We need to solve the equation $\frac{2x}{x-4} = 4 + \frac{8}{x-4}$.

1. Problem Description

We need to solve the equation $\frac{2x}{x-4} = 4 + \frac{8}{x-4}$.

2. Solution Steps

First, we multiply both sides of the equation by $(x-4)$ to eliminate the fractions.

$(x-4) \cdot \frac{2x}{x-4} = (x-4) \cdot (4 + \frac{8}{x-4})$

$2x = 4(x-4) + 8$

$2x = 4x - 16 + 8$

$2x = 4x - 8$

Next, subtract $4x$ from both sides.

$2x - 4x = 4x - 8 - 4x$

$-2x = -8$

Now, divide both sides by $-2$.

$\frac{-2x}{-2} = \frac{-8}{-2}$

$x = 4$

However, we must check if $x=4$ is a valid solution because it makes the denominator $x-4$ equal to zero. If $x=4$, the original equation becomes

$\frac{2(4)}{4-4} = 4 + \frac{8}{4-4}$

$\frac{8}{0} = 4 + \frac{8}{0}$

Since division by zero is undefined, $x=4$ is not a solution. Therefore, there is no solution.

3. Final Answer

NO SOLUTION