Solving the rational equation, we have:
[tex]\begin{gathered} \frac{5}{x+1}+\frac{1}{x-3}=\frac{-6}{x^2-2x-3} \\ \frac{5(x-3)+1(x+1)}{(x+1)(x-3)}=\frac{-6}{(x+1)(x-3)} \\ 5x-15+x+1=-6 \\ 6x-14=-6 \\ 6x=-6+14 \\ 6x=8 \\ x=\frac{8}{6}=\frac{4}{3} \end{gathered}[/tex]
So the solution is x = 4/3 = 1.33
Now let's calculate the values that x cannot be equal, calculating where the denominators of the fractions are equal to zero:
[tex]\begin{gathered} x+1\ne0\to x\ne-1 \\ x-3\ne0\to x\ne3 \end{gathered}[/tex]
So x cannot be equal to -1 or 3.
