Solution: $x \le -3$ OR $x \ge \frac{5}{3}$

Interval Notation: $(-\infty, -3] \cup [\frac{5}{3}, \infty)$

Special Cases:

$|X| < -5$: No solution (absolute value cannot be negative).
$|X| > -5$: All real numbers (absolute value is always greater than a negative number).
$|X| < 0$: No solution (absolute value is non-negative).
$|X| \le 0$: $X=0$ (only 0 has absolute value 0).

A concise way to express ranges of numbers, often used for inequality solutions.

Inequality	Interval Notation	Description
$a < x < b$	$(a, b)$	Open interval; $x$ is between $a$ and $b$, excluding $a, b$.
$a \le x \le b$	$[a, b]$	Closed interval; $x$ is between $a$ and $b$, including $a, b$.
$a < x \le b$	$(a, b]$	Half-open; $x$ is greater than $a$ and less than or equal to $b$.
$a \le x < b$	$[a, b)$	Half-open; $x$ is greater than or equal to $a$ and less than $b$.
$x < a$	$(-\infty, a)$	All numbers less than $a$.
$x \le a$	$(-\infty, a]$	All numbers less than or equal to $a$.
$x > a$	$(a, \infty)$	All numbers greater than $a$.
$x \ge a$	$[a, \infty)$	All numbers greater than or equal to $a$.
All Real Numbers	$(-\infty, \infty)$	The entire number line.

Solving inequalities involving quadratic expressions like $ax^2 + bx + c > 0$.

Find Critical Points: Set the quadratic expression equal to zero and solve for $x$. These are the roots of the quadratic equation.
Plot Critical Points: Place these points on a number line. They divide the line into intervals.
Test Intervals: Pick a test value from each interval and substitute it into the original inequality.
Determine Solution: Identify the intervals where the inequality holds true.

Critical Points: $x^2 - x - 6 = 0 \implies (x-3)(x+2) = 0$. Roots are $x=3$ and $x=-2$.
Intervals: $(-\infty, -2)$, $(-2, 3)$, $(3, \infty)$.
Test Points:
- For $(-\infty, -2)$, pick $x=-3$: $(-3)^2 - (-3) - 6 = 9 + 3 - 6 = 6$. Since $6 > 0$, this interval is a solution.
- For $(-2, 3)$, pick $x=0$: $(0)^2 - (0) - 6 = -6$. Since $-6 \not> 0$, this interval is not a solution.
- For $(3, \infty)$, pick $x=4$: $(4)^2 - (4) - 6 = 16 - 4 - 6 = 6$. Since $6 > 0$, this interval is a solution.
Solution: $x < -2$ OR $x > 3$.
Interval Notation: $(-\infty, -2) \cup (3, \infty)$.

Solving inequalities with rational expressions, e.g., $\frac{P(x)}{Q(x)} > 0$.

Find Critical Points: Set both the numerator and denominator equal to zero and solve for $x$.</