Absolute-value equations solved by splitting them into two related linear equations, one for each sign the inside expression could take. Includes isolating the absolute value first, recognising when the right side is negative and there is no solution, checking both answers, and building an equation from two given solutions using their midpoint and distance.
Subject: Algebra 1 · 65 slides · symbolic lesson
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Title
Algebra 1 · Chapter 6 — Solving and Graphing Linear Inequalities
Solving Absolute-Value Equations
Objectives
Five outcomes, each one you can test yourself on with a pencil and no answer key.
McDougal Littell Algebra 1: Concepts and Skills, Ch. 6 Solving and Graphing Linear Inequalities — Lesson 6.6 Solving Absolute-Value Equations §6.6, pp. 355-360 — the lesson these objectives are drawn from
Warm-up
Lesson 2.2 defined absolute value as distance from zero. This lesson asks which numbers are a given distance away.
Discussion prompt
Which numbers have an absolute value of 8? Mark them on a number line and say why there is more than one.
Hint: Distance is measured in both directions.
Answer:
\[ |x| = 8 \;\Longrightarrow\; x = 8 \;\text{ or }\; x = -8 \]
Both eight and negative eight are eight units from zero, so both qualify. A distance says nothing about direction, which is precisely why an absolute-value equation splits into two.
Concept
An absolute-value equation has the form the absolute value of ax plus b equals c. For c at least zero it is solved by solving two related linear equations, one with c on the right and one with negative c. For c negative it has no solution.
absolute-value equation — An equation containing an absolute-value expression. It is solved by splitting into two linear equations, one for each sign the inside expression could take.
Absolute value always indicates a number that is not negative.
Figure (svg): An absolute-value equation split into its two related equations
McDougal Littell Algebra 1: Concepts and Skills, Ch. 6 Solving and Graphing Linear Inequalities — Lesson 6.6 Solving Absolute-Value Equations §6.6, pp. 355-355
Section
Section 1
Concept
Absolute value measures distance from zero, and every positive distance is reached by two numbers — one on each side. That is why an absolute-value equation normally has two answers.
\[ |x| = 8 \;\Longleftrightarrow\; x = 8 \;\text{ or }\; x = -8 \]
The word or here is exactly the connector from Lesson 6.5.
Figure (svg): Two numbers the same distance from zero on a number line
McDougal Littell Algebra 1: Concepts and Skills, Ch. 6 Solving and Graphing Linear Inequalities — Lesson 6.6 Solving Absolute-Value Equations §6.6, pp. 355-355 — Example 1, Solve an Absolute-Value Equation
Picture it
Two arrows of equal length.
Figure (svg): Two numbers the same distance from zero on a number line
The two answers are symmetric about zero, and they will be symmetric about a different centre once the expression inside the bars is more complicated than a bare x.
Worked example
This is Example 1 from the textbook, both parts.
\[ \text{Solve } \; |x| = 8 \; \text{ and } \; |x| = -10. \]
Take the first
Why: Two numbers are eight units from zero.
\[ x = 8\text{ or } x = -8 \]
State the answer
Why: The equation has two solutions.
\[ 8\text{ and } -8 \]
Take the second
Why: The right side is negative.
\[ | x | = -10 \]
State the answer
Why: A distance is never negative, so no number works.
Figure (svg): Two numbers the same distance from zero on a number line
\[ x = \pm 8 \qquad |x| = -10: \text{ no solution} \]
Verify: substitute both answers into the first equation
Why: The absolute value of eight is eight and the absolute value of negative eight is also eight, so both check. The second equation cannot be checked because there is nothing to substitute — and that absence is the answer.
McDougal Littell Algebra 1: Concepts and Skills, Ch. 6 Solving and Graphing Linear Inequalities — Lesson 6.6 Solving Absolute-Value Equations §6.6, pp. 355-355
Sorting
Look at the number on the right.
Sort into buckets
Sort each equation by its number of solutions.
The number on the right decides the count before any solving begins, which makes it worth checking first — especially for the negative case, where all the work would be wasted.
Worked example
Guided Practice 1 to 3. Two, one and none.
\[ \text{Solve } \; |x| = 6, \quad |x| = 0, \quad |x| = -6. \]
Take six
Why: Two numbers are six from zero.
\[ 6\text{ and } -6 \]
Take zero
Why: Only zero is zero units from zero.
\[ 0\text{ only} \]
Take negative six
Why: No distance is negative.
