Solve an absolute value equation

Explain it like I am five

Suppose you must stand exactly eight blocks from a sign on a straight road. You can stand on its right or its left. An absolute value equation asks for every position with an exact distance. The unknown sits inside the bars, perhaps after several operations. First clear away the operations outside the bars until the distance stands alone. Then ask whether that distance is possible. Positive gives two directions. Zero means standing at the sign. Negative cannot describe a distance. In our rules, A is a nickname for everything inside the bars, and B is the number on the other side. In |2x − 6| = 8, A means 2x − 6 and B means 8.

The equation has only the two inputs where the V crosses height 5.

Reminder

Multiplication beside bars. 3|x + 2| means 3 × |x + 2|. The bars group the whole inside, so finish x + 2 before multiplying.
Equation balance. Subtract or add the same amount to both sides. Dividing an equation by a negative changes no equality sign: −2u = −8 becomes u = 4.
Fractions multiplying x. $2 over 3$ x = 4 becomes x = 4 × $3 over 2$ = 6. The reciprocal $3 over 2$ cancels $2 over 3$ because their product is 1.
Negative substitution. At x = −1, write 2(−1) − 6 = −2 − 6 = −8. Parentheses keep the negative input together.
Answer sets. {−7, 3} means the set containing −7 and 3. {4} has one answer. ∅ and { } have none.

Why it works. For a positive B, only B and −B have distance B from zero. So |A| = B has exactly the same answers as A = B or A = −B. Each of those equations is a branch, one case to solve. They concern the entire inside A, never each term separately. At B = 0, both directions meet at A = 0. At B < 0, no real number has that distance. For A = mx + c, m and c are fixed numbers and m ≠ 0. Undoing addition and multiplication gives one input per branch. Other inside expressions can have different counts.

RuleIsolate |A| = B first. For B > 0, solve A = B or A = −B. For B = 0, solve A = 0. For B < 0, there is no solution. These give two, one, or no inputs when A = mx + c and m ≠ 0. Check every input in the original equation.

The same idea, five ways

Say it

Say: the absolute value of A equals B. A names the whole inside, and B is the number on the other side.

Write it

Find every input whose whole inside has exactly the required distance from zero.

In math

|A| = B
For B > 0: A = B or A = −B
|2x − 6| = 8 gives x = −1 or x = 7
Set notation: {−1, 7}
One answer: {4}
No answers: ∅, also written { }

Find both houses exactly eight blocks from one sign. Zero distance means the sign itself; negative distance cannot occur.

See it

The two crossing inputs, −1 and 7, make |2x − 6| equal 8.

The same idea, other ways

As two directions

An exact positive distance from a landmark places you that far left or that far right. Neither direction can be discarded.

As reversing a machine

Absolute value sends both −5 and 5 to 5. Reversing it must consider both possible inputs. Then reverse the earlier addition or multiplication to recover x.

As a graph crossing

The graph y = |x + 2| crosses y = 5 twice, at −7 and 3. The line y = 0 meets the corner once. A line below zero never meets this distance graph because every output is zero or positive.

Solving the equation means finding the inputs where the two graphs have equal heights.

Why the list is complete

An inside value between −5 and 5 has distance less than 5; an inside value outside that span has distance greater than 5. Only its two boundary values can give distance exactly 5.

B, the number on the other side	Inside equation or equations	If A = mx + c and m ≠ 0
B > 0	A = B or A = −B	Two solutions
B = 0	A = 0	One solution
B < 0	No possible inside value	No solution: ∅

.1B positive

Two opposite positions give the same positive distance. Solve both equations.

For B > 0, |A| = B means A = B or A = −B.
When A = mx + c and m ≠ 0, each inside equation gives one input and the two inputs differ.
For |x| = 3, the answer set is {−3, 3}.

A positive horizontal line crosses the parent V twice.

Worked exampleAn exact three-unit distance

Solve |x| = 3.

A positive horizontal line crosses the parent V twice.

What it asks. Find every number whose distance from zero is exactly 3.

Plan. Choose the position three units left and the position three units right, then check them.

The distance is positive.
Both directions count.

x = 3 or x = −3.These are the two positions three units from zero.
Keep both possible inputs after checking |−3| = 3 and |3| = 3.An equation needs every input that makes it true.

Answer

x = −3 or x = 3
Set: {−3, 3}

Check |−3| = 3 and |3| = 3.

Counterexamples: what it is not, and the tempting wrong moves

✗ Not this: Keep only x = 3 for |x| = 3.

x = −3 also gives |−3| = 3, so one valid input would be missing.

