[tex]\dfrac{5}{x+1}+\dfrac{1}{x-3}=\dfrac{-6}{x^2-2x-3}\\\\\dfrac{5(x-3)}{(x+1)(x-3)}+\dfrac{1(x+1)}{(x+1)(x-3)}=\dfrac{-6}{x^2-2x-3}\ \ \ \ |\text{use distributive property}\\\\\dfrac{(5)(x)+(5)(-3)+(1)(x)+(1)(1)}{(x)(x)+(x)(-3)+(1)(x)+(1)(-3)}=\dfrac{-6}{x^2-2x-3}\\\\\dfrac{5x-15+x+1}{x^2-3x+x-3}=\dfrac{-6}{x^2-2x-3}\\\\\dfrac{(5x+x)+(-15+1)}{x^2-2x-3}=\dfrac{-6}{x^2-2x-3}\\\\\dfrac{6x-14}{x^2-2x-3}=\dfrac{-6}{x^2-2x-3}[/tex]
The fractions with the same denominators are equal if nominators are equal.
Therefore we have the equation:
[tex]6x-14=-6\ \ \ \ |\text{add 14 to both sides}\\\\6x=8\ \ \ \ |\text{divide both sides by 6}\\\\x=\dfrac{8}{6}\\\\\boxed{x=\dfrac{4}{3}}[/tex]
