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

The given equation is

$\frac{4}{x-2} + \frac{1}{2(x-2)} = \frac{3}{4(x-2)}$.

We first find a common denominator for the terms. The least common denominator (LCD) is $4(x-2)$.

Multiply each term by the appropriate factor to get the common denominator.

$\frac{4}{x-2} \cdot \frac{4}{4} + \frac{1}{2(x-2)} \cdot \frac{2}{2} = \frac{3}{4(x-2)}$.

This gives us

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

Now we can add the fractions on the left side:

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

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

Since the denominators are the same, we can equate the numerators, provided $x \neq 2$:

$18 = 3$.

This equation is never true, regardless of the value of $x$.

Therefore, there is no solution to the given equation.

3. Final Answer

No solution.