✓ Instead: x = −3 or x = 3. Set: {−3, 3}.

Tips and tricks

Use the word or between the two input values. You are listing alternatives, not requiring one x to be both numbers.

.2B zero

You have distance zero only when you are at the center. Left and right give the same position, so do not count it twice.

|A| = 0 means A = 0.
Do not list +0 and −0 as different solutions.

The V meets height zero only at input 4.

Worked exampleOne solution

Solve |x − 4| = 0.

The V meets height zero only at input 4.

What it asks. Find the input exactly zero units from 4.

Plan. Set the whole inside to zero, solve for the center, and substitute it back.

Only an inside value of zero has absolute value zero.
Add 4 to both sides to find that input.

Set x − 4 = 0.This asks for the input where the inside vanishes, the only way the distance can be zero.
Add 4 to both sides: x = 0 + 4 = 4.This finds the center input by undoing subtracting 4.
Check |4 − 4| = |0| = 0, then write {4}.The input satisfies the original equation, and the braces list its one answer.

Answer

x = 4
Set: {4}

Check |4 − 4| = |0| = 0.

Counterexamples: what it is not, and the tempting wrong moves

✗ Not this: List x = +4 and x = −4 for |x − 4| = 0.

The ± choice belongs to the inside distance 0, and +0 = −0. At x = −4, the actual distance is |−4 − 4| = 8.

✓ Instead: Set x − 4 = 0. The one input is x = 4, with answer set {4}.

Tips and tricks

When B is zero, solve one inside equation. There is only one zero-distance position.

.3B negative

A negative required distance is impossible. You stop when isolation reveals it.

|A| ≥ 0 for every real A.
If isolation gives B < 0, no input can satisfy |A| = B. The answer set is ∅.

The graph of |3x + 1| + 5 is always at least 5, so it cannot meet height 2.

Worked exampleA distance cannot equal −3

Solve |3x + 1| + 5 = 2.

The graph of |3x + 1| + 5 is always at least 5, so it cannot meet height 2.

What it asks. Find whether any x can make |3x + 1| + 5 equal 2.

Plan. Remove the outside +5, inspect the resulting distance, and stop if it is negative.

Subtract 5 from both sides.
Distance can be zero or positive, but never negative.

Subtract 5 from both sides: |3x + 1| = 2 − 5 = −3.This isolates the bars so the required distance can be inspected.
There is no solution. Write ∅.No value inside the bars produces distance −3; the equation cannot hold for any input.

Answer

No solution
Set: ∅

Check The original left side is at least 0 + 5 = 5, larger than 2 for every x.

Counterexamples: what it is not, and the tempting wrong moves

✗ Not this: Solve 3x + 1 = −3 and keep x = − for |3x + 1| = −3.

At that input the inside is −3, but the bars return 3, not −3.

✓ Instead: Stop when the isolated other side is negative. No solution: ∅.

Tips and tricks

Inspect B after isolation. An outside negative can turn a negative starting number into a positive B.

.4The bars may be on the right

A balanced scale can be read from either side. Swap the complete sides if it helps you see the bars on the left. Then undo the outside operations in the usual order.

The left side equals the right side also means the right side equals the left side. You can swap the two complete expressions.
Swapping sides changes no signs inside either expression.
In an equation, dividing by a negative keeps equality. The inequality flip rule concerns inequality signs.

The original downward V meets output zero at inputs −11 and 1.

Worked exampleA zero on the left

Solve 0 = −|x + 5| + 6.

Both labeled inputs make the original right side zero.

What it asks. Find every input that makes the right side equal zero.

Plan. Swap sides, subtract 6, divide by −1, then solve two inside equations.

Read 0 = −|x + 5| + 6 from the right side to the left side.
Division by a negative keeps the equals sign.
The isolated distance will be 6.

Swap the complete sides: −|x + 5| + 6 = 0.Equal amounts can be written in either order without changing the equation.
Subtract 6 from both sides: −|x + 5| = −6.This removes the addition outside the bars.
Divide both sides by −1: |x + 5| = 6.This cancels the outside minus. Equality stays equality.
Write x + 5 = 6 or x + 5 = −6.The positive distance 6 can come from inside value 6 or −6.
Subtract 5 in each equation: x = 6 − 5 = 1 or x = −6 − 5 = −11.This finds the inputs that produce the two possible inside values.

Answer

x = −11 or x = 1
Set: {−11, 1}

Check At −11 the original right side is −|(−11) + 5| + 6 = −|−6| + 6 = 0. At 1 it is −|1 + 5| + 6 = −6 + 6 = 0.

