Solve the equation. \[ \frac{3 x+1}{x-2}+\frac{4}{x}=\frac{-8}{x^{2}-2 x} \]

To solve the equation \[ \frac{3 x+1}{x-2}+\frac{4}{x}=\frac{-8}{x^{2}-2 x}, \] first, we observe that the right side can be simplified since \(x^2 - 2x = x(x - 2)\). Therefore, we can rewrite the equation as \[ \frac{3x + 1}{x-2} + \frac{4}{x} = \frac{-8}{x(x-2)}. \] Next, we find a common denominator for all the fractions in the equation, which is \(x(x-2)\). We will rewrite the left-hand side with this common denominator: \[ \frac{(3x + 1)x}{x(x - 2)} + \frac{4(x - 2)}{x(x - 2)} = \frac{(3x + 1)x + 4(x - 2)}{x(x - 2)}. \] We can thus rewrite the whole equation as follows: \[ \frac{(3x + 1)x + 4(x - 2)}{x(x - 2)} = \frac{-8}{x(x - 2)}. \] Now, we can equate the numerators: \[ (3x + 1)x + 4(x - 2) = -8. \] Expanding both sides results in: \[ 3x^2 + x + 4x - 8 = -8. \] This simplifies to: \[ 3x^2 + 5x - 8 + 8 = 0, \] which simplifies to \[ 3x^2 + 5x = 0. \] Factoring out \(x\) gives: \[ x(3x + 5) = 0. \] Setting each factor to zero provides the solutions: 1. \(x = 0\), 2. \(3x + 5 = 0 \Rightarrow 3x = -5 \Rightarrow x = -\frac{5}{3}\). Now, we need to verify if these solutions produce any division by zero in the original equation. The original denominators were \(x(x - 2)\) and \(x - 2\). - For \(x = 0\), we have division by zero, so this solution is not valid. - For \(x = -\frac{5}{3}\), neither of the denominators is zero as \(-\frac{5}{3} \neq 0\) and \(-\frac{5}{3} - 2 \neq 0\). Thus, the only valid solution to the given equation is \[ \boxed{-\frac{5}{3}}. \]