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:
- arithmetic mistakes;
- sign mistakes;
- incorrect case setup;
- domain restrictions from other expressions in the equation.
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
- Isolate the absolute-value expression.
- Inspect the value on the other side.
- If it is negative, conclude no real solution.
- If it is zero, set the inside equal to zero.
- If it is positive, create the positive and negative cases.
- Solve each branch.
- Apply any domain or range restrictions.
- 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
- Absolute value measures distance and is therefore always nonnegative.
- \(|A|=k\) with \(k>0\) produces two branches: \(A=k\) or \(A=-k\).
- \(|A|=0\) produces one branch: \(A=0\).
- \(|A|=k\) with \(k<0\) has no real solution.
- Isolate the absolute-value expression before splitting into cases.
- The two-case rule comes directly from the piecewise definition of absolute value.
- \(|x-a|\) represents the distance between \(x\) and \(a\).
- Equal absolute values represent equal distances and can often be solved by case splitting or midpoint reasoning.
- If \(|A|=B\), every solution must satisfy \(B\ge0\).
- Parentheses are essential when forming a negative branch.
- Verification is useful for catching sign errors and enforcing other restrictions.
- Absolute-value equations connect algebraic solving, piecewise functions, geometry, and distance on the number line.
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.