\( \frac{1-n}{n+1} + \frac{n-1}{1-n} = 1\)
(1-n)^2 + (n+ 1)(n - 1)
_________________ = 1 multiply both sides by the fraction denominator
(1 + n) (1 - n)
(1 - n)^2 + (n+ 1) (n - 1) = (1 + n) (1 - n) simplify
1 - 2n + n^2 + n^2 - 1 = 1 - n^2
2n^2 - 2n = 1 - n^2
3n^2 - 2n - 1 = 0
(3n + 1) (n - 1) = 0
Set each factor to 0 and solve for n
3n +1 = 0 n - 1 = 0
3n = -1 n = 1 (reject as this makes an original denominator = 0 )
n = -1/3