Chapter 11 - Section 11.3 - Solving Equations by Using Quadratic Methods - Exercise Set - Page 788: 47

$x=-3$

$\frac{x}{x-1} + \frac{1}{x+1} = \frac{2}{x^2-1}$ $(x-1)(x+1)*\frac{x}{x-1} + (x-1)(x+1)*\frac{1}{x+1} = (x-1)(x+1)*\frac{2}{x^2-1}$ $(x+1)*x+1*(x-1)=2$ $x^2+x+x-1=2$ $x^2+2x-1=2$ $x^2+2x-3=0$ $(x+3)(x-1)=0$ $x+3=0$ $x=-3$ $x-1=0$ $x=1$ $x=1$ cannot be a solution as the first fraction would be $1/0$, making the problem unsolvable.