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

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



-4(v-3)=4v+2(1+v)
We move all terms to the left:
-4(v-3)-(4v+2(1+v))=0
We add all the numbers together, and all the variables
-4(v-3)-(4v+2(v+1))=0
We multiply parentheses
-4v-(4v+2(v+1))+12=0
We calculate terms in parentheses: -(4v+2(v+1)), so:
4v+2(v+1)
We multiply parentheses
4v+2v+2
We add all the numbers together, and all the variables
6v+2
Back to the equation:
-(6v+2)
We get rid of parentheses
-4v-6v-2+12=0
We add all the numbers together, and all the variables
-10v+10=0
We move all terms containing v to the left, all other terms to the right
-10v=-10
v=-10/-10
v=1

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