Counterexamples: what it is not, and the tempting wrong moves

✗ Not this: Changing 0 = −|x + 5| + 6 to 0 = |x + 5| + 6 by moving the minus sign.

Swapping sides does not change the expression. Removing its minus changes the function and makes the equation impossible.

✓ Instead: Write −|x + 5| + 6 = 0, subtract 6, then divide by −1.

Tips and tricks

Write the whole equation in reverse order before doing algebra. Keep every sign attached to its expression.

Strategy: step by step

Get the bars and their complete contents alone. Undo addition or subtraction outside first, then multiplication or division.
Name A, everything inside the bars, and B, the number on the other side. Inspect B before splitting.
For B > 0, write A = B or A = −B. For B = 0, write only A = 0. For B < 0, stop with no solution.
Solve each possible equation by undoing the inside operations. This finds the inputs that produce the required inside values.
Substitute each possible answer into the original equation. Keep only inputs that make both sides equal.
Write the answer as inputs and as a set. {−7, 3} lists two answers. {4} lists one answer. ∅ means the empty set, no answers, and can also be written { }. Braces collect answers; a comma separates them.

Strategy

Solve an absolute value equation

Are the bars alone on one side?

YesInspect B, the number on the other side.

NoUndo outside addition or subtraction, then multiplication or division. If the bars are on the right, swap the complete sides first. Equality stays equality, even when dividing by a negative.

↓

Is B negative?

YesNo distance is negative. Stop with no solution: ∅.

NoCheck whether B is zero.

↓

Is B zero?

YesSet the whole inside A equal to 0 to find the input where the distance vanishes. Solve and check it in the original.

NoB is positive. Solve A = B and A = −B separately, check both inputs, and list the answers in braces.

Isolate the bars by undoing what happens outside them. The resulting B is the actual distance to inspect.
Check whether B is negative, zero, or positive before writing inside equations.
Solve every possible inside equation, then substitute into the original and list the complete answer set.

Worked exampleIsolate before choosing directions

Solve 3|x + 2| − 4 = 11.

The equation has only the two inputs where the V crosses height 5.

What it asks. Find every x that makes the left side equal 11, then write those inputs as a set.

Plan. Undo minus 4, then times 3 outside the bars. Solve two equations for the whole inside and check both inputs.

The outside operations are times 3, then minus 4. Undo them in reverse order.
After isolation, |x + 2| = 5 means x + 2 equals 5 or −5.
A positive distance has a left position and a right position.

Add 4 to both sides: 3|x + 2| − 4 + 4 = 11 + 4, so 3|x + 2| = 15.This removes the outside subtraction while keeping the sides equal.
Divide both sides by 3: |x + 2| = $15 over 3$ = 5.Division cancels the outside multiplication. Now B, the number on the other side, is the true distance 5.
Write x + 2 = 5 or x + 2 = −5.The complete inside must be one of the two numbers whose absolute value is 5.
In the first equation subtract 2 from both sides: x = 5 − 2 = 3.This finds the input that produces inside value 5.
In the second equation subtract 2 from both sides: x = −5 − 2 = −7.This finds the input that produces inside value −5. Both possible directions must be checked.
Substitute 3 and −7 into the original equation and keep both. Write {−7, 3}.Both give original left side 11, so both inputs belong to the answer set.

Answer

x = −7 or x = 3
Set: {−7, 3}

Check At x = 3, 3|3 + 2| − 4 = 3 × 5 − 4 = 15 − 4 = 11. At x = −7, 3|(−7) + 2| − 4 = 3|−5| − 4 = 15 − 4 = 11. The isolated picture also puts both inputs five units from −2.

Ladder: from easy to exam-hard. Press Try it first on any rung to hide its steps and use them as hints.

Rung 1Rung 1: bare bars

Solve |x| = 6.

The V meets height 6 at inputs −6 and 6.

What it asks. Find every number exactly six units from zero.

Plan. Use both directions from zero and check both.

The bars are already alone.
A positive B gives two inside values.

The bars are isolated and B = 6 > 0.There are no outside operations to undo.
x = 6 or x = −6.Both directions have distance six from zero.

Answer

x = −6 or x = 6
Set: {−6, 6}

Check |−6| = |6| = 6.

Rung 2Rung 2: undo an inside shift

Solve |x − 2| = 3.

The two crossing inputs lie three units from center 2.

What it asks. Find every number exactly three units from 2.

Plan. Split the inside into 3 or −3, then add 2 and check both inputs.

Keep x − 2 together as the whole inside.
The center is 2.

