Answer
The simplified form of the expression $\frac{x-2}{x+5}-\frac{x+3}{x-4}$ is $\frac{-7\left( 2x+1 \right)}{\left( x-4 \right)\left( x+5 \right)}$.
Work Step by Step
$\frac{x-2}{x+5}-\frac{x+3}{x-4}$
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 $\left( x-4 \right)\left( x+5 \right)$.
The rational expression is written as a fraction in such a way that the LCD is the denominator of the fraction:
$\frac{x-2}{x+5}=\frac{\left( x-2 \right)\left( x-4 \right)}{\left( x+5 \right)\left( x-4 \right)}$
And
$\frac{x+3}{x-4}=\frac{\left( x+3 \right)\left( x+5 \right)}{\left( x-4 \right)\left( x+5 \right)}$
Since the fractions have a common denominator, add the numerators and the LCD is the denominator:
$\begin{align}
& \frac{x-2}{x+5}-\frac{x+3}{x-4}=\frac{\left( x-2 \right)\left( x-4 \right)}{\left( x+5 \right)\left( x-4 \right)}-\frac{\left( x+3 \right)\left( x+5 \right)}{\left( x+5 \right)\left( x-4 \right)} \\
& =\frac{\left( x-2 \right)\left( x-4 \right)-\left( x+3 \right)\left( x+5 \right)}{\left( x+5 \right)\left( x-4 \right)}
\end{align}$
Simplify the above expression as:
$\begin{align}
& \frac{x-2}{x+5}-\frac{x+3}{x-4}=\frac{x\cdot \left( x-4 \right)-2\cdot \left( x-4 \right)-x\cdot \left( x+5 \right)-3\cdot \left( x+5 \right)}{\left( x+5 \right)\left( x-4 \right)} \\
& =\frac{{{x}^{2}}-4x-2x+8-{{x}^{2}}-5x-3x-15}{\left( x+5 \right)\left( x-4 \right)} \\
& =\frac{{{x}^{2}}-{{x}^{2}}-4x-2x-5x-3x+8-15}{\left( x+5 \right)\left( x-4 \right)} \\
& =\frac{-14x-7}{\left( x+5 \right)\left( x-4 \right)}
\end{align}$
Take out $-7$ from the numerator.
$\frac{x-2}{x+5}-\frac{x+3}{x-4}=\frac{-7\left( 2x+1 \right)}{\left( x-4 \right)\left( x+5 \right)}$
