1+3v+2=5(1-2)+2(-6+4v)

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Solution for 1+3v+2=5(1-2)+2(-6+4v) equation: 



1+3v+2=5(1-2)+2(-6+4v)

We move all terms to the left:

1+3v+2-(5(1-2)+2(-6+4v))=0

We add all the numbers together, and all the variables

3v-(5(-1)+2(4v-6))+1+2=0

We add all the numbers together, and all the variables

3v-(5(-1)+2(4v-6))+3=0



We calculate terms in parentheses: -(5(-1)+2(4v-6)), so:

5(-1)+2(4v-6)

determiningTheFunctionDomain
2(4v-6)+5(-1)

We add all the numbers together, and all the variables

2(4v-6)-5

We multiply parentheses

8v-12-5

We add all the numbers together, and all the variables

8v-17

Back to the equation:

-(8v-17)



We get rid of parentheses

3v-8v+17+3=0

We add all the numbers together, and all the variables

-5v+20=0

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

-5v=-20

v=-20/-5

v=+4

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