Write x − 2 = 3 or x − 2 = −3.A positive distance 3 can come from either whole inside value.
Add 2 in the first equation: x = 3 + 2 = 5.This finds the input three units right of 2.
Add 2 in the second equation: x = −3 + 2 = −1.This finds the input three units left of 2.

Answer

x = −1 or x = 5
Set: {−1, 5}

Check |−1 − 2| = 3 and |5 − 2| = 3, each three units from 2.

Rung 3Rung 3: subtraction from a number

Solve |7 − x| = 3.

Reversing the inside difference keeps this distance graph centered at 7.

What it asks. Find every input whose distance from 7 is exactly 3.

Plan. Set the whole inside equal to 3 and −3, then undo subtraction of x in each equation.

Keep 7 − x together.
|7 − x| = |x − 7| because reversing a difference keeps its distance.

Write 7 − x = 3 or 7 − x = −3.The whole inside has to be one of the two values whose absolute value is 3.
In the first equation subtract 7: −x = 3 − 7 = −4. Divide by −1: x = 4.This finds the input that makes the inside 3. Division keeps equality.
In the second equation subtract 7: −x = −3 − 7 = −10. Divide by −1: x = 10.This finds the input that makes the inside −3.

Answer

x = 4 or x = 10
Set: {4, 10}

Check |7 − 4| = |3| = 3 and |7 − 10| = |−3| = 3. Both inputs are three units from 7.

Rung 4Rung 4: a number multiplying x inside

Solve |2x − 6| = 8.

The graph represents |2x − 6|, and its two crossing inputs give the equation answers.

What it asks. Find every x whose whole inside 2x − 6 has distance 8 from zero.

Plan. Write the inside equal to 8 and −8, then solve and check each equation.

A is 2x − 6, not x alone.
Add 6, then divide by 2, in both branches.

Write 2x − 6 = 8 or 2x − 6 = −8.B = 8 > 0, so the complete inside has two possible values.
In the first equation add 6 to both sides: 2x = 8 + 6 = 14.This removes the inside subtraction while keeping both sides equal.
Divide by 2: x = $14 over 2$ = 7.This finds the input that makes the inside 8.
In the second equation add 6: 2x = −8 + 6 = −2.This removes the same subtraction for the other possible direction.
Divide by 2: x = $−2 over 2$ = −1.This finds the input that makes the inside −8.

Answer

x = −1 or x = 7
Set: {−1, 7}

Check |2 × 7 − 6| = |8| = 8 and |2 × (−1) − 6| = |−8| = 8.

Rung 5Rung 5: isolate two outside operations

Solve 2|x − 1| + 3 = 11.

The original left side 2|x − 1| + 3 reaches height 11 at inputs −3 and 5.

What it asks. Find every x that makes twice the distance from 1, plus 3, equal 11.

Plan. Subtract 3 and divide by 2 before choosing the two inside values.

Undo the outside operations in reverse order.
After isolation, B is 4.

Subtract 3 from both sides: 2|x − 1| = 11 − 3 = 8.This removes the outside addition.
Divide both sides by 2: |x − 1| = $8 over 2$ = 4.The bars now stand alone, revealing the required distance.
Write x − 1 = 4 or x − 1 = −4.The whole inside must have distance 4 from zero.
Add 1 in the first equation: x = 4 + 1 = 5.This finds the input on the right.
Add 1 in the second equation: x = −4 + 1 = −3.This finds the input on the left.

Answer

x = −3 or x = 5
Set: {−3, 5}

Check At −3, 2|(−3) − 1| + 3 = 2|−4| + 3 = 11. At 5, 2|5 − 1| + 3 = 2 × 4 + 3 = 11. Both are four units from 1.

Rung 6Rung 6: compare zero and negative B

Solve |2x − 8| = 0 and |2x − 8| = −1.

The zero-height line meets the V only at input 4.

The negative-height line never meets the distance graph.

What it asks. Solve two separate equations. First find zero distance; then decide whether negative distance is possible.

Plan. For B = 0 solve one inside equation. For B = −1 stop with no solution.

Zero is a possible distance.
Negative one is not a possible distance.

For |2x − 8| = 0, set 2x − 8 = 0.This finds the input where the inside vanishes, the only way to have distance zero.
Add 8 to both sides: 2x = 8.This removes the inside subtraction.
Divide by 2: x = 4. Check |2(4) − 8| = |0| = 0.This isolates the input and confirms it in the original zero-distance equation.
For |2x − 8| = −1, there is no solution: ∅.Every distance is at least zero, so no input gives −1.

