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Solve for \(n\)\(\frac{1-n}{n+1} + \frac{n-1}{1-n} = 1\)

 Jan 8, 2019
 #1
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1 - n    +     n - 1

____         ____    =       1

n + 1          1 - n

 

[ 1 - n] [ 1 - n ] + [ n - 1 ] [ n + 1 ]

__________________________   =  1

[ 1 + n ] [ 1 - n ]

 

[ n^2 - 2n + 1  + n^2 - 1 ]

____________________   =     1

 [ 1- n^2 ]

 

 

[ 2n^2 - 2n ]   =   1 - n^2

 

3n^2 - 2n - 1 =  0      factor

 

(3n  + 1) ( n - 1) = 0

 

Set each factor to 0  and solve for n  and we have that

 

n = -1/3       or    n  = 1

 

Reject the second solution since it makes  an original denominator = 0

 

So

 

n =  -1 / 3   

 

 

cool cool cool 

 Jan 8, 2019

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