The problem is to solve the linear equation $\frac{4x}{3} - \frac{2x-4}{3} + \frac{\frac{3x-5}{4} - \frac{5x-3}{6}}{\frac{4x-3}{9} - \frac{2x-5}{4}} = \frac{2x}{3}$ for $x$.

1. Problem Description

The problem is to solve the linear equation $\frac{4x}{3} - \frac{2x-4}{3} + \frac{\frac{3x-5}{4} - \frac{5x-3}{6}}{\frac{4x-3}{9} - \frac{2x-5}{4}} = \frac{2x}{3}$ for $x$.

2. Solution Steps

First, simplify the left side of the equation.

Combine the first two terms:

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

Next, simplify the numerator of the fraction:

$\frac{3x-5}{4} - \frac{5x-3}{6} = \frac{3(3x-5) - 2(5x-3)}{12} = \frac{9x-15-10x+6}{12} = \frac{-x-9}{12}$

Next, simplify the denominator of the fraction:

$\frac{4x-3}{9} - \frac{2x-5}{4} = \frac{4(4x-3) - 9(2x-5)}{36} = \frac{16x-12-18x+45}{36} = \frac{-2x+33}{36}$

Now, we have the equation:

$\frac{2x+4}{3} + \frac{\frac{-x-9}{12}}{\frac{-2x+33}{36}} = \frac{2x}{3}$

Simplify the fraction:

$\frac{\frac{-x-9}{12}}{\frac{-2x+33}{36}} = \frac{-x-9}{12} \cdot \frac{36}{-2x+33} = \frac{(-x-9)3}{-2x+33} = \frac{-3x-27}{-2x+33} = \frac{3x+27}{2x-33}$

Substitute back into the original equation:

$\frac{2x+4}{3} + \frac{3x+27}{2x-33} = \frac{2x}{3}$

Subtract $\frac{2x+4}{3}$ from both sides:

$\frac{3x+27}{2x-33} = \frac{2x}{3} - \frac{2x+4}{3} = \frac{2x - (2x+4)}{3} = \frac{2x-2x-4}{3} = \frac{-4}{3}$

Cross-multiply:

$3(3x+27) = -4(2x-33)$

$9x+81 = -8x+132$

$17x = 132-81$

$17x = 51$

$x = \frac{51}{17}$

$x = 3$

Now we check if $x=3$ is a valid solution by checking the denominator $\frac{4x-3}{9} - \frac{2x-5}{4}$ and $2x-33$.

$\frac{4(3)-3}{9} - \frac{2(3)-5}{4} = \frac{12-3}{9} - \frac{6-5}{4} = \frac{9}{9} - \frac{1}{4} = 1 - \frac{1}{4} = \frac{3}{4} \neq 0$.

Also, $2x-33 = 2(3)-33 = 6-33 = -27 \neq 0$.

Thus $x=3$ is a valid solution.

3. Final Answer

$x=3$