62+1=2(3z-1)

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Solution for 62+1=2(3z-1) equation: 



62+1=2(3z-1)

We move all terms to the left:

62+1-(2(3z-1))=0

We add all the numbers together, and all the variables

-(2(3z-1))+63=0



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

2(3z-1)

We multiply parentheses

6z-2

Back to the equation:

-(6z-2)



We get rid of parentheses

-6z+2+63=0

We add all the numbers together, and all the variables

-6z+65=0

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

-6z=-65

z=-65/-6

z=10+5/6

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