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

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



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

We move all terms to the left:

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

We add all the numbers together, and all the variables

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

We multiply parentheses

4v+8v-(6(v+2)+4v)-8+6=0



We calculate terms in parentheses: -(6(v+2)+4v), so:

6(v+2)+4v

We add all the numbers together, and all the variables

4v+6(v+2)

We multiply parentheses

4v+6v+12

We add all the numbers together, and all the variables

10v+12

Back to the equation:

-(10v+12)



We add all the numbers together, and all the variables

12v-(10v+12)-2=0

We get rid of parentheses

12v-10v-12-2=0

We add all the numbers together, and all the variables

2v-14=0

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

2v=14

v=14/2

v=7

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