Note the pattern
Why: The right side decides how many answers there are.
Figure (svg): Three columns of how many solutions an absolute-value equation can have
\[ \pm 6; \quad 0; \quad \text{no solution} \]
Verify: explain why zero is the boundary case
Why: Zero is the only number that is its own negative, so the two branches x equals zero and x equals negative zero give the same answer. Every positive right side gives two distinct branches and every negative one gives none, which makes zero the exact dividing line.
McDougal Littell Algebra 1: Concepts and Skills, Ch. 6 Solving and Graphing Linear Inequalities — Lesson 6.6 Solving Absolute-Value Equations §6.6, pp. 355-355
Trap
\[ |x| = 8 \]
Answer x = 8
Why: Eight is the number visible in the equation, and absolute value produces positive results, so a positive answer feels right.
Negative eight is also eight units from zero, so it is equally a solution. Half the answer has been dropped, and it is always the same half.
\[ x = 8 \;\text{ or }\; x = -8 \]
Ask which numbers are that far from zero, in both directions
Why: The bars measure distance, and a distance is achieved on either side.
Substituting the negative answer is what makes it believable: the absolute value of negative eight really is eight.
Faded example
A distance is reached two ways.
Fill in the blanks
|x| = 8 \;\Longrightarrow\; x = 8 \;\text-8\; x = ___
Why: Both eight and negative eight are eight units from zero, so both satisfy the equation. Reporting only the positive one is the standard omission and it always loses exactly the same half.
Elimination
The right side is negative.
Eliminate the wrong options
Which reason is correct?
Survives elimination: A
Why: The bars always produce a value that is zero or positive, so they can never equal a negative number. Option B is worth naming because it confuses the sign of x with the sign of the whole expression — negative values of x are entirely welcome inside the bars.
Socratic
It sits between two answers and none.
Discussion prompt
Explain why an absolute-value equation with zero on the right has exactly one solution rather than two or none. Then say what this means about the two branches in that case.
Hint: Ask what number is its own negative.
Answer:
The two branches are the inside expression equal to zero and equal to negative zero, and those are the same equation, since zero is the only number equal to its own negative. So the two branches collapse into one and produce a single answer.
Geometrically, only one number is zero units from the centre — the centre itself. As the right side grows from zero the two answers separate symmetrically, and as it drops below zero they vanish. Zero is exactly where the two answers meet before disappearing.
Section
Section 2
Concept
If the absolute value of an expression equals c, then the expression itself is c or negative c. Writing both cases and solving each gives the two solutions.
\[ |ax + b| = c \;\Longleftrightarrow\; ax + b = c \;\text{ or }\; ax + b = -c \]
Each case is an ordinary linear equation from Chapter 3.
Figure (svg): An absolute-value equation split into its two related equations
McDougal Littell Algebra 1: Concepts and Skills, Ch. 6 Solving and Graphing Linear Inequalities — Lesson 6.6 Solving Absolute-Value Equations §6.6, pp. 356-356 — Example 2, Solve an Absolute-Value Equation
Picture it
Two equations, one for each sign.
Figure (svg): An absolute-value equation split into its two related equations
The two branches differ only in the sign on the right-hand side. Everything inside the bars is copied unchanged into both.
Worked example
This is Example 2 from the textbook.
\[ \text{Solve } \; |x - 2| = 5. \]
Write the positive case
Why: The inside expression equals five.
\[ x - 2 = 5 \]
Solve it
Why: Add two to each side.
\[ x = 7 \]
Write the negative case
Why: The inside expression equals negative five.
\[ x - 2 = -5 \]
Solve it
Why: Add two to each side.
\[ x = -3 \]
Figure (svg): An absolute-value equation split into its two related equations
\[ x = 7 \;\text{ or }\; x = -3 \]
Verify: check both in the original
Why: Seven minus two is five, whose absolute value is five. Negative three minus two is negative five, whose absolute value is also five. The second check is the one that demonstrates why the negative branch exists.
McDougal Littell Algebra 1: Concepts and Skills, Ch. 6 Solving and Graphing Linear Inequalities — Lesson 6.6 Solving Absolute-Value Equations §6.6, pp. 356-356
Translation
The inside is copied; only the right side changes sign.
