+308 Answer: \(\mathbb{R} - (-1, 2)\) (with what I have, anyway...)
I assume this is what you mean: \((\frac{2-n}{n+1})+(\frac{2n-4}{2-n})\).
Of course, n could be literally anything other than -1 or 2 because there isn't an equality present (maybe you forgot it????? Or maybe I went blind????)
+2668 We have the equation: \({2 - n \over n+1} + {2n - 4 \over 2 - n} = 1\)
Make a common denominator: \({(2 -n)(2-n) \over (n + 1)(2-n)} + {(2n-4)(n+1) \over (2-n)(n+1)} = 1\)
Simplify: \({n^2−4n+4 \over −n^2+n+2} + { 2n^2−2n−4\over -n^2 + n + 2} = 1\)
Add the fractions: \({3n^2 - 6n \over -n^2 + n + 2} = 1\)
Cross multiply: \(3n^2 - 6n = -n^2 + n + 2 \)
Move everything to the left-hand side: \(4n^2 - 7n - 2 = 0\)
Factor: \((4n+1)(n−2)=0\)
So, the solutions are \(-{1 \over 4} \) and \(2\).
But, to verify their validity, we need to plug them back into the original equation.
Doing so, eliminates \(n = 2\), so \(n = \color{brown}\boxed{-{1 \over 4}}\)
