(2-n)/(n+1)+(2n-4)/(2-n)=1

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Solution for (2-n)/(n+1)+(2n-4)/(2-n)=1 equation: 



(2-n)/(n+1)+(2n-4)/(2-n)=1

We move all terms to the left:

(2-n)/(n+1)+(2n-4)/(2-n)-(1)=0



Domain of the equation: (n+1)!=0

We move all terms containing n to the left, all other terms to the right

n!=-1

n∈R





Domain of the equation: (2-n)!=0

We move all terms containing n to the left, all other terms to the right

-n!=-2

n!=-2/-1

n!=+2

n∈R



We add all the numbers together, and all the variables

(-1n+2)/(n+1)+(2n-4)/(-1n+2)-1=0

We calculate fractions


((-1n+2)*(-1n+2))/((n+1)*(-1n+2))+((2n-4)*(n+1))/((n+1)*(-1n+2))-1=0



We calculate terms in parentheses: +((-1n+2)*(-1n+2))/((n+1)*(-1n+2)), so:

(-1n+2)*(-1n+2))/((n+1)*(-1n+2)

We multiply all the terms by the denominator


(-1n+2)*(-1n+2))

Back to the equation:

+((-1n+2)*(-1n+2)))





We calculate terms in parentheses: +((2n-4)*(n+1))/((n+1)*(-1n+2)), so:

(2n-4)*(n+1))/((n+1)*(-1n+2)

We multiply all the terms by the denominator


(2n-4)*(n+1))

Back to the equation:

+((2n-4)*(n+1)))

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