Absolute-value inequalities rewritten as compound inequalities: less than becomes an and statement giving a band around the centre, and greater than becomes an or statement giving two rays. Includes reversing the symbol on the negative branch, checking one value from each region, and reading such inequalities as statements about distance and tolerance.
Subject: Algebra 1 · 65 slides · symbolic lesson
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Title
Algebra 1 · Chapter 6 — Solving and Graphing Linear Inequalities
Solving Absolute-Value Inequalities
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 6.7 Solving Absolute-Value Inequalities §6.7, pp. 361-366 — the lesson these objectives are drawn from
Warm-up
Lesson 6.6 solved absolute-value equations by splitting them in two. This lesson does the same to inequalities, and the connector depends on which way the symbol points.
Discussion prompt
Which numbers have an absolute value less than 3? And which have an absolute value greater than 3? Mark both sets on number lines.
Hint: Think about distance from zero in each case.
Answer:
\[ |x| < 3 \;\Longleftrightarrow\; -3 < x < 3 \qquad |x| > 3 \;\Longleftrightarrow\; x < -3 \;\text{ or }\; x > 3 \]
Close to zero gives a single band around zero, and far from zero gives two rays with a gap. The direction of the symbol decides which shape you get, which is the whole content of this lesson.
Concept
An absolute-value inequality is solved by rewriting it as two related inequalities. A less-than inequality gives two conditions joined by and; a greater-than inequality gives two joined by or.
absolute-value inequality — An inequality containing an absolute-value expression. It is solved by rewriting as a compound inequality — with and for less than, and with or for greater than.
The same rules apply to the or equal to versions.
Figure (svg): The two rewriting rules for absolute-value inequalities
McDougal Littell Algebra 1: Concepts and Skills, Ch. 6 Solving and Graphing Linear Inequalities — Lesson 6.7 Solving Absolute-Value Inequalities §6.7, pp. 361-361
Section
Section 1
Concept
If the absolute value of an expression is less than c, the expression is less than c and greater than negative c. If it is greater than c, the expression is greater than c or less than negative c.
\[ |u| < c \;\Longleftrightarrow\; u < c \;\text{ and }\; u > -c \]
Figure (svg): The two rewriting rules for absolute-value inequalities
McDougal Littell Algebra 1: Concepts and Skills, Ch. 6 Solving and Graphing Linear Inequalities — Lesson 6.7 Solving Absolute-Value Inequalities §6.7, pp. 361-361 — the rules for rewriting less-than and greater-than absolute-value inequalities
Picture it
Different connectors and different second symbols.
Figure (svg): The two rewriting rules for absolute-value inequalities
Both changes on the right-hand branch come from the same source, so remembering that the second branch reverses covers both rules at once.
Worked example
Both rules become obvious once read as sentences.
\[ \text{Describe } \; |x| < 3 \; \text{ and } \; |x| > 3 \; \text{ in words, then in symbols.} \]
Read the first
Why: The distance from zero is less than three.
Say which numbers qualify
Why: Everything between negative three and three.
\[ -3 < x < 3 \]
Read the second
Why: The distance from zero is greater than three.
Say which numbers qualify
Why: Everything below negative three or above three.
\[ x < -3\text{ or } x > 3 \]
Figure (svg): A number line showing why the negative branch reverses
\[ -3 < x < 3 \qquad x < -3 \;\text{ or }\; x > 3 \]
Verify: check that the two sets are complements
Why: Together with the two endpoints, the band and the two rays cover the whole line and never overlap. That is what you would expect, since every number is either within three of zero or not — so the two rules describe opposite sets.
Sorting
Look at the direction of the symbol.
Sort into buckets
Sort each absolute-value inequality by the connector it needs.
Whether the symbol includes or equal to makes no difference to the connector. Only the direction matters, which is why there are two rules rather than four.
Worked example
The four symbols split into two pairs.
\[ \text{Which connector goes with } \; <, \; \le, \; >, \; \ge? \]
Take less than
Why: The expression is trapped between two bounds.
Take less than or equal to
Why: The same, with the bounds included.
Take greater than
Why: The expression is beyond one bound or the other.
Take greater than or equal to
Why: The same, with the bounds included.
