The problem is to solve the inequality $\frac{4}{5}x - (x+3) < \frac{1}{3}(x-1)$.

1. Problem Description

The problem is to solve the inequality $\frac{4}{5}x - (x+3) < \frac{1}{3}(x-1)$.

2. Solution Steps

We need to solve the inequality $\frac{4}{5}x - (x+3) < \frac{1}{3}(x-1)$.

First, we simplify the left side:

$\frac{4}{5}x - x - 3 < \frac{1}{3}(x-1)$

$\frac{4}{5}x - \frac{5}{5}x - 3 < \frac{1}{3}(x-1)$

$-\frac{1}{5}x - 3 < \frac{1}{3}(x-1)$

Now we simplify the right side:

$-\frac{1}{5}x - 3 < \frac{1}{3}x - \frac{1}{3}$

Next, we want to isolate $x$. Add $\frac{1}{5}x$ to both sides:

$-3 < \frac{1}{3}x + \frac{1}{5}x - \frac{1}{3}$

$-3 < \frac{5}{15}x + \frac{3}{15}x - \frac{1}{3}$

$-3 < \frac{8}{15}x - \frac{1}{3}$

Now, add $\frac{1}{3}$ to both sides:

$-3 + \frac{1}{3} < \frac{8}{15}x$

$-\frac{9}{3} + \frac{1}{3} < \frac{8}{15}x$

$-\frac{8}{3} < \frac{8}{15}x$

Now, multiply both sides by $\frac{15}{8}$. Since $\frac{15}{8}$ is positive, we do not need to flip the inequality sign:

$-\frac{8}{3} \cdot \frac{15}{8} < x$

$-\frac{15}{3} < x$

$-5 < x$

Therefore, $x > -5$.

3. Final Answer

$x > -5$