## College Algebra (10th Edition)

$x=\frac{3}{4}$ or $x=\frac{4}{3}$
$\displaystyle \frac{1}{(x-1)^{2}}+\frac{1}{x-1}=12$ We multiply both sides by $(x-1)^2$: $1+(x-1)=12(x-1)^2$ $12(x-1)^2-(x-1)-1=0$ $[4(x-1)+1][3(x-1)-1]=0$ $(4x-4+1)(3x-3-1)=0$ $(4x-3)(3x-4)=0$ $(4x-3)=0$ or $(3x-4)=0$ $x=\frac{3}{4}$ or $x=\frac{4}{3}$