Figure (svg): The solution to Worked example which connector for each symbol shown as a ladder of expressions, one row per algebraic move
\[ <, \le \;\to\; \text{and} \qquad >, \ge \;\to\; \text{or} \]
Verify: check the or equal to versions behave the same
Why: Adding or equal to changes only whether the endpoints are included, and the endpoints do not affect which connector is needed. The connector depends on the direction of the symbol alone.
McDougal Littell Algebra 1: Concepts and Skills, Ch. 6 Solving and Graphing Linear Inequalities — Lesson 6.7 Solving Absolute-Value Inequalities §6.7, pp. 361-361
Trap
\[ |x - 4| < 3 \]
Write x - 4 < 3 or x - 4 > -3
Why: The absolute-value equation in Lesson 6.6 used or, so the habit carries over.
Every number satisfies at least one of those two conditions, so the answer would be all real numbers. Testing one hundred shows it fails the original, since ninety-six is not less than three.
\[ x - 4 < 3 \;\text{ and }\; x - 4 > -3 \;\Longrightarrow\; 1 < x < 7 \]
Use and when the symbol is less than
Why: Being close to a centre requires both bounds at once.
Reading the inequality as a distance statement predicts the connector: within three of four is one band, not two rays.
Faded example
Connector and second symbol.
Fill in the blanks
|u| < c means u < c and u > -c, while |u| > c means u > c or u < -c.
Why: Less than traps the expression between two bounds, requiring both, and greater than puts it beyond one bound or the other, requiring only one. In both rules the second condition has its symbol reversed as well as its number negated.
Elimination
The inequality is the absolute value of x, greater than 5.
Eliminate the wrong options
Which compound inequality is equivalent?
Survives elimination: A
Why: Being more than five from zero means being above five or below negative five. Option D is the complement of the right answer, which is what makes choosing the wrong connector so damaging: you get exactly the numbers the inequality rejects.
Socratic
The rules look arbitrary until you read them as distances.
Discussion prompt
Explain why a small absolute value requires both conditions while a large one requires only one. Then say why no number can satisfy both conditions of the greater-than rule.
Hint: Ask what being close and being far each mean.
Answer:
Being close to a centre means not too far in either direction, so both bounds have to hold at once — a number more than three above four fails, and so does one more than three below it. Being far means far in some direction, and a number cannot be far to the left and far to the right at the same time, so only one condition is needed and only one can hold.
For the greater-than rule the two conditions are being above c and being below negative c, and since c is positive those two regions cannot overlap. That is why the connector must be or: an and reading would give no solutions, which would be wrong for every inequality of this form.
Section
Section 2
Concept
For a less-than inequality, write the two related inequalities joined by and, solve both, and combine them into a single compound statement.
The answer is a bounded segment, as in Lesson 6.4.
Figure (svg): An absolute-value inequality with less than, graphed as a band around the centre
McDougal Littell Algebra 1: Concepts and Skills, Ch. 6 Solving and Graphing Linear Inequalities — Lesson 6.7 Solving Absolute-Value Inequalities §6.7, pp. 362-362 — Example 2, Solve an Absolute-Value Inequality
Picture it
Everything within three of the centre.
Figure (svg): An absolute-value inequality with less than, graphed as a band around the centre
The centre is four because that is what the inside expression subtracts, and the band's half-width is three because that is the number on the right. Both can be read off before solving.
Worked example
This is Example 2 from the textbook.
\[ \text{Solve } \; |x - 4| < 3 \; \text{ and graph the solution.} \]
Write the related inequalities
Why: The inside is less than three and greater than negative three.
\[ x - 4 < 3\text{ and } x - 4 > -3 \]
Solve the first
Why: Add four to each side.
\[ x < 7 \]
Solve the second
Why: Add four to each side.
\[ x > 1 \]
Combine and graph
Why: One band from one to seven, both ends open.
\[ 1 < x < 7 \]
Figure (svg): An absolute-value inequality with less than, graphed as a band around the centre
\[ 1 < x < 7 \]
Verify: test a number inside and one outside
Why: Four gives zero inside the bars, which is less than three, so four is a solution. Nine gives five, which is not less than three, so nine is correctly excluded. Both agree with the band.
