Answer
$\left\{ 3-\sqrt{7},3+\sqrt{7} \right\}$
Work Step by Step
Multiplying both sides of the given equation, $
\dfrac{2}{x}+\dfrac{3}{x-1}=1
,$ by the $LCD=
x(x-1)
$, then,
\begin{array}{l}\require{cancel}
x(x-1)\left( \dfrac{2}{x}+\dfrac{3}{x-1} \right)=\left( 1 \right)x(x-1)
\\\\
(x-1)(2)+x(3)=x(x-1)
\\\\
2x-2+3x=x^2-x
\\\\
-x^2+(2x+3x+x)-2=0
\\\\
-x^2+6x-2=0
.\end{array}
Using $\dfrac{-b\pm\sqrt{b^2-4ac}}{2a}$ (or the Quadratic Formula), the solutions of the quadratic equation, $
-x^2+6x-2=0
,$ are
\begin{array}{l}\require{cancel}
\dfrac{-(6)\pm\sqrt{(6)^2-4(-1)(-2)}}{2(-1)}
\\\\=
\dfrac{-6\pm\sqrt{36-8}}{-2}
\\\\=
\dfrac{-6\pm\sqrt{28}}{-2}
\\\\=
\dfrac{-6\pm\sqrt{4\cdot7}}{-2}
\\\\=
\dfrac{-6\pm\sqrt{(2)^2\cdot7}}{-2}
\\\\=
\dfrac{-6\pm2\sqrt{7}}{-2}
\\\\=
\dfrac{-2(3\pm\sqrt{7})}{-2}
\\\\=
\dfrac{\cancel{-2}(3\pm\sqrt{7})}{\cancel{2}}
\\\\=
3\pm\sqrt{7}
.\end{array}
Upon checking, both solutions satisfy the original equation. Hence, the solutions are $
\left\{ 3-\sqrt{7},3+\sqrt{7} \right\}
.$
