Answer
$(f-g)(x)=\frac{x+2}{x+3}$
$(-\infty,-5)\cup(-5,-3)\cup(-3,\infty)$.
Work Step by Step
The given functions are
$f(x)=\frac{2x-3}{x+5}$ and $g(x)=\frac{x^2-4x-19}{x^2+8x+15}$
Use $(f-g)(x)=f(x)-g(x)$.
$\Rightarrow (f-g)(x)=\frac{2x-3}{x+5}-\frac{x^2-4x-19}{x^2+8x+15}$.
Factor the denominator of the second fraction.
$\Rightarrow x^2+8x+15$
Rewrite the middle term $8x$ as $5x+3x$.
$\Rightarrow x^2+5x+3x+15$
Group the terms.
$\Rightarrow (x^2+5x)+(3x+15)$
Factor each group.
$\Rightarrow x(x+5)+3(x+5)$
Factor out $(x+5)$.
$\Rightarrow (x+5)(x+3)$
Back substitute the factor into the equation.
$\Rightarrow (f-g)(x)=\frac{2x-3}{x+5}-\frac{x^2-4x-19}{(x+5)(x+3)}$.
The LCD of both the fractions is $(x+5)(x+3)$.
$\Rightarrow (f-g)(x)=\frac{(2x-3)(x+3)}{(x+5)(x+3)}-\frac{x^2-4x-19}{(x+5)(x+3)}$
$\Rightarrow (f-g)(x)=\frac{2x^2+6x-3x-9}{(x+5)(x+3)}-\frac{x^2-4x-19}{(x+5)(x+3)}$
$\Rightarrow (f-g)(x)=\frac{2x^2+6x-3x-9-(x^2-4x-19)}{(x+5)(x+3)}$
Simplify.
$\Rightarrow (f-g)(x)=\frac{2x^2+6x-3x-9-x^2+4x+19}{(x+5)(x+3)}$
Add like terms.
$\Rightarrow (f-g)(x)=\frac{x^2+7x+10}{(x+5)(x+3)}$
Factor the numerator.
$\Rightarrow x^2+7x+10$
Rewrite the middle term $7x$ as $5x+2x$.
$\Rightarrow x^2+5x+2x+10$
Group the terms.
$\Rightarrow (x^2+5x)+(2x+10)$
Factor each group.
$\Rightarrow x(x+5)+2(x+5)$
Factor out $(x+5)$.
$\Rightarrow (x+5)(x+2)$
Back substitute the factor into the fraction.
$\Rightarrow (f-g)(x)=\frac{(x+5)(x+2)}{(x+5)(x+3)}$
Cancel common terms.
$\Rightarrow (f-g)(x)=\frac{x+2}{x+3}$
This gives:
$(-\infty,-5)\cup(-5,-3)\cup(-3,\infty)$.