McDougal Littell Algebra 1: Concepts and Skills, Ch. 6 Solving and Graphing Linear Inequalities — Lesson 6.7 Solving Absolute-Value Inequalities §6.7, pp. 362-362
Faded example
The second one reverses.
Fill in the blanks
|x - 4| < 3 \;\Longrightarrow\; x - 4 < 3 \;\text>\; x - 4 -3 ___
Why: The second branch negates the three and reverses the symbol, giving x minus four greater than negative three. Both changes together produce the lower bound of one, and applying only one of them gives a ray rather than a band.
Worked example
Guided Practice 2 and 5. One strict and one inclusive.
\[ \text{Solve } \; |x - 2| < 5 \; \text{ and } \; |x - 2| \le 7. \]
Take the first
Why: The inside is between negative five and five.
\[ -5 < x - 2 < 5 \]
Solve it
Why: Add two throughout.
\[ -3 < x < 7 \]
Take the second
Why: The inside is between negative seven and seven, inclusive.
\[ -7 \le x - 2 \le 7 \]
Solve it
Why: Add two throughout.
\[ -5 \le x \le 9 \]
Figure (svg): The solution to Worked example two from guided practice shown as a ladder of expressions, one row per algebraic move
\[ -3 < x < 7 \qquad -5 \le x \le 9 \]
Verify: check the centre and the half-width of each
Why: Both bands are centred on two, since that is what the inside subtracts, and their half-widths are five and seven — the numbers on the right. Predicting the centre and half-width before solving is a check on the whole computation.
McDougal Littell Algebra 1: Concepts and Skills, Ch. 6 Solving and Graphing Linear Inequalities — Lesson 6.7 Solving Absolute-Value Inequalities §6.7, pp. 362-362
Error analysis
The student solved a less-than absolute-value inequality.
Annotate
On: \( \begin{aligned} |x - 4| &< 3 \\ x - 4 &< 3 \;\text{ and }\; x - 4 < -3 \\ x &< 7 \;\text{ and }\; x < 1 \end{aligned} \)
The negative branch always changes twice: the number is negated and the symbol is reversed. Applying only the first change is the standard error and it produces a ray instead of a band.
Translation
Two conditions joined by and.
Match the pairs
Why: Each band is centred on whatever the inside expression subtracts, with a half-width equal to the number on the right. The last has a centre of zero, since there is nothing subtracted inside the bars.
Prediction
The inequality is the absolute value of x minus 4, less than 3.
Predict first
What are the centre and half-width of the solution band?
Correct: Centre 4, half-width 3.
\[ |x - 4| < 3 \;\Longrightarrow\; 4 - 3 < x < 4 + 3 \]
Why: The number subtracted inside the bars is the centre, and the number on the right is the distance permitted from it. So the band runs from one to seven, which the solving confirms. The fourth option uses the whole width rather than half of it, which is the same confusion as using the full difference in Lesson 6.6.
Socratic
The reversal is not the rule from Lesson 6.2.
Discussion prompt
Explain why the negative branch of a less-than absolute-value inequality has its symbol reversed, given that nothing was multiplied by a negative number. Then say what the reversal is really recording.
Hint: Ask what the two branches are describing.
Answer:
The two branches are not two operations on one inequality; they are two separate conditions describing a band. The upper bound says the expression is below c and the lower bound says it is above negative c — and a lower bound is naturally a greater-than statement. The reversal records that the second branch is a floor rather than a ceiling.
It is not the Lesson 6.2 reversal at all, since nothing has been multiplied or divided by a negative. It is a consequence of what a band is: bounded above by one number and below by another, so the two conditions must point in opposite directions to close it off at both ends.
Section
Section 3
Concept
For a greater-than inequality, write the two related inequalities joined by or, solve each independently, and report both as an or statement.
The two answers cannot be combined into a chain.
Figure (svg): An absolute-value inequality with greater than, graphed as two rays
McDougal Littell Algebra 1: Concepts and Skills, Ch. 6 Solving and Graphing Linear Inequalities — Lesson 6.7 Solving Absolute-Value Inequalities §6.7, pp. 361-362 — Examples 1 and 3, on greater-than absolute-value inequalities
Picture it
Everything more than five from zero.
