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

The simplified form of the expression $\frac{2x}{x-5}+\frac{3}{x+4}$ is $\frac{2{{x}^{2}}+11x-15}{\left( x+4 \right)\left( x-5 \right)}$ .
$\frac{2x}{x-5}+\frac{3}{x+4}$ Solve to obtain the LCD of the denominator of the fractions that is $\left( x-5 \right)$ and $\left( x+4 \right)$. On solving the LCD is $\left( x-5 \right)\left( x+4 \right)$. 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; $\frac{2x}{x-5}=\frac{\left( 2x \right)\left( x+4 \right)}{\left( x-5 \right)\left( x+4 \right)}$ And, $\frac{3}{x+4}=\frac{3\left( x-5 \right)}{\left( x-5 \right)\left( x+4 \right)}$ Since the fractions have a common denominator, add the numerators and the LCD is the denominator: \begin{align} & \frac{2x}{x-5}+\frac{3}{x+4}=\frac{\left( 2x \right)\left( x+4 \right)}{\left( x-5 \right)\left( x+4 \right)}+\frac{3\left( x-5 \right)}{\left( x-5 \right)\left( x+4 \right)} \\ & =\frac{\left( 2x \right)\left( x+4 \right)+3\left( x-5 \right)}{\left( x-5 \right)\left( x+4 \right)} \end{align} Solve the above expression as: \begin{align} & \frac{2x}{x-5}+\frac{3}{x+4}=\frac{2{{x}^{2}}+8x+3x-15}{\left( x-5 \right)\left( x+4 \right)} \\ & =\frac{2{{x}^{2}}+11x-15}{\left( x-5 \right)\left( x+4 \right)} \end{align}