Express $\dfrac{33-8x}{x^2+2x-15}+2$ as a single algebraic fraction. Hence, without using a calculator, solve exactly the inequality $\dfrac{33-8x}{x^2+2x-15}>-2$. Using your answer to part (a), find the set of values of $x$ for which $\dfrac{33-8e^{2x}}{e^{4x}+2e^{2x}-15}>-2$.

1. Combine the expression into a single algebraic fraction.

$\frac{33-8x}{x^2+2x-15}+2$
$= \frac{33-8x+2(x^2+2x-15)}{x^2+2x-15}$
$= \frac{33-8x+2x^2+4x-30}{x^2+2x-15}$
$= \frac{2x^2-4x+3}{x^2+2x-15}$

2. Rewrite the inequality to compare with zero and analyse the sign of the numerator.

$\frac{33-8x}{x^2+2x-15} > -2$
$\frac{33-8x}{x^2+2x-15} + 2 > 0$
$\frac{2x^2-4x+3}{x^2+2x-15} > 0$
Check the discriminant ($\Delta$) of the numerator $2x^2-4x+3$:
$\Delta = (-4)^2 - 4(2)(3) = 16 - 24 = -8$
Since the discriminant is negative ($-8 < 0$) and the coefficient of $x^2$ is positive ($2 > 0$), the quadratic expression is always positive, meaning $2x^2-4x+3 > 0$ for all real $x$.

3. Determine the required condition for the denominator and solve the resulting inequality.

For the rational fraction to be positive when the numerator is strictly positive, the denominator must also be strictly positive:
$x^2+2x-15 > 0$
Factorising the quadratic expression:
$(x+5)(x-3) > 0$
Solving this inequality gives:
$x < -5 \text{ or } x > 3$

1. Apply the result from part (a) by using the substitution $x = e^{2x}$.

From part (a), solving $\dfrac{33-8x}{x^2+2x-15} > -2$ yields the solution $x < -5$ or $x > 3$.

Replacing $x$ with $e^{2x}$ in the inequality gives:
$e^{2x} < -5 \quad \text{or} \quad e^{2x} > 3$

2. Solve the resulting exponential inequalities and reject any invalid cases.

Since the exponential function $e^{2x} > 0$ for all real $x$, the inequality $e^{2x} < -5$ has no real solution.

Solving the valid inequality:
$e^{2x} > 3$
Taking the natural logarithm on both sides:
$2x > \ln 3$
$x > \frac{1}{2}\ln 3$