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

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



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

We move all terms to the left:

(3n)/(n-2)-((3n)/(n-2))=0



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

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

n!=2

n∈R





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

n∈R



We calculate fractions


(-3n^2)/((n-2)*(n-2)))+(-(3n*(n-2))/((n-2)*(n-2)))=0



We calculate terms in parentheses: -(3n*(n-2))/((n-2)*(n-2))), so:

3n*(n-2))/((n-2)*(n-2))

We multiply all the terms by the denominator


3n*(n-2))

We multiply parentheses

3n^2-6n^2

We add all the numbers together, and all the variables

-3n^2

Back to the equation:

-(-3n^2)



We get rid of parentheses

(-3n^2)/((n-2)*(n-2)))+(+3n^2=0

We multiply all the terms by the denominator


(-3n^2)+3n^2*((n-2)*(n-2)))+(=0

We get rid of parentheses

-3n^2+3n^2*((n-2)*(n-2)))+(=0

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