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

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



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

We move all terms to the left:

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

We add all the numbers together, and all the variables

5z-(4-(z+2))-5=0



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

4-(z+2)

determiningTheFunctionDomain
-(z+2)+4

We get rid of parentheses

-z-2+4

We add all the numbers together, and all the variables

-1z+2

Back to the equation:

-(-1z+2)



We get rid of parentheses

5z+1z-2-5=0

We add all the numbers together, and all the variables

6z-7=0

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

6z=7

z=7/6

z=1+1/6

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