Match the pairs
Why: In every case the expression inside the bars is written twice, unchanged, and only the number on the right differs in sign. Negating anything inside the bars is the standard error and it produces an answer that fails the original.
Worked example
Guided Practice 4 to 6. Different insides, same method.
\[ \text{Solve } \; |x + 3| = 5, \quad |x - 3| = 5, \quad |4x - 2| = 6. \]
Take the first
Why: x plus three is five or negative five.
\[ x = 2\text{ or } x = -8 \]
Take the second
Why: x minus three is five or negative five.
\[ x = 8\text{ or } x = -2 \]
Take the third
Why: 4x minus two is six or negative six.
\[ 4 x = 8\text{ or } 4 x = -4 \]
Finish the third
Why: Divide each by four.
\[ x = 2\text{ or } x = -1 \]
Figure (svg): The solution to Worked example three from guided practice shown as a ladder of expressions, one row per algebraic move
\[ \{2, -8\}, \quad \{8, -2\}, \quad \{2, -1\} \]
Verify: find the midpoint of each pair
Why: The pairs average to negative three, three and one half. Each of those is the value making the inside expression zero, which is the centre the two answers are symmetric about — a pattern the last section of this lesson will use in reverse.
McDougal Littell Algebra 1: Concepts and Skills, Ch. 6 Solving and Graphing Linear Inequalities — Lesson 6.6 Solving Absolute-Value Equations §6.6, pp. 356-356
Error analysis
The student solved an absolute-value equation by splitting it.
Annotate
On: \( \begin{aligned} |x - 2| &= 5 \\ x - 2 &= 5 \;\text{ or }\; -x - 2 = 5 \\ x &= 7 \;\text{ or }\; x = -7 \end{aligned} \)
The inside expression is copied unchanged into both branches, and only the number on the right changes sign. Checking the second answer catches this in one line.
Faded example
Negate the right side, not the inside.
Fill in the blanks
|x - 2| = 5 \;\Longrightarrow\; x - 2 = 5 \;\text-5\; x - 2 = -3 \;\Longrightarrow\; x = 7 \;\text___\; x = ___
Why: The negative case sets the inside equal to negative five, giving x equal to negative three. Substituting it gives negative five inside the bars, whose absolute value is five — which is exactly why the branch is there.
Elimination
The equation is the absolute value of 3x plus 1, equals 7.
Eliminate the wrong options
Which split is correct?
Survives elimination: A
Why: The inside is copied and the right side is negated, giving solutions of two and negative eight thirds. Option D is the instructive one: it looks like a second case and simplifies to the first, so it would produce only one of the two answers.
Socratic
The two branches are joined by a word from Lesson 6.5.
Discussion prompt
Explain why the two related equations are joined by or rather than and. Then say what an and version would mean and whether any number could satisfy it.
Hint: Ask what a solution has to do.
Answer:
A number solves the original if the inside expression is five, or if it is negative five — either one produces an absolute value of five. So satisfying one branch is enough, which is exactly what or means.
An and version would require the inside expression to be both five and negative five at once, which no value achieves. So an and reading would give no solutions at all, and the two answers we do find each satisfy exactly one branch and fail the other — which is the clearest possible sign that or is the right connector.
Section
Section 3
Concept
When something is added to or multiplied by the absolute-value expression, undo it first. The equation can only be split once the bars are alone on one side.
Splitting first would attach the outside term to only one branch.
Figure (svg): An absolute-value expression isolated before the equation is split
McDougal Littell Algebra 1: Concepts and Skills, Ch. 6 Solving and Graphing Linear Inequalities — Lesson 6.6 Solving Absolute-Value Equations §6.6, pp. 356-356 — Example 3, which isolates the absolute value before splitting
Picture it
One subtraction, then two branches.
Figure (svg): An absolute-value expression isolated before the equation is split
The plus five belongs to the whole equation rather than to the inside of the bars, so it has to be dealt with before the bars are opened.