Figure (svg): An absolute-value inequality with greater than, graphed as two rays
The gap between the two rays is exactly the band the corresponding less-than inequality would have produced, which is why the two rules give complementary answers.
Worked example
This is Example 1 from the textbook.
\[ \text{Solve } \; |x| > 5 \; \text{ and graph the solution.} \]
Read it as a distance
Why: The distance from zero is more than five.
Write the related inequalities
Why: Greater than five, or less than negative five.
\[ x > 5\text{ or } x < -5 \]
Note the connector
Why: The symbol is greater than, so the parts are joined by or.
Graph
Why: Two rays, open dots at negative five and five.
Figure (svg): An absolute-value inequality with greater than, graphed as two rays
\[ x < -5 \;\text{ or }\; x > 5 \]
Verify: test one number from each region
Why: Negative six gives six, which is greater than five, and so does six. Zero gives zero, which is not, so zero is correctly excluded. Three test numbers cover all three regions of the graph.
McDougal Littell Algebra 1: Concepts and Skills, Ch. 6 Solving and Graphing Linear Inequalities — Lesson 6.7 Solving Absolute-Value Inequalities §6.7, pp. 361-361
Faded example
Or, with the second reversed and negated.
Fill in the blanks
|x + 5| \ge 2 \;\Longrightarrow\; x + 5 \ge 2 \;\text<=\; x + 5 -2 ___
Why: The second branch negates the two and reverses the symbol, giving x plus five at most negative two. Solving both gives x at least negative three or x at most negative seven — two rays around a centre of negative five.
Worked example
This is Example 3 from the textbook.
\[ \text{Solve } \; |x + 5| \ge 2 \; \text{ and graph the solution.} \]
Write the related inequalities
Why: The inside is at least two, or at most negative two.
\[ x + 5 \ge 2\text{ or } x + 5 \le - 2 \]
Solve the first
Why: Subtract five.
\[ x \ge - 3 \]
Solve the second
Why: Subtract five.
\[ x \le - 7 \]
Combine and graph
Why: Two rays with solid endpoints.
\[ x \le - 7\text{ or } x \ge - 3 \]
Figure (svg): The solution to Worked example an inclusive greater-than inequality shown as a ladder of expressions, one row per algebraic move
\[ x \le -7 \;\text{ or }\; x \ge -3 \]
Verify: test the suggested values
Why: The Study Tip recommends ten, five and zero, and negative versions work equally well: negative ten gives five, which is at least two; negative five gives zero, which is not; and zero gives five, which is. Choosing round numbers keeps the substitution trivial.
McDougal Littell Algebra 1: Concepts and Skills, Ch. 6 Solving and Graphing Linear Inequalities — Lesson 6.7 Solving Absolute-Value Inequalities §6.7, pp. 362-362
Trap
\[ x \le -7 \;\text{ or }\; x \ge -3 \]
Compress it to -7 >= x >= -3
Why: Compound answers were written on one line for and inequalities, and the habit carries over.
That chain claims negative seven is at least negative three, which is false, and it means and rather than or. An or answer must stay as two statements.
\[ x \le -7 \;\text{ or }\; x \ge -3 \quad \text{as two statements} \]
Keep the word or, exactly as in Lesson 6.5
Why: The chain notation is reserved for the and case.
Any chain whose left number exceeds its right one is a signal that an or answer has been compressed wrongly.
Translation
Two parts joined by or.
Match the pairs
Why: Each answer is two rays about the centre the inside expression names, with a gap whose half-width is the number on the right divided by any coefficient. The third has a coefficient of three, so its bounds are nine divided by three rather than nine.
Elimination
The solution is x at most -7 or x at least -3.
Eliminate the wrong options
Which numbers fail the inequality?
Survives elimination: A
Why: A number fails an or inequality only by failing both parts, which happens strictly between the two bounds. That gap is exactly the band the corresponding less-than inequality would have produced, which is the connection between this section and the previous one.
Socratic
The gap of one is the band of the other.
Discussion prompt
Explain why the solution of a greater-than absolute-value inequality is exactly what the corresponding less-than one excludes. Then say what happens at the two boundary values.
