3(n+2)-1=8-4(n-3)

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Solution for 3(n+2)-1=8-4(n-3) equation: 



3(n+2)-1=8-4(n-3)

We move all terms to the left:

3(n+2)-1-(8-4(n-3))=0

We multiply parentheses

3n-(8-4(n-3))+6-1=0



We calculate terms in parentheses: -(8-4(n-3)), so:

8-4(n-3)

determiningTheFunctionDomain
-4(n-3)+8

We multiply parentheses

-4n+12+8

We add all the numbers together, and all the variables

-4n+20

Back to the equation:

-(-4n+20)



We add all the numbers together, and all the variables

3n-(-4n+20)+5=0

We get rid of parentheses

3n+4n-20+5=0

We add all the numbers together, and all the variables

7n-15=0

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

7n=15

n=15/7

n=2+1/7

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