Worked example
This is Example 3 from the textbook.
\[ \text{Solve } \; |2x - 7| + 5 = 14. \]
Isolate the absolute value
Why: Subtract five from each side.
\[ | 2 x - 7 | = 9 \]
Write the positive case
Why: 2x minus seven equals nine.
\[ 2 x = 16, x = 8 \]
Write the negative case
Why: 2x minus seven equals negative nine.
\[ 2 x = -2, x = -1 \]
State both answers
Why: Eight and negative one.
\[ 8\text{ and } -1 \]
Figure (svg): An absolute-value expression isolated before the equation is split
\[ x = 8 \;\text{ or }\; x = -1 \]
Verify: check both in the original equation
Why: At eight the inside is nine, so the left side is nine plus five, which is fourteen. At negative one the inside is negative nine, whose absolute value is nine, giving fourteen again. Both checks include the plus five, which is what confirms the isolation step.
McDougal Littell Algebra 1: Concepts and Skills, Ch. 6 Solving and Graphing Linear Inequalities — Lesson 6.6 Solving Absolute-Value Equations §6.6, pp. 356-356
Sorting
The absolute value must stand alone.
Sort into buckets
Sort each equation by whether it can be split immediately.
The multiplication case is worth noticing: dividing both sides by three gives the absolute value of x equal to four, and only then can it be split.
Worked example
Guided Practice 8 and 9. Something added and something subtracted.
\[ \text{Solve } \; |x + 1| + 2 = 4 \; \text{ and } \; |2x - 8| - 3 = 5. \]
Isolate the first
Why: Subtract two from each side.
\[ | x + 1 | = 2 \]
Split and solve
Why: x plus one is two or negative two.
\[ x = 1\text{ or } x = -3 \]
Isolate the second
Why: Add three to each side.
\[ | 2 x - 8 | = 8 \]
Split and solve
Why: 2x minus eight is eight or negative eight.
\[ x = 8\text{ or } x = 0 \]
Figure (svg): The solution to Worked example two from guided practice shown as a ladder of expressions, one row per algebraic move
\[ \{1, -3\} \qquad \{8, 0\} \]
Verify: check the second problem's zero
Why: At x equal to zero the inside is negative eight, whose absolute value is eight, and eight minus three is five. Zero is easy to dismiss as a non-answer, and substituting it shows it is a genuine solution.
McDougal Littell Algebra 1: Concepts and Skills, Ch. 6 Solving and Graphing Linear Inequalities — Lesson 6.6 Solving Absolute-Value Equations §6.6, pp. 356-356
Trap
\[ |2x - 7| + 5 = 14 \]
Write 2x - 7 + 5 = 14 or 2x - 7 + 5 = -14
Why: The equation is split immediately, carrying the plus five along inside.
The plus five was outside the bars, so it is not part of what the absolute value acts on. The first branch gives x equal to eight by coincidence and the second gives a wrong answer.
\[ |2x - 7| = 9 \;\Longrightarrow\; 2x - 7 = \pm 9 \]
Undo everything outside the bars before writing either case
Why: The rule applies only when the absolute value stands alone on its side.
Checking the second answer in the original catches it, since the corrupted branch produces a value that does not satisfy the equation.
Faded example
Undo the outside first.
Fill in the blanks
|2x - 7| + 5 = 14 \;\Longrightarrow\; |2x - 7| = 9 \;\Longrightarrow\; 2x - 7 = \pm 9
Why: Subtracting five isolates the absolute value at nine, and only then do the two branches read 2x minus seven equals nine or negative nine. Splitting before isolating would attach the five to the inside, which is not where it was.
Elimination
The equation is 3 times the absolute value of x, equals 12.
Eliminate the wrong options
What comes first?
Survives elimination: A
Why: Dividing both sides by three gives the absolute value of x equal to four, which then splits into x equals four or negative four. Anything multiplying or added to the bars belongs to the equation, not to the expression inside them.
Socratic
The two branches would look almost right.
Discussion prompt
Explain what goes wrong if you split an equation before isolating the absolute value. Then say what the analogous rule was for solving ordinary equations in Chapter 3.
Hint: Ask what the bars are acting on.
Answer:
The rule says that if the absolute value of an expression equals c, the expression is c or negative c. That statement is about the expression inside the bars alone, so applying it while something else sits outside them means applying it to the wrong thing — the outside term gets swept into the inside where it does not belong.