Hint: Ask what the two inequalities say about the same distance.
Answer:
The two inequalities ask opposite questions about the same distance: is it below c, or above c? Every number has some definite distance from the centre, so it satisfies exactly one of the two — unless its distance is exactly c, in which case it satisfies neither strict version.
At the boundary values the distance is exactly c, so both strict inequalities fail. That is why the strict band and the strict pair of rays leave the two endpoints out of both, and why the or-equal-to versions put each endpoint into exactly one of them. The two solution sets and the two boundary points together account for the whole line.
Section
Section 4
Concept
To check an absolute-value inequality, test one value from each region of the graph. For an or answer that means three values: one from each ray and one from the gap.
The unshaded region is where a wrong connector shows up.
Figure (svg): Three test values, one from each region of an or graph
McDougal Littell Algebra 1: Concepts and Skills, Ch. 6 Solving and Graphing Linear Inequalities — Lesson 6.7 Solving Absolute-Value Inequalities §6.7, pp. 361-362 — the Check panels and the Study Tip on choosing simple values
Picture it
Two pass and one fails.
Figure (svg): Three test values, one from each region of an or graph
The middle test is the informative one. An answer with the wrong connector would accept zero, and testing it is what exposes that in a single line.
Worked example
This is the check accompanying Example 1.
\[ \text{Check the solution } x < -5 \;\text{ or }\; x > 5 \text{ of } \; |x| > 5. \]
Test the left ray
Why: Negative six has an absolute value of six.
\[ 6 > 5,\text{ true} \]
Test the right ray
Why: Six has an absolute value of six.
\[ 6 > 5,\text{ true} \]
Test the gap
Why: Zero has an absolute value of zero.
\[ 0 > 5,\text{ false} \]
Compare with the graph
Why: Both rays shaded, gap unshaded.
Figure (svg): Three test values, one from each region of an or graph
\[ -6: \;\checkmark \quad 6: \;\checkmark \quad 0: \;\times \]
Verify: ask what a single test would have shown
Why: Testing only six would pass under both the correct answer and one with the connector wrong, since six is far from zero either way. Only the gap test distinguishes them, which is why the middle region matters most.
McDougal Littell Algebra 1: Concepts and Skills, Ch. 6 Solving and Graphing Linear Inequalities — Lesson 6.7 Solving Absolute-Value Inequalities §6.7, pp. 361-361
Elimination
The solution claimed is x less than -5 or x greater than 5.
Eliminate the wrong options
Which value best tests it?
Survives elimination: A
Why: The gap is the only region a wrong connector changes, so a value from it is the one that can fail. Zero also makes the arithmetic trivial, which is exactly what the Study Tip recommends.
Worked example
The textbook's Study Tip suggests picking round numbers.
\[ \text{Which values would you use to check } \; |x + 5| \ge 2, \text{ with answer } x \le -7 \;\text{ or }\; x \ge -3? \]
Choose from the left ray
Why: Negative ten is comfortably below negative seven.
\[ x = -10 \]
Choose from the gap
Why: Negative five is between the two bounds.
\[ x = -5 \]
Choose from the right ray
Why: Zero is above negative three and trivial to substitute.
\[ x = 0 \]
Run the three tests
Why: Five, zero and five inside the bars.
Figure (svg): The solution to Worked example choosing simple values shown as a ladder of expressions, one row per algebraic move
\[ -10: \;\checkmark \quad -5: \;\times \quad 0: \;\checkmark \]
Verify: notice why negative five is a good gap value
Why: It sits at the centre of the band, where the inside expression is zero — the smallest possible absolute value. If any number fails a greater-than inequality, the centre does, so it is the strongest test of the gap.
McDougal Littell Algebra 1: Concepts and Skills, Ch. 6 Solving and Graphing Linear Inequalities — Lesson 6.7 Solving Absolute-Value Inequalities §6.7, pp. 362-362
Trap
\[ |x - 4| < 3 \text{, reported as } x < 7 \]
Test 4 and 5, both of which work, and accept the answer
Why: Both are solutions of the original and of the reported answer.
Both also satisfy the incorrect answer, so neither test distinguishes them. Testing zero would expose it: zero gives four inside the bars, which is not less than three, and the reported answer accepts zero.
