\[\begin{array}{rl}(f-g)(x)=f(x)-g(x)&\text{Replace }f\text{ and }g \\ (f-g)(x)=\dfrac{x-3}{x+5}-\dfrac{-5x+7}{x^2+6x+5}&\text{Factor the second denominator} \\ (f-g)(x)=\dfrac{x-3}{x+5}-\dfrac{-5x+7}...\[\begin{array}{rl}(f-g)(x)=f(x)-g(x)&\text{Replace }f\text{ and }g \\ (f-g)(x)=\dfrac{x-3}{x+5}-\dfrac{-5x+7}{x^2+6x+5}&\text{Factor the second denominator} \\ (f-g)(x)=\dfrac{x-3}{x+5}-\dfrac{-5x+7}{(x+5)(x+1)}&\text{Unlike denominators; LCD}=(x+5)(x+1) \\ (f-g)(x)=\color{blue}{\dfrac{(x+1)}{(x+1)}}\color{black}{}\cdot\dfrac{(x-3)}{(x+5)}-\dfrac{-5x+7}{(x+5)(x+1)}&\text{Rewrite each fraction with the LCD} \\ (f-g)(x)=\dfrac{(x+1)(x-3)}{(x+5)(x+1)}-\dfrac{-5x+7}{(x+5)(x+1)}&\text{Multiply the …
