## Elementary and Intermediate Algebra: Concepts & Applications (6th Edition)

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