1+2(3v-2)=-3(5v-1)+5v

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Solution for 1+2(3v-2)=-3(5v-1)+5v equation: 



1+2(3v-2)=-3(5v-1)+5v

We move all terms to the left:

1+2(3v-2)-(-3(5v-1)+5v)=0

We multiply parentheses

6v-(-3(5v-1)+5v)-4+1=0



We calculate terms in parentheses: -(-3(5v-1)+5v), so:

-3(5v-1)+5v

We add all the numbers together, and all the variables

5v-3(5v-1)

We multiply parentheses

5v-15v+3

We add all the numbers together, and all the variables

-10v+3

Back to the equation:

-(-10v+3)



We add all the numbers together, and all the variables

6v-(-10v+3)-3=0

We get rid of parentheses

6v+10v-3-3=0

We add all the numbers together, and all the variables

16v-6=0

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

16v=6

v=6/16

v=3/8

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