Test a value the answer accepts that the original might reject
Why: Zero lies inside x less than seven and outside the true band.
A useful test value is one the candidate answers disagree about, which is the same principle as in Lesson 6.1.
Faded example
The middle region is the informative one.
Fill in the blanks
For |x| > 5 at x = 0: the absolute value of 0 is 0, which is not greater than 5, so 0 is correctly excluded.
Why: Zero fails the inequality, which matches the graph leaving the gap unshaded. An answer that had used and instead of or, or the wrong directions, would have accepted zero — so this one test catches the whole class of connector errors.
Comparison
Fill the blanks from memory before you scroll back.
Comparison matrix
| Less-than answer (a band) | Greater-than answer (two rays) | |
|---|---|---|
| Regions in the graph | three: below, band, above | three: ray, gap, ray |
| Shaded regions | one | two |
| Most informative test | one from outside the band | one from the gap |
Both graphs have three regions, and in each case the test that distinguishes a right answer from a wrong connector comes from a region the answer claims to exclude.
Socratic
It seems more natural to confirm the solutions.
Discussion prompt
Explain why testing a number your answer excludes is more informative than testing one it includes. Then say what kind of error each of the two tests can catch.
Hint: Ask which test can fail.
Answer:
A number inside the solution set usually satisfies the original whether or not the answer is correct, because wrong connectors and wrong directions tend to produce sets that still contain the obvious solutions. A number the answer excludes is a claim that can be falsified, so testing it is where the information is.
The inclusion test catches a boundary in the wrong place or an arithmetic slip in the solving. The exclusion test catches the wrong connector and the wrong direction on a branch, which are the errors specific to this lesson. Running both is cheap, and running only the first is what lets a reversed connector through.
Section
Section 5
Concept
An absolute-value inequality says how far the inside expression is from a centre. The number subtracted inside is the centre, and the number on the right is the permitted or excluded distance.
Every tolerance statement has this shape.
Figure (svg): A tolerance band around a target value
McDougal Littell Algebra 1: Concepts and Skills, Ch. 6 Solving and Graphing Linear Inequalities — Lesson 6.7 Solving Absolute-Value Inequalities §6.7, pp. 361-366 — the distance readings used throughout Examples 1 to 3
Picture it
Fifty grams, give or take two.
Figure (svg): A tolerance band around a target value
One line of absolute-value notation says what would otherwise need two inequalities and a connector, which is why tolerances are almost always written this way.
Worked example
Turning a sentence into an absolute-value inequality.
\[ \text{A bag is acceptable if its weight is within } 2 \text{ grams of } 50. \text{ Write and solve the inequality.} \]
Identify the centre
Why: The target weight is fifty.
\[ \text{centre } 50 \]
Identify the tolerance
Why: Two grams either way.
\[ \text{distance } 2 \]
Write the inequality
Why: Within means less than or equal to.
\[ | w - 50 | \le 2 \]
Solve it
Why: The and rule gives a band.
\[ 48 \le w \le 52 \]
Figure (svg): A tolerance band around a target value
\[ |w - 50| \le 2 \;\Longleftrightarrow\; 48 \le w \le 52 \]
Verify: check the two extremes
Why: At forty-eight the inside is negative two, whose absolute value is two — exactly the tolerance, and the or-equal-to admits it. At fifty-two it is positive two, admitted for the same reason. The two endpoints are the limiting acceptable bags.
Matching
Centre inside, distance on the right.
Match the pairs
Why: Within gives a less-than symbol and a band; more than or at least gives a greater-than symbol and two rays. The last has a negative centre, which appears inside the bars as a plus sign.
Worked example
The complement of a tolerance is a greater-than inequality.
\[ \text{Write the condition for a bag to be rejected, and solve it.} \]
Read the condition
Why: Rejected means more than two grams from fifty.
