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

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



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

We move all terms to the left:

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

We add all the numbers together, and all the variables

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

We get rid of parentheses

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



We calculate terms in parentheses: -(2(3n-1)), so:

2(3n-1)

We multiply parentheses

6n-2

Back to the equation:

-(6n-2)



We add all the numbers together, and all the variables

5n-(6n-2)-1=0

We get rid of parentheses

5n-6n+2-1=0

We add all the numbers together, and all the variables

-1n+1=0

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

-n=-1

n=-1/-1

n=1

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