Solving a Modulus Inequality

A modulus inequality unfolds by the sign of the inequality. $|A| < k$ becomes the band $-k < A < k$; the “greater than” version would split into two outer regions instead. This one is the bounded “between” case.

Solve the inequality $|2x - 1| < 5$. [3]

The working

Step 1, unfold $|2x - 1| < 5$ into a double inequality (because it is “less than”): $-5 < 2x - 1 < 5 \quad \text{(M1)}$

Step 2, add $1$ to all three parts: $-4 < 2x < 6 \quad \text{(M1)}$

Step 3, divide all three parts by $2$: $-2 < x < 3 \quad \text{(A1)}$

So the solution is the single interval $-2 < x < 3$.

Where the marks are won and lost

• “Less than” gives one band. Writing it as two separate regions ($2x - 1 < -5$ or $2x - 1 > 5$) is the “greater than” method and gives the wrong answer here.
• Operate on all three parts together. Adding $1$ and dividing by $2$ apply to the left, middle and right simultaneously.
• Dividing by a positive $2$ keeps the inequality directions. (If you ever divide by a negative, both inequality signs flip.)

Common mistakes

• Solving as $|2x-1| > 5$ (outside regions) by mistake.
• Only operating on the middle and one side, leaving the bounds inconsistent.
• Forgetting the strict inequalities (open interval, endpoints excluded).

Full method: Modulus Equations & Inequalities notes. Topic home: Equations, Inequalities and Graphs pillar.