Solve the equation. \[ \frac{2 x-6}{x}=\frac{x-3}{3} \]
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To solve the equation \[ \frac{2x-6}{x}=\frac{x-3}{3}, \] let's start by cross-multiplying to eliminate the fractions: \[ 3(2x - 6) = x(x - 3). \] Expanding both sides gives: \[ 6x - 18 = x^2 - 3x. \] Rearranging this equation to set it to zero: \[ x^2 - 3x - 6x + 18 = 0 \implies x^2 - 9x + 18 = 0. \] Next, we can factor the quadratic: \[ (x - 3)(x - 6) = 0. \] Setting each factor to zero provides the solutions: \[ x - 3 = 0 \implies x = 3, \] \[ x - 6 = 0 \implies x = 6. \] Now, we should verify if these solutions are valid by substituting them back into the original equation. Both \( x = 3 \) and \( x = 6 \) result in valid equations without leading to division by zero. Thus, the solutions to the equation are: \[ \boxed{3} \quad \text{and} \quad \boxed{6}. \]