(Z-2)/2-(z+1)/6=(3z-3)/6

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Solution for (Z-2)/2-(z+1)/6=(3z-3)/6 equation: 



(-2)/2-(Z+1)/6=(3Z-3)/6

We move all terms to the left:

(-2)/2-(Z+1)/6-((3Z-3)/6)=0

We calculate fractions


(-2Z-2)/()+(-((3Z-3)*2)/()+()/()=0



We calculate terms in parentheses: +(-((3Z-3)*2)/()+()/(), so:

-((3Z-3)*2)/()+()/(

We add all the numbers together, and all the variables

-((3Z-3)*2)/()+1

We multiply all the terms by the denominator


-((3Z-3)*2)+1*()



We calculate terms in parentheses: -((3Z-3)*2), so:

(3Z-3)*2

We multiply parentheses

6Z-6

Back to the equation:

-(6Z-6)



We add all the numbers together, and all the variables

-(6Z-6)

We get rid of parentheses

-6Z+6

Back to the equation:

+(-6Z+6)



We get rid of parentheses

(-2Z-2)/()-6Z+6=0

We multiply all the terms by the denominator


(-2Z-2)-6Z*()+6*()=0

We add all the numbers together, and all the variables

(-2Z-2)-6Z*()=0

We get rid of parentheses

-2Z-6Z*()-2=0

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

-2Z-6Z*()=2

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