Answer
The solutions are $x=\dfrac{3}{2}\pm\dfrac{\sqrt{5}}{2}$
Work Step by Step
$\dfrac{x-1}{x-2}+\dfrac{x}{x-3}=\dfrac{1}{x^{2}-5x+6}$
Factor the denominator of the fraction on the right side of the equation:
$\dfrac{x-1}{x-2}+\dfrac{x}{x-3}=\dfrac{1}{(x-2)(x-3)}$
Multiply the whole equation by $(x-2)(x-3)$:
$(x-2)(x-3)\Big[\dfrac{x-1}{x-2}+\dfrac{x}{x-3}=\dfrac{1}{(x-2)(x-3)}\Big]$
$(x-1)(x-3)+x(x-2)=1$
$x^{2}-3x-x+3+x^{2}-2x=1$
Take all terms to the left side of the equation and simplify:
$x^{2}-3x-x+3+x^{2}-2x-1=0$
$2x^{2}-6x+2=0$
Take out common factor $2$ from the left side:
$2(x^{2}-3x+1)=0$
Take the $2$ to divide the right side:
$x^{2}-3x+1=\dfrac{0}{2}$
$x^{2}-3x+1=0$
Use the quadratic formula to solve this equation. The formula is $x=\dfrac{-b\pm\sqrt{b^{2}-4ac}}{2a}$
In this case, $a=1$, $b=-3$ and $c=1$
Substitute the known values into the formula and evaluate:
$x=\dfrac{-(-3)\pm\sqrt{(-3)^{2}-4(1)(1)}}{2(1)}=\dfrac{3\pm\sqrt{9-4}}{2}=...$
$...=\dfrac{3\pm\sqrt{5}}{2}=\dfrac{3}{2}\pm\dfrac{\sqrt{5}}{2}$
