Answer
The solutions are $x=1$ and $x=7$
Work Step by Step
$\dfrac{3}{x-3}+\dfrac{5}{x-4}=\dfrac{x^{2}-20}{x^{2}-7x+12}$
Factor the denominator of the fraction on the right side of the equation:
$\dfrac{3}{x-3}+\dfrac{5}{x-4}=\dfrac{x^{2}-20}{(x-3)(x-4)}$
Multiply the whole equation by $(x-3)(x-4)$
$(x-3)(x-4)\Big[\dfrac{3}{x-3}+\dfrac{5}{x-4}=\dfrac{x^{2}-20}{(x-3)(x-4)}\Big]$
$3(x-4)+5(x-3)=x^{2}-20$
$3x-12+5x-15=x^{2}-20$
Take all terms to the right side:
$0=x^{2}-20-3x-5x+12+15$
Simplify and rearrange:
$x^{2}-8x+7=0$
Solve by factoring:
$(x-1)(x-7)=0$
Set both factors equal to $0$ and solve each individual equation for $x$:
$x-1=0$
$x=1$
$x-7=0$
$x=7$
The solutions are $x=1$ and $x=7$
