Absolute Value Inequalities

For $a > 0$, these two rules are the foundation of everything:

These rules come directly from the definition of absolute value as distance from zero.

• $|x| < a$ means "$x$ is within distance $a$ of zero" — i.e., $x$ lies between $-a$ and $a$: a single interval.
• $|x| > a$ means "$x$ is more than distance $a$ from zero" — i.e., $x$ lies to the left of $-a$ or to the right of $a$: two separate intervals.

The same rules apply when $x$ is replaced by any expression, e.g. $|2x - 1| < 5$ becomes $-5 < 2x - 1 < 5$.

|x| < a -a < x < a — one interval (between -a and a); |x| > a x < -a or x > a — two intervals (union of two rays)

Pause — copy both fundamental absolute value rules into your book: $|x| < a \Leftrightarrow -a < x < a$ (one interval); $|x| > a \Leftrightarrow x < -a$ or $x > a$ (two separate rays).

We just saw that $|x| < a Leftrightarrow -a < x < a$ and $|x| > a Leftrightarrow x < -a$ or $x > a$. That raises a question: when the expression inside the modulus contains $x$ (like $|2x-3| le 5$), which algebraic technique resolves the absolute value without guessing the sign? This card answers it → case analysis (split on whether the inner expression is $ge 0$ or $< 0$) and squaring both sides (valid when both sides are non-negative).

Method 1: Case Analysis

For expressions like $|2x - 1| > 5$, use two cases based on the sign of the expression inside the bars:

• Case 1: $2x - 1 \ge 0$ (so $|2x - 1| = 2x - 1$). Solve $2x - 1 > 5$, giving $x > 3$. Combined with $x \ge \tfrac{1}{2}$: solution from this case is $x > 3$.
• Case 2: $2x - 1 < 0$ (so $|2x - 1| = -(2x - 1) = 1 - 2x$). Solve $1 - 2x > 5$, giving $x < -2$. Combined with $x < \tfrac{1}{2}$: solution from this case is $x < -2$.
• Final solution: Union of both cases: $x < -2$ or $x > 3$.

Method 2: Squaring Both Sides

When both sides of the inequality are non-negative, we can square both sides without changing the inequality direction. This is particularly useful for inequalities like $|x - 3| < |x + 1|$:

• Square both sides: $(x - 3)^2 < (x + 1)^2$
• Expand: $x^2 - 6x + 9 < x^2 + 2x + 1$
• Simplify: $-6x + 9 < 2x + 1 \Rightarrow 8 < 8x \Rightarrow x > 1$

Case analysis: Case 1 — expression inside 0, so || = (); Case 2 — expression inside < 0, so || = -(); Always intersect each case's solution with the condition used in that case

Pause — copy the case-analysis method into your book: Case 1 (expression $\ge 0$): remove modulus sign directly; Case 2 (expression $< 0$): negate; always intersect each case's solution with the condition defining that case.