=x^2+6x+9. \]

Then

\[ -8x=8, \]

so

\[ x=-1. \]

Because both sides are already nonnegative, squaring preserves equivalence here.

A Faster Rule for Equal Absolute Values

For real expressions \(A\) and \(B\),

\[ |A|=|B| \]

is equivalent to

\[ A=B \]

or

\[ A=-B. \]

For example,

\[ |2x-1|=|x+5| \]

gives two cases:

\[ 2x-1=x+5 \]

or

\[ 2x-1=-(x+5). \]

This preserves both possible sign relationships.

Worked Equal-Absolute-Value Example

Solve

\[ |2x-1|=|x+5|. \]

Case 1:

\[ 2x-1=x+5. \]

Then

\[ x=6. \]

Case 2:

\[ 2x-1=-x-5. \]

Then

\[ 3x=-4, \]

so

\[ x=-\frac{4}{3}. \]

Therefore,

\[ \boxed{x=6,\ -\frac{4}{3}}. \]

Absolute Value and Squaring

For real \(A\) and \(k\ge0\),

\[ |A|=k \]

is equivalent to

\[ A^2=k^2. \]

For example,

\[ |x-4|=3 \]

can be squared:

\[ (x-4)^2=9. \]

Then

\[ x-4=\pm3. \]

This is valid, but the direct two-case rule is usually simpler.

Absolute Value and Distance Between Two Quantities

The expression

\[ |a-b| \]

is the distance between \(a\) and \(b\).

Therefore,

\[ |x-10|=3 \]

means that the distance between \(x\) and \(10\) is \(3\).

The solutions are

\[ x=7 \]

and

\[ x=13. \]

This interpretation is useful in measurement, tolerance, error bounds, and geometry.

Application: Measurement Tolerance

Suppose a machine part is designed to have length \(50\) mm, and the actual length \(x\) differs from the target by exactly \(0.2\) mm.

This is modeled by

\[ |x-50|=0.2. \]

So

\[ x-50=0.2 \]

or

\[ x-50=-0.2. \]

Therefore,

\[ x=50.2 \]

or

\[ x=49.8. \]

Application: Equal Distance

Suppose a point \(x\) on a number line is equally distant from \(-4\) and \(10\).

We can write

\[ |x+4|=|x-10|. \]

The point equidistant from two locations is their midpoint:

\[ x=\frac{-4+10}{2}=3. \]

So

\[ \boxed{x=3}. \]

Worked Example: Isolate First

Solve

\[ 5|x+2|-10=20. \]

Add \(10\):

\[ 5|x+2|=30. \]

Divide by \(5\):

\[ |x+2|=6. \]

Case 1:

\[ x+2=6, \]

so

\[ x=4. \]

Case 2:

\[ x+2=-6, \]

so

\[ x=-8. \]

Therefore,

\[ \boxed{x=-8,\ 4}. \]

Worked Example: No Real Solution

Solve

\[ 3|2x-5|+1=-8. \]

Subtract \(1\):

\[ 3|2x-5|=-9. \]

Divide by \(3\):

\[ |2x-5|=-3. \]

An absolute value cannot equal a negative real number.

Therefore,

\[ \boxed{\varnothing}. \]

Worked Example: One Solution

Solve

\[ 4|3x+6|=0. \]

Divide by \(4\):

\[ |3x+6|=0. \]

So

\[ 3x+6=0. \]

Then

\[ 3x=-6, \]

and

\[ x=-2. \]

There is only one solution because the required distance is zero.

Worked Example: Variable on the Right

Solve

\[ |2x-3|=x+1. \]

The right side must be nonnegative:

\[ x+1\ge0, \]

so

\[ x\ge-1. \]

Case 1:

\[ 2x-3=x+1. \]

Then

\[ x=4. \]

Case 2:

\[ 2x-3=-(x+1). \]

Then

\[ 2x-3=-x-1, \] \[ 3x=2, \] \[ x=\frac{2}{3}. \]

Both candidates satisfy \(x\ge-1\), and both check in the original equation.

Therefore,

\[ \boxed{x=4,\ \frac{2}{3}}. \]

A Reliable Solving Strategy

  1. Isolate the absolute-value expression.
  2. Inspect the value on the other side.
  3. If it is negative, conclude no real solution.
  4. If it is zero, set the inside equal to zero.
  5. If it is positive, create the positive and negative cases.
  6. Solve each branch.
  7. Apply any domain or range restrictions.
  8. Check candidates in the original equation.

Common Errors

Dropping the Absolute-Value Bars

From

\[ |x|=5, \]

writing only

\[ x=5 \]

loses the valid solution \(x=-5\).

Forgetting to Isolate First

For

\[ 2|x-3|+1=9, \]

do not immediately write \(x-3=\pm9\). First isolate the absolute value.

Creating Two Cases When the Right Side Is Zero

If

\[ |A|=0, \]

both “cases” are the same:

\[ A=0. \]

Trying to Solve an Impossible Negative Case

If

\[ |A|=-4, \]

there is no real solution.

Forgetting Parentheses in the Negative Branch

From

\[ |2x-1|=x+5, \]

the negative branch is

\[ 2x-1=-(x+5), \]

not

\[ 2x-1=-x+5. \]

Ignoring Restrictions on the Other Side

If

\[ |A|=B, \]

then every solution must satisfy

\[ B\ge0. \]

Practice

1. Basic Absolute Value

Solve:

\[ |x|=6. \]
Check Answer \[ x=6 \]

or

\[ x=-6. \]

2. Shifted Absolute Value

Solve:

\[ |x-5|=2. \]
Check Answer \[ x-5=2 \]

or

\[ x-5=-2. \]

So

\[ x=7 \]

or

\[ x=3. \]

3. Isolate First

Solve:

\[ 3|x+1|=12. \]
Check Answer \[ |x+1|=4. \]

So

\[ x+1=4 \]

or

\[ x+1=-4. \]

Therefore,

\[ x=3 \]

or

\[ x=-5. \]

4. No Solution

Solve:

\[ |2x-7|=-1. \]
Check Answer

No real solution:

\[ \varnothing. \]

5. One Solution

Solve:

\[ |4x+8|=0. \]
Check Answer \[ 4x+8=0, \]

so

\[ x=-2. \]

6. Multistep

Solve:

\[ 2|3x-1|+4=14. \]
Check Answer \[ 2|3x-1|=10, \] \[ |3x-1|=5. \]

So

\[ 3x-1=5 \]

or

\[ 3x-1=-5. \]

Therefore,

\[ x=2 \]

or

\[ x=-\frac{4}{3}. \]

7. Absolute Values on Both Sides

Solve:

\[ |x-2|=|x+6|. \]
Check Answer

The solution is the midpoint of \(2\) and \(-6\):

\[ x=\frac{2+(-6)}{2}=-2. \]

Key Ideas

Where This Leads

Absolute-value equations introduce a major new solving pattern: one relationship can naturally split into multiple valid cases.

This case-based reasoning will become even more important in absolute-value inequalities, piecewise functions, distance constraints, and equations where different domains require different algebraic rules.