3(n-1)=2(n+4)+6(n-3)

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Solution for 3(n-1)=2(n+4)+6(n-3) equation: 



3(n-1)=2(n+4)+6(n-3)

We move all terms to the left:

3(n-1)-(2(n+4)+6(n-3))=0

We multiply parentheses

3n-(2(n+4)+6(n-3))-3=0



We calculate terms in parentheses: -(2(n+4)+6(n-3)), so:

2(n+4)+6(n-3)

We multiply parentheses

2n+6n+8-18

We add all the numbers together, and all the variables

8n-10

Back to the equation:

-(8n-10)



We get rid of parentheses

3n-8n+10-3=0

We add all the numbers together, and all the variables

-5n+7=0

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

-5n=-7

n=-7/-5

n=1+2/5

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