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2v+4=5(v+1)-3(v+2)
We move all terms to the left:
2v+4-(5(v+1)-3(v+2))=0
We calculate terms in parentheses: -(5(v+1)-3(v+2)), so:We get rid of parentheses
5(v+1)-3(v+2)
We multiply parentheses
5v-3v+5-6
We add all the numbers together, and all the variables
2v-1
Back to the equation:
-(2v-1)
2v-2v+1+4=0
We add all the numbers together, and all the variables
5!=0
There is no solution for this equation
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