3z-2(z+4)=16-3(2z+1)-z+2

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Solution for 3z-2(z+4)=16-3(2z+1)-z+2 equation: 



3z-2(z+4)=16-3(2z+1)-z+2

We move all terms to the left:

3z-2(z+4)-(16-3(2z+1)-z+2)=0

We multiply parentheses

3z-2z-(16-3(2z+1)-z+2)-8=0



We calculate terms in parentheses: -(16-3(2z+1)-z+2), so:

16-3(2z+1)-z+2

determiningTheFunctionDomain
-3(2z+1)-z+16+2

We add all the numbers together, and all the variables

-1z-3(2z+1)+18

We multiply parentheses

-1z-6z-3+18

We add all the numbers together, and all the variables

-7z+15

Back to the equation:

-(-7z+15)



We add all the numbers together, and all the variables

z-(-7z+15)-8=0

We get rid of parentheses

z+7z-15-8=0

We add all the numbers together, and all the variables

8z-23=0

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

8z=23

z=23/8

z=2+7/8

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