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

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



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

We move all terms to the left:

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

We multiply parentheses

5v+(v-2)-(2(v+4)-2(v-3))+15=0

We get rid of parentheses

5v+v-(2(v+4)-2(v-3))-2+15=0



We calculate terms in parentheses: -(2(v+4)-2(v-3)), so:

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

We multiply parentheses

2v-2v+8+6

We add all the numbers together, and all the variables

14

Back to the equation:

-(14)



We add all the numbers together, and all the variables

6v-1=0

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

6v=1

v=1/6

v=1/6

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