Write the inequality
Why: Greater than, so the or rule applies.
\[ | w - 50 | > 2 \]
Write the related inequalities
Why: The inside is above two or below negative two.
\[ w - 50 > 2\text{ or } < -2 \]
Solve
Why: Add fifty to each.
\[ w < 48\text{ or } w > 52 \]
Figure (svg): The solution to Worked example the rejection condition shown as a ladder of expressions, one row per algebraic move
\[ w < 48 \;\text{ or }\; w > 52 \]
Verify: compare with the acceptance condition
Why: The rejected weights are exactly those the acceptance band leaves out, and the two boundary values belong to the acceptance set. Every bag is either accepted or rejected and never both, which is what complementary conditions should do.
Trap
\[ |x + 3| < 5 \]
Say the band is centred on 3
Why: The three is the number visible inside the bars, so it looks like the centre.
The expression x plus three is x minus negative three, so the centre is negative three. Solving confirms it: the band runs from negative eight to two, centred on negative three.
\[ |x - (-3)| < 5 \;\Longrightarrow\; \text{centre } -3, \text{ half-width } 5 \]
Rewrite a plus sign inside as minus a negative before reading the centre
Why: The form is always x minus the centre.
Averaging the two bounds of the solved answer is an independent way to find the centre, and it agrees.
Faded example
The form is x minus the centre.
Fill in the blanks
In |x + 3| < 5 the centre is -3 and the half-width is 5, so the band runs from -8 to 2.
Why: A plus sign inside the bars means the centre is negative, since the standard form subtracts the centre. Averaging the two bounds of the answer gives negative three, confirming it independently.
Hypothesis
Predict before you check.
Predict first
What does |w - 50| <= 0 describe?
Correct: Only w = 50, since no distance can be less than zero.
\[ |w - 50| \le 0 \;\Longrightarrow\; w = 50 \qquad |w - 50| < 0: \;\text{ no solution} \]
Why: An absolute value is never negative, so at most zero means exactly zero, which forces w to equal fifty. Physically it is a specification with no tolerance at all — the machine must hit the target exactly. Had the symbol been strictly less than zero, no weight would qualify and the answer would be no solution.
Socratic
The same thing could be written as two inequalities.
Discussion prompt
Explain what the absolute-value form of a tolerance gives you that the two-inequality form does not. Then say when you would convert to the two-inequality form anyway.
Hint: Ask what each form makes visible.
Answer:
The absolute-value form shows the target and the tolerance as two separate numbers, each meaning something in the situation — a nominal weight and an allowed error. The two-inequality form shows the limits instead, and the target and tolerance have to be recovered by averaging and halving.
You convert when you need the actual limits: to set a machine's cut-off, to check a particular measurement quickly, or to graph the set. So the absolute-value form is how a specification is stated and the compound form is how it is applied — which is exactly the relationship between the two halves of this lesson.
Comparison
Fill the blanks from memory before you scroll back.
Comparison matrix
| Less than | Greater than | |
|---|---|---|
| Connector | and | or |
| Shape of the graph | one band | two rays with a gap |
| Distance reading | within d of the centre | more than d from the centre |
| The second branch | reversed and negated | reversed and negated |
Only the last row is the same in both columns. Everything else flips, and all the flips follow from the direction of the original symbol.
Pattern
Whether the symbol points left or right, the same five moves cover it.
Step two before step three matters: knowing which shape to expect makes a wrong connector visible as soon as the answer appears.
OpenStax Intermediate Algebra 2e, §2.7 Solve Absolute Value Inequalities §2.7
Check
Less than gives a band.
Check your understanding
Solve |x - 3| < 4.
Answer: A
Why: The two related inequalities are x minus three less than four and greater than negative four, giving a band from negative one to seven centred on three. Testing three gives zero inside the bars, which is less than four.
Check
Greater than gives two rays.
Check your understanding
Solve |x + 2| >= 6.
Answer: A
Why: The related inequalities are x plus two at least six or at most negative six, giving x at least four or at most negative eight. The centre is negative two and the gap has half-width six.
Check
The centre is what is subtracted.
Check your understanding
Where is the solution band of |x + 7| < 2 centred?
Answer: A
Why: The expression x plus seven is x minus negative seven, so the centre is negative seven. The band runs from negative nine to negative five, whose average is negative seven.
Real world
This is Exercise 42's situation. Water shot upward from a fountain has velocity v equal to negative 32t plus 40 feet per second after t seconds, and you want the times when its speed exceeds 24 feet per second in either direction.
