Modulus inequalities can be solved by thinking of modulus as distance on the number line.
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.