This mathematics problem involves applying core mathematical principles and formulas. Below you will find a complete step-by-step solution with detailed explanations for each step, helping you understand not just the answer but the method behind it.
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To find the set of real values of x for which |(x+3)/(x-5)| = (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 ≠ 0 x ≠ 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. (x+1)/(x-5) 0 We can analyze the signs of the numerator and denominator: • If x+1 0 x -1 • If x-5 > 0 x > 5 We can use a sign table for (x+1)/(x-5): | Interval | x+1 | x-5 | (x+1)/(x-5) | |---------------|-------|-------|-------------------| | x < -1 | - | - | + | | -1 x < 5| + | - | - | | x > 5 | + | + | + | So, the condition (x+1)/(x-5) 0 holds for x -1 or x > 5. This is the valid domain for our solutions. Step 3: Solve the equation |(x+3)/(x-5)| = (x+1)/(x-5). We can rewrite the left side as (|x+3|)/(|x-5|). (|x+3|)/(|x-5|) = (x+1)/(x-5) We need to consider two cases based on the sign of x-5. Case 1: x-5 > 0 x > 5. In this case, |x-5| = x-5. The equation becomes: (|x+3|)/(x-5) = (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 0 x -3. Since we are in the interval x > 5, the condition x -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 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 x < 5. In this case, |x-5| = -(x-5) = 5-x. The equation becomes: (|x+3|)/(-(x-5)) = (x+1)/(x-5) (|x+3|)/(5-x) = (x+1)/(x-5) Multiply both sides by (x-5): (|x+3|)/(-(x-5)) · (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 -1. So, we are looking for solutions in the interval x -1. Subcase 2.1: x+3 0 x -3. Combining with x -1, this means we are in the interval -3 x -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 x < -3. Combining with x -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: |(-2+3)/(-2-5)| = |(1)/(-7)| = |-(1)/(7)| = (1)/(7) RHS: (-2+1)/(-2-5) = (-1)/(-7) = (1)/(7) Since LHS = RHS, x=-2 is correct. The set of real values of x for which the equation holds is \-2\.