7(n+3)=3(n-4)+5

Solution for 7(n+3)=3(n-4)+5 equation:

7(n+3)=3(n-4)+5

We move all terms to the left:

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

We multiply parentheses

7n-(3(n-4)+5)+21=0


We calculate terms in parentheses: -(3(n-4)+5), so:

3(n-4)+5

We multiply parentheses

3n-12+5

We add all the numbers together, and all the variables

3n-7

Back to the equation:

-(3n-7)


We get rid of parentheses

7n-3n+7+21=0

We add all the numbers together, and all the variables

4n+28=0

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

4n=-28

n=-28/4

n=-7