We need to solve the equation $\frac{2x}{x-3} = 4 + \frac{6}{x-3}$ for $x$.

1. Problem Description

We need to solve the equation $\frac{2x}{x-3} = 4 + \frac{6}{x-3}$ for $x$.

2. Solution Steps

First, we multiply both sides of the equation by $(x-3)$ to eliminate the fractions.

$(x-3) \cdot \frac{2x}{x-3} = (x-3) \cdot (4 + \frac{6}{x-3})$

This simplifies to:

$2x = 4(x-3) + 6$

Expanding the right side gives:

$2x = 4x - 12 + 6$

Combining like terms on the right side:

$2x = 4x - 6$

Subtract $4x$ from both sides:

$2x - 4x = 4x - 6 - 4x$

$-2x = -6$

Divide both sides by $-2$:

$\frac{-2x}{-2} = \frac{-6}{-2}$

$x = 3$

However, we must check if $x=3$ is a valid solution. If we plug $x=3$ into the original equation, we get:

$\frac{2(3)}{3-3} = 4 + \frac{6}{3-3}$

$\frac{6}{0} = 4 + \frac{6}{0}$

Since division by zero is undefined, $x=3$ is not a valid solution.

Therefore, the equation has no solution.

3. Final Answer

(D) NO SOLUTION