Answer
$\{1, 7\}$
Work Step by Step
$\displaystyle \frac{3}{x-3}+\frac{5}{x-4}=\frac{x^{2}-20}{x^{2}-7x+12}$
The denominator on the RHS can be factored:
Two factors of 12 that add to -7 are -4 and -3
$\displaystyle \frac{3}{x-3}+\frac{5}{x-4}=\frac{x^{2}-20}{(x-3)(x-4)} \qquad $ ... multiply with LCD = $(x-3)(x-4)$
We want to write the equation as $ ax^{2}+bx+c=0$
First, we exclude any x that yields 0 in a denominator
$ x\neq 3,4$
$3(x-4)+5(x-3)=x^{2}-20$
$3x-12+5x-15=x^{2}-20$
$8x-17==x^{2}-20$
$0=x^{2}-8x+7$
We factor this trinomial by finding two factors of $7$ that add to $-8$; they are $-7$ and $-1$.
$(x-7)(x-1)=0$
$ x=7,\qquad x=1$
The solution set is
$\{1, 7\}.$
