9(n-3)-6=2(n+4)+29

Solution for 9(n-3)-6=2(n+4)+29 equation:

9(n-3)-6=2(n+4)+29

We move all terms to the left:

9(n-3)-6-(2(n+4)+29)=0

We multiply parentheses

9n-(2(n+4)+29)-27-6=0


We calculate terms in parentheses: -(2(n+4)+29), so:

2(n+4)+29

We multiply parentheses

2n+8+29

We add all the numbers together, and all the variables

2n+37

Back to the equation:

-(2n+37)


We add all the numbers together, and all the variables

9n-(2n+37)-33=0

We get rid of parentheses

9n-2n-37-33=0

We add all the numbers together, and all the variables

7n-70=0

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

7n=70

n=70/7

n=10