The problem asks to solve the equation $\frac{2}{3}(3x-5) - \frac{3}{5}(2x-3) = 3$ for $x$.

1. Problem Description

The problem asks to solve the equation $\frac{2}{3}(3x-5) - \frac{3}{5}(2x-3) = 3$ for $x$.

2. Solution Steps

First, distribute the fractions into the parentheses:

$\frac{2}{3}(3x-5) - \frac{3}{5}(2x-3) = 3$

$\frac{2}{3}(3x) - \frac{2}{3}(5) - \frac{3}{5}(2x) + \frac{3}{5}(3) = 3$

$2x - \frac{10}{3} - \frac{6x}{5} + \frac{9}{5} = 3$

Now, combine the $x$ terms and the constant terms:

$2x - \frac{6x}{5} = \frac{10x}{5} - \frac{6x}{5} = \frac{4x}{5}$

$-\frac{10}{3} + \frac{9}{5} = -\frac{50}{15} + \frac{27}{15} = -\frac{23}{15}$

So the equation becomes:

$\frac{4x}{5} - \frac{23}{15} = 3$

Add $\frac{23}{15}$ to both sides:

$\frac{4x}{5} = 3 + \frac{23}{15} = \frac{45}{15} + \frac{23}{15} = \frac{68}{15}$

$\frac{4x}{5} = \frac{68}{15}$

Multiply both sides by 5:

$4x = \frac{68}{15} \times 5 = \frac{68}{3}$

Divide both sides by 4:

$x = \frac{68}{3} \div 4 = \frac{68}{3} \times \frac{1}{4} = \frac{68}{12} = \frac{17}{3}$

3. Final Answer

$x = \frac{17}{3}$