It is the same principle as undoing operations from the outside in, which Chapter 3 used for every multi-step equation. The absolute value is the outermost structure on its side, so everything wrapped around it must be peeled off before it can itself be opened.
Section
Section 4
Concept
Substituting both solutions into the original equation confirms them. Before solving, the number on the right predicts how many answers there will be.
The count applies after the absolute value has been isolated, not before.
Figure (svg): Three columns of how many solutions an absolute-value equation can have
McDougal Littell Algebra 1: Concepts and Skills, Ch. 6 Solving and Graphing Linear Inequalities — Lesson 6.6 Solving Absolute-Value Equations §6.6, pp. 355-356 — the rule for c less than zero and the Study Tip on checking solutions
Picture it
The right side decides.
Figure (svg): Three columns of how many solutions an absolute-value equation can have
The prediction is worth making before solving, since a negative right side means the whole split can be skipped and the answer written down immediately.
Worked example
The textbook's Study Tip recommends substituting each answer.
\[ \text{Check that } 8 \text{ and } -1 \text{ solve } \; |2x - 7| + 5 = 14. \]
Substitute eight
Why: Sixteen minus seven is nine.
\[ | 9 | + 5 \]
Simplify
Why: Nine plus five is fourteen.
Substitute negative one
Why: Negative two minus seven is negative nine.
\[ | - 9 | + 5 \]
Simplify
Why: The absolute value is nine, so the total is fourteen.
Figure (svg): Both solutions of an absolute-value equation substituted back
\[ 8: \;\checkmark \qquad -1: \;\checkmark \]
Verify: notice which check uses the absolute value
Why: The first gives a positive value inside the bars and would work without them. The second gives negative nine, and only the bars turn it into nine — so the second check is the one that tests whether the negative branch was handled correctly.
McDougal Littell Algebra 1: Concepts and Skills, Ch. 6 Solving and Graphing Linear Inequalities — Lesson 6.6 Solving Absolute-Value Equations §6.6, pp. 356-356
Prediction
Isolate first, then look at the right side.
Predict first
How many solutions does |x + 5| - 3 = -3 have?
Correct: One, since isolating gives |x + 5| = 0.
\[ |x + 5| - 3 = -3 \;\Longrightarrow\; |x + 5| = 0 \;\Longrightarrow\; x = -5 \]
Why: Adding three to both sides gives an isolated right side of zero, so the two branches coincide and the single answer is x equal to negative five. The third option looks at the right side before isolating, which is exactly the mistake the count rule warns against — the rule applies to the isolated equation.
Worked example
Guided Practice 7 and a negative case.
\[ \text{How many solutions do } \; |3x - 2| = 0 \; \text{ and } \; |x - 4| = -2 \; \text{ have?} \]
Look at the first right side
Why: Zero, so the two branches coincide.
Solve it
Why: 3x minus two equals zero, so x is two thirds.
\[ x = \frac{2}{3} \]
Look at the second right side
Why: Negative, so no distance can match it.
State the second answer
Why: There is nothing to solve.
Figure (svg): The solution to Worked example predict the count first shown as a ladder of expressions, one row per algebraic move
\[ x = \tfrac{2}{3}; \quad \text{no solution} \]
Verify: confirm the single answer is genuinely single
Why: Both branches of the first equation read 3x minus two equals zero, since negative zero is zero, so they give the same answer rather than two. Writing both branches out once is worth doing to see that they coincide rather than assuming it.
McDougal Littell Algebra 1: Concepts and Skills, Ch. 6 Solving and Graphing Linear Inequalities — Lesson 6.6 Solving Absolute-Value Equations §6.6, pp. 356-356
Trap
\[ |x - 4| = -2 \]
Split into x - 4 = -2 or x - 4 = 2 and solve
Why: The splitting rule is applied automatically without looking at the right side.
\[ x = 2 \;\text{ or }\; x = 6 \quad \text{(neither works)} \]
Substituting two gives the absolute value of negative two, which is two rather than negative two. Both reported answers fail the original.
The right side is negative, so there is no solution.
Look at the isolated right side before splitting anything
Why: A negative there ends the problem immediately.
Checking the answers would have caught it too, which is why the Study Tip's advice to substitute is worth following even when the solving felt routine.
Faded example
The bars turn a negative into a positive.
