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

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Solution for 3(n+4)=2(n-1) equation: 



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

We move all terms to the left:

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

We multiply parentheses

3n-(2(n-1))+12=0



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

2(n-1)

We multiply parentheses

2n-2

Back to the equation:

-(2n-2)



We get rid of parentheses

3n-2n+2+12=0

We add all the numbers together, and all the variables

n+14=0

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

n=-14

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