## College Algebra 7th Edition

$\frac{3\pm\sqrt{6}}{3}$
$\frac{1}{x}+\frac{2}{x-1}=3$ We multiply through by $(x-1)x$: $(x-1)+2(x)=3(x)(x-1)$ $x-1+2x=3x^{2}-3x$ $0=3x^{2}-6x+1$ Next we use the quadratic formula ($a=3, b=-6, c=1$): $x= \frac{-b\pm\sqrt{b^{2}-4ac}}{2a}=\frac{--6\pm\sqrt{(-6)^{2}-4*3*1}}{2*3}=\frac{6\pm\sqrt{36-12}}{6}=\frac{6\pm\sqrt{24}}{6}=\frac{6\pm 2\sqrt{6}}{6}=\frac{3\pm\sqrt{6}}{3}$