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

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

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

We move all terms to the left:

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

We multiply parentheses

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


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

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

We multiply parentheses

2n+(n+1)-6

We get rid of parentheses

2n+n+1-6

We add all the numbers together, and all the variables

3n-5

Back to the equation:

-(3n-5)


We add all the numbers together, and all the variables

3n-(3n-5)-5=0

We get rid of parentheses

3n-3n+5-5=0

We add all the numbers together, and all the variables

=0

n=0/1

n=0