Math101 · Independent Ontario learning library
Math101learn.math101.caAbsolute Value Inequalities
Absolute value inequalities describe distance ranges. For $k>0$, $|u|<k$ means $-k<u<k$ (inside a band), while $|u|>k$ means $u<-k$ or $u>k$ (outside the band).
These inequalities model allowable error, measurement tolerance, and exclusion zones. Their geometry makes compound inequalities and set unions meaningful.
Intuition and core definition
Absolute value inequalities describe distance ranges. For $k>0$, $|u|<k$ means $-k<u<k$ (inside a band), while $|u|>k$ means $u<-k$ or $u>k$ (outside the band). Inclusive symbols produce inclusive endpoints. Negative or zero bounds require separate logical analysis.
Notation, language, and conditions
$|x-a|\le k$ describes the closed interval $[a-k,a+k]$ for $k\ge0$. $|x-a|>k$ describes two rays $(-\infty,a-k)\cup(a+k,\infty)$. “And” corresponds to intersection between bounds; “or” corresponds to the union of outside regions.
Why this idea matters
Absolute-value inequalities describe points inside or outside a distance band, connecting compound inequalities to geometric intervals.
A dependable method
- Isolate the absolute-value expression and inspect the bound.
- Translate a “less than” comparison into a compound AND inequality.
- Translate a “greater than” comparison into two OR inequalities.
- Solve each part, preserving endpoint inclusion.
- Graph and test a centre point plus points inside and outside the boundaries.
Worked example
Representations and interpretation
On a number line, $|x-a|$ is distance from $a$. A small-distance inequality shades between two boundary points; a large-distance inequality shades away from them. The graph of $y=|x-a|$ compared with $y=k$ shows the same regions.
Reasoning about variations
If $|x|<-1$, there are no solutions because absolute value is never negative. If $|x|>-1$, every real number works. At bound zero, $|x|\le0$ gives only $x=0$, while $|x|>0$ excludes zero.
Common mistakes
How to check your work
- Test the centre, an endpoint, and an outside point in the original inequality.
- Read the graph as a distance statement.
- Compare interval notation, compound notation, and shading for agreement.
Practice
- Solve $|x+2|<5$.
- Solve $|2x|\ge6$.
- Solve $|x-1|<-4$.
Answers and brief solutions
Show answers
- $-7<x<3$ $-5<x+2<5$, then subtract $2$ throughout.
- $x\le-3$ or $x\ge3$ $2x\le-6$ or $2x\ge6$.
- No solution A nonnegative distance cannot be less than $-4$.
Synthesis and transfer
A manufacturing tolerance of less than a fixed deviation produces an intersection between two bounds, while a minimum-separation rule produces two outward rays.
A target mass of $250$ g with tolerance $3$ g gives $|m-250|\le3$, or $247\le m\le253$. The word “within” creates the inside of a distance band, so the compound statement uses an intersection. By contrast, requiring a reading to be at least $3$ units from the target gives $m\le247$ or $m\ge253$, two outward regions joined by a union. Endpoint inclusion follows from whether equality is permitted. Plotting the centre and radius before manipulating symbols makes the AND/OR structure visible and provides a check against memorized rules that are easily reversed.
Related topics
Teaching and accessibility note
Try it yourself
Hints are part of learning. Open one whenever it makes the next step feel possible.
Solve $|x+2|<5$.
- $-5<x+2<5$, then subtract $2$ throughout.
End of lesson
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