3/(n+4)=-3/(n-2)

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Solution for 3/(n+4)=-3/(n-2) equation:



3/(n+4)=-3/(n-2)
We move all terms to the left:
3/(n+4)-(-3/(n-2))=0
Domain of the equation: (n+4)!=0
We move all terms containing n to the left, all other terms to the right
n!=-4
n∈R
Domain of the equation: (n-2))!=0
n∈R
We calculate fractions
3n/((n+4)*(n-2)))+(-(-3*(n+4))/((n+4)*(n-2)))=0
We calculate terms in parentheses: -(-3*(n+4))/((n+4)*(n-2))), so:
-3*(n+4))/((n+4)*(n-2))
We multiply all the terms by the denominator
-3*(n+4))
We multiply parentheses
-3n-
We add all the numbers together, and all the variables
-3n
Back to the equation:
-(-3n)
We get rid of parentheses
3n/((n+4)*(n-2)))+(+3n=0
We multiply all the terms by the denominator
3n+3n*((n+4)*(n-2)))+(=0

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