Answer
$\emptyset$
Work Step by Step
In factored form, the given equation, $
\dfrac{3}{x-3}-\dfrac{2}{x-2}=\dfrac{3}{x^2-5x+6}
,$ is equivalent to
\begin{align*}
\dfrac{3}{x-3}-\dfrac{2}{x-2}&=\dfrac{3}{(x-2)(x-3)}
.\end{align*}
Multiplying both sides by the $LCD=
(x-2)(x-3)
,$ the equation above is equivalent to
\begin{align*}
(x-2)(x-3)\left(\dfrac{3}{x-3}-\dfrac{2}{x-2}\right)&=\left(\dfrac{3}{(x-2)(x-3)}\right)(x-2)(x-3)
\\\\
(x-2)(3)+(x-3)(-2)&=3(1)
\\
3x-6-2x+6&=3
\\
(3x-2x)+(-6+6)&=3
\\
x&=3
.\end{align*}
If $x=3$ is substituted in the original equation, the first term, $\dfrac{3}{x-3}$, results to a division by zero (which is not allowed). Thus, $x=3$ is NOT a solution to the original equation. Hence, there is no solution to the given equation.
