5) \( \frac{2}{x-3}+\frac{2}{x}=\frac{1}{2 x} \)

To solve the equation \( \frac{2}{x-3}+\frac{2}{x}=\frac{1}{2 x} \), we start by finding a common denominator for the left-hand side, which is \( x(x-3) \). Rewriting each fraction: \[ \frac{2x}{x(x-3)} + \frac{2(x-3)}{x(x-3)} = \frac{2x + 2(x-3)}{x(x-3)} = \frac{2x + 2x - 6}{x(x-3)} = \frac{4x - 6}{x(x-3)} \] Now, for the right-hand side \( \frac{1}{2x} \), we will rewrite it with a common denominator of \( 2x(x-3) \): \[ \frac{1}{2x} = \frac{x-3}{2x(x-3)} \] Equating both sides gives us: \[ \frac{4x - 6}{x(x-3)} = \frac{x-3}{2x(x-3)} \] Cross-multiplying results in: \[ 2(4x - 6) = (x - 3) \] Expanding this leads to: \[ 8x - 12 = x - 3 \] Now we rearrange the equation: \[ 8x - x = -3 + 12 \] That simplifies down to: \[ 7x = 9 \Rightarrow x = \frac{9}{7} \] Thus, the solution to the equation is: \[ x = \frac{9}{7} \]