Answer
$x=1\pm\dfrac{\sqrt{6}}{3}$
Work Step by Step
$\dfrac{1}{x}+\dfrac{2}{x-1}=3$
Multiply the whole equation by $x(x-1)$:
$x(x-1)\Big(\dfrac{1}{x}+\dfrac{2}{x-1}=3\Big)$
$(x-1)+(2x)=3x(x-1)$
$x-1+2x=3x^{2}-3x$
Take all terms to the right side:
$0=3x^{2}-3x-2x-x+1$
$3x^{2}-6x+1=0$
Use the quadratic formula to solve this equation. The formula is $x=\dfrac{-b\pm\sqrt{b^{2}-4ac}}{2a}$. For this equation, $a=3$, $b=-6$ and $c=1$.
Substitute the known values into the formula and simplify:
$x=\dfrac{-(-6)\pm\sqrt{(-6)^{2}-4(3)(1)}}{2(3)}=\dfrac{6\pm\sqrt{36-12}}{6}=...$
$...=\dfrac{6\pm\sqrt{24}}{6}=\dfrac{6\pm2\sqrt{6}}{6}=\dfrac{6}{6}\pm\dfrac{2\sqrt{6}}{6}=1\pm\dfrac{\sqrt{6}}{3}$