\( \frac{x+1}{x-1}=\frac{-9}{x+3}+\frac{8}{x^{2}+2 x-3} \) One or more solutions:

To solve the equation: \[ \frac{x+1}{x-1} = \frac{-9}{x+3} + \frac{8}{x^{2}+2x-3} \] **Step 1: Factor the Denominator** Notice that \( x^2 + 2x - 3 \) factors as \( (x+3)(x-1) \). Rewrite the equation: \[ \frac{x+1}{x-1} = \frac{-9}{x+3} + \frac{8}{(x+3)(x-1)} \] **Step 2: Find a Common Denominator** The common denominator for all terms is \( (x+3)(x-1) \). Rewrite each term with this denominator: \[ \frac{(x+1)(x+3)}{(x+3)(x-1)} = \frac{-9(x-1)}{(x+3)(x-1)} + \frac{8}{(x+3)(x-1)} \] **Step 3: Eliminate the Denominator** Since \( (x+3)(x-1) \neq 0 \), multiply both sides of the equation by \( (x+3)(x-1) \): \[ (x+1)(x+3) = -9(x-1) + 8 \] **Step 4: Expand and Simplify** Expand the left side and simplify the right side: \[ x^2 + 4x + 3 = -9x + 17 \] Bring all terms to one side: \[ x^2 + 13x - 14 = 0 \] **Step 5: Solve the Quadratic Equation** Use the quadratic formula: \[ x = \frac{-13 \pm \sqrt{13^2 - 4(1)(-14)}}{2(1)} = \frac{-13 \pm \sqrt{225}}{2} = \frac{-13 \pm 15}{2} \] This gives two potential solutions: \[ x = 1 \quad \text{and} \quad x = -14 \] **Step 6: Check for Valid Solutions** Substitute \( x = 1 \) into the original equation: \[ \frac{2}{0} \quad \text{(undefined)} \] Thus, \( x = 1 \) is not a valid solution. Substitute \( x = -14 \) into the original equation: \[ \frac{-13}{-15} = \frac{9}{-11} + \frac{8}{165} \] Simplifying both sides: \[ \frac{13}{15} = \frac{143}{165} \] Both sides are equal, confirming that \( x = -14 \) is a valid solution. **Final Answer:** \( \boxed{-14} \)