We are asked to solve the equation $\frac{2}{3}(3x-5) - \frac{3}{5}(2x-3) = 3$ for $x$.

1. Problem Description

We are asked to solve the equation $\frac{2}{3}(3x-5) - \frac{3}{5}(2x-3) = 3$ for $x$.

2. Solution Steps

First, we distribute the fractions into the parentheses:

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

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

Next, we want to combine the $x$ terms and the constant terms.

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

To combine the $x$ terms, we need a common denominator. The common denominator of 1 and 5 is 5, so we have:

$2x = \frac{10}{5}x$

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

To combine the constant terms, we need a common denominator. The common denominator of 5 and 3 is 15, so we have:

$\frac{9}{5} = \frac{27}{15}$

$\frac{10}{3} = \frac{50}{15}$

$\frac{27}{15} - \frac{50}{15} = -\frac{23}{15}$

So now our equation looks like:

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

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

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

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

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

Multiply both sides by $\frac{5}{4}$:

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

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

$x = \frac{17}{1} \cdot \frac{1}{3}$

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

3. Final Answer

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