Fill in the blanks
x = -1 \text9 |2x - 7| + 5: \quad |-9| + 5 = 14 + 5 = ___
Why: The inside comes out as negative nine and the bars turn it into nine, so the total is fourteen and the answer checks. This is the check that actually tests the negative branch, since the positive branch would work even without the bars.
Elimination
The equation is |x - 1| + 6 = 4.
Eliminate the wrong options
How should you judge the number of solutions?
Survives elimination: A
Why: Subtracting six gives an isolated right side of negative two, so no number satisfies the equation. The count rule is about the isolated form, which is why isolating comes before any judgement as well as before any splitting.
Socratic
The two branches are solved the same way.
Discussion prompt
Explain what a check of the negative branch's answer tests that a check of the positive branch's does not. Then say what kind of error would pass the first check and fail the second.
Hint: Ask which check uses the bars.
Answer:
The positive branch's answer makes the inside expression positive, so the bars change nothing and the check would pass even in an equation with no bars at all. The negative branch's answer makes the inside negative, so the check genuinely exercises the absolute value.
An error in the sign of the negative branch — negating something inside the bars, or forgetting to negate the right side — produces a wrong second answer while leaving the first untouched. So checking only the first answer would pass, which is exactly why both are worth substituting.
Section
Section 5
Concept
Given two numbers, an absolute-value equation with those solutions is built from their midpoint and their common distance from it: the absolute value of x minus the midpoint equals the distance.
\[ |x - \text{midpoint}| = \text{distance} \]
The two solutions are always symmetric about the midpoint.
Figure (svg): Two solutions on a number line with their midpoint and common distance marked
McDougal Littell Algebra 1: Concepts and Skills, Ch. 6 Solving and Graphing Linear Inequalities — Lesson 6.6 Solving Absolute-Value Equations §6.6, pp. 357-357 — Example 4, Write an Absolute-Value Equation
Picture it
Two equal jumps from the centre.
Figure (svg): Two solutions on a number line with their midpoint and common distance marked
The midpoint is the average of the two numbers and the distance is half their difference. Both are read off the picture directly.
Worked example
This is Example 4 from the textbook.
\[ \text{Write an absolute-value equation whose solutions are } 7 \text{ and } 15. \]
Find the midpoint
Why: The average of seven and fifteen is eleven.
\[ \text{midpoint } 11 \]
Find the distance
Why: Each is four units from eleven.
\[ \text{distance } 4 \]
Write the equation
Why: The absolute value of x minus eleven equals four.
\[ | x - 11 | = 4 \]
Check both solutions
Why: Seven and fifteen both give four inside the bars, up to sign.
Figure (svg): Two solutions on a number line with their midpoint and common distance marked
\[ |x - 11| = 4 \]
Verify: solve the equation forwards
Why: Splitting gives x minus eleven equals four or negative four, so x is fifteen or seven — the two numbers asked for. Running the construction forwards is the natural check, and it uses the method from earlier in the lesson.
McDougal Littell Algebra 1: Concepts and Skills, Ch. 6 Solving and Graphing Linear Inequalities — Lesson 6.6 Solving Absolute-Value Equations §6.6, pp. 357-357
Translation
Midpoint inside, distance outside.
Match the pairs
Why: The midpoint is the average of the two solutions and the distance is half their difference. The third pair is Guided Practice 10, and the fourth shows that the method works unchanged when one solution is negative.
Worked example
This is Example 5 from the textbook.
\[ \text{Miniature poodles stand from } 10 \text{ to } 15 \text{ inches at the shoulder. Write an equation with these as solutions.} \]
Find the midpoint
Why: The average of ten and fifteen is twelve and a half.
\[ \text{midpoint } 12.5 \]
Find the distance
Why: Each is two and a half units away.
\[ \text{distance } 2.5 \]
Write the equation
Why: The absolute value of x minus 12.5 equals 2.5.
\[ | x - 12.5 | = 2.5 \]
Interpret it
Why: The heights differing from 12.5 inches by exactly 2.5 inches.
Figure (svg): The solution to Worked example the poodle heights shown as a ladder of expressions, one row per algebraic move
\[ |x - 12.5| = 2.5 \]
Verify: say what the equation describes and what it does not
Why: It has exactly the two extreme heights as solutions, and says nothing about the heights in between — a poodle of thirteen inches does not satisfy it. Describing the whole range rather than its two endpoints needs an inequality, which is Lesson 6.7.
