Inequalities with a modulus Cambridge IGCSE Additional Mathematics revision
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In plain words
|x| < 4 says "x is less than 4 away from zero": between −4 and 4. |x| > 4 says "x is more than 4 away from zero": below −4 or above 4. Every modulus inequality is one of these two pictures.
5 things to know
- |ax + b| < c means −c < ax + b < c. The solution is a single stretch.
- |ax + b| > c means ax + b < −c or ax + b > c. The solution is two separate stretches.
- The same holds with ≤ and ≥.
- When both sides are moduli, as in |ax + b| ≤ |cx + d|, square both sides. Both are non-negative, so the inequality is kept, and a quadratic inequality is left.
- A graph of both sides shows the answer too: find where one graph is below the other.
Worked example
Solve |2x − 1| < 5.
- −5 < 2x − 1 < 5.
- Add 1 throughout: −4 < 2x < 6.
- Divide by 2: −2 < x < 3.
Worked example
Solve |x − 1| ≤ |2x + 3|.
- Square both sides: x² − 2x + 1 ≤ 4x² + 12x + 9.
- Rearrange: 0 ≤ 3x² + 14x + 8, which is (3x + 2)(x + 4) ≥ 0.
- The critical values are −4 and −2/3. The parabola is above the axis outside them: x ≤ −4 or x ≥ −2/3.
Tips and tricks
- "Less than" gives one region between two values. "Greater than" gives two regions. Getting this the wrong way round is the usual error.
- Squaring is only safe when both sides are certainly not negative, as two moduli are.
It lands in your notebook with its questions as flashcards.
Inequalities with a modulus: 5 questions and answers
These are the quiz’s questions. Do the quiz first, then come back here for the ones that got you.
What is the solution of |x| < 4?
x is less than 4 away from zero.
What is the solution of |x| > 2?
x is more than 2 away from zero, on either side.
What is the solution of |x − 3| ≤ 1?
−1 ≤ x − 3 ≤ 1.
What is the solution of |2x| ≥ 6?
2x ≤ −6 or 2x ≥ 6.
What is the solution of |x + 1| < 3?
−3 < x + 1 < 3.
Quiz
5 questions
Tap an answer and you’ll see straight away whether it’s right, and why.
Worksheet
3 questions, 10 marks. Write your answers on paper, then check them.
Inequalities with a modulus
Cambridge IGCSE Additional Mathematics 0606 · 10 marks · papermunch.org
Name ______________________________ Date ______________
Solve |3x + 2| ≤ 8.[3]
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−10/3 ≤ x ≤ 2. −8 ≤ 3x + 2 ≤ 8, so −10 ≤ 3x ≤ 6.
Solve |x − 5| > 2.[3]
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x < 3 or x > 7. x − 5 < −2 or x − 5 > 2.
Solve |2x − 1| < |x + 1|.[4]
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0 < x < 2. Squaring gives 4x² − 4x + 1 < x² + 2x + 1, so 3x² − 6x < 0, so 3x(x − 2) < 0.
Answers: Inequalities with a modulus
- 1. −10/3 ≤ x ≤ 2. −8 ≤ 3x + 2 ≤ 8, so −10 ≤ 3x ≤ 6.
- 2. x < 3 or x > 7. x − 5 < −2 or x − 5 > 2.
- 3. 0 < x < 2. Squaring gives 4x² − 4x + 1 < x² + 2x + 1, so 3x² − 6x < 0, so 3x(x − 2) < 0.



