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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