The modulus \(|x-a|\) means the distance between \(x\) and \(a\).

So modulus inequalities describe numbers that are either within a distance or outside a distance.

For \(a>0\):

\[ |x| < a \quad \Longleftrightarrow \quad -a < x < a \]

\[ |x| \le a \quad \Longleftrightarrow \quad -a \le x \le a \]

\[ |x| > a \quad \Longleftrightarrow \quad x < -a \text{ or } x > a \]

\[ |x| \ge a \quad \Longleftrightarrow \quad x \le -a \text{ or } x \ge a \]

If the modulus is centred at \(a\), use the same idea:

\[ |x-a| < b \quad \Longleftrightarrow \quad a-b < x < a+b \]

\[ |x-a| > b \quad \Longleftrightarrow \quad x < a-b \text{ or } x > a+b \]

These work because \(|x-a|\) is the distance from \(x\) to \(a\).

Solve:

\[ |x-2| < 5 \]

This means the distance between \(x\) and 2 is less than 5:

\[ -5 < x-2 < 5 \]

\[ -3 < x < 7 \]

Solve:

\[ |2x+1| \ge 7 \]

Split into two cases:

\[ 2x+1 \ge 7 \quad \text{or} \quad 2x+1 \le -7 \]

\[ 2x \ge 6 \quad \text{or} \quad 2x \le -8 \]

\[ x \ge 3 \quad \text{or} \quad x \le -4 \]

Solve:

\[ |3x-4| \le 8 \]

Write as a double inequality:

\[ -8 \le 3x-4 \le 8 \]

Add 4 throughout:

\[ -4 \le 3x \le 12 \]

\[ -\frac{4}{3} \le x \le 4 \]

If the right-hand side is negative, be careful.

\[ |x| < -3 \]

has no solution, because modulus can never be negative.

\[ |x| \ge -3 \]

is true for all real \(x\), because modulus is always at least 0.

• \(|x-a| < b\): values inside the interval
• \(|x-a| \le b\): values inside, including endpoints
• \(|x-a| > b\): values outside the interval
• \(|x-a| \ge b\): values outside, including endpoints