3(n-4)+4(n-5)=5(n+2)+16

Solution for 3(n-4)+4(n-5)=5(n+2)+16 equation:

3(n-4)+4(n-5)=5(n+2)+16

We move all terms to the left:

3(n-4)+4(n-5)-(5(n+2)+16)=0

We multiply parentheses

3n+4n-(5(n+2)+16)-12-20=0


We calculate terms in parentheses: -(5(n+2)+16), so:

5(n+2)+16

We multiply parentheses

5n+10+16

We add all the numbers together, and all the variables

5n+26

Back to the equation:

-(5n+26)


We add all the numbers together, and all the variables

7n-(5n+26)-32=0

We get rid of parentheses

7n-5n-26-32=0

We add all the numbers together, and all the variables

2n-58=0

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

2n=58

n=58/2

n=29