Solve the equation. \[ \frac{7 x+1}{x-2}+\frac{3}{x}=\frac{-6}{x^{2}-2 x} \]

To solve the equation \[ \frac{7 x + 1}{x - 2} + \frac{3}{x} = \frac{-6}{x^2 - 2x}, \] we start by rewriting the right-hand side. Notice that \( x^2 - 2x \) can be factored: \[ x^2 - 2x = x(x - 2). \] So we can express the right side as: \[ \frac{-6}{x(x - 2)}. \] Now, our equation becomes: \[ \frac{7 x + 1}{x - 2} + \frac{3}{x} = \frac{-6}{x(x - 2)}. \] Next, we find a common denominator for the left-hand side, which will be \( x(x - 2) \). We rewrite each fraction: 1. For \( \frac{7x + 1}{x - 2} \): \[ \frac{7x + 1}{x - 2} \cdot \frac{x}{x} = \frac{(7x + 1)x}{x(x - 2)} = \frac{7x^2 + x}{x(x - 2)}. \] 2. For \( \frac{3}{x} \): \[ \frac{3}{x} \cdot \frac{x - 2}{x - 2} = \frac{3(x - 2)}{x(x - 2)} = \frac{3x - 6}{x(x - 2)}. \] Now, combine these fractions: \[ \frac{7x^2 + x + 3x - 6}{x(x - 2)} = \frac{7x^2 + 4x - 6}{x(x - 2)}. \] So now our equation reads: \[ \frac{7x^2 + 4x - 6}{x(x - 2)} = \frac{-6}{x(x - 2)}. \] Since the denominators are the same, we set the numerators equal to each other: \[ 7x^2 + 4x - 6 = -6. \] Adding 6 to both sides gives: \[ 7x^2 + 4x - 6 + 6 = 0, \] which simplifies to: \[ 7x^2 + 4x = 0. \] Factoring out \( x \): \[ x(7x + 4) = 0. \] Setting each factor to zero gives us: 1. \( x = 0 \). 2. \( 7x + 4 = 0 \) ⟹ \( 7x = -4 \) ⟹ \( x = -\frac{4}{7} \). Now we have two potential solutions: \( x = 0 \) and \( x = -\frac{4}{7} \). However, we need to check for extraneous solutions, especially since the original equation has denominators. \( x = 0 \) makes the \( \frac{3}{x} \) undefined, so we discard it. Thus, the only solution is: \[ \boxed{-\frac{4}{7}}. \]