Answer
The simplified form of the expression $\frac{2x}{x-5}+\frac{3}{x+4}$ is $\frac{2{{x}^{2}}+11x-15}{\left( x+4 \right)\left( x-5 \right)}$
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Work Step by Step
$\frac{2x}{x-5}+\frac{3}{x+4}$
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:
$\begin{align}
& \frac{2x}{x-5}+\frac{3}{x+4}=\frac{2{{x}^{2}}+8x+3x-15}{\left( x-5 \right)\left( x+4 \right)} \\
& =\frac{2{{x}^{2}}+11x-15}{\left( x-5 \right)\left( x+4 \right)}
\end{align}$