McDougal Littell Algebra 1: Concepts and Skills, Ch. 6 Solving and Graphing Linear Inequalities — Lesson 6.6 Solving Absolute-Value Equations §6.6, pp. 357-357
Trap
Write an absolute-value equation with solutions 7 and 15.
The numbers are 8 apart, so write |x - 11| = 8
Why: Eight is the visible gap between the two numbers, so it looks like the distance in the equation.
That equation has solutions three and nineteen, since each is eight units from eleven. The distance in the equation is from the midpoint to each solution, which is half the gap.
\[ |x - 11| = 4 \]
Halve the difference to get the distance from the midpoint
Why: The two solutions are on opposite sides of the centre, so the gap between them is twice the distance.
Solving the finished equation forwards catches this immediately, and it takes two lines.
Faded example
Average, then halve the difference.
Fill in the blanks
\text11 7 \text4 15: \quad \text___ ___, \; \text___ ___ \;\Longrightarrow\; |x - 11| = 4
Why: The midpoint eleven goes inside the bars, subtracted from x, and the distance four goes on the right. Using the whole difference of eight instead would give an equation whose solutions are three and nineteen.
Hypothesis
Predict before you check.
Predict first
What absolute-value equation has 6 as its only solution?
Correct: |x - 6| = 0.
\[ |x - 6| = 0 \;\Longrightarrow\; x - 6 = 0 \;\Longrightarrow\; x = 6 \]
Why: With both solutions at six the midpoint is six and the distance is zero, so the equation is the absolute value of x minus six equals zero. That matches the count rule from earlier: a right side of zero gives exactly one solution. The third option has solutions six and negative six rather than six alone.
Socratic
The construction is the lesson run backwards.
Discussion prompt
Explain why the number subtracted inside the bars is the midpoint of the two solutions. Then say what would change if the equation were written with a plus sign inside instead.
Hint: Ask what the inside expression measures.
Answer:
The expression x minus a measures the distance from x to a, so setting its absolute value equal to d asks for the numbers exactly d away from a. Those numbers are symmetric about a, which makes a their midpoint — the construction is just this statement read in reverse.
A plus sign inside would be x minus negative a, so the centre would be negative a rather than a. Writing the absolute value of x plus three equals five gives solutions two and negative eight, which are symmetric about negative three — so a plus sign inside means a negative centre, and reading it as a positive one is a standard slip.
Comparison
Fill the blanks from memory before you scroll back.
Comparison matrix
| Linear equation (Chapter 3) | Absolute-value equation | |
|---|---|---|
| Usual number of solutions | one | two |
| How it is solved | undo operations to isolate the variable | isolate the bars, then split into two cases |
| When there is no solution | when the variables cancel to a false statement | when the isolated right side is negative |
The second row is why the third differs: an absolute-value equation can fail before any solving begins, simply because a distance was asked to be negative.
Pattern
Whether the equation has one term outside the bars or several, the same five moves cover it.
Step two costs a glance and can end the problem immediately, which is why it belongs before the splitting rather than after.
OpenStax Intermediate Algebra 2e, §2.7 Solve Absolute Value Inequalities §2.7
Check
Copy the inside, negate the right.
Check your understanding
Solve |x - 4| = 6.
Answer: A
Why: The two cases are x minus four equals six, giving ten, and x minus four equals negative six, giving negative two. Checking negative two: negative two minus four is negative six, whose absolute value is six.
Check
Isolate before judging.
Check your understanding
How many solutions does |x + 2| + 7 = 3 have?
Answer: A
Why: Subtracting seven gives an isolated right side of negative four, and no absolute value is negative, so no number satisfies the equation.
Check
Midpoint inside, distance outside.
Check your understanding
Which equation has 3 and 11 as its solutions?
Answer: A
Why: The midpoint of three and eleven is seven and each is four units away, so the equation is the absolute value of x minus seven equals four. Solving it forwards gives eleven and three.
Real world
A bag of crisps is labelled 50 grams. The factory's machines fill each bag to within 2 grams of that weight, and quality control pulls out the bags that are exactly at the limit for inspection.
