6(3v+1)+3v+1=6(v-2)-4v

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Solution for 6(3v+1)+3v+1=6(v-2)-4v equation: 



6(3v+1)+3v+1=6(v-2)-4v

We move all terms to the left:

6(3v+1)+3v+1-(6(v-2)-4v)=0

We add all the numbers together, and all the variables

3v+6(3v+1)-(6(v-2)-4v)+1=0

We multiply parentheses

3v+18v-(6(v-2)-4v)+6+1=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

2v-12

Back to the equation:

-(2v-12)



We add all the numbers together, and all the variables

21v-(2v-12)+7=0

We get rid of parentheses

21v-2v+12+7=0

We add all the numbers together, and all the variables

19v+19=0

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

19v=-19

v=-19/19

v=-1

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