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

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



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

We move all terms to the left:

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



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

4+2(n-1)

determiningTheFunctionDomain
2(n-1)+4

We multiply parentheses

2n-2+4

We add all the numbers together, and all the variables

2n+2

Back to the equation:

-(2n+2)



We get rid of parentheses

3n-2n-2-1=0

We add all the numbers together, and all the variables

n-3=0

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

n=3

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