Discussion prompt
Write an absolute-value equation whose solutions are the two limiting weights, solve it, and say what it describes. Then say what equation would describe a target of 50 grams with no tolerance at all.
Hint: The midpoint is the target and the distance is the tolerance.
Answer:
\[ |w - 50| = 2 \;\Longrightarrow\; w = 52 \;\text{ or }\; w = 48 \]
The two solutions are the heaviest and lightest bags the tolerance permits. The equation describes only those two extremes, not the acceptable range between them — for that you would need an inequality, which is the subject of Lesson 6.7.
With no tolerance at all the equation is the absolute value of w minus fifty equals zero, whose only solution is fifty. That is the count rule's middle case: a right side of zero collapses the two branches into one, and physically it says the machine must hit the target exactly.
Commit first
Answer, then rate your confidence honestly. Confident and wrong is the combination worth finding.
Predict first
What are the two cases of |x - 2| = 5?
Correct: x - 2 = 5 or x - 2 = -5.
\[ x = 7 \;\text{ or }\; x = -3 \]
\[ |-3 - 2| = |-5| = 5 \;\checkmark \]
Why: The expression inside the bars is copied unchanged into both cases, and only the number on the right changes sign. The first option negates the x inside, which gives negative seven — and negative seven minus two is negative nine, whose absolute value is nine rather than five. Checking the second answer catches every version of this error in one line, which is why the Study Tip recommends substituting both.
Explain it
They can solve linear equations and have just met bars around an expression.
Discussion prompt
In no more than four sentences, explain why absolute-value equations have two answers and how to find both. Then tell them the two things to check before starting.
Hint: Distance, in two directions.
Answer:
A usable answer: the bars measure distance from zero, and you can be a given distance away in either direction — so the thing inside the bars could be the number on the right or its negative. Write both of those as ordinary equations, keeping the inside exactly as it is and changing only the sign on the right, then solve each one.
Before starting, make sure the bars are alone on their side, undoing anything added to or multiplying them first. Then look at the number left on the right: if it is negative there is no solution at all, and you can stop.
Exit ticket
Name the weakest spot before you close the deck. That is the one worth ten minutes tonight.
Predict first
Which of these would you least want to be handed cold on a quiz tomorrow?
Correct: Whichever you picked is the right answer — and each one has a specific fix.
Why: Both cases are fixed by writing the two branches side by side before solving either. Negating the right side is fixed by copying the inside expression twice, unchanged, before touching anything. Isolating is fixed by checking that nothing sits outside the bars. Building an equation is fixed by taking the average for the midpoint and half the difference for the distance, then solving forwards to check. Pick yours and do five of that kind tonight rather than twenty mixed ones.
Connect it up
Do this on paper. It is worth more than rereading the slides.
Draw it
At the top of a page solve one absolute-value equation that needs isolating first, showing the isolation as its own step and then the two branches side by side in two columns. Substitute both answers back into the original underneath, and circle the check that produced a negative value inside the bars. In the middle, write three equations differing only in the number on the right — one positive, one zero and one negative — and beside each write how many solutions it has and why. To the right, draw a number line with your two solutions marked, mark the midpoint between them, draw the two equal jumps, and write the equation you would build from that picture, checking it matches the one you started with. Finally, in the margin, write the two things to check before splitting any absolute-value equation.
The equation you build from the number line should be the same as the one you solved at the top, once both are isolated. If the distance differs by a factor of two, the whole gap was used instead of half of it.
Recap
Five things, and the second is where the sign goes in the wrong place.
| If the question says | Your first move is |
|---|---|
| Solve |x - 2| = 5 | Write both cases, negating only the right side |
| Something is added outside the bars | Isolate the absolute value first |
| The right side is negative | Report no solution |
| The right side is zero | Expect exactly one solution |
| Write an equation with these solutions | Use the midpoint and half the difference |
Lesson 6.7 replaces the equals sign with an inequality. The absolute value of an expression being less than a number becomes a compound and inequality, and being greater than one becomes a compound or inequality — which is why the last two lessons came first.
McDougal Littell Algebra 1: Concepts and Skills, Ch. 6 Solving and Graphing Linear Inequalities — Lesson 6.6 Solving Absolute-Value Equations §6.6, pp. 355-360 — everything on these slides traces back here
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