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solve for n: $\frac{1-n}{n+1} + \frac{n-1}{1-n} = 1$

 Jun 12, 2019
 #1
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\( \frac{1-n}{n+1} + \frac{n-1}{1-n} = 1\)

 

(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

 

 

 

cool cool cool

 Jun 12, 2019

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