Without using a calculator, solve $\frac{30-11x}{x^2-9} \le -2.$ Solve $(a-3bx^2)e^{ax-bx^3} < 0,$ where $a$ and $b$ are positive constants.

1. Rearrange the inequality to form a single rational fraction on one side with zero on the other.

\frac{30-11x}{x^2-9} + 2 \le 0 \
\frac{30-11x+2(x^2-9)}{x^2-9} \le 0 \
\frac{30-11x+2x^2-18}{x^2-9} \le 0 \
\frac{2x^2-11x+12}{x^2-9} \le 0

2. Factorise both the numerator and the denominator to identify all critical values.

\frac{(2x-3)(x-4)}{(x-3)(x+3)} \le 0 \
\text{Critical values are } x = -3, \frac{3}{2}, 3, \text{ and } 4.

3. Use a sign test or number line to determine the intervals where the fraction is negative or zero, taking care to exclude values that make the denominator zero.

\text{Using a number line to test the sign of } \frac{(2x-3)(x-4)}{(x-3)(x+3)}:\

• \text{For } x < -3, \text{ sign is } +\
• \text{For } -3 < x < \frac{3}{2}, \text{ sign is } -\
• \text{For } \frac{3}{2} < x < 3, \text{ sign is } +\
• \text{For } 3 < x < 4, \text{ sign is } -\
• \text{For } x > 4, \text{ sign is } +\

\text{Since we require the expression to be } \le 0, \text{ we take the intervals with negative signs.}\
\text{The roots of the numerator } (x = \frac{3}{2}, x = 4) \text{ are included.}\
\text{The roots of the denominator } (x = -3, x = 3) \text{ result in an undefined expression and must be excluded.}\
\text{Solution: } -3 < x \le \frac{3}{2} \text{ or } 3 < x \le 4.

1. Eliminate the exponential factor by applying its positivity property.

Since $e^{ax-bx^3} > 0$ for all real values of $x$, dividing both sides by $e^{ax-bx^3}$ does not change the inequality sign:
$a - 3bx^2 < 0$

2. Solve the resulting quadratic inequality.

$3bx^2 > a$
Since $b$ is a positive constant, we can divide by $3b$ without flipping the inequality sign:
$x^2 > \frac{a}{3b}$
Since $a$ and $b$ are positive, $\frac{a}{3b} > 0$. Solving the quadratic inequality gives:
$x < -\sqrt{\frac{a}{3b}} \quad \text{or} \quad x > \sqrt{\frac{a}{3b}}$