Inequality sign changing when multiplying with a unknown variable? This is the inequality:
$$\frac{(2x-1)}{(3-5x)} < 2$$
Next step is:
$$\frac{(2x-1)}{(3-5x)} \cdot (3-5x) < 2\cdot (3-5x)$$
Now is the sign $<$ or $>$?
The end solution is $x < \frac{7}{12}.$

In general, when multiplying an inequality by an unknown quantity, we can't tell whether the sign changes or not. To solve the problem, then, we must consider both cases, i.e. whether the unknown quantity is positive or negative. In this example, we would do the following:
CASE 1: $3-5x > 0$
\begin{align*}
\frac{2x-1}{3-5x} (3-5x) < 2(3-5x)
\quad \Rightarrow \quad x < \frac{7}{12}
\end{align*}
CASE 2: $3-5x < 0$
\begin{align*}
\frac{2x-1}{3-5x} (3-5x) > 2(3-5x)
\quad \Rightarrow \quad x > \frac{7}{12}
\end{align*}
Note: I have left out the case $3-5x = 0$ because it is in the denominator of a fraction. So, there are in fact two solutions: we require $3-5x>0$ and $x<\frac{7}{12}$, OR we require $3-5x<0$ and $x>\frac{7}{12}$. 
However, we can simplify this, since $3-5x>0$ implies $x<\frac{3}{5}$, and similarly $3-5x<0$ implies $x>\frac{3}{5}$. Then since $\frac{3}{5} > \frac{7}{12}$, the two possibilities become:
\begin{align*}
x < \frac{3}{5} \text{ and } x < \frac{7}{12}
\quad \Rightarrow \quad x < \frac{7}{12} \\
x > \frac{3}{5} \text{ and } x > \frac{7}{12}
\quad \Rightarrow \quad x > \frac{3}{5} \\
\end{align*}
So, the solution set is $x < \frac{7}{12}$ OR $x > \frac{3}{5}$.
