Equations, Inequalities and Graphs · 0606 Topic 4

Modulus Equations & Inequalities |ax+b|

Teacher Rig, IGCSE Add Math tutor

Written by Teacher Rig

8 years teaching IGCSE Add Math · Updated 12 June 2026

Modulus equations and inequalities are casework exercises: each modulus hides two possibilities, and the marks go to students who surface both on paper.

Equations: ax+b=k\lvert ax + b \rvert = k

For k>0k > 0, split explicitly:

3x2=7\lvert 3x - 2 \rvert = 7 \to 3x2=73x - 2 = 7 or 3x2=73x - 2 = -7 \to x=3x = 3 or x=53x = -\frac{5}{3}

The written “or”-split line is the method mark. For k=0k = 0 there’s one solution; for k<0k < 0, none, and stating “no solutions, since a modulus cannot be negative” can be the entire answer.

Two moduli: f(x)=g(x)\lvert f(x) \rvert = \lvert g(x) \rvert, square both sides

2x1=x+4\lvert 2x - 1 \rvert = \lvert x + 4 \rvert \to (2x1)2=(x+4)2(2x - 1)^2 = (x + 4)^2 \to 4x24x+1=x2+8x+164x^2 - 4x + 1 = x^2 + 8x + 16 \to 3x212x15=03x^2 - 12x - 15 = 0 \to x24x5=0x^2 - 4x - 5 = 0 \to x=5,x=1x = 5, x = -1

Squaring kills both moduli at once and feeds a standard quadratic. The ±\pm split works too but doubles the cases; squaring is the safer exam route.

Modulus = expression: solve, then check

x3=2x\lvert x - 3 \rvert = 2x \to x3=2xx - 3 = 2x gives x=3x = -3; (x3)=2x-(x - 3) = 2x gives x=1x = 1. Check x=3x = -3: RHS=6<0\text{RHS} = -6 < 0, but a modulus can’t equal a negative, reject, with the reason written. Check x=1x = 1: 2=2\lvert -2 \rvert = 2x=1x = 1

The rejection sentence is a mark, and unchecked invalid solutions are a headline examiner complaint.

Inequalities: two unpackings

For k>0k > 0:

  • ax+b<k\lvert ax + b \rvert < k \to one sandwich: k<ax+b<k-k < ax + b < k
  • ax+b>k\lvert ax + b \rvert > k \to two regions: ax+b>kax + b > k or ax+b<kax + b < -k

2x35\lvert 2x - 3 \rvert \ge 5 \to 2x352x - 3 \ge 5 or 2x352x - 3 \le -5 \to x4x \ge 4 or x1x \le -1

Write two-region answers as two inequalities joined by “or”, never chained. If you blank on which unpacking is which, sketch y=ax+by = \lvert ax + b \rvert against y=ky = k (the graph method); the picture settles it in five seconds. The squaring method also works for inequalities between two moduli: x1<x4\lvert x - 1 \rvert < \lvert x - 4 \rvert \to square, solve the resulting quadratic inequality.

Common mistakes

  • Only the positive case solved, half the solutions, every time
  • ”<” and ”>” unpackings swapped
  • Modulus-equals-expression answers left unchecked
  • Squaring used without collecting to zero before solving
  • Final answers chained into impossible inequalities

Full topic context: Equations, Inequalities & Graphs notes · the graphical side: graphs of modulus functions.

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