## Algebra: A Combined Approach (4th Edition)

$x=-11$ and $x=1$
$\dfrac{x+1}{x+3}=\dfrac{x^{2}-11x}{x^{2}+x-6}-\dfrac{x-3}{x-2}$ 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 $x=1$