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

To solve the equation \[ \frac{3 x+4}{3}-\frac{4 x}{x-4}=x, \] first, we will find a common denominator for the fractions. The common denominator is \(3(x - 4)\). Rewriting the left side of the equation with this common denominator: \[ \frac{(3x + 4)(x - 4)}{3(x - 4)} - \frac{4x \cdot 3}{3(x - 4)} = x. \] This simplifies to: \[ \frac{(3x + 4)(x - 4) - 12x}{3(x - 4)} = x. \] Now, we'll clear the fraction by multiplying both sides by \(3(x - 4)\) (note that \(x \neq 4\) to avoid division by zero): \[ (3x + 4)(x - 4) - 12x = 3x(x - 4). \] Expanding both sides: Left side: \[ (3x + 4)(x - 4) = 3x^2 - 12x + 4x - 16 = 3x^2 - 8x - 16, \] so \[ 3x^2 - 8x - 16 - 12x = 3x^2 - 20x - 16. \] Right side: \[ 3x(x - 4) = 3x^2 - 12x. \] Setting them equal gives: \[ 3x^2 - 20x - 16 = 3x^2 - 12x. \] Now, subtract \(3x^2\) from both sides: \[ -20x - 16 = -12x. \] Next, add \(20x\) to both sides: \[ -16 = 8x. \] Dividing by \(8\) yields: \[ x = -2. \] To verify, substitute \(x = -2\) back into the original equation: Left side: \[ \frac{3(-2)+4}{3}-\frac{4(-2)}{-2 - 4} = \frac{-6 + 4}{3} - \frac{-8}{-6} = \frac{-2}{3} - \frac{4}{3} = -2. \] Right side: \[ -2. \] Both sides are equal, so the solution is confirmed as: \[ \boxed{-2}. \]