Answer

|2x − 8| = 0: x = 4; set {4}
|2x − 8| = −1: no solution; set ∅

Check At x = 4 the inside is zero. Every other input has nonnegative absolute value too, so none gives −1.

Rung 7Rung 7: a fraction multiplying x inside

Solve | $2 over 3$ x − 1| = 3.

The two crossing inputs give |x − 1| = 3.

What it asks. Find every x that makes the entire inside exactly three units from zero.

Plan. Set the inside equal to 3 and −3. Add 1, then multiply by the reciprocal in each equation.

$2 over 3$ x means $2 over 3$ × x.
Multiplying $2 over 3$ by $3 over 2$ gives 1.
Keep exact fractions until the calculation finishes.

Write $2 over 3$ x − 1 = 3 or $2 over 3$ x − 1 = −3.The positive B = 3 allows two whole inside values.
In the first equation add 1: $2 over 3$ x = 4.This removes the inside subtraction.
Multiply both sides by $3 over 2$ : x = 4 × $3 over 2$ = $12 over 2$ = 6.The reciprocal cancels the multiplier of x and finds the first possible input.
In the second equation add 1: $2 over 3$ x = −2.This removes the same subtraction in the other branch.
Multiply both sides by $3 over 2$ : x = −2 × $3 over 2$ = − $6 over 2$ = −3.The reciprocal finds the input that produces the negative inside value.

Answer

x = −3 or x = 6
Set: {−3, 6}

Check At 6 the inside is × 6 − 1 = 4 − 1 = 3. At −3 it is × (−3) − 1 = −2 − 1 = −3. Their absolute values both equal 3.

Rung 8Rung 8: a negative outside multiplier and fractional answer

Solve −2|3x − 1| + 5 = −3.

The isolated graph |3x − 1| meets height 4 at −1 and .

What it asks. Find every x that makes the whole expression equal −3.

Plan. Remove +5, divide by −2, then solve the two inside equations and check the fractional input exactly.

An equation keeps its equals sign when divided by a negative.
The isolated distance becomes positive 4.
The coefficient is the number multiplying a letter or the bars. Here −2 multiplies the bars and 3 multiplies x.

Subtract 5 from both sides: −2|3x − 1| = −3 − 5 = −8.This removes the outside addition before inspecting the distance.
Divide both sides by −2: |3x − 1| = $−8 over −2$ = 4.This cancels the outside multiplier. Equality stays equality; no inequality sign is present to flip.
Write 3x − 1 = 4 or 3x − 1 = −4.The positive B = 4 allows two whole inside values.
In the first equation add 1: 3x = 5. Divide by 3: x = $5 over 3$ .These moves find the input that produces inside value 4. The fraction is exact.
In the second equation add 1: 3x = −3. Divide by 3: x = −1.These moves find the input that produces inside value −4.

Answer

x = −1 or x = $5 over 3$
Set: {−1, $5 over 3$ }

Check At , 3 × − 1 = 5 − 1 = 4. At −1, 3(−1) − 1 = −3 − 1 = −4. Both give original left side −2 × 4 + 5 = −8 + 5 = −3.

Counterexamples: what it is not, and the tempting wrong moves

✗ Not this: Split 3|x + 2| − 4 = 11 into 3(x + 2) − 4 = 11 or 3(x + 2) − 4 = −11.

The second equation gives 3(x + 2) = −7, then x = − − 2 = −. Substitution gives 3|− + 2| − 4 = 3 × − 4 = 3, not 11. The outside arithmetic changed the required distance.

✓ Instead: First add 4 and divide by 3 to get |x + 2| = 5. Then split x + 2 = 5 or x + 2 = −5, giving {−7, 3}.

✗ Not this: |x| = −5 gives x = −5.

Substitution gives |−5| = 5, which does not equal −5.

✓ Instead: The equation has no solution. Its answer set is ∅.

✗ Not this: Write {∅} for no answers.

The braces would describe a set containing the empty set as an object. They would not describe an empty collection of numerical answers.

✓ Instead: Write ∅ or { } for no answers. Write {4} when the one numerical answer is 4.

Tips and tricks

Box the isolated |A| = B before choosing cases. Inspect B there, rather than the number in the starting equation.
Two answers: list both in braces, smaller first. One answer: use one number in braces. No answers: use ∅.
The textbook may write the second case as −A = B. Multiplying both sides by −1 gives A = −B, so the two versions have the same answers.
Memory cue: alone, then two roads. First isolate the bars; take both roads only when B is positive.

Trap. Splitting before isolating, or keeping only the positive branch. In 3|x + 2| − 4 = 11, the needed distance is 5, not 11. Box the isolated equation, then choose its branches.