2007 Cyprus MO/Lyceum/Problem 8

Problem

If we subtract from $2$ the inverse number of $x-1$ , we get the inverse of $x-1$ . Then the number $x+1$ equals to

$\mathrm{(A) \ } 0\qquad \mathrm{(B) \ } 1\qquad \mathrm{(C) \ } -1\qquad \mathrm{(D) \ } 3\qquad \mathrm{(E) \ } \frac{1}{2}$

$2-\frac1{x-1}=\frac1{x-1}$

$2=\frac2{x-1}$

Multiplying out and solving, we get that $x = 2$ , so $x+1=3\Longrightarrow\mathrm{ D}$ .