## Precalculus (6th Edition) Blitzer

$\{-5, -1\}.$
First, exclude any x that yields 0 in a denominator $x\neq 3,-3$ $\displaystyle \frac{2x}{x-3}+\frac{6}{x+3}=\frac{-28}{x^{2}-9} \qquad$ ... multiply with LCD = $(x-3)(x+3)$ $2x(x+3)+6(x-3)=-28\qquad$ ... write as $ax^{2}+bx+c=0$ $2x^{2}+6x+6x-18=-28$ $2x^{2}+12x+10=0\qquad$ ... divide with 2 $x^{2}+6x+5=0$ We factor this trinomial by finding two factors of 5 that add to 6; they are +5 and +1. $(x+1)(x+5)=0$ $x=-1,\qquad x=-5\qquad$(neither was excluded) The solution set is $\{-5, -1\}.$