N+(n+2)=3(n+4)-27

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Solution for N+(n+2)=3(n+4)-27 equation: 



+(N+2)=3(N+4)-27

We move all terms to the left:

+(N+2)-(3(N+4)-27)=0

We get rid of parentheses

N-(3(N+4)-27)+2=0



We calculate terms in parentheses: -(3(N+4)-27), so:

3(N+4)-27

We multiply parentheses

3N+12-27

We add all the numbers together, and all the variables

3N-15

Back to the equation:

-(3N-15)



We get rid of parentheses

N-3N+15+2=0

We add all the numbers together, and all the variables

-2N+17=0

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

-2N=-17

N=-17/-2

N=8+1/2

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