## College Algebra (11th Edition)

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