2z-1+(3z-2)=5-(2z-1)

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Solution for 2z-1+(3z-2)=5-(2z-1) equation: 



2z-1+(3z-2)=5-(2z-1)

We move all terms to the left:

2z-1+(3z-2)-(5-(2z-1))=0

We get rid of parentheses

2z+3z-(5-(2z-1))-2-1=0



We calculate terms in parentheses: -(5-(2z-1)), so:

5-(2z-1)

determiningTheFunctionDomain
-(2z-1)+5

We get rid of parentheses

-2z+1+5

We add all the numbers together, and all the variables

-2z+6

Back to the equation:

-(-2z+6)



We add all the numbers together, and all the variables

5z-(-2z+6)-3=0

We get rid of parentheses

5z+2z-6-3=0

We add all the numbers together, and all the variables

7z-9=0

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

7z=9

z=9/7

z=1+2/7

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