6(2+v)-v=4+2(v+1)

Solution for 6(2+v)-v=4+2(v+1) equation:

6(2+v)-v=4+2(v+1)

We move all terms to the left:

6(2+v)-v-(4+2(v+1))=0

We add all the numbers together, and all the variables

6(v+2)-v-(4+2(v+1))=0

We add all the numbers together, and all the variables

-1v+6(v+2)-(4+2(v+1))=0

We multiply parentheses

-1v+6v-(4+2(v+1))+12=0


We calculate terms in parentheses: -(4+2(v+1)), so:

4+2(v+1)

determiningTheFunctionDomain
2(v+1)+4

We multiply parentheses

2v+2+4

We add all the numbers together, and all the variables

2v+6

Back to the equation:

-(2v+6)


We add all the numbers together, and all the variables

5v-(2v+6)+12=0

We get rid of parentheses

5v-2v-6+12=0

We add all the numbers together, and all the variables

3v+6=0

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

3v=-6

v=-6/3

v=-2