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

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Solution for 6(n-1)=4(n+2)-n equation: 



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

We move all terms to the left:

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

We multiply parentheses

6n-(4(n+2)-n)-6=0



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

4(n+2)-n

We add all the numbers together, and all the variables

-1n+4(n+2)

We multiply parentheses

-1n+4n+8

We add all the numbers together, and all the variables

3n+8

Back to the equation:

-(3n+8)



We get rid of parentheses

6n-3n-8-6=0

We add all the numbers together, and all the variables

3n-14=0

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

3n=14

n=14/3

n=4+2/3

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