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

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



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

We move all terms to the left:

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

We add all the numbers together, and all the variables

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

We multiply parentheses

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



We calculate terms in parentheses: -(3(v-1)+2), so:

3(v-1)+2

We multiply parentheses

3v-3+2

We add all the numbers together, and all the variables

3v-1

Back to the equation:

-(3v-1)



We add all the numbers together, and all the variables

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

We get rid of parentheses

5v-3v+1+4=0

We add all the numbers together, and all the variables

2v+5=0

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

2v=-5

v=-5/2

v=-2+1/2

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