3(2n-5)+n=3(n-3)-(n+1)

Solution for 3(2n-5)+n=3(n-3)-(n+1) equation:

3(2n-5)+n=3(n-3)-(n+1)

We move all terms to the left:

3(2n-5)+n-(3(n-3)-(n+1))=0

We add all the numbers together, and all the variables

n+3(2n-5)-(3(n-3)-(n+1))=0

We multiply parentheses

n+6n-(3(n-3)-(n+1))-15=0


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

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

We multiply parentheses

3n-(n+1)-9

We get rid of parentheses

3n-n-1-9

We add all the numbers together, and all the variables

2n-10

Back to the equation:

-(2n-10)


We add all the numbers together, and all the variables

7n-(2n-10)-15=0

We get rid of parentheses

7n-2n+10-15=0

We add all the numbers together, and all the variables

5n-5=0

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

5n=5

n=5/5

n=1