Absolute Value Equations & Inequalities: Distance Models

Absolute value measures distance from zero

For a real number x, |x| is its non-negative distance from zero. Thus |−7|=7, |7|=7 and |0|=0. Absolute value is not an instruction to remove a negative sign anywhere in a longer expression.

Evaluate the expression inside the bars first. For example |3−8|=|−5|=5, while |3|−|8|=−5. The location of the bars changes the operation.

Shift the centre of the distance

The expression |x−a| measures distance from a. In |x−2|=4, the points are 4 units to either side of 2: x=−2 and x=6.

For |x+3|, the centre is −3 because x+3=x−(−3). Draw the centre before marking endpoints to avoid the same sign reversal that occurs in vertex form.

Count zero, one or two solutions

For |u|=k, a positive k gives u=k or u=−k. If k=0, the two cases coincide at u=0. If k<0, there is no real solution because distance cannot be negative.

Solve any algebra inside u after choosing the valid cases. Check both answers in the original equation, especially if you first move constants or divide by coefficients.

Solve inside-distance inequalities

For k>0, |x−a|<k means a−k<x<a+k. The point is closer than k to the centre. Using ≤ includes both endpoints.

For |x−3|≤2, the solution is 1≤x≤5. Test the centre 3, an endpoint 1 and an outside point 0. They explain the filled endpoints and shaded interval.

Solve outside-distance inequalities

For k>0, |x−a|>k means x<a−k or x>a+k. There are two separate rays; the word “or” is essential. Using ≥ includes the endpoints.

For |x+1|>3, the centre is −1 and the boundaries are −4 and 2. Solutions lie below −4 or above 2. Values between the boundaries are too close to the centre.

Isolate the bars before splitting cases

For 2|x−1|+3=11, subtract 3 and divide by 2 to get |x−1|=4. Then x−1=4 or −4, giving x=5 or −3.

Do not split the original equation into x−1=11 and x−1=−11. The outer multiplication and addition are part of the equation and must be undone first.

Handle negative and zero bounds directly

Every absolute value is ≥0. Thus |x|<0 has no solutions, |x|≥0 has all real solutions and |x|≤0 has only x=0. These boundary cases need reasoning rather than a memorised positive-bound template.

If multiplying or dividing an inequality by a negative number, reverse the inequality sign. Isolating the absolute value first usually makes this easy to see.

A solution record that shows the meaning

Record the centre, distance condition, endpoints, whether boundaries are included and one test point from each region. Give the final answer in words, inequality or interval form as your course requests.

A real-world tolerance can use this model: a length within 0.2 cm of 10 cm satisfies |L−10|≤0.2. State units and the context’s allowed domain. This guide uses simple linear expressions inside the bars; equations with several absolute-value terms need additional case analysis.

See the reasoning

Worked examples and checks

Try, explain, check

12 questions with explained answers

Try each task before opening its explanation. These are original BananaLearn questions, not official exam or past-paper questions. Where several answers can work, the explanation gives the criteria.

1. Evaluate |−12|.

   Check answer and reasoning

   12
   It is a distance from zero.

2. Evaluate |2−9|.

   Check answer and reasoning

   7
   Evaluate inside first.

3. Solve |x|=5.

   Check answer and reasoning

   x=−5 or 5
   Both are 5 units from zero.

4. Solve |x|=0.

   Check answer and reasoning

   x=0
   Only zero has zero distance.

5. Solve |x|=−5.

   Check answer and reasoning

   No real solution
   Distance cannot be negative.

6. Solve |x−1|=3.

   Check answer and reasoning

   x=−2 or 4
   Move 3 units either side of 1.

7. Solve |x+4|≤2.

   Check answer and reasoning

   −6≤x≤−2
   The centre is −4 and the boundaries are included.

8. Solve |x−2|>1.

   Check answer and reasoning

   x<1 or x>3
   Use the two outside rays.

9. Solve 3|x|=12.

   Check answer and reasoning

   x=−4 or 4
   Divide by 3 before splitting cases.

10. Is |x|≥0 true for every real x?

    Check answer and reasoning

    Yes
    All absolute values are non-negative.

11. A length is within 0.5 cm of 8 cm. Give its interval.

    Check answer and reasoning

    7.5≤L≤8.5 cm
    The wording includes the boundary tolerance here.

12. Does |a+b| always equal |a|+|b|?

    Check answer and reasoning

    No
    For a=3,b=−3, the left side is 0 and the right is 6.

A checking record to use in your notebook

Copy these headings and fill them for one worked task, then for a fresh attempt. Use your own evidence rather than copying the model answer.

Question or sentence

What am I being asked to explain?

Model or evidence

Which diagram, rule or passage detail supports my method?

Work

What steps connect the question to the answer?

Independent check

Can I substitute, recount, trace a path or check the text?

Next attempt

What changes in a fresh example?

Choose the next useful step

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Sources, level and update

Original BananaLearn explanations, examples, illustrations and practice. Source facts checked 7 November 2026. Suggested levels are learning guides, not a claim that every board or textbook uses the same sequence.

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