Without using a calculator, solve the inequality
$\frac{3x^2 + 9x - 8}{x - 1} < 2.$ Hence solve the following inequality
$\frac{3e^{2x} + 9e^x - 8}{e^x - 1} < 2,$ Hence solve the following inequality
$\frac{3 + 9x - 8x^2}{x - x^2} < 2.$

1. Rearrange the inequality to compare with zero and express the expression as a single algebraic fraction.

$\frac{3x^2 + 9x - 8}{x - 1} - 2 < 0$
$\frac{3x^2 + 9x - 8 - 2(x - 1)}{x - 1} < 0$

2. Expand and simplify the numerator.

$\frac{3x^2 + 9x - 8 - 2x + 2}{x - 1} < 0$
$\frac{3x^2 + 7x - 6}{x - 1} < 0$

3. Factorise the quadratic numerator to identify the critical values.

$\frac{(3x - 2)(x + 3)}{x - 1} < 0$

The critical values are $x = \frac{2}{3}$, $x = -3$, and $x = 1$.

4. Perform a sign test (or use a number line) to determine the intervals that satisfy the strict inequality.

Testing the intervals partitioned by the critical values:

• For $x < -3$, expression is $\frac{(-)(-)}{(-)} < 0$ (Satisfied)
• For $-3 < x < \frac{2}{3}$, expression is $\frac{(+)(-)}{(-)} > 0$
• For $\frac{2}{3} < x < 1$, expression is $\frac{(+)(+)}{(-)} < 0$ (Satisfied)
• For $x > 1$, expression is $\frac{(+)(+)}{(+)} > 0$

Thus, the solution is $x < -3 \text{ or } \frac{2}{3} < x < 1$.

1. Solve the base inequality from part (i) to establish the solution range.

Solving $\frac{3x^2 + 9x - 8}{x - 1} < 2$:
$\frac{3x^2 + 9x - 8}{x - 1} - 2 < 0$
$\frac{3x^2 + 7x - 6}{x - 1} < 0$
$\frac{(3x - 2)(x + 3)}{x - 1} < 0$
Critical values are $x = -3$, $x = \frac{2}{3}$, and $x = 1$.
By testing the intervals, the solution to the inequality is:
$x < -3$ or $\frac{2}{3} < x < 1$.

2. Substitute $e^x$ into the established inequality.

The given inequality $\frac{3e^{2x} + 9e^x - 8}{e^x - 1} < 2$ has the exact same structure as the inequality in part (i), with $x$ replaced by $e^x$.
Using the result from (i), we can deduce:
$e^x < -3 \quad \text{or} \quad \frac{2}{3} < e^x < 1$

3. Solve for $x$, discarding any invalid intervals.

Since $e^x > 0$ for all real values of $x$, the inequality $e^x < -3$ has no real solution.
For the remaining inequality $\frac{2}{3} < e^x < 1$, taking the natural logarithm on all parts gives:
$\ln\left(\frac{2}{3}\right) < x < \ln(1)$
$\ln\left(\frac{2}{3}\right) < x < 0$

1. Identify the substitution that relates the target inequality to the inequality solved in part (i).

Let $y = \frac{1}{x}$. Substituting $y$ into the inequality from part (i) yields:
$\frac{3\left(\frac{1}{x}\right)^2 + 9\left(\frac{1}{x}\right) - 8}{\frac{1}{x} - 1} < 2$
Multiplying the numerator and the denominator on the left-hand side by $x^2$ (assuming $x \neq 0$):
$\frac{3 + 9x - 8x^2}{x(1 - x)} < 2$
$\frac{3 + 9x - 8x^2}{x - x^2} < 2$
This precisely matches the target inequality, meaning the solution can be found by replacing the variable in part (i)'s solution with $\frac{1}{x}$.

2. Apply the substitution to the solution derived in part (i).

From part (i), the solution to $\frac{3y^2 + 9y - 8}{y - 1} < 2$ is:
$y < -3 \quad \text{or} \quad \frac{2}{3} < y < 1$
Substituting $y = \frac{1}{x}$ into these solution ranges:
$\frac{1}{x} < -3 \quad \text{or} \quad \frac{2}{3} < \frac{1}{x} < 1$

3. Solve the newly formed inequalities for $x$ and combine them for the final answer.

For the first inequality:
$\frac{1}{x} < -3 \implies \frac{1}{x} + 3 < 0 \implies \frac{1 + 3x}{x} < 0$
By analyzing the signs of the numerator and denominator, this gives $-\frac{1}{3} < x < 0$.

For the second inequality:
$\frac{2}{3} < \frac{1}{x} < 1$
Since all terms are strictly positive, taking the reciprocal of all parts reverses the inequality signs:
$\frac{3}{2} > x > 1 \implies 1 < x < \frac{3}{2}$

Combining both ranges gives the final solution:
$-\frac{1}{3} < x < 0 \quad \text{or} \quad 1 < x < \frac{3}{2}$