Solve equations and inequalities with moduli
Split modulus equations at sign changes and combine inequalities with interval or sign analysis.
Cambridge IGCSE Additional Mathematics 0606, 2028–2030 syllabus, Version 1. Paper 1 is non-calculator; Paper 2 requires a scientific calculator. These original lessons provide self-checked topic practice, not a Cambridge grade or a timed paper. Full-paper assessment and human specialist review are still outstanding.
Before you begin
- Cambridge assumes knowledge of IGCSE Mathematics or an equivalent course. Check the domain before using an inverse or logarithm, and retain exact values where the question asks for them.
Work on paper. Try each question before revealing help, and explain why your method works.
Quick readiness check
What two equations replace |A| = c when c > 0?
What you will learn
- Solve equations involving linear and quadratic moduli
- Solve modulus inequalities algebraically or graphically
- State strict and non-strict solution intervals correctly
A modulus equation can have multiple branches
For |A| = c with c ≥ 0, solve A = c or A = −c. For |A| = |B|, solve A = B or A = −B, then check each candidate in the original equation.
For a quadratic inside a modulus, first identify where the quadratic is positive or negative. A sketch is often the clearest way to manage the reflected pieces.
Translate inequalities carefully
For c > 0, |A| ≤ c is equivalent to −c ≤ A ≤ c. For |A| < c, endpoints are excluded. For |A| ≥ c, solve A ≤ −c or A ≥ c.
When the expression inside the modulus is quadratic, combine the resulting inequalities with its graph or a sign table. Keep endpoint equality consistent with the original symbol.
Pause and explain
What interval is equivalent to |x − 2| ≤ 5?
Worked example
Solve |2x − 3| = 5 and |x − 1| < 3.
Show the worked solution
- For the equation, solve 2x − 3 = 5 or 2x − 3 = −5.
- These give x = 4 or x = −1.
- For the inequality, write −3 < x − 1 < 3 and add 1 throughout.
Answer Equation solutions x = −1 or 4; inequality solution −2 < x < 4.
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 + 1| = 8.
Give me a hint
Use both signs before solving.
Compare my reasoning
- 3x + 1 = 8 gives x = 7/3.
- 3x + 1 = −8 gives x = −3.
- Substitute both values into the original modulus equation.
x = 7/3 or x = −3.
Look for these in your work
- Both branches solved
- Fraction simplified
- Solutions checked
Solve |x − 4| ≥ 2.
Give me a hint
A value is at least 2 units from 4 on either side.
Compare my reasoning
- Write x − 4 ≤ −2 or x − 4 ≥ 2.
- Solve each branch.
- Keep the union of the two regions.
x ≤ 2 or x ≥ 6.
Look for these in your work
- Correct inequality branches
- Both regions solved
- OR retained
Solve |x² − 4| ≤ 5 and explain why no extra interval appears from the lower bound.
Give me a hint
Convert the modulus inequality to a double inequality.
Compare my reasoning
- Write −5 ≤ x² − 4 ≤ 5.
- Add 4: −1 ≤ x² ≤ 9.
- Since x² ≥ 0 automatically, only x² ≤ 9 restricts x.
−3 ≤ x ≤ 3.
Look for these in your work
- Double inequality correct
- Non-negative square used
- Endpoints included
Common mistakes
- Using only the positive branch of a modulus equation
- Including endpoints for a strict inequality
- Combining alternative branches with AND instead of OR
What can you explain now?
Solve |x| = 4.
Compare with the explanation
x = −4 or x = 4
Both positive and negative inputs have modulus 4.
After trying it yourself, choose your next review. This is your self-assessment.
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Make it stick
For a modulus equation and a modulus inequality, explain how you choose signs and endpoints.
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.