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3(v-1)=5v+3-2(-3v-3)
We move all terms to the left:
3(v-1)-(5v+3-2(-3v-3))=0
We multiply parentheses
3v-(5v+3-2(-3v-3))-3=0
We calculate terms in parentheses: -(5v+3-2(-3v-3)), so:We get rid of parentheses
5v+3-2(-3v-3)
determiningTheFunctionDomain 5v-2(-3v-3)+3
We multiply parentheses
5v+6v+6+3
We add all the numbers together, and all the variables
11v+9
Back to the equation:
-(11v+9)
3v-11v-9-3=0
We add all the numbers together, and all the variables
-8v-12=0
We move all terms containing v to the left, all other terms to the right
-8v=12
v=12/-8
v=-1+1/2
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