3(z-2)=-7(z+2)+28

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Solution for 3(z-2)=-7(z+2)+28 equation:



3(z-2)=-7(z+2)+28
We move all terms to the left:
3(z-2)-(-7(z+2)+28)=0
We multiply parentheses
3z-(-7(z+2)+28)-6=0
We calculate terms in parentheses: -(-7(z+2)+28), so:
-7(z+2)+28
We multiply parentheses
-7z-14+28
We add all the numbers together, and all the variables
-7z+14
Back to the equation:
-(-7z+14)
We get rid of parentheses
3z+7z-14-6=0
We add all the numbers together, and all the variables
10z-20=0
We move all terms containing z to the left, all other terms to the right
10z=20
z=20/10
z=2

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