Solve modulus inequalities using intervals and valid sign cases
Translate a distance bound into an inside or outside interval and split a variable comparison where its inner expression changes sign.
Section 2.1 · Algebra
Exam scope and assessment notes
Pure Mathematics 2 (Paper 2, AS only) and Pure Mathematics 3 (Paper 3, A Level) · Shared Algebra foundations, Mathematics 9709, 2026–2027. Paper 1 knowledge is assumed. These modulus and polynomial requirements occur in both sections 2.1 and 3.1. Paper 3 additionally requires partial fractions and rational-power binomial expansions; follow its separate lessons for those methods. The AS route using Papers 1 + 2 cannot be carried forward to A Level. Both sections use linear modulus graphs; nonlinear y = |f(x)| and y = f(|x|) graphs are excluded. Polynomial dividends have degree at most 4 and divisors are linear or quadratic. Written practice is self-checked.
Before you begin
- Solve linear inequalities and distinguish strict from inclusive endpoints.
- Interpret the two branches of a linear modulus.
Work on paper. Try each question before revealing help, and explain why your method works.
Quick readiness check
Which interval represents |x − 2| < 3?
What you will learn
- Solve strict and inclusive modulus bounds with correct endpoint treatment.
- Solve an inequality with a variable comparison by intersecting each result with its sign-case interval.
A distance bound gives an inside or outside interval
For b > 0, |x − a| < b means a − b < x < a + b: the input is less than b units from a. With ≤, include both endpoints. By contrast |x − a| > b means x < a − b or x > a + b; the two outside intervals are joined by ‘or’, not ‘and’. With ≥, include the endpoints. A positive coefficient can first be factored out as a scale on the distance.
For |3x − 6| < 9, divide the modulus scale by 3 to get |x − 2| < 3, so −1 < x < 5. The endpoints are excluded because equality gives exactly 9. A non-positive bound needs separate care: |u| < 0 has no solutions, |u| ≤ 0 requires u = 0, and |u| ≥ 0 holds for all real u where the expression is defined. Do not apply a positive-distance interval rule to a negative bound.
Intersect each branch with the interval where it applies
For 2x + 5 < |x + 1|, split at x = −1. When x ≥ −1, the right side is x + 1, so the inequality gives x < −4, which has no overlap with that branch interval. When x < −1, the right side is −x − 1, so 3x < −6 gives x < −2, entirely within its branch. The complete solution is x < −2, with the boundary excluded.
Squaring an inequality is order-preserving only under suitable non-negative sign conditions. Here the left side is negative for some valid inputs, so squaring both sides blindly can discard solutions. Keep the original comparison, solve each linear branch and combine the valid intersections. Verify a sample from each reported interval and test boundary equality to decide whether endpoints belong.
Pause and explain
For the branch x ≥ −1 in 2x + 5 < |x + 1|, solving gives x < −4. What remains?
Worked example
Solve 2x + 5 < |x + 1| over the reals.
Show the worked solution
- For x ≥ −1, use |x + 1| = x + 1: the resulting x < −4 has no overlap with x ≥ −1.
- For x < −1, use |x + 1| = −x − 1: the inequality gives x < −2, valid in this branch.
- At x = −2 both sides equal 1, so the strict comparison excludes the boundary; combine to get x < −2.
Answer x < −2.
Try a topic test
Choose Easy, Normal or Difficult. Save your answers and open worked review after finishing.
Test this topic →From guided practice to a new situation
Write your answer and reasoning first. Use a hint only if you are stuck. The model solution is for self-checking; your written work is not automatically marked.
Solve |3x − 6| < 9 and mark the boundary inputs on a number line.
Give me a hint
Use −9 < 3x − 6 < 9 and divide the full compound inequality by positive 3.
Compare my reasoning
- Add 6 to get −3 < 3x < 15.
- Divide by 3 to get −1 < x < 5.
- Use open endpoints at −1 and 5 because both give modulus 9 rather than a value below 9.
−1 < x < 5; both endpoints excluded.
Look for these in your work
- Apply each operation to all three parts of the inequality.
- Use open endpoints for the strict comparison.
Solve |2x + 1| ≥ 5, stating whether each endpoint is included.
Give me a hint
The two signed cases are 2x + 1 ≤ −5 or 2x + 1 ≥ 5.
Compare my reasoning
- The negative branch gives 2x ≤ −6, so x ≤ −3.
- The positive branch gives 2x ≥ 4, so x ≥ 2.
- Combine with ‘or’; both endpoints give modulus 5 and are included.
x ≤ −3 or x ≥ 2; both endpoints included.
Look for these in your work
- Use two outside intervals joined by ‘or’.
- Retain equality at both boundary values.
For d(t) = 3t − 12 metres on 0 ≤ t ≤ 6 seconds, find the full time interval when the reference distance |d(t)| is at most 3 m.
Give me a hint
An ‘at most’ distance bound is inclusive; intersect its interval with the stated time domain.
Compare my reasoning
- Use −3 ≤ 3t − 12 ≤ 3.
- Adding 12 and dividing by 3 gives 3 ≤ t ≤ 5.
- This interval lies within [0, 6]; both endpoint distances equal 3 m, so both times are included.
3 ≤ t ≤ 5 seconds, with both endpoints included.
Look for these in your work
- Use an inclusive inside-distance interval.
- Check both boundary values and the allowed time domain.
Common mistakes
- Use ‘or’ for two outside intervals, not an impossible ‘and’ condition.
- Intersect a branch solution with the interval where that branch is valid.
- Keep strict endpoints open and do not square a signed comparison without justification.
What can you explain now?
Solve |x − 2| > 3.
Compare with the explanation
x < −1 or x > 5.
The input must lie more than 3 units from 2, so use the two outside intervals with strict endpoints.
After trying it yourself, choose your next review. This is your self-assessment.
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Make it stick
Contrast inside and outside modulus intervals, then explain why a sign-case answer must be intersected with its branch condition.
If you needed the solution, close it and solve the problem again from a blank page. A correct answer is stronger when you can explain the reason for each step.