Discussion prompt
Write the condition as an absolute-value inequality, solve it, and interpret the answer. Then say what the gap in the graph represents physically.
Hint: Speed in either direction is the absolute value of the velocity.
Answer:
\[ |-32t + 40| > 24 \;\Longrightarrow\; -32t + 40 > 24 \;\text{ or }\; -32t + 40 < -24 \]
\[ -32t > -16 \;\text{ or }\; -32t < -64 \;\Longrightarrow\; t < 0.5 \;\text{ or }\; t > 2 \]
The water is moving faster than twenty-four feet per second for the first half second, while it is still rising quickly, and again after two seconds, once it has been falling long enough to build up speed. Both parts required a reversal when dividing by negative thirty-two, exactly as in Lesson 6.5.
The gap from half a second to two seconds is the period around the top of the arc, where the water is slowing, stopping and beginning to fall — moving at twenty-four feet per second or less throughout. Absolute value is what lets one inequality ask about speed regardless of direction, which is precisely the question a fountain designer would ask.
Commit first
Answer, then rate your confidence honestly. Confident and wrong is the combination worth finding.
Predict first
What does |x - 4| < 3 become as a compound inequality?
Correct: x - 4 < 3 and x - 4 > -3.
\[ 1 < x < 7 \]
\[ \text{check } x = 100: \; |96| = 96, \text{ not} < 3 \]
Why: A less-than absolute value traps the inside expression between two bounds, so both conditions are required and the second has its symbol reversed as well as its number negated. The first option uses or, and every number satisfies at least one of those two conditions, so it would give all real numbers — testing one hundred exposes that immediately, since ninety-six is not less than three. The fourth option is the answer to the greater-than version, which is the complement of the truth.
Explain it
They can solve absolute-value equations and are using or for every inequality.
Discussion prompt
In no more than four sentences, explain when to use and and when to use or, without asking them to memorise a rule. Then give them the one test that catches a wrong connector.
Hint: Read the inequality as a sentence about distance.
Answer:
A usable answer: read the bars as distance from a centre. If the distance has to be small, the number is close to the centre on both sides at once, so you get one band and the word and. If the distance has to be large, the number is far off to one side or the other, so you get two rays and the word or.
To check, pick a number from the region your answer leaves out and substitute it into the original. If it turns out to be a solution after all, your connector is the wrong way round — and zero is usually the easiest number to try.
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: The connector is fixed by reading the inequality as a distance sentence: close gives a band, far gives two rays. The second branch is fixed by making both changes together — negate the number and reverse the symbol. The centre is fixed by rewriting a plus inside as minus a negative. The test value is fixed by choosing from the region your answer excludes, usually zero. 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
Divide a page in two. On the left, solve one less-than absolute-value inequality: write both related inequalities with the second one's reversal marked, solve them, graph the band, and write the distance sentence it corresponds to. On the right, do the same for the greater-than version of the same expression with the same number, graphing the two rays and writing its distance sentence. Underneath both, mark on each graph the region the other one covers, and write one sentence saying why the two answers are complements. In the lower half, write a real tolerance of your own as an absolute-value inequality, solve it into a band, and mark the two limiting values. Finally, in the margin, write three test values for your greater-than answer — one from each region — with the verdict of each.
Your two graphs should together cover the whole line, meeting only at the two boundary points. If they overlap anywhere or leave a wider gap, one of the two connectors is the wrong way round.
Recap
Five things, and the first decides the shape of everything that follows.
| If the question says | Your first move is |
|---|---|
| |x - a| < c | Write two conditions joined by and |
| |x - a| > c | Write two conditions joined by or |
| Within c of a | Write |x - a| <= c |
| There is a plus inside the bars | The centre is negative |
| Check your answer | Test a value from the excluded region |
Lesson 6.8 closes the chapter by moving inequalities into two variables. The solutions become a whole half-plane rather than a set of points on a line, and the boundary is a line rather than a point.
McDougal Littell Algebra 1: Concepts and Skills, Ch. 6 Solving and Graphing Linear Inequalities — Lesson 6.7 Solving Absolute-Value Inequalities §6.7, pp. 361-366 — everything on these slides traces back here
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