#### Answer

The simplified form of the expression $\frac{2a+b}{a-b}-\frac{4}{3a-3b}$ is $\frac{6a+3b-4}{3\left( a-b \right)}$.

#### Work Step by Step

$\frac{2a+b}{a-b}-\frac{4}{3a-3b}$
Factor the expressions in the denominator:
$3a-3b=3\left( a-b \right)$
Solve to obtain the LCD of denominator of the fractions that is $a-b$ and$3\left( a-b \right)$.
On solving the LCD is$3\left( a-b \right)$.
The rational expression is written as a fraction in such a way that the LCD is the denominator of the fraction. Consider, the individual terms of the provided rational terms;
$\frac{2a+b}{a-b}=\frac{\left( 2a+b \right)\left( 3 \right)}{3\left( a-b \right)}$
And,
$\frac{4}{3a-3b}=\frac{4}{3\left( a-b \right)}$
Since the fractions have a common denominator, subtract the numerators and the LCD is the denominator.
$\begin{align}
& \frac{2a+b}{a-b}-\frac{4}{3a-3b}=\frac{\left( 2a+b \right)\left( 3 \right)}{3\left( a-b \right)}-\frac{4}{3\left( a-b \right)} \\
& =\frac{\left( 2a+b \right)\left( 3 \right)-\left( 4 \right)}{3\left( a-b \right)} \\
& =\frac{6a+3b-4}{3\left( a-b \right)}
\end{align}$