Let's analyze and solve the given equation:

$\frac{4}{x+1} - \frac{4}{x^2 - 2x - 3} = \frac{1}{3 - x}$

Step 1: Factorize the quadratic term

The quadratic term in the second denominator $x^2 - 2x - 3$ can be factorized as:

$x^2 - 2x - 3 = (x - 3)(x + 1)$

Thus, the equation becomes:

$\frac{4}{x+1} - \frac{4}{(x-3)(x+1)} = \frac{1}{3 - x}$

Step 2: Express $3 - x$ as $-(x - 3)$

Notice that $3 - x$ can be rewritten as $-(x - 3)$. So, the right-hand side becomes:

$\frac{1}{3 - x} = -\frac{1}{x - 3}$

Now, the equation looks like:

$\frac{4}{x+1} - \frac{4}{(x-3)(x+1)} = -\frac{1}{x-3}$

Step 3: Combine the terms on the left-hand side

We need a common denominator on the left-hand side. The common denominator is $(x - 3)(x + 1)$, so:

$\frac{4}{x+1} = \frac{4(x - 3)}{(x - 3)(x + 1)}$

Thus, the left-hand side becomes:

$\frac{4(x - 3)}{(x - 3)(x + 1)} - \frac{4}{(x - 3)(x+1)} = \frac{4(x - 3) - 4}{(x - 3)(x + 1)}$

Simplifying the numerator:

$4(x - 3) - 4 = 4x - 12 - 4 = 4x - 16$

So, the left-hand side is now:

$\frac{4x - 16}{(x - 3)(x + 1)}$

Step 4: Set up the equation

Now the equation becomes:

$\frac{4x - 16}{(x - 3)(x + 1)} = -\frac{1}{x - 3}$

Step 5: Multiply both sides by $(x - 3)$

To eliminate the denominator $x - 3$, we multiply both sides by $(x - 3)$ (but be careful with the restriction $x \neq 3$):

$\frac{4x - 16}{x + 1} = -1$

Step 6: Solve for $x$

Multiply both sides by $x + 1$:

$4x - 16 = -(x + 1)$

Expand the right-hand side:

$4x - 16 = -x - 1$

Now, collect like terms:

$4x + x = -1 + 16$

$5x = 15$

Finally, solve for $x$:

$x = 3$

Step 7: Verify the solution

However, $x = 3$ is not a valid solution because it would make the denominators $(x - 3)$ zero. Therefore, the solution $x = 3$ must be rejected.

Final Conclusion:

There is no valid solution for this equation, since $x = 3$ is not allowed.

Would you like more details on any step or have further questions? Here are 5 related questions for practice:

1. How do you factor quadratic expressions like $x^2 - 2x - 3$?
2. What happens if a solution makes a denominator zero in a rational equation?
3. How can you rewrite fractions with common denominators?
4. Why do we reject certain solutions in rational equations?
5. How do we handle negative signs when manipulating rational equations?

Tip: Always check your solutions in the original equation to ensure they don't violate any restrictions (like making a denominator zero).