5. \( \frac{2}{x+4}-\frac{1}{x}=\frac{2}{3 x} \)
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To solve the equation \( \frac{2}{x+4}-\frac{1}{x}=\frac{2}{3 x} \), we first find a common denominator for the left side, which is \( x(x+4) \). Rewriting each term, we get: \[ \frac{2x}{x(x+4)} - \frac{(x+4)}{x(x+4)} = \frac{2}{3x} \] Combining the left side: \[ \frac{2x - (x+4)}{x(x+4)} = \frac{2}{3x} \] This simplifies to: \[ \frac{x - 4}{x(x+4)} = \frac{2}{3x} \] Now, we can cross-multiply: \[ 3x(x - 4) = 2(x(x + 4)) \] Expanding both sides: \[ 3x^2 - 12x = 2x^2 + 8x \] Rearranging gives: \[ 3x^2 - 12x - 2x^2 - 8x = 0 \] Which simplifies to: \[ x^2 - 20x = 0 \] Factoring out \( x \): \[ x(x - 20) = 0 \] Thus, \( x = 0 \) or \( x = 20 \). Since \( x = 0 \) would make the original equation undefined, we discard it. Therefore, the solution is: \[ \boxed{20} \]
