We are given the equation $\frac{4}{x-2} + \frac{1}{2(x-2)} = \frac{3}{4(x-2)}$ and we need to solve for $x$.

1. Problem Description

We are given the equation $\frac{4}{x-2} + \frac{1}{2(x-2)} = \frac{3}{4(x-2)}$ and we need to solve for $x$.

2. Solution Steps

First, we find a common denominator for all terms. The common denominator is $4(x-2)$. We rewrite each fraction with the common denominator:

$\frac{4}{x-2} = \frac{4 \cdot 4}{4(x-2)} = \frac{16}{4(x-2)}$

$\frac{1}{2(x-2)} = \frac{1 \cdot 2}{2(x-2) \cdot 2} = \frac{2}{4(x-2)}$

$\frac{3}{4(x-2)}$ is already in the desired form.

Now we rewrite the equation with the common denominator:

$\frac{16}{4(x-2)} + \frac{2}{4(x-2)} = \frac{3}{4(x-2)}$

Since all terms have the same denominator, we can add the numerators on the left side:

$\frac{16 + 2}{4(x-2)} = \frac{3}{4(x-2)}$

$\frac{18}{4(x-2)} = \frac{3}{4(x-2)}$

Now, we can multiply both sides of the equation by $4(x-2)$, provided that $x \ne 2$:

$4(x-2) \cdot \frac{18}{4(x-2)} = 4(x-2) \cdot \frac{3}{4(x-2)}$

$18 = 3$

This is a contradiction, which means there is no solution for $x$.

3. Final Answer

No solution.