## calyne show me how to solve for x when 1/(x+1)^2 = [x/(x+1) - 2]/(x-1) 2 years ago 2 years ago

1. satellite73

bunch of algebra, i will try to write it

2. calyne

ik thanks

3. satellite73

$\frac{x}{x+1}-2=\frac{-x-2}{x+1}$ is a start

4. satellite73

so right hand sides is $\frac{-x-2}{(x+1)(x-1)}$

5. calyne

wait how did you start out like that

6. satellite73

$\frac{1}{(x+1)^2}=\frac{-x-2}{(x+1)(x-1)}$ $\frac{1}{x+1}=\frac{-x-2}{x-1}$ $x-1=(x+1)(-x-2)$ $x-1=-x^2-3x-2$ $x^2+3x+1=0$

7. satellite73

$\frac{x}{x+1}-2=\frac{x}{x+1}-\frac{2(x+1)}{x+1}$ etc

8. calyne

still don't get thte first step talk about moronic what the flutter