## College Algebra (11th Edition)

$$\frac{x}{x-1}-\frac{1}{x+1}=\frac{2}{x^2-1}$$ $$((x+1)(x-1))(\frac{x}{x-1}-\frac{1}{x+1})=((x+1)(x-1))(\frac{2}{x^2-1})$$ $$x(x+1)-(x-1)=2$$ $$x^2+x-x+1=2$$ $$x^2-1=0$$ $$(x+1)(x-1)=0$$ $$x=-1,1$$ Both -1 and 1 are extraneous solutions. There is no solution.