Answer
$x=-11$ and $x=1$
Work Step by Step
$\dfrac{x+1}{x+3}=\dfrac{x^{2}-11x}{x^{2}+x-6}-\dfrac{x-3}{x-2}$
Factor the denominator of the second fraction:
$\dfrac{x+1}{x+3}=\dfrac{x^{2}-11x}{(x+3)(x-2)}-\dfrac{x-3}{x-2}$
Multiply the whole equation by $(x+3)(x-2)$:
$(x+3)(x-2)\Big[\dfrac{x+1}{x+3}=\dfrac{x^{2}-11x}{(x+3)(x-2)}-\dfrac{x-3}{x-2}\Big]$
$(x+1)(x-2)=x^{2}-11x-(x-3)(x+3)$
$x^{2}-x-2=x^{2}-11x-x^{2}+9$
Take all terms to the left side of the equation and simplify it by combining like terms:
$x^{2}-x-2-x^{2}+11x+x^{2}-9=0$
$x^{2}+10x-11=0$
Solve this equation by factoring:
$(x+11)(x-1)=0$
The two solutions are:
$x=-11$ and $x=1$
The original equation is not undefined for neither of the values found. So the answer is:
$x=-11$ and $x=1$
