2(6z-)=32

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Solution for 2(6z-)=32 equation:



2(6z-)=32
We move all terms to the left:
2(6z-)-(32)=0
We add all the numbers together, and all the variables
2(+6z)-32=0
We multiply parentheses
12z-32=0
We move all terms containing z to the left, all other terms to the right
12z=32
z=32/12
z=2+2/3

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