Solve the inequality

For the fractions to be defined, the denominators cannot be zero.

$$x - 5 \neq 0 \implies x \neq 5$$

Since the left-hand side is an absolute value, it must be non-negative. Therefore, the right-hand side must also be non-negative.

$$\frac{x+1}{x-5} \ge 0$$

We can analyze the signs of the numerator and denominator:

• If $x+1 \ge 0 \implies x \ge -1$
• If $x-5 > 0 \implies x > 5$

We can use a sign table for $\frac{x+1}{x-5}$:

Interval	$x+1$	$x-5$	$\frac{x+1}{x-5}$
$x < -1$	$-$	$-$	$+$
$-1 \le x < 5$	$+$	$-$	$-$
$x > 5$	$+$	$+$	$+$

So, the condition $\frac{x+1}{x-5} \ge 0$ holds for $x \le -1$ or $x > 5$.
This is the valid domain for our solutions.

We can rewrite the left side as $\frac{|x+3|}{|x-5|}$.

$$\frac{|x+3|}{|x-5|} = \frac{x+1}{x-5}$$

We need to consider two cases based on the sign of $x-5$.

Case 1: $x-5 > 0 \implies x > 5$.
In this case, $|x-5| = x-5$.
The equation becomes:

$$\frac{|x+3|}{x-5} = \frac{x+1}{x-5}$$

Since $x-5 > 0$, we can multiply both sides by $x-5$:

$$|x+3| = x+1$$

Now, we solve this equation for $x > 5$.
Subcase 1.1: $x+3 \ge 0 \implies x \ge -3$.
Since we are in the interval $x > 5$, the condition $x \ge -3$ is always satisfied.
So, $x+3 = x+1$.

$$3 = 1$$

This is a contradiction, so there are no solutions in this subcase.

Subcase 1.2: $x+3 < 0 \implies x < -3$.
This condition $x < -3$ contradicts the condition $x > 5$. So, there are no solutions in this subcase.
Therefore, there are no solutions when $x > 5$.

Case 2: $x-5 < 0 \implies x < 5$.
In this case, $|x-5| = -(x-5) = 5-x$.
The equation becomes:

$$\frac{|x+3|}{-(x-5)} = \frac{x+1}{x-5}$$ $$\frac{|x+3|}{5-x} = \frac{x+1}{x-5}$$

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

$$\frac{|x+3|}{-(x-5)} \cdot (x-5) = x+1$$ $$-|x+3| = x+1$$ $$|x+3| = -(x+1)$$

Now, we solve this equation for $x < 5$. We must also satisfy the domain condition $x \le -1$. So, we are looking for solutions in the interval $x \le -1$.

Subcase 2.1: $x+3 \ge 0 \implies x \ge -3$.
Combining with $x \le -1$, this means we are in the interval $-3 \le x \le -1$.
In this range, $x+3 = -(x+1)$.

$$x+3 = -x-1$$ $$2x = -4$$ $$x = -2$$

This solution $x = -2$ lies within the interval $[-3, -1]$.
Thus, $x = -2$ is a valid solution.

Subcase 2.2: $x+3 < 0 \implies x < -3$.
Combining with $x \le -1$, this means we are in the interval $x < -3$.
In this range, $-(x+3) = -(x+1)$.

$$-x-3 = -x-1$$ $$-3 = -1$$

This is a contradiction, so there are no solutions in this subcase.

The only solution found is $x = -2$.
We verify this solution with the original equation:
LHS: $\left|\frac{-2+3}{-2-5}\right| = \left|\frac{1}{-7}\right| = \left|-\frac{1}{7}\right| = \frac{1}{7}$
RHS: $\frac{-2+1}{-2-5} = \frac{-1}{-7} = \frac{1}{7}$
Since LHS = RHS, $x=-2$ is correct.

The set of real values of $x$ for which the equation holds is $\boxed{\{-2\}}$.