To find the set of real values of $x$ for which $\left|\frac{x+3}{x-5}\right| = \frac{x+1}{x-5}$, we need to consider the conditions for the equation to be defined and then solve it. Step 1: Determine the domain of the equation. For the fractions to be defined, the denominators cannot be zero. $$ x - 5 \neq 0 \implies x \neq 5 $$ Step 2: Analyze the condition for the right-hand side to be non-negative. 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. Step 3: Solve the equation $\left|\frac{x+3}{x-5}\right| = \frac{x+1}{x-5}$. 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. Step 4: Combine the results. 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 $\boxed{\{-2\}}$.